PHILOSOPHIC  PRIICTIGAL  MATHEMATICS 


DESIGNED    FOR   THE   USE    OF 


ACCOUNTANTS,  MERCHANTS,  BUSINESS  MEN. 

PRIVATE  LEARNERS,  HIGH  GRADE  COMMERCIAL  COLLEGES 

AND   NORMAL  SCHOOLS. 

THE  PHILOSOPHIC  SYSTEM 

Is   Presented  and  Applied  to  the   Solution   of  COMMERCIAL,  FINANCIAL,  MECHANICAL  and 

Professional  Calculations  and  Problems,  covering 
the  various  fields  of  Business  Life. 


IT  SOLVES  AND  DISCUSSES  HUNDREDS  OF  PRACTICAL  QUESTIONS— COMPLEX 

AND  COMPOUND-NEVER  BEFORE  PRESENTED  IN  ANY 

SIMILAR  TREATISE. 


It  Sparkles  with  New  Thought,  Advanced  Methods,  and  Higher  Work.    It  presents  the  Rarest 

Gems  of  the  Science  of  Practical  Mathematics,  and  it  Teaches  that  a  NEW  TRUTH 

is  better  than  an  OLD  ERROR,  ha.vever  ancient  or  renowned. 

By    GEO.   SOULE, 

"Practical  and  Consulting  Accountant,  Lecturer  on  Commercial  Sciences,  Tresident  of  Soule  Commercial 
College  and  Literary  Institute,  Fetl-jw  of  the  Institute  of  Accounts  of  U^ew  York,  cMemher  of  the 
Associated  Accountants  of  U^ew  Orleans,  and  Director  tn  and  Tresident  of  various  Corporations. 
Author  of  iSoule's  '^Intermediate  Thilosophic  Arithmetic,"  ^^Contractions  in  Slumbers," 
'^Thilosophic  Drill  Problems,^'  ^^ Analytic  and  Philosophic,  Commercial  and  Exchange  Calculator," 
''tMamial  ofAudiling,"  and  "d^ew  Science  and  Practice  ofAccuunts.^' 

Sixth  Edition,  Revised  and  Enlarged  1910. 


PUBLISHED  BY  THE  AUTHOR. 
NEW  ORLEANS.     '    '.  ''' 


^ 


Entered  According  to  Act  of  Congress  in  the  Years  1872-1895-1898-1901-1905-1910 

By    GEO.    SOULE. 

In  the  Office  of  the  Librarian  op  Congress,  at  Washington 

ALL  RIGHTS   RESERVED. 


COMPOSITION  ELECTROTYPCD 

BV  BY 

MAUUS   4   HOFCUINE:.  N.  O.    ENO.  &   ELEO.   00. 

PRINTED  AND   BOUND 
BY 
SEARCY    &    PFAFF.    LTD. 


PREFACE. 


Twenty-four  centuries  ago  the  renowned  Grecian  mathematician,  Pythagoras,  an- 
nounced to  the  world  his  first  great  discovery  in  Mathematics,  and  since  that  eventful  time, 
the  science  of  Mathematics  has  been  by  master  minds  enriched  with  new  discoveries  until, 
as  a  science,  it  stands  the  equal  of  any.  and  is  as  unerring  in  its  results  as  the  laws  which 
govern  the  planets  in  their  revolutions  through  the  realms  of  space.  But,  while  as  a  science, 
Mathematics  stands  without  a  peer,  and  in  general  usefulness,  without  a  superior,  the  first, 
and  l>y  far  the  most  generally  used  subdivision  of  the  science.  Arithmetic,  is  uniformly  pre- 
sented in  the  text  books  of  the  day,  by  methods  which  are  generally  admitted  to  be  more  or 
less  defective  in  logical  reasons  for  the  operations,  and  in  clear  elucidations  of  the  subjects. 
In  addition  to  these  defects,  the  various  treatises  on  the  subject,  almost  without  an 
exception,  are  deficient  in  practical  problems,  and  are  by  far  too  brief  and  of  too  low  a 
grade  to  meet  the  commercial  and  business  demands  of  this  age  of  progress. 

With  a  view,  therefore,  to  remedy  in  a  measure  the  imperfections  named,  the  author 
presents  a  thousand  page  work  on  Philosophic  Practical  Mathematics,  replete  with  all  known 
abbreviated  methods  and  with  practical  problems  covering  the  various  fields  of  trade, 
finance,  mechanics  and  business  all  of  which  are  fully  elucidated  by  operations,  philosophic 
reasoning  and  pertinent  discussions.  Thus  the  work  supplies  a  long-felt  want,  and  is  of 
great  value  as  a  book  of  reference  and  authority  for  all  classes  of  business  men  and 
practical  book-keepers.  For  private  learners,  teachers,  commercial  colleges  and  literary 
institutions,  it  is  believed  to  have  no  equal  and  to  possess  the  crown  of  superiority. 

The  Science  of  Numbers,  or  Practical  Arithemetic,  is  one  of  the  queenly  products  of 
the  human  mind.  With  man,  it  has  come  down  the  evolving  centuries,  compounding  itself 
with  his  thought  and  occupation,  and  unfolding  its  beauties  as  the  human  race  advanced 
on  the  planes  of  civilization.  Receiving  new  inspirations  and  contributions  from  profouncj 
thinkers  as  the  ages  passed  by.  it  has  attained  to  the  dignity  of  a  queenly  science  before 
whose  throne  the  logic  of  mind  and  the  industries  of  the  world  worship.  The  erudite 
Hindoo  and  the  learned  Egj'ptian  communed  at  its  infant  shrine.  Before  it  Thales  andl 
Pythagoras  reasoned  with  uncovered  heads :  at  its  feet  Plato  laid  the  genius  of  his  great 
mind ;  and  Archimedes  and  Aristotle  employed  their  eloquence  and  philosophy  in  unfold- 
ing its  mysteries.  Newton  and  La  Place  adorned  it  with  modern  thought,  and  Pestalozzi 
leavened  it  with  the  great  principle  of  analytic  reason. 

But  while  the  mathematicians  of  ancient  fame  were  enthusiastic  regarding  numbers, 
and  spent  much  time  in  fanciful  contemplation  and  speculation  upon  their  properties,  they 
did  but  little  for  the  advancement  of  practical  arithmetic.  Their  numerical  recreation  and 
labors  were  largely  with  magic  squares  and  circles,  while  their  philosophic  talents  were 
bestowed  largely  upon  the  noble  science  of  geometry.  This  science  was  so  highly  esteemed 
by  Plato  that  over  the  door  of  his  lecture  hall  he  wrote:  "Let  no  one  enter  here  who  is 
ignorant   of  geometry." 

Thus,  until  the  time  of  Recorde,  isoo  to  1545:  Ramus.  1517  to  1572;  Stevinus,  1550 
to  1585;  Newton,  1642  to  1727;  Leibnitz,- 1646  to  1716;  and  later  Pestalozzi,  1746  to  1827, 
arithmetic  did  not  receive  the  attention  due  to  its  merits,  and  which  modern  business, 
methods  now  demand. 

The  Pestalozzian  process  of  analytic  reasoning  has  been,  during  the  past  seventy 
years,  utilized  by  more  than  a  hundred  authors.  Colburn,  Emerson,  Hobart,  Stewart, 
McCormick.  Ray,  Greenleaf,  Dodd,  Perkins.  Brooks,  Davies,  Robinson,  Thompson  and  all 
authors  since  1810  have  used  in  various  modified  forms,  more  or  less,  the  analytic  methods 
of  Pestalozzi.  and  thus  they  have  moved  on  with  the  evolving  world  and  performed  a 
measure  of  efficient  missionarj'  work  in  the  pagan  realms  of  practical   mathematics. 

The  analytic  method  of  Pestalozzi  is  the  foundation  of  the  golden  Philosophic  Sys- 
tem presented   in  this  and  the  author's  other  works  on  the   Science   of   Numbers. 

In  the  higher  evolution  of  this  system,  the  author  of  this  work  employs  and  combines 
the  three  great  processes  of  acquiring  knowledge  and  eliciting  truth,  viz :  comparison, 
analysis  and  synthesis.  These  three  processes  constitute  the  trinity  of  mathematics,  and  by 
an  ingenious  use  of  them,  the  author  is  believed  to  have  advanced  the  science  to  loftier 
planes  and  to  more  rational  methods  than  were  ever  before  achieved.  In  no  ancient  or 
modern  work  on  numbers  have  comparison,  analysis  and  synthesis  been  woven  into  such  an 
extended  chain  of  logical  and  philosophic  reasoning  as  is  herein  presented. 


r^nfT;«4i 


IV  PREFACE. 

By  the  Philosophic  System,  all  the  difficult  scientific  statements  of  true  proportion  are 
avoided,  all  the  ohjections  raised  against  the  purely  analytic  method  are  obviated,  the  some- 
what entangling  cause  and  effect  method  is  surpassed,  and  that  monstrosity  of  a  process 
which  "considers  whether  the  answer  is  to  be  greater  or  less  than  the  third  term "  js 
anathematized  and  consigned  to  the  shades  of  the  dark  age's  of  arithmetic. 

By  the  Philosophic  System,  all  the  arbitrary  rules  that  overload  the  faculty  of 
memory  and  prevent  the  expansion  of  the  higlier  faculties  of  causality  and  comparison  are 
abandoned,  and  the  reasoning  organs  of  the  mind  are  brought  into  action,  thereby  capaci- 
t:'.ling  tlie  learner  not  only  to  produce  the  results  of  problents,  but  to  observe  fine  distinctions, 
reason  logically,  and  deduce  correctly.  Thus  like  geometry,  it  serves  to  enlarge  the  pow- 
ers of  the  mind  and  to  qualify  for  high  planes  of  usefulness,  not  only  in  the  fields  of  mathe- 
matics, but  in  all  the  vocations  of  life. 

The  Pliilosophic  System  is  believed  to  be  the  most  valuable  improvement  yet  made  to 
impart  a  thorough  knowledge  of  the  principles  of  numbers  and  to  capacitate  the  student 
ti)  utilize  the  same  in  the  practical  afifairs  of  business  life.  But  notwithstanding  its 
superiority  and  the  fact  that  its  advocates  include  many  of  the  most  profound  mathematic.il 
minds,  yet.  like  every  other  improvement  or  discovery  in  education,  commerce,  art,  or  science, 
it  has  some  opponents  and  is  regarded  with  inditiference  by  those  who  are  satisfied  with  the 
non-progressive   and   non-reasoning   methods   of   past   ages. 

For  forty-eight  years  the  author  has  labored  with  tongue  and  pen  in  the  develop- 
ment of  the  Philosophic  System  of  Arithmetic,  and  has  tested  its  superior  merits  in  the 
school  room  and  lecture  hall  with  over  18,000  students;  and  from  a  full  knowledge  of 
its  advantages,  he  conscientiously  assures  his  co-laborers  in  the  mathematical  field  of  edu- 
cation, that  a  more  thorough  knowledge  of  the  science  of  numbers,  can  be  imparted,  and  in 
far  less  time,  by  this  system  than  by  the  usual  methods  and  systems  of  work. 

It  stands  without  a  peer  in  the  annals  of  mathematics,  and  it  is  believed  that  the  day 
is  not  far  distant  when  its  banner  will  wave  in  triumph  from  every  spire,  pinnacle  and 
dome  of  the  Temple   of   Practical   Mathematics. 

The  method  of  disseminating  or  imparting  arithmetical  knowledge,  that  is  in  almost 
luiiversal  use  by  the  great  army  of  authors  and  teachers,  is  the  rule  method,  or  in  other 
words,  the  non-reasoning  method.  By  this  method  the  principles  involved'  are  generally 
lost  sight  of,  and  results  without  reasons  are  sought  for.  Being  fully  convinced  of  this 
fact  and  knowing  that  the  best  results  are  to  be  attained  only  by  thoroughly  understanding 
the  principles  of  the  science,  the  author  has  in  this  treatise  labored  most  earnestly  to 
dethrone  the  venerable  absurdities  that  have,  as  he  believes,  obscured  the  science  of  num- 
bers for  ages. 

The  entire  work  sparkles  with  the  rarest  gems  of  the  science  of  numbers  and 
teaches  that  a  new  truth  is  better  than  an  old  error,  and  that  facts  and  reasons  are  better 
than  fallacious  theories,  however  ancient  or  renowned. 

To  obviate  a  formal  introduction  to  the  work,  the  subject  matter  usually  classed 
under  that  head  has  been  presented  in  remarks  and  discussions  throughout  the  work,  in 
connection  with  the  various  topics  treated.  By  reason  of  this  arrangement,  the  remarks 
and  discussions  have  been  made  more  pertinent,  and  will  be  read  at  the  time  when  the 
learner  most  needs  them. 

The  special  features  of  the  work  and  its  many  points  of  superior  merit  are  so  num- 
erous and  the  number  of  subjects  treated  is  so  great  as  to  preclude  the  propriety  of  giving 
them  space  in  the  preface.  A  cursory  look  through  the  alphabetical  index  following  the 
preface,  will  give  full  information  thereof. 

In  the  preparation  of  the  work,  the  -author  has  been  guided  b,y  an  experience  of 
forty-eight  years  as  a  teacher  and  lecturer  on  the  science  of  numbers  and  as  a  pr.actical 
accountant  and  commercial  lawyer  in  daily  professional  intercourse  with  all  classes  of  busi- 
ness men;  and  many  of  the  intricate  computations  that  have  been  submitted  to  him  for 
adjustment  are  herein  presented  and  fully  elucidated.  These  problems  and  the  solution 
thereof  constitute  a  special  and  most  valuable  feature  of  the  work. 

In  concluding  the  preface,  the  author  desires  to  extend  his   thanks  to: 

1.  Miss  Carrie  McGuigin.  his  faithful  amanuensis,  for  six  years  of  efficient  service 
in  the  preparation  of  the  copy. 

2.  Mr.  J.  M.  Buchee,  thirty-five  years  his  sincere  friend  and  co-worker  m  the  cause 
of  practical  education,  for  his  cheerful  assistance  rendered  in  proof-reading  and  in  work- 
ing problems.  To  him  is  to  be  credited  the  preparation  of  Involution  and  Evolution,  Square 
and  Cube  Root  Arithmetical  and  Geometrical  Progressions  and  Annuities.  His  discus- 
sions and  criticisms  on  various  matters  in  connection  with  the  work  have  been  valuable  and 
will  be  kindly  remembered.  ,    „     , ,  , 

3.  To  his  sons,  Albert  Lee  Soule,  Edward  Everett  Soule  and  Frank  Soule,  teachers 
in  the  Soule  College,  for  assistance  in  proof-reading  and  in  working  problems.  To  A.  L. 
Soule.  is  to  be  credited  the  work  of  preparing  Continued  Fractions,  Savings  Banks,  Per- 
mutations, Combinations.  Probabilities  and  Chances. 


^^P 


PREFACE.  V 

In  presenting  this  work  to  the  pubhc,  the  great  tribunal  that  is  to  pass  judgment 
upon  it,  the  author,  with  the  deepest  convictions  of  the  superior  merit  and  vakie  of  the 
philosophic  system,  indulges  the  hope  that  it  may  prove  acceptable  and  he  asks,  that  in 
befialf  of  justice,  in  the  interest  of  progress  and  the  science  of  numbers,  a  diecision  may 
not  be  rendered  until  a  thorough  examination  shall  have  been  made  and  its  merits  contrasted 
with  those  of  other  systems.  , 

GEO.   SOULE. 


PREFACE    TO    EDITION    OF    1910. 


In  1872  the  Authorpublished  a  work  entitled  Soul<5's  Analytic  and  Philosophic  Com- 
mercial and  Exchange  Calculator,  which  comprised  about  one-half  the  matter  contained 
in  the  present  work.  Five  years'  labor  was  devoted  to  this  treatise  and  it  was  received 
with  so  much  favor  by  Business  Men,  Accountants  and  Learners,  that  the  Author  was 
encouraged  to  enlarge  the  work  and  to  include,  elucidate  and  discuss  man.v  practical 
subjects  and  problems  from  various  lields  of  business  mathematics,  that  were  not  pre- 
sented in  the  first  work.  The  new  and  enlarged  work  was  given  to  the  public  in  1S95 
under  the  name  of  Soul^'s  Philosopliic  Practical  Alathematics.  This  work  was  revised 
and  a  new  edition  published  in  18118.  In  extent  of  subjects,  in  material,  in  original  and 
in  philosophic  solutions  and  discussion  of  problems,  covering  the  whole  held  of  Practical 
Matlieraatics,  this  work  is  believed  to  be  far  in  advance  of  any  similar  publication.  In 
1901  another  edition  was  published  with  several  new  problems.  In  1905  a  5th  edition 
was  pulilished  with  some  additional  new  matter. 

The  present  edition,  1910,  has  been  revised  and  enriched  with  the  presentation  and 
solution  of  many  new  practical  problems  drawn  from  different  lines  of  trade,  finance, 
mechanics,  agriculture,  general  business,  etc.  Large  numbers  of  these  problems  have 
been  presented  to  the  Author,  for  solution  by  accountants  from  all  sections  of  the  coun- 
try, and,  as  they  are  more  or  less  complex  coming  from  all  kinds  of  business,  they  are 
gems  of  varied  hue  in  the  crown  of  Practical  Mathematics. 

The  brilliantconquests  of  commerce,  the  amazing  magnitude  of  finance  and  exchange 
and  the  higher  demands  of  business,  have  produced  a  need  for  more  contracted  and 
rapid  systems  for  ordinary  computations,  and  for  a  higher  knowledge  of  the  more  intricate 
and  complex  practical  mathematical  problems  that  are  daily  occurring  in  business 
affairs.  To  meet  this  demand  this  work  has  been  prepared  and  revised  from  time  to 
time.  It  is  now  tendered  to  the  public  in  its  improved  and  complete  form,  with  the 
confident  hope  that  it  will  prove  valuable  to  those  for  whom  it  has  been  written  and 
published. 

GEO.  SOULE. 

New  Orleans,  May  14th,  1910. 


ALPHABETICAL    INDEX 


A 

PAGES 

Abbreviations  am''  Signs 3S-39 

Abstract  Fractional  Numbers,    to  Mul- 
tiply         ISS 

Abstract    Number    26 

Abstract   Numbers,   Reasoning   for  the 

Division    of    200 

Acceptances  and  Notes,  Discounting  of  551 
Accident  Insurance,  Definition  of  ....  508 
Account,  Bank,    Savings    and    Savings 

Banks     669 

Account,  Joint    Sales    on    663 

Account  of  Purchase,  or  an  Invoice,  De- 
finition of   480 

Account  Sales,    Definition    of    480 

Account    Sales,   Equation   of    643 

Account    Sales,   on   Joint    647 

Accounts    and  Notes  Bearing  Interest, 

Equation    of    649 

Accounts,  Compound  Equation  or  Aver- 
age  of    636 

Accounts    Current     and     Interest     Ac- 

■counts,  and  Cash  Balances   653 

Accounts    Current     and     Interest     Ac- 
counts, in  English  Money 661 

Accounts    Current     and     Interest    Ac- 
counts,  Problems   in    653-663 

Accounts     Current    and     Interest    Ac- 
counts  with   Brokers    705 

Accounts,  Equation  of,  Problems  in     628-646 

Acute  Angle,  Definition  of   337 

Addition     40-62 

Addition  and  Subtraction  Tables  . . .  41-42 
Addition    by    Casting    Out    Nines    and 

Elevens,   Proof  of   152 

Addition  by    Grouping    48 

Addition  by  One  of  the  Properties  of 

9     55 

Addition,  Civil   Service   Method    46 

Addition,  Contracting    iMethods    58 

Addition,  Drill    Exercises    43-58 

Addition,  General    Directions    for    ... .       47 

Addition,  Horizontal     49 

Addition     of     Compound     Denominate 

Numbers     267-268 

Addition    of   Decimals    223 

Addition     of  Dollars  and  Cents   52 

Addition    of   Fractional    Numbers    ....       51 

Addition  of  Fractions    175-179 

Addition  of  Fractions       by       a       New 

Method 179 


TAGES 

Addition   of  Metric   Numbers    936 

Addition  of    Several    Columns    in    One 

Operation     53 

Addition,  Problems     46-62 

Addition,  Proof  of 47,  152 

Addition   Tables   41 

Addition,  Vicenary    System    56-57 

Adjustment  of  Interest,  Definition  of. .  627 
Adjustment    of    Interest    on     Partners' 

Accounts    845 

Adjustment  of  Fire  Insurance  Losses.  512 
Adjustment  of  Losses,  Problems  . .     523-527 

Adjustment   of   the   Calendar 237 

Ad  Valorem  Duty,  Definition  of  500 

Agent,  an  Insurance,  Definition  of  ....  511 

Aliquot,  Definition    of 147 

Aliquot  Parts  of  1000   (Percentage)    . .  436 

Aliquots,   Table    of    94 

Alligation     789 

Alligation,  Alternate,   Definition   of 7S9 

Alligation,  Analytic    Method    793 

Alligation,  Linking   Method    792 

Alligation,  Loss  and  Gain  Method    . . .  792 
Alligation,  Medial    or    Medial     Propor- 
tion,   Definition    of    789 

Alligation,  Problems    in 792-802 

Allowance,  Definition   of   502 

Alloy,    Definition    of 233 

American    Experience    Table    of    Mor- 
tality      541 

American  Express  Company  Travelers' 

Checks     715 

American      Weights      and      Measures 

Reduced   to   Metric   941 

Amount,   Definition   of    547 

Amount  or  Final  Value    920 

Amusing  and  Curious  Questions  ......  974 

Analysis,   Quantitative   Chemical    467 

Angle,  Definition     of     252 

Angles,    Definition    of 337 

Angular  or  Circular  Measure   252 

Annual  Payments  of  Principal  and  In- 
terest,   Equal     588 

Annual   Premium   Rates  Table    539 

Annual,  Semi-Annual     and     Quarterly 

Interest    607 

Annual,  Simple     and      Compound     In- 
terest  Compared    615 

Annuity,  A   Certain,   Definition   of 920 

Annuity,  A  Contingent,   Definition  of.... 920 

Annuity,  A   Perpetual.   Definition  of...  920 

Annuity,  Forborne  or  in   Arrears    ....  920 

Annuity  Policy,   Definition  of   533 


ALPHABETICAL   INDEX. 


VII 


PAGES 

Annuity    Tables    925-926-927 

Annuities    920 

Annuities    Problems    in    921-927 

Annuities,  Remarks  on    921 

Antecedents  of  Proporticn   310 

Apothecaries'  Fluid  Measure   251 

Apothecaries'  Weight   250 

Appendix    977 

Application    of    Eleven    Check    Figure 

Method   of   Locating   Errors    .....     156 
Application     of    Eleven    Check    Figure 
Method  of  Proving  Transfers    and 

Postings     156 

Application  of  the  Lever   420 

Application  of   the    Screw    431 

Application  of  the  Wedge   429 

Applications    of    Geometrical    Progres- 
sion   to    Compound    Interest     and 

Compound   Discount 917 

Applications  of  Square  Root  .........     SSS 

Appraiser,  The,  Definition  of 502 

Approximate    Contractions,   Practical..     131 

Arabic    Notation    30 

Arbitration,  Compound,     Definition     of 

736 

Arbitration   of   Exchange    736 

Arbitration  of  Exchange,   Problems  in 

736-739 

Arbitration  of  Foreign  Exchange    ....     774 
Arbitration,  Sim|)le,   Definition   of   ....     736 

Arc.  Definition   of   252 

Are,  The  Definition  of 930 

Area 370 

Area,  Definition  of   243-33S 

Areas  and  Segments  of  a  Circle,  Table 

of 353-354 

Argentine  Republic,  Exchange  on   ....     771 
Arithmetic,  Complement  of  a  Number.       26 

Arithmetic,  Definition    of    27 

Arithmetic,  Philosophic,    System   of    . .       26 
Arithmetic,  Supplement   of  a  Number.       26 

Arithmetic,  Units   of   Orders    26 

Arithmetical   Progression    901 

Arithmetical   Progression,  Problems  in 

901-907 

Arms    of    a    Lever    420 

Arpent     .• 254 

Art,  Definition    of    25 

Articles  of  Partnership,  or  Articles  of 

Agreement    S06 

Artificers'     or     Mechanics'     Work     in 

Surfaces 360 

Assayers'    Weight    251 

Assessment,   Definition    of    677 

Assessment  Roll,  Definition  of   4S9 

Assessor,  Definition  of   4S9 

Assets,    Definition    of    4S7 

Assets  or  Resources    Definition  of  ....     4S7 
Assets,  Resources  or  Effects,  Definition 

of SOS 


PAGES 

Assignee,  Definition  of 487 

Assignment.  An,  Definition  of  4S7 

Associations,  Building  and  Loan   977 

Associations,  Building  and  Loan,  Prob- 
lems   in    9S9-1004 

Associations,    Co-operative    or    Benevo- 
lent   Insurance    535 

Associations,  National   Building   983 

Assumed     Due     Date     or     Focal     Date 

Definition    of    627 

Assured  or  Insured,  The,  Definition  of.. 509 

Astronomical   Day    236 

Austrian  Exchange    767 

Average,  An  Adjuster,  Definition  of...  527 

Average  Clause,   Definition  of    513 

Average,  General   and    Particular    ....  527 
Average,  General  and  Particular,  Prob- 
lems         528-531 

Average,  General,    Definition    of    527 

Average  Investment,  Definition  of  ....  809 

Average  or  Compound  Equation   627 

Average  or  Equation  of  Payments  and 

Accounts     627 

Average  or  Simple  Equation   627 

Average,  Particular,  Definition  of   ....  527 

Average,  Partnership,  Definition  of  . . .  861 
Average,  Partnership,    Compound,  Det. 

of     S61 

Average,  Partnership,    Simple,     Defini- 
tion of   861 

Average   Problems,   Cotton    .......     868-871 

Average   Storage,   Definition   of    778 

Averaging  Sales  in  Commission   867 

Avoirdupois  or  Commercial  Weight  . .  249 

Axiom,    Definition    of    26 

Axis  of   a   Sphere 370 

Axle   and   Wheel    422 

B 

Balance   of   Trade    711 

Balances,    Accounts    Current    and    In- 
terest  Accounts    and    Cash    653 

Bank   Accounts,    Savings    and    Savings 

Banks     669 

Bank  and  True  Discount  Compared. . .  604 

Bankers'  and  Merchants'  Discount....  576 
Bankers'     and     Merchants'     Discount, 

General   Directions    578 

Bankers'     and     Merchants'      Discount, 

Problems    in 577-582 

Bankers'  Interest  on  Daily  Balances..  665 
Bankers,  Method    of     Calculating     In- 
terest      571 

Bank  Notes,  National     230 

Bank  Notes,  United   States    230 

Bankruptcy     487 

Bankruptcy    Problems    488 

Bankrupt   or   Insolvent,  Definition   of.  .  487 

Banks,   Savings,  Definition  of   669 


A 
B 

C 

D 

E 

F 

G 

H 
I 

J 
K 

L 
M 

N 


10t> 

147 
30S 
334 
337 
370 

221 

220 

220 

)2-266 

256 

256 

941 


Q 
R 
S 
T 
U 

Y 

W 

Y 


VIII 


ALPHABETICAL  INDEX. 


PAGES 

Basis  of  All     Interest  Computations. . .  .560 

Basis  of  Life  Insurance    ^j;" 

Belgium,  Exchange  on    • • 

Benevolent   or   Co-operative    Insurance 

Associations     

Benevolent  or   Co-operative   Insurance, 

Definition   of    ••• °"^ 

Bill  and    Notes,    Promissory,     Indorse-     ^^^ 

ments    on     .■.'■■■■; 71  c 

Bill,  A  Short  Sight,  Definition  of   ... .     71b 


PAGES 
Brokers,  Bankers,  Financiers,  Definition 

of  Words  and  Phrases  Used  by     6S2-GS4 

Budget,   The,   Definition   of    4S9 

Building     ^61 

Building  and  Loan  Associations  Si ' 

Building  and  Loan  Associations,  Prob- 
lems   in    989-1004 

Building  and  Loan  Associations,  Table 


of 


97S 


Bill,  Face  of,  Definition  of   . 

Bill   of    Lading   Bills    or   Documentary 

Exchange,  Definition   of    '1° 

Bills    and   Drafts,   Parties   to    •  &^4 

Bills   and   Invoices    • f^l 

Bills,  Bankers,   Definition  of   '  |o 

Bills,  Clear,   Definition    of '|» 

Bills,  Commercial,  Definition  of   ab 

Bills,  Definition    of    ' 

Bills  of  Exchange   V;,"',"  rr. 

Bills     of  Exchange.  Drafts  and  Checks  554 

Bills  of  Exchange,  Foreign 'la 

Bills  of  Exchange,  Foreign,  Set  ot  ....  ;ii^ 

Bills,  Prime,   Definition  of   '^^ 

Binary    Scale    ; " ' 

Black,  S.  L.,  The  Method  of  Calculating  ^^^ 

Interest     ggg 

Board  Foot     •  ■  o  092 

Board   Measure    • "  'i„. 

Boards,  Measurement  of ^ 

Body  or  Solid,  Definition  of   . . . .  •  -  •  ■  •  ^* ' 

Body    Solid  or  Volume,  Definition  of..  6it 
Bonded  Warehouses,  Definition  ^ot^  .  .^.^-^    ^^^ 

„      : .V '. . . ."'  679 

Bonds    g-jg 

Bonds  and  Stock   •  • •  ^^ 

Bonds    and  Stocks,  Problems  m...     6S4-709 

Bonds,  Coupon,   Definition   of. bis 

Bonds,  Interest  on  United  States   ....  569 

Bonds,  Registered,  Definition  of  b^^ 

Bonds,  U.  S ,5^ 

Books  and  Paper  2S4 

Book,  Bill     • 54g 

Borrowers  and  Lenders • 

Borrowing  Members,  Definition  of   . . .  9^8 

Borrowing   Methods   in   Subtraction  66 

Botto^iry  and  Respondentia,  Definition  ^^^ 

of     76S 

Brazilian    Exchange    •         „ 

Breakage   and   Leakage    ^"-"^i 

Brief   History   of   the   Year    ..._....•  •^-     -^' 

Broken  or  Crooked  Line,  Definition  of     336 
Brokerage  and  Commission   . . . . . .  •  •  •  • 

Brokerage    and    Commission    P-'ol^^^^^^^.^gg 

Broker,' An  Insurance,  Definition  of...  511 

Broker.  Customhouse,  Definition  of...  50- 
Brokers,    Accounts    Current     and     In- 

terest  Accounts  with   <  "'^ 


Building  Associations,    National     '982 

Building  Materials   Bill   291 

Bullion,    Definition    of    230 

Business,  Customhouse    500 

Business,  Customhouse    Problems  .     502-507 
Business    Customs    and    Special    Laws.. 556 
Business  Customs  in  Buying  and  Sell- 
ing  Exchange    ''^'J. 

Butter  and  Cheese  Bill    287 

c 

Calculating     Interest,     General     Direc- 
tions   for    563 

Calendar     •  •• jj^ 

Calendar,    Adjustment  of   ^^' 

Canada    Money    •  ■  ^^* 

Cancellation     •     ^''^""^ 

Cancellation,  Definition  of    lol 

Cancellation,  General    Directions   for. .  ibi 

Cancellation   ot   Policies,   Definition   of  5U 

Cancellation    Synopsis   for   Review 163 

Candy    Bill     Vno  q77 

Capital,  Definition    of    bUM-yM 

Capital,    Net  or  Present  Worth,  Defini- 

tion    of    •  ^°^ 

Capital  Stock  or  Stock  of  a  Stock  Com- 

pany.   Definition   of    6((> 

Capitation  Tax  or  Poll  Tax,  Definition 

of     *^^ 

Carpenter's  Time  Sheet  290 

Carpeting    Floors    •  •^*''* 

Cash   Balances,  Accounts  Current  and 


653 


661 


Interest   Accounts 
Cash   Balances,  Accounts   Current  and 

Interest      Accounts,      in      English 

Money    • 

Cash   Balances,  Accounts  Current  and 

Interest  Accounts,  Problems  in     6ou-bb,i 
Cash    Capital  Stock,  or  Fictitious  Divi- 

dend    Forced,    Definition    of b7b 

Cash  Dividend,  Definition  of   61S 

Cash   Notes    J^-i:-;,"     7,0 

Cash  Notes  and  Exchange  Combined..     7i^ 

Cash    Notes,  Problems   in    582-59U 

Cash    Notes,  with     Interest,     Commis- 

sion  and  Brokerage  Combined   ...     584 

Cash  Storage,  Definition  of   777 

Cash  Value  or  Proceeds  of  Notes   ....     bw 

Centigrade   Thermometer    ^'* 

Centimeter,  Definition  of    • ^■'•L 


ALPHABETICAL  INDEX. 


IX 


PAGES 

Centrifugal    Sugar   Bill    293 

Certain  Aunuity,  a  Definition  of 920 

Certificate  of  Stock,  Definition  of   676 

Certificates,    Refunding,    Definition    of...6Sl 

Certificates,    Silver    230 

Chain    Measure     2-14 

Chance  or   Probabilities    945 

Chance  or    Probabilities,    Problems    in 

946-94S 

Changing  One  Scale  to  Another   ....     2S-29 

Character    Bill     942 

Check  or  Key  Figure   152 

Ch: ck  or  Key  Figure  System   155 

Checks,    American  Express   Co.    Trav- 
elers      715 

Checks  and   Drafts,  Bills  of  Exchange  554 

Cheese  and  Butter  Bill   287 

Chemical   Analysis,   Quantitative    467 

Chili,  Exchange  on   771 

Chinese    Exchange    774 

Chord,   Definition  of    252 

Circle.  Definition    of    252-338 

Circle,     of  a  Regular  Polygon,  to  Find 

Radius    of    357 

Circular  or  Angular  Measure    252 

Circle,   Table   of   Areas   and    Segments 

353-354 

Circle,  To  Find  the  Area  of  Segment  of  355 
Circle,  To  Find  the  Chord  of  Segment 

of    356 

Circulating   Decimals     217 

Circumference,    Definition    of    252 

Civil  or  Common  Year  236 

Civil  or  Legal  Day,  Definition  of   236 

Civil  Service  Method  of  Acjdition   ....  46 

Clandestine  Stock,  Definition  of 679 

Classification  of  Fractions   165 

Classification    of   Lumber    3S9 

Clause,  Average,  Definition  of   513 

Clause,  Iron  Safe  and  75  Per  Cent....  513 
Clause,    Lightning,    Definition    of    ....  513 
Clause,  Three-quarter    Loss,    or    Coun- 
try   Definition  of   514 

Clause,  Three-fourths  Value,  Definition 

of     514 

Clearance,  a  Definition  of 501 

Clear   Bills,    Definition    of    716 

Cloth   Measure    243 

Coal  and  Wood  Bill   2S7 

Coal    Gauging    409 

Co-efficient    of    Purity    (Sugar)     467 

Coin,  Weight  of   233 

Coins,  Table  of  Foreign    743 

Co-insurance,     75     Per     Cent      Clause, 

Definition    of    513 

Cold   Storage,  Definition  of   777 

Collection   of   Interest-bearing   Notes..  557 
Collection  of  Notes  that  Bear  Interest 

and    Eventual    Interest    580 


PAGES 

Collector  or  Receiver  of  Taxes,  Defini-  • 

;ion    of    4S9 

Colombia,  Exchange  on   771 

Commendam,    Partners    in,    Definition 

of     S04 

Commercial   Bills,   Definiiion   of    716 

Commercial    Instruments    of    Writing, 

Miscellaneous    551 

Commercial  or  Avoirdupois  Weight...  249 
Commercial      or      General       Partners, 

Definition    of     803 

Commercial    or   Trading    Partnerships, 

Definition   of    S04 

Commercial    Par   of    Exchange,    Defini- 
tion of   7ie 

Commercial   Partnership,   Definition  of  803 

Commission     and    Brokerage    4S0 

Commission    and    Brokerage    Problems 

4S1-4S6 

Commission,    Averaging   Sales   in    867 

Commission,  Definition   of    480 

Common     Denominator,    Definition    of 

172-173 

Common   Divisor,   Definition   of    147 

Common   Fractions,   Definition    of    165 

Common  Multiple,  Definition  of   147 

Common   or   Civil    Year    236 

Combinations   or  Choice    943 

Combinations  or  Choice,    Problems    in 

943-944 

Companies,    Insurance,    Definition   of..  511 

Companies,   Life   Insurance    532 

Companies,  Loan  and  Investment   ....  675 
Company     a   Mixed  or   Stock   and   Mu- 
tual   Insurance,    Definition    of    ...  511 
Company,     a  Mutual  Insurance,  Defini- 
tion   of    511 

Company,    a    Stock    Insurance,    Defini- 
tion of  511 

Comparative  Expectancy  of  Life,  Table  537 

Comparative  Mortality  Table    538 

Comparative     Values      of     Gold      and 

Silver    233 

Comparison,    Definition    of    20 

Comparison   of   Measurements    413 

Comparison   of   Thermometers    274 

Complement   of   a    Number,   .-"irithmet- 

ical     26 

Complement,  to  Multiply  by lOS 

Complex    and    Expert    Interest    Prob- 
lems       5S8 

Complex  Decimal,  Definition  of   215 

Complex   Fractions,   Definition   of    ....  166 
Compound   Arbitration,  Definition  of..  736 
Compound    Denominate    Numbers..     229-255 
Compound  Denominate  Numbers,      Ad- 
dition   of    267-268 

Compound  Denominate  Numbers,    Divi- 
sion  of    273-276 


c 

D 

E 
F 

G 

H 

I 

J 

K 

L 

M 

N 


147 
SOS 
334 
33T 
370 

221 

220 


220 

52-266 

256 

256 

941 


Q 
R 

S 

T 
U 
V 

w 

Y 


ALPHABETICAL   INDEX. 


TAGES 

Compound  Denominate  Numbers,    Sub- 
traction   of    269-270 

Compound    Denominate   Numbers,    Mis- 
cellaneous   Problems   in    277-2S2 

Comijound  Denominate  Numbers,    Mul- 
tiplication   of    271-272 

Compound       Discount     and       Present 

Worth     918 

Compound  Equation  or  Average,  Defin- 
ition  of    627 

Comiiound     Equation     or     Average     of 

Accounts     636 

Composite  Factor,  Definition   ot   146 

Compound    Fraction,    Definition   of. . . .  165 

Compound  Interest     609 

Compound  Interest  and  Compound  Dis- 
count, Applications  of  Geometrical 

Progression    to    917 

Compound  Interest  Problems   in    .  .      609-616 

Compound  Interest,    Tables    611-612 

Compound    Numbers     26 

Composite  Numbers,  Definition  of 146 

Compound   Partnership  Average,  Defin- 
ition   of 861 

Compound   Partnership  Average,  Prob- 
lems   in    S61-S67 

Compound    Proportion,    Definition    ot.  .  310 

Compound   Proportion,   Problems   in...  312 

Compound     Pulleys     425-426 

Compound    Rartio     308 

Compound,  Simple      and      Annual      In- 
terest   Compared    615 

Computations,   Rasis   of   All   Interest..  560 
Computations    of    Freight    in     English 

Money    753 

Computing       Interest,       a       Universal 

Formula    for     561 

Computing  Interest,    Bankers'     Method  571 
Computing  Interest,  Different  Methods 

of    571 

Computing  Interest,    The    S.    L.    Black 

Method     574 

Computing  Interest,     36%     Method 573 

Costa  Rica,  Exchange  on    771 

Cost  of  Different  Articles  of  Imported 

Goods     752 

Cotton  Average  Problems   S68-S71 

Cotton    Bill     2SS 

Coupon  Bonds,  Definition   of 679 

Concrete  or  Denominate  Numbers  ....  26 

Conditions    or    Premises   of    a  Problem  26 

Conditions,    Special   with   Settlement. .  872 

Cone,    Definition    of 370 

Conjoined    Proportion    312-333 

Connecticut    System    of    Partial     Pay- 
ments       618 

Co-operative    or    Benevolent   Insurance 

Associations     535 

Co-operative   or  Benevolent  Insurance, 

Definition    of     509 


PACES 

Consequents   of   Proportion    310 

Consignee,    Definition    of    480 

Consignment,   Definition    of    480 

Consignor,   Definition   of    480 

Consolidation  of  Corporations   703 

Consols,  Definition    of     680 

Consols,  English,  Definition  of   680 

Contents  or  Solidity  ot  a  Solid    370 

Contents  or  Volume,  Definition  of....  247 
Contingent  Annuity,  a  Definition  of  . .  920 
Contingent     Reserve,     or     Redemption 

Fund,  Definition  of   678 

Continued    Fractions     949 

Continued  Fractions,  Problems  in.  949-950 
Contracted  Interest  Divisors,  Table  of  561 
Contracted  Methods,  Drill  Problems..  115 
Contracted   Methods  ot  Finding  Check 

or  Key  Figure  152 

Contracting   Interest   Questions,   Philo- 
sophic   System    of    566 

Contracting    Methods    in    Vicenary    Ad- 
dition            58 

Contractions    In    Division    138-142 

Contraction    in    Multiplication    SS-121 

Contractions  in  Multiplication  of  Frac- 
tions           190-197 

Contractions  in  Practical  Problems  96-101 
Contractions    in    Subtracting     Dollars 

and    Cents    68 

Contractions,       Practical      Approxima- 
tive          191 

Conventional    Interest,    Definition    of. .     547 
Corporation  ot  Stock  Company,  Defini- 
tion   of    676 

Credit,  Letters  of,  Definition  ot  714 

Credit  or  Debit  of  Time,  Definition  of     627 

Creditor,    Definition    of    487 

Creditors,  Preferred,  Definition  of  .  .  487 
Credit  or  Time  Storage,  Definition  of  777 
Crooked  or  Broken  Line,  Definition  ot  336 
Cross  or  Simultaneous  Multiplication     88-93 

Cuban   Exchange    771 

Cube,  Definition    of    247-370 

Cube   Root    891 

Cube  Root,  Problems  in   899-900 

Cubic   Measure    247 

Cubic   Meter  or   Stere    930 

Curious    and    Amusing   Questions    ....     974 

Currency,  Definition    of    230-493 

Currency,  French,   Interest   on    570 

Currency,  German,   Interest   on    571 

Currency,  Gold,    Silver   and   Uncurrent 

Money    493 

Currency,  Gold,    Silver   and   Uncurrent 

Money,    Problems    in 493-499 

Currency    Sixes.    U.   S.    Pacific   R.   R., 

Definition    ot    681 

Curved    Line.   Definition    of    336 

Curved    Surface,   Definition   of    337 

Customhouse   Broker    502 


ALPHABETICAL   INDEX. 


XI 


PACES 

Customhouse    Business    500 

Customhouse    Business,    Prol^lems. .  .502-507 

'Custom  Houses,  Definition  of 500 

Customs,   in   Buying    and    Selling    Ex- 
change       717 

Customs   or   Duties,   Definition   of   ....  500 

Customs,  Special  Laws  and  Business..  556 

Cylinder,   Definitioa  of    270 

D 

Daily  Balances,  Bankers'  Interest  on..  665 

Daily  Balance  Method  of  Storage   ....  779 

Date  Line,  Internatioual    307 

Dates,  to  Find  the  Interval  of  Time  Be- 
tween   Two    296-300 

Day,  Definition   of    236 

Day,  Derivation    of    239 

Day,  Loss    or    Gain    of    by    Traveling 

Around   the    World    306 

Day  on   which   an   Event   Did   or   May 

Occur     299 

Day  or   Night,    to    Find    Length   of    ..  297 

Daj  s  of  Grace,  Definition  of   549 

Davs  of  the  Week,   Derivation  of    240 

jjdbenture,   Insurance  or   Investment     ..535 

Debenture   Stock    677 

Debtor,   Definition  of    4S7 

Decagon     33S 

Decimal   Fractions    216-228 

Decimal  Fractions,  Definition  of 1G5 

Decimal   Point   (.),   Definition   of   216 

Decimal  Scale  Changed  to  Other  Scales 

2S-29 

Decimals,  Addition    of    223 

Decimals,  Circulating     217 

Decimals,  Division    of     226-22S 

Decimals,  Multiplication   of    225 

Decimals,  Reduction    of    220-221 

Decimals,  Subtraction    of    224 

Decimals,  Table    of    218 

Deduction  of  Lumber   3S9 

Definition    25 

Definition  of  Abstract  Number 26 

Definition  of  a    Demonstration    26 

Definition  of  an   Axiom    26 

Definition  of  a  Problem     26 

Definition  of  Arithmetic    27 

Definition  of  Art     25 

Definition  of  a    Theorem    26 

Definition  of  Comparison   26 

Definition  of  Figures    27 

Definition  of  Formula     26 

Definition  of    Notation     30 

Definition  of  Number     25 

Definition  of   Numeration    30 

Definition  of  Philosophy    26 

Definition  of  Quantity    25 

Definition  of    Radix     27 

Definition  of  Science    25 


P.^GES 

Definitions    25-32 

Definitions   and   Terms   in    Partnership 

Settlements    SOS 

Definitions  Concerning  Ratio  308 

Definitions   in   .Addition    40 

Definitions   in   JIuUiplicatiou    75 

Definitions   in   Subtraction    64 

Degree,    (Circular)    Definition   of 252 

Degree    (Geographical)    Definition  of.  .  242 

Demonstration,  Definition   of   26 

Demonstration   of  Subtraction    65-66 

Denmark,   Exchange  on    773 

Denominate  Fractions,  Definition    of  . .  259 
Denominate  Fi-actions,  Reduction  of    259-261 

Denominate   Numbers,  Definition  of    .  .  229 
Denominate    Numbers,    Miscellaneous, 

Problems    in    277-282 

Denominate  Numbers,  Reduction  of    256-25S 

Denominate  or  Concrete   Number    ....  26 

Denominator,    Common     172-173 

Denominator,   Definition   of    165 

Denominator,  Least   Common    172 

Deposits,   Definition    of    669 

Derivation  of  the   Days   of  the   Week.  240 

Derivation  of  the  Names  of  the  Months  240 
Derivation  of  Year,       Month,       Week, 

Da.v,    etc 239 

Deviation    in    Weight   of   Coin,    Allow- 
ance   Made    for    233 

Diagonal    337 

Diagonal  and  Ullage  Rods,  Use  of .  . . .  413 
Diagrams  Showing  Division  of  Govern- 
ment   Lands     246 

Diameter,    Definition    of    252 

Diameter  of  a  Sphere   370 

Diamond   Weight    251 

Different    Methods    of    Computing    In- 
terest       571 

Difference  of    Latitude    301 

Difference  of   Longitude    307 

Difference  of    Longitude    between    two 
places  When  Difference  of  Time  is 

Given     304 

Difference  of  Time  between  two  places 
When    Difference    in    Longitude    is 

Given     305 

DiflScult     and     Practical     Question     in 

Renewing     589 

Directions  for  Addition   47 

Directions   for   Division    129 

Directions  for  Division  of  Fractions   . .  201 

Directions    for   Multiplication    85 

Directions  for  Subtraction    67 

Directions,    General,     for    Calculating 

Interest  563 

Direct  Tax  or  Property  Tax,  Definition 

of     489 

Discharge,   Definition   of    487 

Discount,  Bankers'    and    Merchants'...  576 


D 

E 
F 

G 

H 
I 


J 
K 

L 
M 

N 


100 

147 
30S 
334 
33T 
370 

221 

220 

220 

52-266 

256 

256 

941 


Q 
R 
S 

T 
U 
V 

w 

Y 


XII 


ALPHABETICAL  INDEX. 


PAGES 

Discount,  Bankers'      and      Merchants', 

General  Directions  in   578 

Discount,  Banliers'      and      Mercliants', 

Problems   in    577-582 

Discount,       Compound      and      Present 

Worth     918 

Discount    Day,   Definition    of    550 

Discount,    Definition   of    493 

Discount  Notes   that   Bear  Interest....     5S0 

Discount  on   Time  Drafts    732 

Discounting  Notes,  Definition  of 550 

Discounting  Notes  and  .Acceptances..  551 
Discounting  of  Notes  Bearing  Interest  558 
Discount,  True   and    Ijanlv   Compared..      604 

Discount,  True,  Definition  of   596 

Discount,  True,   Problems   in    596-604 

Dishonoring,   Definition   of    550 

Dissolution  of  Partnerships,   Definition 

of     806 

Distribution   of  Prof.t,  Plans  of   983 

Dividend,  Definition   of    487-534 

Dividends     678 

Dividends,   Definition   of    979 

Divisibility   of   Numbers    147 

Division    122-141 

Division  and  Proportional  Division...  876 
Division  by  Casting  Out  the  Nines  and 

Elevens,   Proof  of    154 

Division,    Contractions   in    138-142 

Division,  Definitions    in    122-123 

Division,  Fractional   Numbers    124 

Division,  General   Directions   for    129 

Division,  Long    129 

Division,  Logic   of    122 

Division,  Method   of   Reasoning    197 

Division,  Miscellaneous  Problems  in     135-137 
Division  of  Abstract  Numbers,  Reason- 
ing  for   the    200 

Division     of     Compound      Denominate 

Numbers     273-276 

Division    of    Decimals     226-228 

Division  of  Fractions    197-205 

Division  of  Metric  Numbers   939 

Division,  Principles    of     123 

Division,  Questions   in    197 

Division,  Projiortional     883 

Division,  Short     128 

Divisor,  a  Philosophic  Method  of  Using 

the   Factors   of   the   Interest    562 

Divisor,  Common     147 

Divisor,  Definition   of    147 

Divisor,  Greatest  Common    147 

Divisors,    Definition   of    123-147 

Divisors,    Interest    559 

Divisors,    Table      of      Contracted      In- 
terest          561 

Divisors,    Table  of  Interest    560 

Documentary  Exchange  or  Bill  of  Lad- 
ing Bills    716 

Dollars  and  Cents,  Addition  of   52 


PAGES 

Dollars  and  Cents,  Multiplication  of  . .  84 
Dollars  and  -Cents    Subtraction  of    . .     67-68 

Dollars,    Standard    Silver    230 

Doyle's   Rule    389 

Draft,  Definition  of   502 

Drafts   and    Bills,    Parties    to    554 

Drafts  and  Checks,  Bills  of  Exchange.  554 

Drawback,    Definition    of     502 

Drawing   Notes,   Points   in 559 

Drill  Exercise  in  Mensuration  of  Solids 

417-418 

Drill  Exercises,    Addition    43-58 

Drill  Problems  in   Contracted   Methods  115 
Drill  Problems  in   Simultaneous   Multi- 
plication       90 

Drug    Bill    286 

Dry    Goods    Bill    294 

Dry  Measure   248 

Duodecimal    Scale    27 

Duty,  Ad  Valorem,  Definition  of 500 

Duty,    Specific,    Definition    of    500 

Duty,  Tonnage,   Definition   of    501 

Duties,   Definition   of    501 

Duties,  Excise,   Definition  of   502 

Duties   or  Customs,   Definition  of    500 

E 

Earliest  Date,   Definition   of    627 

Earnings,  Gross,  Definition  of 677 

Earnings,  Net;  or  Net  Profits,  Defini- 
tion   of    678 

Effects,  Resources  or  Assets,  Definition 

of     80S 

Eggs  and  Poultry  Bill   287 

Eleven    Check    Figure  .  Method,   Errors 

not   Detected    by    153 

Elevens,    Excess    of    154 

Elevens,  Properties   of   150-156 

Ellipse,   Definition   of    339 

Elucidation  of  Vicenary   System   56 

Endless   Screw    431 

Endorsement,    Open    Policy,    Definition 

of    510 

Endowment   Policy,   Definition   of    ....  533 

Engineers'  and  Surveyors'  Measure   . .  244 

English   Consols,   Definition  of   680 

English   Exchange    740 

Englisli  Exchange,   Problems  in    . .     745-758 

English  Goods,  to   Import   and    Mark..  457 

EngliFh  Monetary  Pound,  History  of.  .  740 

English  Money     234 

Englisli  Money,  Computation  of  Freight 

in     753 

English    Money    in    Accounts    Current 

and  Interest  Accounts    661 

English  Money,  Interest  on    569 

English   Money,  Table    234 

English  Money,  to  Compute    Per    Cent 

on    449 


ALPHABETICAL  INDEX. 


XIII 


PAGES 

English  System  of  N'umeration  31 

Entry  Ports,  Definition  of  501 

Equal    Annual    Payments   of    Principal 

and    Interest    5SS 

Equated  Time,  Definition  of   627 

Equation,   Compound   or  Average,    De- 
finition of   627 

Equation     Method,     Accounts    Current 

and   Interest    Account    653 

Equation     of  Account  Sales   643 

Equation    of  Accounts  and  Notes  Bear- 
ing   Interest    6-19 

Equation   of  Accounts,  Problems  in     62S-646 

Equation  of  Notes   651 

Equation    of   Rent   Notes    ,.  .  G4S 

Equation    or      Average      of      Accounts, 

Compound     636 

Equation,  Simple    or    Average,    Defini- 
tion  of    627 

Equation  or  Average  of  Payments  and 

Accounts     627 

Equivalency  and  Relationship  of  Num- 
bers      311 

Equivalents,   Tables   of    931-933 

Errors    Not    Detected    by    the    Eleven 

Check   Figure   Method    153 

Errors     Not     Detected     by     the     Nine 

Check   Figure   Method    153 

Estate,  Real,  or  Real  Property,  Defini- 
tion   of    4S9 

Estates,  Settling  of   S74 

Even    Numbers    25,  146 

Evil   Effects  of  Usury  Laws    545 

Evolution    885 

Excess  of  Elevens,  or  the  Check  or  Key 

Figure    154 

Exchange    710 

Exchange,  Arbitration    of    736 

Exchalnge,  Austrian     767 

Exchange,  Bills    of 713 

Exchange,  Bills  of.  Drafts  and  Checks  554 

Exchange,  Bills  of.   Foreign,   Set  of. . .  713 

Exchange,    Brazilian     768 

Exchange,  Business    Customs    in    Buy- 
ing  and   Selling   717 

Exchange,  Cash  and  Notes  Combined.  732 

Exchange,  Chinese    774 

Exchange    Cuban    771 

Exchange,  Documentary,     or     Bill     of 

Lading  Bills    716 

Exchange,  Foreign,  Arbitration  of  ....  774 

Exchange,  Foreign  Bills   of    715 

Exchange,  French    759 

Exchange,  German     764 

Exchange,  Importance    of    712 

Exchange,  Japanese 772 

Exchange     on      Belgium,      Switzerland, 

Italy,    Spain,    Greece    and    Finland  764 


PAGES 

Exchange,  on    Honduras,     Costa   Rica, 
Guatemala,     Nicaragua,     Salvador 

and   Columbia    771 

Exchange,  on    Mexico   and   the   Argen- 
tine   Republic    771 

Exchange,  on  Peru,  Ecuador  and  Chili  771 

Exchange,  on    Portugal    770 

Exchange,  on    Sweden,     Norway     and 

Denmark     773 

Exchange,  Origin  of   711 

Exchange,  Problems    in    717-735 

Exchange,   Rate   of    744 

Exchange  Russian 771 

Exchange,  The     Commercial     Par     of. 

Definition   of    71G 

Exchange,  The  Intrinsic  Par  of.  Defini- 
tion  of    716 

Exchange,  The   Par  of.  Definition  of. .  716 

Exchange,  The  Rate  of,  Definition  of.  716 

Exchanges,   Stock,  Definition  of   6S1 

Excise   Duties,   Definition  of   502 

Excise  or  Indirect  Tax,  Definition  of. .  489 

Exercises  in  Addition,    Drill    43-45 

Exercises  in  Division    of    Fractions    ..  201 

Exercises  in  Factoring     157 

Exercises   in   Finding    Check    or    Key 

Figure    155 

Exercises  in  Notation     and      Numera- 
tion         33-34 

Exercises,  Oral,  in  Partnership  Settle- 
ments      818 

Expectancy  of  Life,  Comparative  Table 

of    537 

Experience  Table  of  Mortality,  Ameri- 
can      541 

Expert    and     Complex    Interest    Prob- 
lems       5S8 

Express    Company    Travelers'    Checks, 

American     715 

Extension,    Definition    of    241 

Extraction  of  the  Square  Root   SS5 

Extremes  of  Proportion 310 


F 


Face  of  a  Note,  Definition  of 549 

Face  of  Bill,  Definition  of  739 

Factoring     157-160 

Factoring,  Synopsis    for   Review    160 

Factors    Definition  of  146 

Factors  of  the  Interest  Divisor,  a  Phil- 
osophic Method  of  using  the    ....  562 

Factors,  Prime   and    Composite    146 

Factory  Time   Sheet    290-291 

Fahrenheit   Thermometer   274 

Fertilizers     467 

Fictitious,  or  Forced,  or  Capital  Stock 

Dividend,   Definition    of    678 


E 
F 

G 

H 
I 

J 
K 

L 
M 

N 


Q 


147 
30S 
334 
337   D 

370 


221 

220 

220 

52-266 

256 

256 

941 


s 

T 
U 
V 

w 

Y 


XIV 


ALPHABETICAL  INDEX. 


PAGES 

Fidelity    or    Guarantee    Insurance,    De- 
finition of   509 

Figures    and    Geometrical     Definitions 

336-33S 

Figures,  Definition    o£    27 

Figures,  Order  ot   29-30 

Figures,  Value    of    27 

Final  Value,  or  tlie  Amount 920 

Financiers,   Brokers   and   Bankers,   De- 
finition    of     Words     and     Phrases 

used    by     6S2-6S4 

Finland,    Exchange    on    764 

Fire   Insurance,   Definition   of    508 

Fixed   Pulley    424 

Floating  Policy,  a  Definition  of 514 

Focal  Date  or  Assumed  Due  Date,  De- 
finition   of    627 

Foot  of  Different  Nations  Compared...     253 
Forced    Cash    Capital    Stock    or    Ficti- 
tious  Dividend,   Definition   of    ....     67S 
Foreign  Bills  of  Exchange,    Face    and 

Character  of    715 

Foreign  Bills  of  Exchange,  Set  of 713 

Foreign  Coins,  Table  of   743 

Foreign  Exchange,  Arbitration  of    774 

Forfeited   Stock,  Definition  of   677 

Formation    of    Partnerships     Definition 

of    805 

Formula,    Definition   of    26 

Formula     for     Computing     Interest,    a 

Universal    561 

Formula  of  Statements    739 

Fractional    Numbers    in    Division    ....     124 
Fractional    Numbers,    To    Multiply   Ab- 
stract          188 

Fractional   Unit    25,  164 

Fraction,   Definition   of    164 

Fractions     164-168 

Fractions,  Addition    of     175-179 

Fractions,    Classification    of    165 

Fractions,   Continued    949 

Fractions,  Continued,  Problems  in.    .949-950 

Fractions,  necimal     216-22S 

Fractioas,  General    Principles    of 166 

Fractions,  Greatest   Common   Divisor. .     213 

Fractions,  Like,   Definition  of   175 

Fractions,  Reduction      of     Denominate 

259-261 

Fractions,  Subtraction    of    1S0-1S2 

Fractions,  Synopsis  tor  Review   166 

Fractions,  Unlike,  Definition  of   175 

Framing    361 

Free  List,  The  Definition  of   501 

Freight  Bill    289 

Freight,    Computations    of    in   English 

Money     753 

French  Currency,  Interest  on   570 

French  Exchange     759 

French  Exchange,  Problems  in 759-764 

French  Money    235 


PAGES 

French  Money,  Table     235 

French   (Old)   System  of  Measures 253 

French   Rentes,   Definition   of    6S0 

French   System   of   Numeration    30 

Frustrum  of  a  Cone     370 

Frustrum  of  a  Pyramid   370 

Fulcrum,   Definition   of 419 

Funded  Loan  of  1891,  Definition  of  . .  680 
Funded  Loan  of    1907,    Definition    of..  681 
Fund,  Reset  ve,  or  Contingent,    or  Re- 
demption,   Definition    of    678 

Fund,  The  Reserve,  Definition  of   ....  534 

G 

Gains  and  Losses   808 

Gains  and  Losses,   Sharing  of    ....     S03-S61 

Gain,  Net,   Definition  of   809 

Gain   or  Loss  of  a   Day  by    Traveling 

Around  the  World  306 

Gallon,   Definition   of    247 

Gauging  Barrels,  Casks,  etc 413 

Gauging  Coal    409 

General  and  Commercial  Partners,  De- 
finition   of    803 

General  and   Particular  Average    527 

General  and   Particular  Average   Prob- 
lems          528-531 

General  Average,   Definition    of    527 

General   Directions   for  Addition    47 

General  Directions    for      Addition      of 

Decimals     223 

General  Directions  for  Addition  of  Frac- 
tions       177 

General    Directions    for    Bankers'    and 

Merchants'    Discount    578 

General   Directions   for  Calculating  In- 
terest      ■ 563 

General    Directions    for    Cancellation..  162 

General  Directions  for  Division    129 

General  Directions  for  Division  of  Deci- 
mals      228 

General  Dinections    for      Division      of 

Fractions     201 

General  Directions  for  Finding  Greatest 

Common  Divisor  of  Fractions  ....  213 
General    Directions    for   Finding    Least 

Common   Multiple  of  Fractions     214-215 
General     Directions     for     Finding    the 

Greatest  Common  Divisor    158 

General    Directions     for     Finding    the 

Least  Common  Multiple    159 

General  Directions  for    Multiplication.  85 
General  Directions  for  Multiplication  of 

Fractions     225 

General  Directions  for  Proportion   ....  313 
General  Directions  for   Reducing   Com- 
mon Fractions  to  Decimals    222 

General  Directions  for   Reducing   Deci- 
mals to  Common  Fractious    221 


ALPHABETICAL    INDEX. 


XV 


PACKS 

General  Directions  for    Subtraction...  67 
General  Directions    for    Subtraction    of 

Decimals     224 

General  Directions  for    Subtraction    of 

Fractions     181 

General    Partnership     S  "> 

Ge':eral  Principles    of   Fractions    166 

General  Principles  of  Proportion     ....  310 

General  Principles  of  Ratio     309 

Geometrical    Definitions    and    Figures 

336-33S 

Geometrical  Progression     90S 

Geometrical  Progression,  Applications 
of  to  Compound  Interest  and  Com- 
pound  Discount    917 

Geometrical  Progression,    Problems    in 

908-919 

German  Currency,  Interest  on  571 

German    Exchange    764 

German  Exchange,  Problems  in   ...     7G,')-767 

German  Money     236 

German  Money   Table    236 

Glazing  and   Painting   366 

Glucose,  Ratio,  Definition  of   467 

Gold  and  Silver,  Comparative    Values.  233 

Gold  and  Silver,  Value    of    2.34 

Gold  Coin,   Definition   of    230 

Gold,   Currency,   Silver  and   Uncurrent 

Money    493 

Gold,    Currency,   Silver  and   Uncurrent 

Money,    Problems    in 493-499 

Goods,    Insuring    526 

Goods   Shipped,  Invoice  of   295 

Government  Lands,  United   States    . . .  245 

Grace,   Days   of.   Definition   of    549 

Grain  and  Hay  Bill   292 

Grain  Warehouse.  Definition  of 777 

Grain,  Weight  of.  Per  Bushel    235 

Gram,    The    Definition    of    930 

Greatest  Common  Divisor     158 

Greatest  Common   Divisor,  Definition  of  147 

Greatest  Common  Divisor  of  Fractions  213 

Greece,  Exchange  on   764 

Grocery   Bills    2S3-2S4 

Gross  Earnings,  Definition  of   677 

Gross  Weight,   Definition    of    502 

Grouping,    Addition    by    49 

Guarantee,    Definition    of    4S0 

Guaranteed  Stock,  Definition  of   677 

Guarantee  or  Fidelity  Insurance,  De- 
finition  of   509 

Guatemala,   Exchange   on    771 

H 

Hardware    Bill    292 

Hay  and   Grain  Bill    292 

Health   Insurance,   Definition  of    508 

Heptagon     338 

Hexagon    338 


PAr.i:s 

Higher  Terms,  Definition  of   167 

History  of  Interest   543 

History    of     Tlie      English      Moinetary 

Pound     740 

History   of   the    Metric    System    ......  928 

Honduras,   Exchange  on    771 

Horizontal  .\ddition '  '  4;) 

Horizontal  Subtraction     6.S 

Houses,   Custom,   Dcfmition   of    ...  500 

How  Numbers  are  Expressed    25 

Human  Life,  The   Probability  of   .....  542 

Hypotenuse,  Definition  of  ]  337 

I 

Immovable  Proiierty,  Valued  Policies  on  52(> 

lm|)erfect  Number,  Definition  of 14S 

Importance    of    Exchange     712 

Importance   of   Life    Insurance    ...  532 

Importance    of    Multiplication    77 

Imported  Goods,   Cost  of  Different' Ar- 

cles   of   752 

Improper  Fraction,   Definition  ot   .  ....  165 

Inclined    Plane    427 

Income  Tax,  U.  S.,  Definition  o(  .....  .  489 

Indirect   or   Excise   Tax,   Definition   of.  489 

Indorsement,    An,    Definition    of    550 

Indorsements  on  Promissory  Notes  and 

Bills    550 

Indorser,  an.  Definition   of   55(1 

Installment,  Definition  of '.  677 

Installment  Dividend,  Definition  of   . . !  678 

Installment   Plan,   Definition   of    981 

Installments,  Partial  Payments,  or  Pay- 
ments   by     ] 61G 

Insolvency,    Definition   of    809 

Insolvent  or  Bankrupt,  Definition  of  .' .  487 

Inspection   of  Lumber    389 

Instruments  of  Writing,  Miscellaneous 

CommeriJial     551 

Insurance     508 

Insurance,  Accident,  Definition  of   508 

Insurance,  Agent,   An,    Definition   of...  511 
Insurance,  a  Mixed   or  Stock  and   .Mu- 
tual  Co.,  Definition  of   511 

Insurance,  a   Mutual   Company,   Defini- 
tion   of    511 

Insurance    Associations,      Co-operative 

or  Benevolent   53s 

Insurance,  Basis  of  Life   536 

Insurance    Broker,   .\n.   Definition   of. .  511 

Insurance    Companies,  Definition  of   ..  511 
Insurance    Companies,    Life,   Definition 

of    532 

Insurance,  Co-operative  or  Benevolent, 

Definition    of    509 

Insurance,  Fire,  Definition  of   508 

Insurance,  Guarantee    or    Fidelity,    De- 
finition   of    509 

Insurance,  Health,  Definition  of 508 


G 

H 
I 

J 
K 

L 
M 
N 


117 
?,<>% 
334 
337 
370 

Q 
R 

221 

S 

220 

220 

62-266 

256 

256 

T 
U 

941 

V 

w 

Y 

XVI 


ALPHABETICAL  INDEX. 


Forei 

( 

Forei 

Forei 

Forei 

Forf( 

Forn 

( 

Forn 

Forn 

1 

Forn 

Frac 

Frac 

Frac 
Frac 
Frac 
Frac 
Frac 
Frac 
Frac 
Frat 
Frac 
Frac 
Frac 
Frac 

Frac 
Fra, 
Frai 
Frai 
Fret 
Frei 
Frei 

Frei 
Fre 
Fre' 
Fie 


PACKS 

Insurance    Importance  of  Value  of....  514 

Insurance,  Life     532 

Insurance,  Life,  Definition   of 50S 

Insurance,  Life,  Problems    510-542 

Insurance,  Live   Stock, .  Definition   of.  50S 

Insurance,  Marine,  Defijition   of    50S 

Insurance   or   Investment,    Debenture..  535 

Insurance,  Personal,  Definition  of  .'...  508 

Insurance,    Problems     515-527 

Insurance,  Propert.v,   Definition    of    ...  508 

Insurance,  River,    Definition   of    508 

Insurance,  Scrip,   Definition  of    512 

Insurance,  Stock   Company   a    511 

Insurance,  The  Subject  of  the.  Defini- 
tion   of    510 

Insurance,  Transit,    Definition    of    ....  508 

Insured  or  Assured,  The   509 

Insurer,   The    509 

Insuring    Good      526 

Integer,  Definition  of    146 

Interest 543 

Interest    Accounts,    Accounts    Current 

and    Cash    Balances    653 

Interest    Accounts,     Accounts     Current 
and     Cash     Balances,     in     English 

Money    661 

Interest   Accounts,    Accounts    Current 
and    Cash    Balances,    Problems    in 

653-663 

Interest,  Adjustment    of,    Definition    of  627 
Interest    and  Eventual  Interest,  to  Col- 
lect Notes  that  Bear  580 

Interest  and  Premiums,  Plans  of  Charg- 
ing          980-'9Sl 

Interest    and    Principal,    Equal    Annual 

Payments   of    589 

Interest,  a    New    Table     of     Computa- 
tions       576 

Interest,  Annual,      Semi-annual       and 

Quarterly 607 

Interest,  a  Universal  formula  for  Com- 
puting       561 

Interest,  Bankers'    Method,    or    60    day 

or   6'"r    Method    571 

Interest  Bearing    Notes,    Collection    of  557 

Interest  Bearing  Notes,  Discounting  of  558 

Interest,  Compound     609 

Interest,  Compound,  Problems  in..     609-616 
Interest,  Compound,    Simple    and    An- 
nual,  Compared    61 5 

Interest  Computations,  Basis  of  All...  560 
Interest,  Conventional,  Definition  of   .  .  547 
Interest,  Different    Methods     of     Com- 
puting      571 

Interest  Divisor,  a  Philosophic  Method 

of   Using   the    factors    of   the    562 

Interest  Divisor,    Definition   of    559 

Interest  Divisors     559 

Interest  Divisors,  Contracted   Table  of  561 

Interest  Divisors,   Table   of    560 


PAGES 

Interest,  Equation     of    Accounts     and 

Notes    Bearing    649 

Interest,  General  Directions  for  Calcu- 
lating      563 

Interest  in  Advance  Plan,  Definition  of  981 

Interest,  Laws,  Table  of  548 

Interest,  Legal   Definition   of    547 

Interest,  Methods,     Accounts     Current 

and  Interest  Account   653 

Interest   on   Daily   Balances,   Bankers. .  .665 

Interest    on    English    Money     569 

Interest    on    French    Currency    570 

Interest    on  German  Currency    571 

Interest    on   Interest   Payments    595 

Interest  on    Partners'     Accounts,     Ad- 
justment  of    845 

Interest   on  United  States  Bonds    ....  569 
Interest    Problems,    Complex    and    Ex- 
pert       588 

Interest,  Problems   in    562-576 

Interest,  Questions,    Peculiar     567 

Interest,  Questions,     The      Philosophic 

System  of  Contracting  566 

Interest    Received   and    Paid,   To   Find 

the  Rate  of   592 

Interest,  Simple,   Definition   of    547 

Computing    574 

Interest    Tables,   Compound    611-612 

Interest,  Term,  Definition  of   669 

Interest    The    History   of    543 

Interest,  The  Rate  of  543 

Interest,  The    S.    L.    Black    Method    of 

Interest,  367p   Method   of   Computing. .  573 

Interest,  To  Discount  Notes  that  Bear  580 

Internal  Revenue,  Definition  of   502 

International   Date   Line    307 

Interval  of  Time  Between  Two  Dates, 

To    Find    the    29G-300 

Intrinsic  or  Liquidation  Value  of  Stock, 

Intrinsic  Par  of  Exchange,  Definition  of  716 

Inventorying  or  Taking  Stock  of  Goods  527 

Definition    of     677 

Inverse    Ratio    308 

Investment  and  Loan  Companies   675 

Investment,    Average,    Definition    of...  809 

Investment,   Definition  of   S09 

Investment,  Net,  Definition  of   739-809 

Investment  or   Savings   Policy,  a  Ton- 
tine, Definition  of   533 

Investment,  Whole   Amount   of.   Defini- 
tion   of    739 

Invoice  or  Account  of  Purchase,  Defini- 
tion   of    480 

Invoice  or  Manifest,  an,  Definition  of.  .  500 

Invoices   and   Bills    283-295 

Involution    863 

Involution,    Problems   in    864 

Italy,  Exchange  on    764 


ALPHABETICAL   INDEX. 


XV 11 


Japanese    Exchange    772 

Jettison,   Definition   of   52S 

Joint  Account,  Sales    on     047-663 

Joint  Life  Policy,  Definition  of   533 

Iv 

Key  Figure    152 

Key  Figure    System    155 

Kilogram,  or  Kilo,  Definition   oi 933 

Kilometer,   Definition  of    931 

Kinds    of    Fractions    164 

Knot,  Definition  of    242 


Lands,  United  States  Government 245 

Latitude,  Definition    of    301 

Latitude,  Difference   of    301 

Laws   and   Business  Customs,   Special.  556 

Laws  of  Interest,  Table  of   548 

Laws,  Usury,   Evil  Effects  of   545 

Laying  Floors   361 

Lead   Pipe  and   Sheet  Lead   Table    ...  363 

Leakage  and  Breakage,  Definition  of. .  502 

Leap   Year    236 

Least    Common    Denominator,      Defini- 
tion   of    . .  ■. 1T2 

Least  Common   Multiple   159 

Least   Common   Multiple,  Definition   of  147 
Least   Common    Multiple    of   Fractions 

214-215 

Ledger   Accounts    156 

Legal   Interest.  Definition  of   547 

Legal  or  Civil  Day,  Definition  of 236 

Legal  Tender,  Definition  of   230 

Lenders   and   Borrowers    546 

Letters  of  Credit,  Definition  of   714 

Lever,  Definition  of   419 

Liabilities,    Defiintion    of    4S7-S08 

Licenses  and  Taxes,  Definition  of 489 

Licenses  and  Taxes,  Problems   490-492 

License,   Definition   of    489 

Life,  Comparative    Expectancy   Table..  537_ 

Life,  Human,  The  Probability  of   542" 

Life  Insurance 53.. 

Life  Insurance    Basis    of    536 

Life  Insurance  Companiss    532 

Life   Insurance,  DefinitioQ*  of    508 

Life  Insurance,  Importance   of    532 

Life     Insurance,  Problems    540-542 

Life  Policy,  Joint     533 

Life  Policy,  Limited    Payment,    Defini- 
tion of   533 

Life  Policy,  Ordinary   Definition   of    . .  533 

Li.sjhtning  Clause,  The  Definition  of   . .  513 


I'AGF.S 

Like   Numbers    25 

Limited   Partners   in   Louisiaua,  Defini- 
tion   of    : S04 

Limited  Partnership,  Definition   of    ...     <S05 

Linking  Method  of  Alligation    792 

Line,    Definition    of    241-336 

Linear  Measure,    340-346 

Linear  Measure,  Miscellaneous  Notes.  243 
Linear  Measure,  Miscellaneous    Units.     243 

Linear  Measure,   Table    243 

Line,  or   Linear   Measure    241 

Liquidation     or     Intiinsic      Value      of 

Stocks,    Definition    of     677 

Liquid  or  Wine   Measure   247 

List,  Free,  Definition  of 501 

List   Prices   and   Trade   Discounts    ....     45S 

Liter,  The,  Definition  of   930 

Live  Stock   Insurance,   Definition   of.  . .     508 

Loading    or    Margin 534 

Loan  and  Building  .■Associations   977 

Loan  and  Building  .Associations,  Prob- 
lems   in    989-1004 

Loan  and  Investment  Companies  ....  675 
Loaning  Money,  or  Premium  Plans..  9S3 
Loan  of  1891,  Funded,  Definition  of,.  680 
Loan  of  1904,  Funded,  Definition  oi..  681 
Loan  of  1907,  Funded,  Definition  of. .  681 
Loan  of  1925,  Funded     Definition     of..     6S1 

Local  Value  of  Figures   27 

Logical   System   of  Multiplication    77 

Logic   of   Division    122 

Logs,  Measurement  of   3S4 

Long    Division     129 

Longitude,  Difference    of    302 

Longitude  and  Time    303-307 

Longitude,  Definition    of    302 

Long  Ton,   Definition  of    502 

Loss  and  Gain  Method  of  Alligation   .  .     792 

Losses,   Adjustment   of    512 

Lessee,    Adjustments    of     Problem^    m.. 

523-527 

Losses  and   Gains    808 

Losses  and  Gains.   Sharing  of    ....     803-S61 

Loss,  Net,  Definition  of  S09 

Loss   or   Gain    of   a   Day   in   Traveling 

Around   the   World    306 

Lower   Terms,   Definition   of    167 

Lowest  Terms,  Definition  of    167 

Lumber  and  Board  Measure   388-392 

Lumber    Bill     289 

Lumber,  Measurement    of    384 

M 

Manifest  or   Invoice,   an.   Definition   of  500 

Manufacturing  on  Shares  and  ToUage.  7S5 

Margin  or  Loading,  Definition  of 534 

Marine    Insurance     508 

Mariner's    Measure    242 


J 

K 

L 
M 
N 


^  147 

.  30S 

334 

337 

370 

221 

220 

220 

62-266 

256 

256 

941 


Q 
R 
S 
T 
U 
V 

w 

Y 


XVIII 


ALPHABETICAL    INDEX. 


PACKS 

Market,  or  Surrendered  Value,  Defini- 
tion   of    534 

Market  Value  of  Stocks    Definition  of.  C77 

Marking  Goods   451 

Maturity  of  a  Note,  Definition  of   ....  54D 

Means,  in  Proportion    310 

Mean   Solar   Day,   Definition   of    236 

Measure    Linear    340-346 

Measurement  of  Logs,  Timber,  Lumber 

and    Boards    3S4 

Measurement    of    Segments,     Remarks 

on    3S2 

lAeasure  of  Time   236 

Measures     Definition    of    223 

Measures  of  Extension    241 

Measures  of  Weight    249-255 

Measures,    Weights    and    Values,    Com- 
pared       254 

Measure    Tables     231 

Mechanical  Efficiency  of  the  Screw  .  . .  431 

Mechanical    Powers    419-433 

Mechanics'  or  Artificers'  Work  in  Sur- 
faces       360 

Medial    Alligation   or     Medial     Propor- 
tion      789 

Medial    Proportion    333 

Members,  Borrowing,  Definition  of   .  . .  97S 

Members,  Definition   of    977 

Members.  Depositing,  Definition  of  . . .  978 

Mensuration     336-418 

Mensuration  of  Surfaces    346-380 

Mensuration     of     Surfaces,     Practical 

Problems   in    357-380 

Mensuration  of  Solids 370-418 

Merchandise    Insurance — Points    to    be 

Specially    Observed     526 

Merchandise  Sales  on  Joint  Account..  647 

Merchants'  and   Bankers'   Discount....  576 
Merchants'     and     Bankers'     Discount, 

General  Directions  for  578 

Merchants'     and     Bankers'     Discount,  . 

Problems  in    577-582 

Merchants'     System     of     Partial     Pay- 
ments   617 

Meridian,  Definition  of  302 

Meter,  Definition    of    931 

Meter,  Standard  Definition  of   928 

Methods  of  Accounts   Current  and   In- 
terest  Account    653 

Methods    of    Computing    Interest,     Dif- 
ferent      571 

Methods    of   Determining   the    Gain   or 

Loss  of  a  Business   809 

Methods    of   Reading   Metric    Numbers  931 
Method  of  Reasoning  in  Division    ....  197 
Method  of  Reasoning  in  Multiplication  183 
Methods  of  Some  New   Orleans   Build- 
ing and   Loan   Associations    991 

Metric  Numbers,  Addition   of   936 

Metric  Numbers,    Division   of    939 


P.\GES 

Metric  Numbers,  Reduction   of    935 

Metric  Numbers,  Subtraction  of 937 

Metric  Numbers,  to  Write  and  Read..     934 

Metric  System,  History  of   928 

Metric  System  of    Weights    Measures.     928 
Metric  System       of        Weights        and 

Measures,   Problems    in 934-942 

Metric    Tables    931 

Metric  Weights  and  Measures  Reduced 

to   American    940 

Mexico,    Exchange    on    771 

Millimeter,  Definition  of   931 

Mint  or  Troy  Weight   249 

Minuend    64 

Minute,   (Geographical)   Definition  of. .     242 
Miscellaneous  Commercial  Instruments 

of    Writing    551 

Miscellaneous    Oral  Exercises   in   Divi- 
sion   of   Fractions    201 

Miscellaneous    Practical    Questions    in 

Percentage    470-479 

Miscellaneous  Problems.     951-974,    1014-102.5 
Miscellaneous  Problems   in   .\dditions     59-62 
Miscellaneous  Problems  in  Denominate 
and     in     Compound     Denominate 

Numbers    277-282 

Miscellaneous  Problems  in  Division   .135-137 
Miscellaneous  Problems  in  Division  of 

of    Fractions    202 

Miscellaneous   Problems   in   Multiplica- 
tion         85-88 

Miscellaneous   Problems   in    Multiplica- 
tion  of   Fractions    189 

Miscellaneous  Problems  in  Square  and 

Cube  Root   899-900 

Miscellaneous  Problems  in  Subtraction     71-73 
Miscellaneous   Problems   Involving  Ad- 
dition,   Subtraction,    Multiplication 

and   Division     143-145 

Miscellaneous  Problems   Involving   Ad- 
dition,   Subtraction,    Multiplication 
and  Division  of  Fractions   ....     206-212 
Miscellaneous  Units  of  Linear  Measure     243 

Misrepresentation,    Definition    of    512 

Mixed   Decimal,  Definition   of    216 

Mixed  Number,  Definition  of   165 

Mixed   or   Stock  and   Mutual   Ins.    Co., 

Definition   of    511 

Monetary  Pound,  English,  History  of. .     74a 

Monetary  Unit  of  France     235 

Monetary  Unit  of    Germany    236 

Monetary  Unit  of    Great    Britain    234 

Monetary  Unit  of  the  United    States..     234 
Monetary  Units  of  Different  Ages,  Na- 
tions  and   Peoples    230-231 

Money,  Canada    234 

Money,   Definition   of    230 

Money,  English    234 

Money,  English,   Interest   on    569 

Money,  French    235 


ALPHABETICAL   INDEX. 


XIX 


PAGES 

Money,  Gold,  Silver  and  Unciirrent  . .     493 
Money,  Gold,    Silver     and     Uncurrent, 

Problems     493-499 

Money  of  the  World,  Unification  of   . .     776 

Money,  Uncurrent,   Definition   of    493 

Money,  United    States    232 

Month,    Derivation    of    239 

Months,  Derivation  of  Names  of 240 

Mortalitv,  American    Experience   Table 

of    ■ 541 

Mortality    Table,   Comparative    53S 

Mortgage,    Definition    of     6S0 

Movable    Pulley    425 

Multiple,  Common     147 

Multiple,  Definition  of    147 

Multiple,  Least    Common    147 

Multiple  Units  of   the   Metric  System.     930 

Multiples,  Table  of   339 

Multiplicand     '75 

Multiplication 75-121 

Multiplication    by   Aliquots    94-101 

Multiplication    by     Caistlug     Out     the 

Nines    or    Elevens,  Proof  of 154 

Multiplication,  by   Complement    lOS 

Mutiplication    by  Factors    84 

Multiplication,  by    Squaring    Numbers.     118 

Multiplication,    Contractions  in    S8-121 

Multiplication,  Definition  of   75,  1S3 

Multiplication,  Drill    Problems    90 

Multiplication,    General    Directions    for       S5 

Multiplication,    Importance    of    77 

Multiplication    of     Abstract     Numbers, 

with    Reasons    80 

Multiplication    of    Compound     Denomi- 
nate   Numbers    271-272 

Multiplication   of   Decimals    225 

Multiplication  of  Dollars  and  Cents...       S4 

Multiplication   of   Fractions    183-196 

Multiplication    of    Fractions.    Contrac- 
tions  in    190-197 

Multiplication  of  Metric  Numbers   938 

Multiplication,    Philosophic    or    Logical 

System     7'^ 

Multiplication,  Practical   Definition   of.     183 

Multiplication,     Principles   of    75 

Multiplication.  Proof  of    75,154 

Multiplication,  Simultaneous  or  Cross    88-93 

Multiplication,    Table    to    25x25    76 

Multiplication  with  Naughts  on  Right.       83 

Multiplier     '75 

Multiplying  Fractional  Numbers  to  the 

Nearest    Unit    191 

Mutual   Insurance   Company,  a.  Defini- 
tion of   SIX 

N 

.National    Bank    Notes    230 

National   Building   Associations    9S2 

Nautical  Mile,  Definition  of  242 


PAGES 

Naval  Officer,  Definition  of   501 

Negotiable   Paper,   Definition  of    549 

Net  Capital  or  Present  Worth,  Defini- 
tion of   809 

Net  Earnings  or  Net  Profits,  Definition 

of     678 

Net  Gain,  Definition  of   S09 

Net  Investment.    Definition   of    739-809 

Net  Loss,  Definition  of  809 

Net  Plan.  Definition  of   981 

Net  Proceeds,   Definition  of   480 

Net  Weight,   Definition   of 502 

New  Hampshire  System  of  Partial  Pay- 
ments   618 

New   Orleans  Building    and    Loan    As- 
sociations, Methods   Used   by    ....  991 
New   Table   in   Interest  Computations.  576 

Nicaragua,    Exchange   on    771 

Nickel,   The    232 

Night  or  Day,  to  Find  Length  of  297 

Nine  Check  Figure  Method,  Errors  not 

Detected    by    153 

Nines,   Properties   of   150-156 

Nominal  or  Ostensible  Partner,    Defini- 
tion of   805 

Nominal  or  Par  Value  of  Stock,  Defini- 
tion of   677 

Nonagon    338 

Non-Forfeiting  Policies,  Definition  of. .  534 
Non-Participating    Policies,     Definition 

of    511 

Non-Trading   Partners,   Definition   of..  804 

Norway,  Exchange  on    773 

Notation  and   Numeration    33-34 

Notation,   Arabic    30 

Notation,   Definition   of    30 

Notation  of  Decimals   216 

Notation,   Roman    30 

Notation,  Roman   System  of    31 

Note,  a  Negotiable  Promissory,  Defini- 
tion of  549 

Note    Discounting  of.  Definition  of....  550 

Note,  Parties  to,  Definition  of   549 

Notes,   a  Difficult   and  Practical   Ques- 
tion in  Renewing   589 

Notes    and    Bills,   Promissory,   Indorse- 
ments   on    550 

Notes  and   Acceptances,   Discounting. .  551 
Notes   and   .Accounts   Bearing  Interest, 

Equation   of    649 

Notes    Bearing    Interest   and    Maturing 
in     Successive     and     Consecutive 

Months,  to  Find  the  Value  of 590 

Notes  Bearing  Interest,  to  Discount...  558 

Notes,  Cash   582 

Notes,  Cash  and  Exchange,  Combined.  732 

Notes,  Cash,  Problems  in   582-590 

Notes,  Equation   of    651 

Notes,  Forms  of   552-554 

Notes,  Interest-Bearing,  Collection  of..  557 


N 


.  IRG 

.  147 

.  30J 

.  334 

.  337 

.  370 

221 

.  220 

.  220 
62-266 

.  256 

.  256 

,  941 


Q 
R 

S 

T 

U 

Y 

W 

Y 


XX 


ALPHABETICAL   INDEX. 


PAGI':S 

Notes,  Points  in  Drawing  559 

Notes,     Promissory      and      Negotiable 

Paper     549 

Notes,  Rent,  Equation  of  648 

Notes  That  Bear  Interest  and  Eventual 

Interest,  to  Collect   580 

Notes  That  Bear  Interest,  to  Discount.  580 

Note,  The  Face  of,  Definition  of   549 

Note,  The  Maturity  of,  Definition  of.  . .  549 
Note,  The  Proteeds  or  Cash  Value  of. 

Definition  of    550 

Number,   Abstract    26 

Number',  Arithmetical  Complement  of.  26 

Number,   Arithmetical   Supplement  of.  26 

Number,    Compound    26 

Number,  Composite,  Definition  of 146 

Number,   Definition   of    25 

Number,   Denominate   or   Concrete    ...  26 

Number,  Imperfect,   Definition   of   ....  146 

Number,  Perfect,  Definition  of  146 

Number,  Powers  of  a    147 

Number,  Prime,  Definition  of   146 

Number,  Reciprocal  of  a  147 

Numbers,  Divisibility  of   147 

Numbers,   How   Expressed    25 

Numbers,  Like    25 

Numbers,  Odd  and  Even  25,  146 

Numbers,   Properties  of  146-149 

Numbers,  Reading   30 

Numbers,  Reduction  of  Denominate     256-258 
Numbers      Relationship     and     Equiva- 
lency       311 

Numbers,  Scale  or  System  of 27 

Numbers,   Unlike    25 

Number,  The  Unit  of  a  25 

Numeration  and  Notation   33-34 

Numeration,    Definition   of    '0 

Numeration,   English   System    31 

Numeration,  French  System   30 

Numeration  of  Decimals   219 

Numerator,  Definition  of   165 

o 

Oblate    Spheroid    370 

Obtuse  Angle,  Definition  of  337 

Octagon     338 

Odd    Numbers    25,146 

Officer,   Naval,  Definition  of    501 

Ohio  or  Dayton  Plan,  Definition  of 980 

Oils   and   Paints   Bill    291 

Old  French  and  Spanish  Measures...  253 
Onen    Endorsement    Policy,    Definition 

of     510 

Opening  Entry,  to  Change  Single  Entry 

Books   to   Double   Entry,   and   Use 

the   Single   Entry  Ledger   854 

Open   New  Books    From    an    Old    Set 

Which  Has  Been  Incorrectly  Kept.  S55 

Open  Policy,  An,  Definition  of  510 


P.\GES 

Operation,  Definition  of  26 

Operations    in    Mixed    Numbers,    Frac- 
tions, Parenthesis  or  Vinculum     211-212 
Operations    Showing    Change    of    One 

Scale  to  Another   28-29 

Order  of  Figures    29-30 

Orders,  Arithmetical  Units  of 26 

Ordinary  Life  Policy,  Definition  of....  533 

Ordinary    Partners,    Definition    of 804 

Oral  and  Written  Problems  in  Division  198 
Oral    Exercises   in   Partnership    Settle- 
ments      Sli 

Origin  of  Dollar  Sign   (?)    38 

Origin   of   Exchange    711 

Ostensible     or     a     Nominal     Partner, 

Definition  of   805 

P 

Painting   and    Glazing    366 

Paints  and  Oils  Bill   291 

Paper  and  Books   254 

Paper,   Negotiable,   Definition   of   549 

Paralellogram,   Definition   of    338 

Parenthesis  and  Vinculum,  Use  of   . . .  134 

Par  of  Exchange,  Definition  of 716 

Par    of    Exchange,    The     Commercial, 

Definition   of    716 

Par      of      Exchange,      The      Intrinsic, 

Definition   of    716 

Par  or  Nominal  Value  of  Stock,  Defini- 
tion of   677 

Partial   Payments  or  Payments  by   In- 
stallments       616 

Partial  Payments,  Problems  in   ...     619-626 
Partial     Payments,     The     Connecticut 

System     618 

Partial  Payments,  The  Merchants  Sys- 
tem       617 

Partial  Payments,  The  New  Hampshire 

System     618 

Partial    Payments,   The   United    States 

System    616 

Partial  Payments,    The    Vermont    Sys- 
tem       617 

Particular  and  General  Average   527 

Particular  Average,  Definition  of   ....  527 
Particular  Partnership,  Definition  of     804-805 

Parties  to  a  Note,  Definition  of  549 

Parties  to  Bills  and  Drafts   ^54 

Partition    Proportion    332 

Partner,    Silent,    Secret,     Sleeping     or 

Dormant,    Definition   of    805 

Partner,  Nominal  or  Ostensible,  Defini- 
tion   of    805 

Partners'  Accounts,  Adjustment  of  In- 
terest   on    S45 

Partners,      Commercial      or      General, 

Definition   of    803 

Partners  in  Commendam,  Definition  of  804 


A. 


ALPHABETICAL   INDEX. 


XXI 


PAGES 

Partners,  Limited,  in  Louisiana,  Defini- 
tion of   804 

Partners,   Non-Trading,   Definition   of. .  S04 

Partners,  Ordinary,  Definition  of S04 

Partners,   Special  Names  Given  to....  S05 

Partnersliip    SOS 

Partnersliip   and  Partnership  Average.  S61 
Partnersliip  Average,  Compound,  Defin- 
ition of 861 

Partnersliip   Average,    Definition    of. . .  S61 
Partnership    Average,    Simple,    Defini- 
tion of   S61 

Partnership,  Commercial,  Definition  of  S0.3 

Partnership,  General,  Definition  of....  805 

Partnership,    Limited,    Definition   of. . .  844 

Partnership  Settlements   808 

Partnership    Settlements,    Terras    and 

Definitions    SOS 

Partnership,  Universal,    Definition  of    S04-S05 
Partnerships,    Articles    of    or    Articles 

of   Agreement    806 

Partnerships,  Dissolution   of    806 

Partnerships,    Formation   of    805 

Partnerships    in    Louisiana,    Definition 

of     803 

Partnerships     in     the     Common     Law 

States,  Definition  of   804 

Partnerships,   Trading   or   Commercial, 

Definition  of    804 

Paving   Yards   and   Walks    360 

Payments   and   Accounts,   Equation   or 

Average  of  627 

Payments,  Interest  on  Interest   595 

Peculiar     Contractions     in     Multiplica- 
tion   of    Fractions    193 

Peculiar   Interest   Questions    567 

Percentage    434-479 

Percentage,  Definitions    Concerning   . .  434 
Percentage,      Miscellaneous      Practical 

Questions    470-479 

Percentage    Tables     436 

Per  Cent.  Definition   of    434 

Perfect  Number,  Definition  of   146 

Permanent  Plan,  Definition  of 980 

Permutations     944 

Pentagon     33S 

Perpetual  Policies,  Definition  of 510 

Perpetuity,    or    a     Perpetual     Annuity, 

Definition    of   920 

Personal   Insurance,  Definition  of   508 

Peru,  Exchange  on   771 

Philosophic  Method  of  Using  the  Fac- 
tors of  the  Interest  Divisor  562 

Philosophic    Solution    26 

Philosophic  System  of  Arithmetic 26 

Philosophic  System  of  Contracting  In- 
terest   Questions    566 

Philosophic   System   of   Multiplication.  77 

Phlloeophy,  Definition  of   26 


PAGES 

Phrases  and  Words  Used  by  Financiers, 
Banliers    and    Brokers,     Definition 

of    682-684 

Pipe  and   Sheet  Lead  Table   363 

Plane,  Figures,  Definition  of   337 

Plane,  Surface,    Definition    of    337 

Plans    of   Building   and    Loan   Associa- 
tions         979-9S1 

Plans   of   Charging   Premiums   and   In- 
terest          980-981 

Plans  of  Distribution  of  Profit   983 

Plans   of  Loaning   Money  or   Premium 

Plans    983 

Plans  of  Withdrawal    9S3 

Plantation    Sygar   Bill    288 

Plasterers'  Work   365 

Plumbers'    Work    363 

Point,  Definition  of  336 

Points  in   Drawing  Notes   559 

Points    to    be    Specially    Observed    in 

Merchandise   Insurance    526 

Policies,     Cancellation     of.     Definition 

of     511 

Policies,  Floating,  Definition  of 514 

Policies,    Non-Participating,    Definition 

of    511 

Policies,  Transferring  of.  Definition  of.  511 

Policy,  An  Endowment,  Definition  of. .  533 

Policy,  Annuity,  Definition  of   533 

Policy,     A     Non-Forfeiting,     Definition 

of     534 

Policy,  a   Valued,   Definition   of    509 

Policy,  Joint  Life,  Definition  of   533 

Policy,  Limited   Payment,  Definition  of  533 

Policy,  Open,   Definition   of    510 

Policy,  Open    Endorsement,    Definition 

of    510 

Policy,  Ordinary  Life,  Definition  of...  '533 
Policy,  Reserve  Endowment,  Definition 

of    533 

Policy,  Term,  Definition  of 533 

Policy,  The   Definition  of    509 

Policy,  Tontine  Savings  or  Investment, 

Definition  of    533 

Policy,   Trust  Certificate,   Definition  of  533 

Policy.  Value  of,  Definition   of   534 

Poll    Tax   or    Capitation    Tax,     Defini- 
tion of   4S9 

Polygon,  Definition  of 337 

Polygons,  Talkie  of   350 

Ports  of  Entry,  Definition  of 501 

Portugal,  Exchange  on 770 

Poultry   and   Eggs  Bill    287 

Pound,   History   of  the  English   Mone- 
tary   740 

Powers  of  a  Number  147 

Practical  Approximative   Contractions.  191 

Practical  Definition    of    Multiplication.  183 

Practical    Problems    278-282 


.   lBt> 

.  147 
.  30J 

Q 

.  334 

.  337 

R 

.  370 

221 

S 

.  220 

\     220 

T 

262-266 

.  256 
.  256 

U 

c 

.  941 

V 

w 

Y 

XXII 


ALPHABETICAL  INDEX. 


PAGES 

Practical     Problems,     Contractions     in 

96-lUl,   276-2S3 

Practical   Problems   in   Mensuration   of 

Solids    393-416 

Practical  Questions  in   Percentage.      470-479 
Preference  or  Preferred   Stock,  Defini- 
tion of   677 

Preferred    Creditors,    Definition   of. . . .     4S7 
Premises   or  Conditions   of  a   Problem       26 

Premium,  Definition  of    493 

Premium  or  Discount,  Definition   of.  .  .     978 

Premium   Rates,   Table   Showing    539 

Premium,  Return,  Definition  of   510 

Premium,  The  Rate  of,    Definition  of    510-534 
Present    Worth     and    Compound    Dis- 
count         918 

Present  Worth,   Definition  of   596 

Present  Worth,  or  Net  Capital 809 

Present  Worth  or  Value,  Definition  of.     920 

Prime   Bills,   Definition  of    716 

Prime  Factor,  Definition  of   146 

Prime  Number,  Definition  of 146 

Principal    and   Interest,    Equal    Annual 

Payments   of' 588 

Principal,   Definition  of   546 

Principal   Meridian,   Definition   of   ....     245 

Principle  of  Addition   40 

Principle   of    Subtraction    of   Fractions     ISO 

Principles  of  Cancellation    161 

'Principles  of  Division     123 

Principles  of  Division  of  Decimals  ....     227 

Principles  of  Multiplication    75 

Principles  of  Subtraction    64 

Printing    Paper     254 

Probabilities   or   Chance    945 

Probabilities      or      Chance,      Problems 

in     946-948 

Probability  of  Human  Life    542 

Problem,    Definition    of    26 

Problem,  Premises  or  Conditions  of   .  .       26 

Problem,  Solution  of  a    26 

Problems  in  Accounts  Current  and  In- 
terest   Accounts    G53-663 

Problems  in  Addition    46-62 

Problems  in  Adjustment  of  Losses. .  .523-527 

Problems  in   Alligation    792-802 

Problems  in  Annuities    921-927 

Problems  in  Arbitration  of  Ex- 
change         736-739 

Problems  in  Arithmetical    Progression 

901-907 

Problems    in    Bankers    and    Merchants 

Discount     577-582 

Problems     in     Bankers'  -  Interest     on 

Daily    Balances    665-668 

Problems  in  Bankruptcy   4S8 

Problems  in  Building  and  Loan  Associ- 
ations          989-1004 

Problems  in  Cash  Notes 582-590 


PAGES 

Problems  in  Combinations  or 

Choice     943-944 

Problems   in   Commission    and   Broker- 
age          481-486 

Problems    in    Compound    Interest.  .      609-616 
Problems  in  Continued  Fractions.  .     949-950 

Problems,   Cotton   Average    8GS-S71 

Problems     in     Curious     and     Amusing 

Questions     974-975 

Problems     in     Customhouse     Business 

502-507 

Problems  in  English  Exchange    .  . .     745-758 
Problems  in  Equation  of  Accounts.     628-646 

Problems  in   Exchange    717-735 

Problems  in  French  Exchange 759-764 

Problems  in  Geometrical  Pro- 
gression          908-919 

Problems    in    General    and    Particular 

Average     528-531 

Problems    in   German    Exchange...     765-767 
Problems  in  Gold,  Currency,  Silver  and 

Uncurrent  Money    493-49& 

Problems  in  Insurance   515-527 

Problems    in    Interest    562-576 

Problems    in    Interest,     Complex     and 

Expert     588 

Problems  in  Involution    884 

Problems  in   Life  Insurance   540-542 

Problems  in  Linear  Measure 340-346 

Problems  in  Loan  and  Investment  Com- 
panies         675 

Problems  in  Metric  System  of  Weights 

and    Measures    934-942 

Problems  in  Multiplication    75-104 

Problems  in   Partnership   Settle- 
ments         810-882 

Problems  in   Proportion    317-335 

Problems  in  Savings    Banks    and    Sav- 
ings Bank  Accounts    6G9-674 

Problems  in  Square    and     Cube     Root 

899-900 

Problems  in  Stocks  and  Bonds 6S4-709 

Problems  in   Storage    779-7SS 

Problems  in  Subtraction    71-73 

Problems  in  Taxes  and  Licenses  ....490-492 

Problems   in   True   Discount    596-604 

Problems   involving   Addition,   Subtrac- 
tion, Multiplication  and  Division  of 

Fractions     206-212 

Problems  Miscellaneous  ..951-974,  1014-1025 

Proceeds,  Net,  Definition  of   480 

Proceeds    or    Cash    Value     of     Notes, 

Definition   of    550 

Proceeds,    The    Definition    of 547 

Produce   Bill    293 

Produce,   Weight    of   Per   Bushel    ....     255 

Product,   Definition   of    75 

Product  Method,  Accounts  Current  and 

Interest  Account    653- 


ALPHABETICAL   INDEX. 


XXIII 


PAGES 

Profit  and   Loss    450 

Profit  or  Gain   451 

Profits,  Net,  or  Net  Earnings,  Defini- 
tion of     678 

Progression,  Arithmetical   901 

Progression,    Geometrical    90S 

Progression,  Geometrical  Applications 
of  to  Compound  Interest  and  Com- 
pound   Discount    917 

Prolate   Spheroid    370 

Promissory  Note,  Negotiable,  Defini- 
tion of   549 

Promissory  Notes  and  Bills,  Indorse- 
ments  on    550 

Promissory      Notes      and      Negotiable 

Paper   549 

Proof  of  Addition   47,  152 

Proof  of  Addition  by  Casting  Out  the 

Nines  and  Elevens    '. 152 

Proof  of  Division   by   Casting   Out   the 

the  Nines  or  Elevens   154 

Proof  of  Multiplication   75,  154 

Proof  of  Multiplication  by  Casting  Out 

the  Nines  or  Elevens   154 

Proof  of  Subtraction   64,  153 

Proof  of  Subtraction    by    Casting    Out 

the   Nines   or  Elevens    153 

Proper  Fraction,  Definition  of 165 

Properties  of  Nines   and   Elevens    ..150-156 

Properties  of  Numbers   146-149 

Property  Insurance    Definition  of   ....     508 

Property,  Personal,  Definition  of   489 

Property,  Real,  or  Real  Estate,  Defini- 
tion of   

Property  Tax,  or  Direct  Tax,  Defini- 
tion of   

Property,  Valued  Policies  on  Immov- 
able      

Proportion 310 

Proportion,  Medial  or  Medial  Alliga- 
tion          789 

Proportion,  Problems    in     317-335 

Proportion,  Solutions    in     312-316 

Proportional    Division     876 

Proportional    Division    and    Division...     871 

Protesting,    Definition    of    550 

Pulley     424 

Purchase,  an  Invoice  or  Account  of.  De-. . 

finition    of     480 

Pure  or  Simple  Decimal,  Definition  of  216 
Puzzling  Problem  in  Percentage  ....  453 
Pyramid,  Definition   of    370 


489 


489 


526 


Q 


Quadrant,  Definition  of 252 

Quadrilateral  or  Rectangular  Solid   .  . .  370 

Quantitative   Chemical    Analysis    467 

Quantity.   Definition   of    25 


PAGES 

Quarterly,  Seml-Annual  and  Annual  In- 
terest      607 

Quartenary  Scale,  Reduced  to  Decimal  28 

Questions  in   Division   197 

Questions  in   Percentage    47C-479 

Questions,    Curious    and    Amusing    ...  974 

Questions,  Peculiar   Interest    567 

Questions,  Philosophic  System  of  Con- 
tracting Interest    566 

Quinary  Scale,  Reduced  to  Decimal..!  28 

Quotient    ^23 

R 

Radius,   Definition  of   252 

Radius  of  a   Sphere 370 

Radix,    Definition    of ' . .  27 

Range    (Land)    Definition  of    .....  245 

Rate  of  Exchange,     744 

Rate  of  Exchange,   Definition   of    ."."..".  716 

Rate  of  Interest   543 

Rate  of  Interest  Received  and  Paid  . '.  562 
Rate  of  Premium,  The,  Definition  of.. 

510-534 

Rates,  Annual  Premium  Table  539 

Rates  of  Storage  Charges,  Definition  of  777 

Rate.s,   Short,   Definition  of    510 

Ratio     308-309 

Ratio,  Definition   of    338 

Read  and  Write  Metric  Numbers   934 

Reading   Numbers    30 

Reading   and   Writing   Numbers    33-34 

Real   Property  or   Real   Estate,   Defini- 
tion of   489 

Reasoning  for  the  Division  of  Abstract 

Numbers     2OO 

Reaumur    Thermometer    274 

Receipts,  Forms  of  551-552 

Receipts,    Storage   or    Warehouse,    Re- 
ceipts, Definition  of   777 

Receiver  or  Collector  of  Taxes,  Defini- 
tion of  489 

Reciprocal   of  a  Fraction    166 

Reciprocal  of  a  Number    147 

Reciprocal  of  a  Ratio  30S 

Reciprocal    Proportion    334 

Rectangle.   Definition   of 337 

Rectangular  or   Quadrilatral   Solid    ...  370 
Reducing   Common    Fractions   to   Deci- 
mals       221 

Reducing  Decimals   to  a  Common   De- 
nominator      220 

Reducing   Decimals    to   Common    Frac- 
tions       220 

Reduction,    Ascending    262-266 

Reduction,  Definitions    Concerning    ...  256 

Reduction,  Descending    256 

Reduction     of     .-American     to     Metric 

Weights  and   Measures    941 


Q 
R 


U 


W 
Y 


XXIV 


ALPHABETICAL   INDEX, 


PACES 

Reduction  of  Decimals   220-221 

Reduction  of  Denominate  Numbers     25G-25S 
Reduction    of     Denominate     Fractions. . 

259-261 

Reduction  of  Fractions   167-174 

Reduction  of  Metric    Numbers    935 

Reduction     of     Metric     to     American 

Weiglits  and   Measures    940 

Refined   Sugar  Bill    289 

Refunding  Certficates,  Definition  of   . .  6S1 

Registered  Bonds,  Definition  of   ......  6S0 

Reinsurance    511 

Relationship  and  Equivalency  of  Num- 
bers       311 

Remarks  in  Proportion    315 

Remarks   on    Annuities    921 

Renewing  Notes,  a  Difficult  and   Prac- 
tical   Question    in    589 

Rent  and  Services,  Bill  for   290 

Rentes,  French,  Definition  of   6S0 

Rent  Notes,  Equation  of   648 

Repetends.  Del^nition  of   217 

Reserve.     Contingent     or     Redemption 

Fund,  Definition  of   678 

Reserve  Endowment   Policy    Definition 

of     533 

Reserve   Fund,   The,   Definition   of    ...  534 

Reserve,    The  Definition  of   534 

Resources,    Assets    or    Effects,    Defini- 
tion   of    808 

Resources   or   Assets,   Definition   of. . .  4S7 
Respondentia  and  Bottomry,  Definition 

of    512 

Return   Premium,   Definition   of    510 

Revenue,   Internal,   Definition  of    502 

Rhombus,  Definition  of    338 

Rice  Bill   2S8 

Right  Angle,  Definition  of  337 

Right-Angled  Triangle,  Definition  of  . .  337 

Risk,   Definition    of    510 

River  Insurance,  Definition  of 508 

Roll,  Assessment,   Definition  Of   4S9 

Roman  Characters,    Table    of    32 

Roman    Notation    30 

Roman  Svstem  of  Notation   31 

Root,    Cube    891 

Roofs,   Slating  and   Shingling    364 

Root,  Snuare,  Extraction  of  SS5 

Round  Timber   3S5-3S8 

R.  R.  Currency  Sixes,  U.  S.  Pacific,  De- 
finition   of    '^'^'^ 

Russian   Exchange    7''1 

s 

Sales,  an  Account,  Definition  of 4S0 

Sales,  Equation  of  Account    643 

Sales  on  Joint  Account   (547-663 

Salvador,    Exchange   on    "771 

Salvage,  Definition  of   615 


PAGES 

Savings   Banks  and  Savings  Bank  Ac- 
counts         669 

Savings    Banks     Definition   of    669 

Savings   or   Investment   Policy,  a   Ton- 
tine,  Definition  of   533 

Saw   Log,   Standard    3S4-3SS 

Scale    Beam    420 

Scalene   Triangle,  Definition  of    338 

Scale  or  System  of  Numbers  -    27 

Schedule,   Definition   of   487 

Science,   Definition   of    !..       25 

Screw    431 

Scrip,   Insurance,   Definition   of    512 

Scrip,  Definition  of   534 

Seasons,    The    239 

Section,  Definition  of  j;4o 

Sector,    Definition   of    252 

Segments  and  Areas  of  a  Circle,  Table 

of    ....■ 353-354 

Segment,   Definition   of    252 

Segments,    Remarks    on    Measurement 

of    382 

Semi-annual,     Annual     and     Quarterly 

Interest    607 

Semi-Circumference,  Definition  of   252 

Serial    Plan,   The   Definition   of    979 

Services,  Bills  for  289-290 

Settlements,    Partnership    808 

Settlements    with    Special    Conditions.     872 

Settling  Estates    874 

Seventy-five     Per     Cent    Co-Insurance 

Clause,  Definition  of  513 

Sextant,  Definition  of  252 

Shares  and  Tollage    Manufacturing  on     785 

Shares,   Definition   of   978 

Sharing  Gains  and  Losses  803-861 

Shingling  and  Slating  Roofs   364 

Shoe    Bills    286 

Shoemakers'    Measure    242 

Short   Cuts   in   Division    138-142 

Short  Cuts  in  Multiplication 88-121 

Short   Division    128 

Short  Rates,   Definition   of   510 

Short  Sight  Bills,  Definition  of   736 

Sign  of  a  Ratio  308 

Sign,   Definition   of    252 

Sign  of   Per   Cent 434 

Sign   of  Proportion    310 

Signs  and  Abbreviations    3S-39 

Signs  and  Symbols  36-37 

Silent,    Secret,    Sleeping    or    Dormant 

Partner,   Definition   of    805 

Silver  and  Gold  Comparative  Values..     233 

Silver  and    Gold,   Value    of    234 

Silver  Certificates     230 

Silver  Coin 232 

Silver,   Currency,   Gold   and  Uncurrent 

Money    493 

Silver,   Currency,   Gold   and  Uncurrent 

Money,   Problems   in    493-499 


ALPHABETICAL  INDEX. 


XXV 


Silver   Dollars     Standard    230 

Simple,  Annual  and  Compound  Interest 

Compared     615 

Simple   Arbitration,   Definition   of    ....     736 

Simple  Decimal,  Definition  of  216 

Simple    Equation    or    Average,    Defini- 
tion of   627 

Simple   Fraction,   Definition   of    165 

Simple   Interest,  Definition  of   547 

Simple  Partnership  Average   861 

Simple   Partnership  Average  Problems 

861-S63 

Simple   Proportion,  Definition   of 310 

Simple    Proportion,   Problems   in 311 

Simple    Ratio    303 

Simple    Value    27 

Simultaneous  or  Cross  Multiplication     SS-93 

Sizes  of  Paper   254 

Slating  and  Shingling  Roofs   364 

Slaters'    Bill    295 

Smuggling,  Definition  of  501 

Solar  Day,   Definition  of   236 

Solid,  Body  or  Volume,  Definition  of. .     337 

Solid,  Definition    of    241-247 

Solid  or  Cubic  Measure  247 

Solidity  or  Contents  of  a  Solid   370 

Solutions   in   Proportion    312-316 

Solution    of  a  Problem    26 

Solution,  Philosophic    26 

Solution,  Statement    or    Operation    ....     26 

Spain,   Exchange  on    764 

Spanish   (Old)   System  of  Measures   . .     253 
Special  Laws  and  Business  Customs  in 

Interest  and  Discount 556 

Special  Names  Given  to  Partners  ....     805 

Specific   Duty,   Definition   of    500 

Sphere,   Definition   of    370 

Spheroid,   Oblate   and   Prolate    370 

Square,  Definition    of    243-337 

Square  of  Earth,  Definition   of   247 

Square  or  Surface  Measure   243 

Square  Root,  Applications    of    888 

Square  Root.  Extraction   of   SS5 

Square  Root,  Problems    in    899-900 

Sip'.pring    Numbers    US 

Standard   Board    3SS 

Standard   Meter,   Definition   of    92S 

Standard   Silver   Dollars    230 

Standard  Unit  of  Circular    Measure.  .     252 

Standard  Unit  of  Troy   Weight    249 

State  and  United  States  Leagues   9S2 

Statements,  Formula  of  739 

Statement  Line   161,  184 

Statement  Solution,  Definition  of   26 

Steelyard    421 

Stere  or  Cubic  Meter,  Definition  of . . . .     930 
Stock  and  Mutual  Insurance  Company, 

a  Mixed,  Definition  of  511 


P.\GES 

Stock,  Capital,  Dividend,  a  Forced 
Cash  or  Fictitious  Dividend,  De- 
finition  of    678 

Stock,  Certificates   of.   Definition   of...  676 

Stock,  Clandestine    679 

Stock  Company  or  Corporation,  Defini- 
tion   of    676 

Stock  Company,  The  Stock  or  the  Capi- 
tal   Stock   of.   Definition   of    676 

Stock,  Debenture,    Definition    of    677 

Stock,  Exchanges     681 

Stock,  Forfeited,   Definition   of    677 

Stock,  Guaranteed,   Definition  of   677- 

Stock  Insurance  Company,  A,  Defini- 
tion   of    511 

Stock,  Live,  Insurance,  Definition  of. .  508 
Stock    Preference  or  Preferred,  Defini- 
tion of  677 

Stock-taking    or     Inventorying   Goods 

Definition   of    527 

Stock,  The     Intrinsic     or     Liquidation 

Value  of,   Definition  of    677 

Stock,  The  Market  Value  of   677 

Stock,  The   Par  or  Nominal   Value  of. 

Definition   of    677 

Stock,  to  find  the  Margin  to  Carry 704 

Stock.  Watered,  Definition  of   678 

Stock,   Value   of   591 

Stocks    and    Bonds    676 

Stocks  and  Bonds,  Problems  in   . .     084-709 

Stocks,   Definition  of    676 

Storage    777 

Storage,  Average,   Definition  of   778 

Storage,  Cash,   Definition   of    777 

Storage,  Cold,    Definition   of    777 

Storage,  Credit  or  Time,  Definition  of  777 

Storage,  Daily   Balance  Method    779 

Storage,  Problems    in    779-788 

Storage,  Receipts  or  Warehouse  Re- 
ceipts,  Definition  of   777 

Storage,  The  Rates  of  777 

Store-Keeper,  The,  Definition  of   502 

Straight  Line,   Definition   of    336 

Subject  of  Insurance.  The    510 

Sub-Partner,  Definition  of 805 

Subtraction     64-74    ■ 

Subtraction,  Borrowing  Method    66 

Subtraction,  Borrowing  Method,  Im- 
proper      66 

Subtraction   by  Addition    66 

Subtraction   by   Casting   Out  the  Nines  C 

or  Elevens,  Proof  of   153 

Subtraction    by    Complement    of    10..  69 

Subtraction.  Commencing  at  the  Left.  69      -p 

Subtraction,  Demonstration  of 65-66       ' 

Subtraction,  Dollars  and  Cents    67-68 

Subtraction,  General  Directions  for   . .  67     M 

Subtraction,  Horizontal    68 

Supplement  of  a  Number,  Arithmetical  26 

w 

Y 


XXVI 


ALPHABETICAL   INDEX. 


PAGES 

Subtraction  of  Compound  Denominate 

Numbers     2G9-270 

Subtraction    of   Decimals    224 

Subtraction   of  Fractions    1S0-1S2 

Subtraction   of  Metric  Numbers   937 

Supplement  of  Number,   Multiply  by..     108 
Subtraction,  Miscellaneous    Problems     71-73 

Subtraction,  Proof    of     64-153 

Subtraction  Table    64 

Sub-Multiple  Units  of  the   Metric   Sys- 
tem          930 

Sucrose  Multiplier   467 

Sugar  Bills    2S5,  288,  289,  293 

Sugar  Cane,  to  find  tlae  Value  Per  Ton     466 

Surface,  Definition  of   241,  243,  337 

Surface  or  Square  Measure   243 

Surplus,  Definition  of    .512-534 

Surrendered   or   Market   Value,   Defini- 
tion   of    534 

Surveyor,  The,  Definition  of    501 

Surveyors'  and    Engineers'    Measure..     244 

Surveyors'  Square    Measure    245 

Sweden,    Exchange    on     '773 

Switzerland,   Exchange  on    764 

Symbols    and    Signs    36-37 

Synopsis  for  Review  ....35,  63,  74,  121,  142 

149,  IGO,   163,  166. 
Synopsis  for  Review  of  Cancellation  . .     163 
Synopsis   for    Review   of   Factoring    ..     ItiO 

Synopsis  for  Review  of  Fractions   166 

System  of  Addition     Vicenary    56 

System  of  Check   or   Key    Figure    155 

System   of   Contracting   Interest   Ques- 
tions,  The    Philosophic    566 

System  of  Multiplication,    Philosophic.       77 

System  of  Notation     31 

System  of  Notation  .Roman   31 

System   of    Numeration    30-31 

System  of  Numbers    27 

System  or  Scale  of  Numbers   27 

T 

Table,    American   Experience    of   Mor- 
tality      541 

Table,     Annual     Premium     Rates 539 

Table,     Comparative      Expectancy      of 

Life     537 

Table,     Comparative     Mortality   538 

Table  Comparing    Weights,    Measures 

and    Values    254 

Table,  Multiplication     Te 

Table    of   Aliquots    9* 

Table  of  Areas    and    Segments     of    a 

Circle    353-354 

Table  of  Assayers*   Weight    251 

Table  of  Apothecaries   Weight    250 

Table  of  Avoirdupois  Weight   250 

Table  of  Centuries     299 


PAGES 
Table  of  Chain  or  Engineers'  and  Sur- 
veyors'   Measure    244 

Table  of  Circular    Measure    252 

Table  of  Cloth    Measure    243 

Table  of  Contracted    Interest    Divisors  561 

Table  of  Cubic  or  Solid   Measure    ....  247 

Table  of  Diamond   Weight   251 

Table  of  Dry    .Measure     248 

Table  of  English    Money     234 

Table  of     Foreign  Coins   743 

Table  of  Fluid    Measure    251 

Table  of  French    Money    235 

Table  of  German    Money    236 

Table    of   Government    Land    Measure.  245 

Table  of  Interest    Divisors     560 

Table  of  Interest    Laws     548 

Table  of  Linear    Measure     242 

Table  of  Liquid   or  Wine   Measure....  247 

Table  of  Longitude  and  Time   303 

Table  of  Mariners'    Measure     242 

Table      of      Miscellaneous      Units      of 

Linear   Measure    243 

Table  of    Miscellaneous    Weights     and 

Measures     255 

Table  of  Months     299 

Table  of  Multiples     339 

Table  of   New   Interest  Computations.  576 
Table  of  Old  French  Cubic  and  Square 

Measure     253 

Table  of  Old   French   Measure    253 

Table  of  Old   Spanish   Measure    253 

Table  of  Polygons     350 

Table  of  Roiuan   Characters    32 

Table   of    IGths    and    32nds    and    Their 

Equivalents    in    Decimals    870 

Table  of  Square  or  Surface  Measure..  244 

Table  of  Squares   and   Cubes    118 

Table  of  Surveyors'    Square    Measure.  245 
Table     of     Thirty-seconds     and     Their 

Equivalent  Values  in  Decimals    192,  870 

Table   of  Time   Measure    239 

Table  of   Troy   Weight    249 

Table  of  Unearned   Premiums   522 

Table  of  United  States  Money   231 

Table  of  United  States    Coin    232 

Table    of   Whole    Numbers    and   Deci- 
mals       218 

Tables,  Addition    41 

Tables,  Addition   and   Subtraction   ....  41 

Tables,  Compound  Interest 611-612 

Tables,    Measure    231 

Tables,  Metric     931 

Tables    of  Annuity    925-926-927 

Tables   of  Building  and  Loan  Associa- 
tions       978 

Tables  of  Equivalents   932-933 

Tables  of  Trade    Discounts     464-466 

Tables.    Percentage    436-454 

Table  Showing  in  How  Many  Years  a 

given  Principal  Will  Double  itself  604 


ALPHABETICAL  INDEX. 


XXVII 


PAGES 

Table  Showing  the  Price  of  Single  arti- 
cle When  the  Dozen  Price  is  Given     192 
Table  Showing  Time  in  Days  Between 

Two   Dates    29S 

Table,  Subtraction     64 

Taking  Stock  or  Inventorying  Goods..     527 

Tare,   Definition   of    502 

Tariff,  Definition  of  501 

Taxes  and    Licenses     489 

Taxes  and  Licenses,  Problems  in..      490-492 
Taxes,  Collector  or  Receiver  of.  Defini- 
tion of  489 

Taxes,  Definition    of     4S9 

Tax,  Indirect  or  Excise,  Definition  of  489 
Tax,  Poll  or  Capitation,  Definition  of.  4S9 
Tax,  Property  or  Direct,  Definition  of.     489 

Tax,  U.    S.   Income,   Definition   of    489 

Term  Policy,  Definition  ot    53.3 

Term,  The  Interest,  Defiintion  of   669 

Terminate  Decimal,  Definition  of   i'lb 

Terminating  Place,  The,  Definition  of.  iiT'J 
Terms   and   Definitions   in   Partnership 

Settlements    808 

Terms  of  a  Ratio   308 

Terms  of   Proportion    310 

The  Appraiser,  Definition  of   502 

Theorem,    Definition    of    26 

Thermometers,    Comparison    of    274 

The  Surveyor,  Definition  of  501 

Thirty-six  Per  Cent  Method  of  Comput- 
ing   Interest    573 

Three-Fourths  Value  Clause,  Definition 

of    514 

Three-Quarter  Loss,  or  Country  Clause, 

Definition    of     514 

Timber,  Measurement    of    384 

Timber,  Round     385-388 

Time  and  Longitude   303-307 

Time,  Definition    of    236-546 

Time  Drafts,   Discount   on    732 

Time  Exchange    730 

Time,  Measure   of    236 

Time    Measure,    Table    of    239 

Time  or  Credit  Storage,  Definition  of.     777 

Time    Sheets    290-291 

Time,  The  Debit  or  Credit  af,  Defini- 
tion   of    627 

Time,  The  Equated,  Definition  ot 627 

Time,    to    Find    the    Interval    Between 

Two  Dates 296-300 

Tobacco   Bills    2Su 

To  Change  One  Scale  to  Another  . .     28-29 

To  Multiply  by  Complement   108 

To  Read  or  Numerate  Numbers 33 

To  Verify  Notation  or  Writing   33 

To  Write  or  Notate  Numbers   33 

Tollage  and  Shares,  Manufacturing  on    785 

Ton,  Long,  Definition  of   502 

Tonnage  Duty,  Definition  of 501 

Tonneau  or  Ton,  Definition  of 933 


PAGES 
Tontine  Savings  or  Investment  Policy, 

Definition    of    533 

Township.    Definition    of    245 

Trade,  Balance  of   711 

Trade  Discounts   aud    List   Prices    ....     458 

Trade  Discounts,  Tables    of    464-466 

Trade    Dollar    233 

Trading   or   Commercial    Partnerships, 

Definition   of    804 

Transferring    Policies,    Definition    of.  .     511 

Transit  Insurance,   Definition   of   508 

Trapezium,    Definition    of     338 

Trapezoid.    Definition    of    338 

Travelers'    Checks,    American    Express 

Co 715 

Triangles,  Definition  of   338 

Trinary    Scale     27 

Troy  or    Mint    Weight    249 

Troy  W^eight,   Definition    of    249 

True  and  Bank  Discount  Compared  . . .     604 

True  Discount     596 

True  Discount,   Definition   of   696 

True  Discount,  Problems  in   596-604 

Trust  Certificate   Policy,  Definition  of    533 

u 

Ullage    Rods,    Use    of    413 

Uncurrent  Money,  Definition  of   493 

Uncurrent  Money,  Gold,   Currency  and 

Silver    493 

Uncurrent  Money,  Gold,  Currency  and 

Silver    Problems    493-499 

Unearned  Premiums,  Table  of   522 

Unification  of  the  Money  of  the  World     776 

Unit    Definition   of    25,164 

Unit,  Fractional     25 

Unit    of    a    Number    25 

Unit  of  Dry    Measure     248 

Unit  of  Measure  for  Solids   370 

Unit  of  Time    236 

Unit  of  Troy    Weight 249 

Unit  of  Wine   Measure    247 

United   States  and  State  Leagues   9S2 

United   States   Bank   Notes    230 

United  States  Bonds     680 

United  States  Bonds,  Interest    on     569 

United  States  Government  Lands   ....     245 

United   States   Money    231-232 

United  States  System   of    Partial    Pay- 
ments          616 

Units,    Monetary,    of    Different    Ages, 

Nations  and  Peoples 230-231 

Units  of  Orders,  Arithmetical   26 

Universal   Formula   for   Computing   In- 
terest         561 

Universal  Partnership,    Definition  of   804-805 

Unlike    Numbers    25 

Usury,  Definition    of    547 

Usury  Laws,  Evil   Effects  of   545 


T 
U 
V 

w 

Y 


XXVIII 


ALPHABETICAL  INDEX. 


PAGES 

V.  S.  Income  Tax,  Definition  of   489 

U.  S.  Pacific  R.  R.  Currency  Sixes,  De- 
finition   of    6S1 

U.  S.  Stoclvs  and  Bonds  in  Europe  ....  707 

V 

Value,  Definition    of    229 

Value,  Final   or  Amount,   Definition   of  920 

Value,  Local     27 

Value    of   a   Fraction    16G 

Value  of  a  Policy,  Definition  of 534 

Value  of  English  Money  in  U.  S.  Mone- 
tary  Unit    235 

Value    of    Figures    27 

Value  of  Gold  and  Silver  234 

Value   or   Importance   of  Insurance    . .  514 
Value  of  Notes    Bearing    Interest    and 
Maturing   in    Successive   and    Con- 
secutive   Months    590 

Value  of  Notes,  Proceeds  or  Cash,  De- 
finition of   550 

Value   of    Stock    591 

Value  of  Stock,    Intrinsic    or    Liquida- 
tion,   Definition    of    677 

Value  of  Stock,    Par   or    Nominal,    De- 
finition   of    677 

Value,  or  Present  Worth,  Definition  of  920 

Value,  Simple     27 

Value,    The     Surrendered    or    Market, 

Definition    of    534 

Values,    Measures    and   Weights,    Com- 
pared      254 

Values   of   Gold   and    Silver,   Compara- 
tive      233 

Valued    Policies    on    Immovable    Prop- 
erty,   Definition   of    525 

Valued  Policy,  Definition  of  509 

Vermont  System  of  Partial  Payments.  617 
Vicenary  Addition,    Contracting    Meth- 
ods   58 

Vicenary  Scale,    Changed    to    Decimal  29 
Vicenary    System    of    Addition,    Eluci- 
dated          56-57 

Vinculum  and  Parenthesis,  Use  of 134 

Volume  or  Contents,  Definition  of 247 

Volume,  Solid  or  Body   337 

\v 

Wainscoting    361 

Walks  and  Yards,  Paving 360 


PAGES 

Warehouses,  Bonded,  Definition  of... 

501-502,  777 

Warehouse,  Grain,   Definition   of    777 

Warehouse  Receipt  or  Storage  Receipt, 

Definition   of    777 

Watered  Stock,  Definition  of 678 

Wedge 429 

Week,  Day  of,  on  Which  an  Event  Did 

or  May  Occur   299 

Week,   Derivation   of    239 

Weight,  Definition    of    229 

Weight,  Gross,    Definition    of    502 

Weight,    Measures   of    249-255 

Weight    Net,   Definition  of    502 

Weight  of  Coin  233 

Weight   of    Grain     and     Produce,     Per 

Bushel    255 

Weights  and  Measures,  Metric  System 

of     928 

Weights,  Measures  and  Values  Com- 
pared       254 

Wheel  and  Axle   422 

Whole    .\mount    of    Investment    739 

Wine   Measure    247 

Withdrawal,   Plans   of 9S3 

Withdrawals,   Definition   of    979 

Wood    and    Coal    Bill    287 

Words  and  Phrases  Used  by  Finan- 
ciers, Bankers  and  Brokers,  Defini- 
tion  of    6S2-6S4 

World,  Unification  of  the  Money  of  the  776 
Worth,    Present,    and    Compound     Dis- 
count      918 

Worth,  Present  or  Net  Capital   809 

Worth,  The  Present,  Definition  of 596 

Write  and  Read  Metric  Numbers   ....  934 

Writing   and   Reading   Numbers    33-34 

Writing,    Commercial    Instruments    of  551 

Writing   of   Decimals    219 

Written  and  Oral  Problems  in  Division  198 
Written   Problems   in   Partnership  Set- 
tlements   810 


Yard,  Definition  of   241 

Yards  and  Walks,  Paving   360 

Year,  Brief  History  of  237 

Year,  Definition    of    235 

Year.  Derivation   of    239 


Souk's  Philosophic  Practical  Mathematics. 


e^finitions. 


^^^^ 


1.  A  Defiuition  is  the  meaning  or  import  of  a  word  or  words  expressed  by 
other  words. 

2.  Science  is  cdassifled  knowledge. 

3.  Art  IS  the  practical  application  of  the  principles  of  science,  according  to 
prescribed  methods. 

4.  Quantity  is  anything  that  can  bo  increased  or  diminished. 

5.  A  Unit  is  a  single  thing  of  whatsoever  deuomiuatiou  or  nature,  as  one 
orange,  one  jjound,  etc. 

6.  A  Number  is  a  unit  or  a  collection  of  like  units. 

7.  Like  Numbers  are  those  which  express  units  of  the  same  kind.  Thus: 
Five  apples  and  six  apples,  seven  pounds  and  nine  pounds,  are  like  numbers. 

8.  Unlike  Numbers  are  those  which  express  units  of  different  kinds.  Thus: 
Four  pounds,  seven  hours,  ten  peaches,  are  unlike  numbers. 

9.  Odd  and  Even  Numbers.  An  odd  number  is  one  that  cannot  be 
divided  by  2  without  a  remainder,  as  1,  3,  5,  7,  35,  479,  etc.  An  EVEN  NUMBER  is 
one  that  can  be  divided  by  2  without  a  remainder,  as  2,  6, 18,  54,  530,  etc. 

10.  Tlie  Unit  of  a  Number  is  one  of  the  collection  of  units  forming  that 
number.  Thus  the  unit  of  twelve  hats  is  one  hat;  of  Jive  is  one;  o?  four  ijounds, 
one  pound. 

11.  The  Unit  is  the  universal  basis  of  numbers  and  the  foundation  of 
arithmetic.  From  wniij/ arise  two  distinct  classes  of  number :  1.  Integers,  2.  Frac- 
tions. The  first  class.,  Integers,  has  its  origin  in  the  multiplication  of  the  unit: 
and  the  second  class.  Fractions,  results  from  the  division  of  the  unit.  The  first  is 
synthetical,  the  second  is  analytical. 

12.  A  Fractional  Unit  is  one  of  the  equal  parts  into  which  any  integral 
unit  is  divided.  If  the  integral  unit  is  divided  into  two  equal  parts,  each  is  called 
a  half;  if  into  three,  each  is  called  a  third;  if  into  four,  each  is  called  a  fourth; 
and  so  on  according  to  the  number  of  i^arts  into  which  the  integral  unit  is  divided. 
See  Fractions,  page  164. 

13.  Numbers  are  expressed  by  words,  figures,  or  letters. 

(25)  Y 

W 
Y 


26  ,     soule's  philosophic  practical  mathematics.  * 

H.  An  Abstract  Number  is  one  in  which  the  kind  of  unit  or  quantity  is 
not  designated.    Thus :  Three,  four,  five,  etc. 

15.  A  Denominate  or  Concrete  Number  is  one  in  which  the  kind  of  unit 

is  designated.     Thus:  Two  pounds,  iive  yards,  nine  dollars,  etc 

16.  A  Compound  Number  is  a  denominate  number  expressed  in  two  or 
more  denominations.  Thus:  Five  years,  four  months,  and  eight  days;  two  miles, 
five  furlongs,  and  ten  rods  ;  two  yards,  two  feet,  and  five  inches. 

17.  The  Arithmetical  Units  of  Orders  are  1,  10, 100, 1000,  10000,  etc. 

18.  An  Arithmetical  Complement  of  a  Number  is  the  difference  between 
the  number  and  a  unit  of  the  next  higher  order.  Thus:  3  is  the  arithmetical  com- 
plement of  7  ;  26  is  the  arithmetical  complement  of  74 ;  19  is  the  arithmetical  com- 
l)]ement  of  981. 

19.  An  Arithmetical  Supplement  of  a  Number  is  the  difference  between 
the  number  and  a  unit  of  the  same  order.  Thus  :  7  is  the  arithmetical  supplement 
■of  17 ;  12  is  the  arithmetical  supplement  of  112. 

20.  A  Problem  is  a  qtiestiou  proposed  or  given  for  solution,  or  it  is  a  ques- 
tion requiring  some  unknown  result  from  stated  conditions. 

21.  The  Premises  or  Conditions  of  a  problem  are  the  known  facts  and 
truths,  therein  given  as  a  basis  or  predicate,  for  the  solution. 

22.  A  Solution  of  a  problem  is  the  process  of  determining  the  required 
result  or  demonstration. 

23.  A  Philosophic  Solution  is,  in  this  work,  a  full  numerical  statement 
.showing,  step  by  stc]),  how  the  result  of  a  problem  is  obtained,  with  a  logical  reason 
for  each  conclusion  reached  in  the  solution.  This  process  of  solution  and  reasoning 
from  the  premises  and  conditions  of  all  problems  constitutes  the  IPhilosopJiie 
System  of  Arithmetic. 

24:.  An  Axiom  is  a  self-evident  truth.  Axioms  are  the  laws  which  govern 
the  logical  mind  in  processes  of  reasoning. 

25.  A  Theorem  is  a  statement,  the  truth  or  falsity  of  which  is  to  be  deter- 
niined  by  i-easoning. 

26.  Demonstration  is  a  process  of  reasoning  by  which  the  truth  of  a 
theorem  is  proved. 

27.  Comparison  is  considering  the  relations  between  numbers;  the  rela- 
tions between  the  premises  and  conditions  of  problems  and  the  required  result. 

Note.— CoHywrison  is  tlie  true  metbod  of  investigation — it  is  the  roy.al  road  to  tlie  fields  of 
unknown  trntli  and  liuowledge. 

The  comparison  of  two  numbers  as  to  tlieir  unit  value,  gives  ratio ;  the  compariaon  of  two 
ratios  gives  proportion  ;  and  the  compariaon  of  several  numbers  which  differ  by  a  commou  ratio  con- 
stitutes progression. 

28.  A  Solution  Statement,  or  an  Operation,  is  a  statement  of  the  figures 
employed  in  solving  a  ])roblcni. 

29.  A  Formula  is  the  expression,  by  symbols,  of  general  principles  appli- 
cable to  the  o])erations  of  particular  problems. 

30.  Philosophy  is  the  knowledge  of  phenomena  as  explained  by,  and  resolved 
into,  causes  and  reasons,  powers  and  laws. 


^  DEFINITIONS.  if 

31.  Arithmetic  is  the  science  of  numbers ;  or  to  define  it  more  extencledly, 
it  is  that  braucli  of  inatbematics  which  treats  of  the  principles,  properties  and  rela- 
tions of  numbers  when  expressed  by  the  aid  of  figures,  either  singly  or  combined. 
These  principles  and  relations  of  numbers,  combined  ^rith  the  facts  relating  to 
problems,  are  applied,  by  the  reasoning  powers  of  man,  to  the  solution  of  all  numer- 
ical problems  of  business  affairs  and  of  practical  life. 

32.  Figures. — Figures  in  arithmetic,  are  characters  used  to  represent  num- 
bers.   The  ten  Arabic  figiu'es  which  we  use,  are 

Kaught  or  Cipher.     One     Two     Three     Four     Fire     Six     Seven     Eight     Nine 

0  123456789 

By  properly  combining  these  ten  figures,  all  jjossible  numbers  may  be  repre- 
sented. 

The  1,  2,  3,  4,  5,  6,  7,  8,  and  9  are  sometimes  called  digits.  They  are  also 
called  the  significant  figures,  because  each  signifies  a  number  when  alone. 

The  naught  (0)  is  so  called,  because  by  itself  it  does  not  signify  or  express 
any  number.    It  expresses  number  only  when  used  in  connection  with  other  figures. 

33.  Talue  of  Figures. — The  value  of  a  figure  is  its  power  to  express 
quantity. 

Figures  have  two  values  :     1.  A  simple  value.    2.  A  local  value. 

34.  The  Simple  Yalue  of  a  figure  is  the  quantity  exjiressed  when  standing 
alone,  or  in  the  unit's  place.  Thus,  when  we  write  3,  independently  of  other  figures, 
or  in  the  unit  column  of  numbers,  it  has  only  a  simple  value  expressing  3  units. 

35.  The  Local  Talue  of  a  figure  is  the  quantity  expressed  when  standing 
to  the  left  of  other  figm-es.  Thus,  3  in  34  expresses  a  quantity  of  thirty  units ;  and 
3  in  345,  expresses  tliree  hundred  units. 

The  local  value  depends  upon  the  scale  or  system  of  numbers  employed,  and 
the  location  in  the  scale. 

36.  The  Radix  is  the  number  of  units  of  one  order  which  it  takes  to  make 
one  of  the  next  higher  order ;  thus  the  radix  of  the  common  system  is  10 ;  of  the 
quinary  system  (not  in  use),  it  is  5. 

37.  A  Scale  or  System  of  Numbers,  in  Arithmetic,  is  a  succession  of 
units,  increasing  and  decreasing  according  to  some  established  custom  in  the  opera- 
tions of  numbers.  Thus  there  is  the  binary,  the  trinary,  etc.,  etc.,  and  the  decimal^ 
the  duodecimal,  etc.,  etc. 

38.  In  these  four  named  different  scales,  the  value  of  four  ones  (1111)  would 
be  as  follows :  In  the  binary  scale  or  system,  the  first  1  on  the  right  is  1 ;  the  second 
1  is  2;  the  third  1  is  4;  and  the  fourth  1  is  8 ;  making  15  altogether. 

39.  In  the  trinary  system  or  scale,  the  first  1  on  the  right  is  1 ;  the  second 
1  is  3  ;  the  third  1  is  9 ;  and  the  fourth  1  is  27  ;  making  40  altogether. 

40.  In  the  decimal  scale,  the  first  1  is  1 ;  the  second  1  is  10 ;  the  third  1  is 
100;  and  the  fourth  1  is  1000 ;  making  in  all  1111. 

41.  In  the  duodecimal  scale,  the  first  1  is  1 ;  the  second  1  is  12 ;  the  third  1 
is  144;  and  the  fourth  1  is  1728 ;  making  in  all  1885. 

From  the  foregoing,  we  see  that  the  system  of  scale  derives  its  name  from 


28  SOUI.e's    PHILOSOrillC    PRACTICAL    MATHEMATICS.  * 

tlio  Radix  or  R(ttio  of  vahu',  yiven  to  each  succeeding  figure  from  the  right  toward 
the  left. 

42.  The  Decimal  Scale  or  System  is  one  in  which  the  Radix  or  law  of 
iucrease  and  decrease  is  always  ten.  This  system  is  in  general  use,  and  derives  its 
name  from  the  Latin  word  decern,  which  means  ten. 

In  any  scale,  tlie  number  of  figures  or  characters  required,  including  the  0,  is 
the  same  as  the  radix. 

TO  CHANGE  ONE  SCALE  TO  ANOTHER. 

43.  The  following  elucidations  will  show  the  maune'r  of  reducing  one  scale  to 
the  notation  of  another  : 

1.  Reduce  2436  of  the  Decimal  Scale  to  the  notation  of  the  Quinary  Scale. 

OPERATION. 
2436  —  5  =  iSlfres  and  1  unit  over. 
487  fives  -H  5  ^  Q7 Jives  oi  fivis  or  25's  and  2  fives  over. 
97  25'8  -^  5  =  19  fives  <>(  ftvcs  o[  fives  or  125's  and  2  25's  over. 
19  125's  ^5  =  3  625's  and  i  125's  over. 
Hence  2436  of  the  Decimal  Scale  =  34221  of  the  Quinary  Scale. 

There  is  no  conventional  method  of  reading  numbers  written  in  other  scales 
than  the  decimal.  This  number  34221,  Quinary  Scale,  would  be  read,  3  C25's,  4  125'8, 
2  25's,  2  5's  and  1  unit. 

2.  Reduce  34221  of  the  Quinary  Sqale  to  the  Decimal  Scale. 

FIRST   OPEnATION.  .SECOXD   OPKRATIOX 

BY    SIMPLE    KEDUCTION. 

1  tinit  of  the  Quinary  Scale  =       1    unit  of  the  Decimal  Scale.  625'3    125'3    25's    S's      units, 

2  5's      "        "        "        "       =     10    units      ..  "  "  3  4  2        2  1 

2  25's    "         "         "         "        =     50       •'  "  "  " 
4  125's  "        •'        "         "       =  500      "           "            "            " 

3  625'3  "        "        "        "       =1875      "  "  "  " 

Hence  34221  of  the  Quin- 
ary Scale    -      -     -      =2436  units  of  the  Decimal  Scale. 


5 


19 

125'3 

5 

97 

25's 

5 

487 

5'3 

5 

Scale. 


2436    units. 
3.    Reduce  2031  of  the  Quaternary  Scale  to  the  notation  of  the  Decimal 


FIRST   OPERATION. 


1  unit  of  the  4  scale  =:     1  unit  of  the  10  scale. 
3  fours  "     "    4      "     =  12  units      "       10     " 
Osixteens  "     "4      "     =    0  units      "       10     " 

2  sixty-fours  "     "    4      "    =128  units      "      10     " 

Hence  2031  "     "    4      "    =141  units      "      10     " 


SECOND     OPERATIOy 
BY    SIMPLE     REDUCTION. 

64's        16's        4'3 

units. 

2 
4 

0 

3 

1 

8 

16'8 

4 

35 

4-3 

4 

141 

units. 

DEFINITIONS.  2Q 

4.  lleduce  141  of  the  Decimal  Scale  to  tbe  Quaternary  Scale. 

OPERATION. 

141  —  4  =  35  4's  and  1  unit  over. 
35  4's  -r-  4  =  8  16's  and  3  4's  over. 
8  16's  —  4  =  2  64's  and  0  16's  over. 

Hence  141  of  the  Decimal  Scale  =  2031  of  the  Quaternary  Scale. 

This  2031  of  the  Quaternary  Scale  is  read,  2  64's,  0  16's,  3  4'3  and  1  unit. 

5.  Reduce  4893  of  the  Decimal  Scale  to  the  notation  of  the  Yicenary  or 
Vigesimal  Scale  and  reverse  the  work. 

Before  this  reduction  can  be  performed,  we  must  adopt  20  characters  to  repre- 
sent 20  figures  in  the  Vicenary  Scale.  To  do  this  in  the  plainest  manner,  let  us  use 
the  10  figures  of  the  Decimal  Scale  and  the  fi^rst  10  letters  of  the  English  alphabet, 
for  the  required  additional  numbers. 

10  11  12  13  14  15  IG  17  18  19 
Thus,  12345G789  abcdefghijO 

Here  we  have  20  characters,  each,  except  the  naught,  representing  a  single 
number  or  integer  in  the  Vicenary  Scale.  The  numerical  power  of  the  letters  is 
shown  by  the  small  figures  i)laced  directly  above  them. 

Having  the  required  figures,  we  make  the  reduction  by  the  following 

OPERATION. 

4893  —  20  =  244  20's  and  13  (d)  units  over. 
244  20's  —  20  =  12  (c)  400's  and  4  20's  over. 

Hence  4893,  Decimal  Scale,  =  12  400's  4  20's  and  13  units.  Or,  expressed  in  the  figures 
adopted  as  above,  it  equals  c  4  d. 

REVERSE   OPERATION. 


(12) 

(13) 

c    4 

d 

20 

244    20's 

or,  thus : 

d 

20 

4    20's 

c  400's 

4893  units. 

=  13  units. 

=  80      " 

=  4800      " 

4893      " 

44.  Order  of  Figures. — The  successive  places  occupied  by  figures  are 
called  orders.  Thus  in  the  Decimal  System,  a  figure  in  the  first  place  is  called  a 
figure  of  the_^rs^  order,  or  of  the  order  of  units  ;  a  figure  in  the  second  place  is  a 
figure  of  the  second  order,  or  of  the  order  of  tens  ;  in  the  third  place,  of  the  third 
order,  or  of  the  order  of  hundreds,  and  so  on ;  each  figure  next  to  the  left  belonging 
to  a  distinct  order,  the  unit  of  which  is  tenfold  the  size  or  value  of  a  unit  of  the 
order  of  the  figure  on  its  right. 


30  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

45.  From  the  above,  we  see  1st:  That  teii  units  of  any  order  in  a  number, 
in  the  Decimal  System,  make  one  unit  of  the  next  higher  order. 

lid.  That  moving  a  figure  one  place  to  the  left,  increases  its  representative 
value  tenfold. 

3d.  That  moving  a  figure  one  jilace  to  the  right,  decreases  its  representative 
value  tenfold. 

46.  Notation  is  a  method  of  writing  numbers.  There  are  two  systems,  the 
Arabic  and  the  Roman. 

47.  By  the  Arabic  notation,  numbers  are  expressed  or  written  by  figures. 
This  system  is  in  general  use  and  is  so  called  because  it  was  introduced  into  Europe 
by  the  Arabians,  iu  the  10th  century. 

48.  By  the  Roman  notation,  numbers  are  expressed  or  written  by  letters. 
This  system  is  now  used  chiefly  to  number  chapters  and  divisions  of  books.  It  is 
so  called  because  it  was  used  by  the  ancient  Komans. 

49.  Numeration  is  naming  the  places  which  figures  occupy. 

50.  Beading  Numbers  is  expressing  their  value  orally. 

51.  There  are  two  systems  of  numeration,  or  reading  numbers, — the  French 
and  the  English. 

The  French  system  is  the  one  in  general  use  in  the  United  States  and  on 
the  Continent  of  Euro^ie. 
The  English  system  is  that  generally  used  in  England  and  in  English  Provinces. 


si 

a   oj  fcB 

p-st; 

. 

o  '^ 

cc               :« 

CO 

GO 

ii£ 

S   cS 

2          a          = 

0 

.2           .                               '3 

S:2  ,. 

=            o           o 

"     .        !»             ..;             m             a 

P  =  p 

i 

H 

w  5 

f  Octillio 
illious. 

f  Septilli 
itillions. 

f  Sextilli 
tillions. 

.5 .0 

5-2 

f  Quadril 
idrillions 

f  Trillion 
lions. 

f  Billions 
[ions. 

f  Millioni 
lions. 

f  Thousa 
lusands. 

III 

«  O*  3) 

!^  1  »    . 

ff)      0      □      « 

>C 

0-5       c  t-   .  0  v^ 

.  "0  -5  '^■ 

is.i  1 

■c  ^  ii  2 
=  2  £  s 

© 

CO 

'=  '£  ~  '=  Z  "5  =  2       _ 

i-rll 

2  ■•/:  X   -  "^  —  -  ■-'^  ■-  3  '^  S  S  *  ^  3 

ccS 

H 

03 

s  ^ 

1c   bx 

«; 

^ 

C  H  0  a  H  'O:  S  H  X'  ffi  H  w 
3  8  4,5  G  1,3  3  8. 17  9. 

'S  H  5'S  H  H  K  H  5  S  H  ?^  K  H  H  K 
6  0  4,  9  3  2,  4  8  7,  2  6  3, 1  9  6,  5 

6  8. 

N** 

>-.'C 

7            7 

"  ?  3  H 

VI 

Oj 

^ 

^^ v^_/^-^^^-..^s^^ 

— '•• .J — ' 

^--^/ '-.w»^^«^v-*'^^«.^\- V ''-^-V '^ ■ 

-V ' 

p  0  0  '^ 
2  -  .2    »>- 

ss 

-1  ® 

_o 

•Period  Period  of  Period 

of  Period  of  Period  of     Period      Period      Period  Period  of    Period 

i  S2  i 

u 

o  ij; 

of            Sep-          Sex. 

Quin. 

(Juad.          of               of              of          Thou- 
rillions.  Trillions.  Billions.  Millions,    sands.       V 

dred  eighty-four  octillions,  five  hundred  E 
thirty-eight  sextillions,  one  hundred  sev 

of 

"^  H  ^  a 

^  4^ 

[V,  Ci-i 

^    O 
® 

^.2 

o 

OctillioDS  tillions.    tillions.     tillions. 

Tlio  number  is  read  two  hun^ 
one  septillions,  three  hundred 

nits. 

iixty- 
enty- 

0  0  § 
®    2    w    =3 

1 1  §  s 

■"  "  i  •§ 

o 

nine  fjuintilllons,  six  hundred  four  quadrillions,  nine  hundred  thirty-two 

"o  's  -3  a 

"C 

trillions,  four  hundred  eighty-seven  billions,  two  hiindred  sixty-three  millions, 

CI  ft 

«5 

a 

one  hundred  ninety-six  thousand,  five  hundred  sixty -eight. 

5  ««  „- 

H    1    „-  g 

S 

ill 

DEFINITIONS.  3 1 


S^-      ES  §  §  §  I^SjS- 

1^      4:1  ;S   .  .2  .2  i3£-:a^,| 

,    &,.  =    (T^^  E-io  na  r^§  gSfe^^ 

s  «^  s;  ^  <t;  §  2  ,x  ^s  ^s  ^=5  II a ^1 

5  s||  ^o-ll  ^H  -3  ^s  .  ^               12:I| 

1 115  illii  lill  fill  iill  1^        lilSl 

^s^-Hjc^-c-i  Bg^tij  H|jMg  Hgr^sg  h|          ic=t; 

«  ^1^  ^.  2  =^2  2'-SS°'-SS  "So^'^iS  'SS^'-SfS  c2.             fe-ll  =  r 


OS' 


•^   r.j::    O'Ch    O    :/j-    O    ■~±:    z    :f±.    Z    =J-    Z    :A±    O   =±-    O    :/  ^  __.   £f  6,  =  H 


,2-2j 


3f/c§_^   28456  1,33817  9,  G0493  2,  4:8720  3,  19G56  8.  S.rJ  ^ 

g  I  ElDj^  ^ V ^ V ^ V ^ V " V ^giisi 

S  _  sj       Period  of  Quad-    Period  of  Trillions.  Period  of  Billions.  Period  of  Millions.  Period  of  Units.     tfr=  -S  2  " 

£  2   r/;  rilUons.  .3  J     =3 

H  -r-.  +J  The  number  i.s  read  two  hundred  eighty-four  thou.sand  five  hundred  sixty-'i"i^  J.2 

M  2  one  quadrillions,  three  hundred  thirty-eight  thousand  one  hundred  seventy-  S^  S  ^  J  2 

5^  £  EJo  nine  trillions,  six  hundred  four  thousand  nine  hundred  thirty-two  billion's,  t.^t'^'^a. 

\a  P'Z  four  hundred  eighty-seven  thousand  two  hundred  sixty-three  millions,  oneKS"-=s| 

J^rSi  hundred  mnety-six  thousand  five  hundred  sixtv-eight.                                              1 1  =  Sw 

THE  EOMAN  SYSTEM  OF  NOTATION. 

54.  In  tbe  Eomau  system  of  notation  the  letter  I  represents  one  ;  Y,  five  ; 
X,  ten  ;  L,  fifty  ;  C,  one  hxdidrcd  ;  D,  five  hundred,  and  M,  one  thousand. 

The  intermediate  and  succeeding  numbers  are  expressed  according  to  tlie 
following  principles : 

1st.  Every  time  a  letter  is  repeated,  its  value  is  repeated ;  thus  II  represents 
two  ;  XX  represents  twenty  ;  CCCC  represents /o??r  hundred. 

2d.  When  a  letter  is  placed  after  one  of  greater  value,  the  sum  of  their 
values  is  the  number  expressed.  Thus,  VI  expresses  six;  XVII  exjiresses  seven- 
teen. 

3d.  When  a  letter  is  placed  before  one  of  greater  value  the  difference  of  their 
values  is  the  number  expressed.    Thus,  IV  expresses _/om>-  ;  XL  expres.ses /orf;/. 

4th.  When  a  letter  is  placed  between  two  letters  of  a  greater  value,  it  is 
combined  with  the  one  following  it.  Thus,  XIX  expresses  nineteen;  CXLVI 
expresses  one  hundred  forty-six. 

5th.  A  dash  or  bar  placed  over  a  letter  multiplies  its  value  by  1000.  Thus^ 
V  expresses  5000 ;  X  expresses  10,000 ;  L  expresses  50,000 ;  C  expresses  100,000 ; 
D  expresses  500,000 ;   M  expresses  1,000,000. 

6th.  In  like  manner,  a  double  or  a  treble  dash  placed  over  a  letter  midtiplies 
its  value  by  1000  tico  times  for  the  double  dash,  and  three  times  for  the  treble  dash. 
Thus,  G  expresses  100,000,000;  M  expresses  1,000,000,000;  0  expresses  100,000,000,000; 
SI  expresses  1,000,000,000,000. 


32 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TABLE  OF  liOMAN  CHARACTERS. 


I 

one. 

XXV 

twenty-five. 

II 

two. 

XXVI 

twenty-six. 

III 

three. 

XXVII 

twenty-seven. 

IV 

four. 

XXVIII 

twenty-eight. 

V 

five. 

XXIX 

twenty-nine. 

VI 

sis. 

XXX 

thirty. 

VII 

seven. 

XL 

forty. 

VIII 

eight. 

L 

fifty. 

IX 

nine. 

LX 

sixty. 

X 

ten. 

LXX 

seventy. 

XI 

eleven. 

LXXX 

eighty. 

XII 

twelve. 

XC 

ninety. 

XIII 

thirteen: 

C 

one  hundred. 

XIV 

fonrteen. 

CC 

two  hundred. 

XV 

fifteen. 

CCC 

three  hundred. 

XVI 

sixteen. 

cccc 

four  hundred. 

XVII 

seventeen. 

D 

five  hundred. 

XVIII 

eighteen. 

DC 

six  hundred. 

XIX 

nineteen. 

DCC 

seven  hundred. 

XX 

twenty, 

DCCC 

eight  hundred. 

XXI 

twenty-one. 

DCCCC 

nine  hundred. 

XXII 

twenty-two. 

M 

one  thousand. 

XXIII 

twenty-three. 

MM 

two  thousand. 

XXIV 

twenty-four. 

MDCCCXCIV^ 

=1894. 

MDCLXVII 

DXLII 

=     10006G7. 
=     1426542. 

MCCCCXXVM 

mFiSSiDcOxSmclmcxiv 

=     1003721151114. 

Note. — It  may  he  proper  to  remark  that  the  methods  used  by  the  ancient  Romans  to  express 
large  numbers  were  varied  and  somewhat  difl'erent  from  the  method  now  used,  and  as  here  explained. 
Because  of  the  complex  system  of  the  Roman  numerals  and  the  great  difficulty  in  using  them,  in  the 
r^tudy  of  arithmetic  and  in  business  calculations,  the  Romans  themselves  used  a  mechanical  contri- 
vance called  an  Abacus.  This  Abacus  consisted  of  a  table  or  board  with  compartments  or  grooves 
representing  a  different  value  to  be  given  to  the  counters  or  pebbles  placed  therein.  In  later  years, 
the  Abacus  was  made  with  wires  on  which  were  placed  movable  counters,  each  wire  representing  a 
unit  of  a  different  v.alue.     And  in  this  form,  the  Abacas  is  still  used  by  some  Europeans. 

The  Roman  notation  is  chiefly  used  in  numbering  chapters  aud  sections  in  books,  marking 
dials  of  clocks  and  watches,  aud  in  writing  physicians'  prescriptions. 


otation  and  Kumeration, 


^ 


—  OR 


Writing  and  Reading  Numbers. 


55.  To  "Write,  or  Notate  Numbers,  begin  at  the  left  and  write  the  figiu-es 
of  each  period  in  tlioir  proper  places,  filliiig-  the  vacant  orders,  if  any,  with  naughts. 

50.  To  Bead  or  Numerate  Numbers,  begin  at  the  riglit  and  point  off  the 
number  into  jieriods  of  three  figures  each.  Theu  commence  at  the  left  aud  read  iu 
succession  each  period  with  its  name. 

57.  To  Verify  the  Notatiou,  or  Writiug,  numerate  the  number  and  see 
if  it  agrees  with  the  number  given. 

EXERCISES    I^"   iS'OTATIOX   AND   NUMERATIOJf. 

58.  Write  seven ;  nine;  ten;  fifteen;  thirty-two;  eighty-nine;  ninety-nine; 
one  hundred  nine. 

"Write  six  hundred  twenty-two;  twenty;  one  hundred  twenty-two;  thirty- 
seven. 

Write  four;  forty;  forty-four;  four  hundred;  four  hundred  four ;  four  hiin- 
dred  forty-four. 

"Write  tM'o;  ten;  two  hundred  eight;  two  hundred  twelve  ;  three  hundred; 
four  hundred  four;  five  hundred  fifty-four;  six  hundred;  seven  hundi-ed;  eight 
hundred;  nine  hundred;  teii  hundred  ten. 

Write  one  thousand  one;  two  thousand  two  liundred  one;  three  thousand 
thirty-one;  forty-four  hundred  forty;  five  thousand  five  hundred  five;  sixty  hun- 
dred sixty;  eight  thousand  nine  hundred  eighty-eight;  twelve  thousand  two; 
twenty-three  thousand  forty-five ;  thirty  thousand  fifty-four, 

59.  Kead  the  following  numbers : 

307     1328    34540     1424G2     81562111     220      8123      19703      770077  40404040 

170    2813     10010    555055    27624555    202      7890      88888      809010      3333333333333 
200    3218    40101     606060     11111110    322      9087      90909      100107      1201050607082 

60.  Write  the  following  numbers : 

1.  Write  1  unit;  1  ten;  1  hundred;  1  unit  aiul  1  ten;  1  hundred,  1  ten  and 
1  unit. 

2.  Write  1  unit,  2  tens,  and  three  huuiheds ;  4  units,  5  tens,  and  6  hundreds; 
7  hundreds,  8  tens,  and  9  units. 

3.  Write  0  units,  1  ten,  and  0  huiulreds ;  4  units,  0  tens,  0  hundreds  aud  4 
thousands. 

(33) 


34  SOULE  S    PHILOSOrHIC    PRACTICAL    MATHEMATICS.  * 

4.  Three  liuudred  forty-one  tbousand,  twenty-two. 

5.  Sixty-live  million,  one  luiudredtliiity-two  thonsand,  tliree  linndred  eighty- 
seven. 

C.     Twelve  billion,  sisteen  million,  forty-three  thonsand,  one  hundred  eleven. 

7.  Nine  hnndred  thonsand,  three  hnndred  fifty. 

8.  Six  million,  one  hundred  .sixty-nine  thonsand,  four  liuiidred  thirty-seven. 

9.  Twenty-two  billion,  one  hundred  three  million,  five  hundred  seventy-six 
thousand,  one  hundred  two. 

10.  One  hundred  two  trillion,  one  hundred  twenty-five  million,  four  hundred 
three. 

11.  Write  208  million,  18  thousand,  1  unit. 

12.  Write  10  billion,  8  million,  103  thousaud,  eleven. 

13.  Write  200  sextillion,  1  quintilliou,  100  quadrillion,  10  trillion,  111  billion, 
1  million,  10  thousand,  10  units. 

14.  Write  97  million,  14 ;  5  thousaud,  5. 

15.  Write  eleven  thousand,  11  hundred,  11 ;  16  thousaud,  16  hundred,  16. 

16.  Five  billion,  fourteen  million,  nine. 

17.  Two  hundred  one  million,  twenty  thousaud. 

18.  Seventy-seven  trillion,  seventy-six. 

19.  Eight  thousand,  nine  hundred,  ninety. 

20.  Four  hundred  twenty  million,  one. 

21.  Thirty  thousand  million,  thirty. 

22.  Three  hundred  eleven  octillion. 

23.  Four  hundred  two  vigiutillion. 

24.  Fill  all  the  orders  of  figures  with  9's  from  the  order  of  units  to  the  order 
of  octillious,  both  inclusive,  point  oft",  and  read  the  same  accordiug  to  the  French 
and  the  English  systems  of  numeration. 

61.  Write  in  figures  the  following  numbers,  and  numerate  them  according 
to  the  English  system  of  numeration : 

1.  Twelve  hundred  forty-one  million,  one  hundred  twenty-three  thousand 
five  hundred  fourteen. 

2.  Four  hundred  nineteen  thousand  one  hundred  fifty-two  million,  twenty-one 
thousand  forty-seven. 

3.  Fifty-three  billion,  two  hundred  twelve  thousand  twenty-six  million, 
seventy-five  thousand  three  hundred  eighty -four.  Ans.    53,212026,075384. 

4.  1342  trillion,  11122  billion,  14  million,  19  units. 

5.  1  quadrillion,  1  trillion,  1  million,  1. 

62.  Kead  the  following  numbers : 
XXXIV. 
XXXV. 
XXXVI. 
XLIX. 

Write,  in  the  Komau  System  of  Notation,  the  following  numbers : 

9, 12,  14,  37,  49,  83,  108,  519, 1519, 14704,  88976,  13140363,  1001001001. 


LI. 

C. 

PCG. 

DC. 

MMDCLI. 

LXI. 

CI. 

CD. 

DLXIX. 

MMMXC. 

LXIX. 

ex. 

CDXIV. 

M. 

MMCCX. 

XC. 

CL. 

D. 

MC. 

MMCMMDX, 

SYNOPSIS    FOR    REVIEW. 


SY^oj^sjs  :por  r^vi:b\v. 


63.    Define  the  foUowiug  words  and  phrases : 


3. 
4. 

5. 
6. 


9. 

10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 

19. 

20. 
21. 
22. 
23. 
24. 
25. 
26. 


A  Defiuitiou.  28. 
Science. 

Art.  29. 

Quantitj-.  30. 

A  Unit.  31. 

A  Xumber.  32. 

Like  Numbers.  33. 

IJnlike  Numbers.  34. 

Odd  and  Even  Numbers.  35. 

The  Unit  of  a  Number.  36. 

The  Unit.  37. 

A  Fractional  Unit.  38. 

Numbers.  39. 

An  Abstract  Number.  40. 
A  Denominate  or  Concrete  Number.  41. 

A  Compound  Number.  42. 
The  Arithmetical  Units  of  Orders.  43.- 
An  Arithmetical  Complement  of  a  44. 

Number.  45. 
An  Arithmetical  Sui^plemeut  of  a  40. 

Number.  47. 

A  Problem.  48. 

The  Premises  or  Conditions.  49. 

A  Solution.  50. 

A  Philosophic  Solution.  51. 

An  Axiom.  52. 

A  Theorem.  53. 

Demonstration.  54. 
Comparison. 


A  Solution  Statement,  or  an  Opera- 
tion. 
A  Formula. 
Philosophy. 
Arithmetic. 
Figures. 

Value  of  Figures. 
The  Simjile  Value. 
The  Local  Value. 
The  Eadix. 

A  Scale  or  System  of  Numbers. 
The  Binary  Scale. 
The  Triuary  Scale. 
The  Decimal  Scale. 
The  Duodecimal  Scale. 
The  Decimal  Scale  or  System. 
To  Change  one  Scale  to  Another, 
Oixler  of  Figures. 
Decimal  System  of  Figures. 
Notation. 
Arabic  Notation. 
Koman  Notation. 
Numeration. 
Eeading  Numbers. 
Two  Systems  of  Numeration. 
French  System  of  Numeration. 
English  System  of  Numeration. 
The  Eoman  System  of  Notation. 


igns  arid  Syrmlbols. 


64.  Signs  and  Symbols  are  used  to  abridge  arithmetical  operations.  They 
also  indicate  some  relationship  existing  among  numbers,  aud  what  operation  is  to 
be  performed. 

The  signs  in  general  use  in  Arithmetic  are  as  follows : 

j_     y     ^     —     M     or ...      .      1(52      ,/     '     1°    20    30 

65.  The  perpendicular  or  Greek  Cross,  (  +  )  is  the  sign  of  Addition;  it  is 
called  i^Zms,  and  is  read  j>/!(s,  or  and.  It  means  more,  and  indicates  that  the  num- 
bers between  which  it  is  placed  are  to  be  added.  Thus,  7  +  9  is  read  7  plus  9,  and 
means  that  7  and  9  are  to  be  added.  When  used  after  a  uumber,  thus,  5  +,  whicl> 
is  read  5  plus,  it  means  5  and  a  small  excess. 

66.  The  horizontal  line  ( — )  is  the  sign  of  Subtraction,  and  is  called  minus. 
It  means  less,  and  indicates,  when  jilaced  between  two  numbers,  that  the  one  that 
follows  it  is  to  be  taken  from  the  one  before  it.    Thus,  7 — 3  equals  4. 

67.  The  obliqiie  or  Saint  Antlrew's  Cross,  ( X )  is  the  sign  of  Multiplication. 
It  is  read  multiply  hy,  or  times.  It  indicates  that  the  numbers  between  which  it  is 
placed  are  to  be  midtlplied  together.  Thus,  7  x  5  is  read,  7  multiplied  by  5,  or  5 
times  7. 

68.  The  horizontal  line  with  a  point  above  and  a  point  below  it,  (-r)  is  the 
sign  of  Division.  It  is  read  divided  hy.  It  indicates,  when  i^laced  between  two 
numbers,  that  the  one  before  it  is  to  be  measured  or  divided  by  the  one  after  it. 
Thus  124-4  equals  3. 

69.  The  parallel  horizontal  lines  (=)  are  the  sign  of  Equality.  It  is  read 
equals  or  is  equal  to.  It  indicates  that  the  quantities  between  which  it  is  placed  are 
equal.  Thus,  5-|-3=12 — 4.  A  statement  of  this  kind  is  called  an  equation,  because 
the  quantity  of  5+3  is  equal  to  12 — i. 

70.  The  (  )  or is  the  sign  of  Aggregation.     The  first  is  the  Parenthesis, 

the  second  is  the  Vinculum.  They  are  both  used  for  the  same  purpose.  They 
indicate  that  the  numbers  within  the  parenthesis  or  below  the  vinculum,  are  to  be 
considered  as  one  quantity.     Thus,  10  —  (.j+4)=7,  or  16  —  5+4=7.     See  page  134. 

71.  The  single  point  (.)  is  the  Decimal  sign.  It  indicates  that  the  num- 
bers which  follow  it  are  tenths,  hundredths,  etc.  Thus,  .5,  .05  are  read  5  tenths,  5 
hundredths. 

72.  The  ( : )  is  the  sign  of  Ratio.    It  is  read,  is  to  or  the  ratio  of. 

73.  The  ( : : )  is  the  sign  of  Proportion.  It  is  read,  as,  or  equal.  Thus,  3  :  6 
:  :  5  :  10  is  read  3  is  to  G  as  5  is  to  10;  i.  e.  3  is  such  a  part  of  6  as  5  is  a  part  of  ten. 

(36) 


^  SIGN'S    AND    SYMBOLS.  37 

7i.  The  (10- )  is  the  sign  of  lit  volution.  TUe  small  figure  to  the  right  and  the 
toi>  of  the  number  imlicates  the  power  to  whieli  the  number  is  to  be  raised.  Thus, 
8-  indicates  that  8  is  to  be  raised  to  the  second  iwwer  or  takeu  as  a  factor  twice. 
Thus,  8x8=04^.  8''  indicates  that  the  third  power  of  8  is  required.  Thus, 
8x8x8=512. 

75.  The  radical  sign  (  •/  )  is  the  sign  of  Evolution.  It  indicates  that  some 
root  of  a  number  is  to  be  found.  Thus,  -/'jl  indicates  that  the  square  root  of  04:  is 
to  be  found.  ^/')V2  indicates  that  the  cuhc  root  of  512  is  required.  V^oyO  indicates 
that  the  fourth  root  of  4090  is  to  be  extracted.  The  small  figure  within  the  branches 
of  the  radical  sign  is  called  the  index  and  indicates  what  root  is  required. 

Other  signs  of  evolution,  now  often  used,  are  as  follows : 
Thus,  250-  indicates  that  tlie  square  root  is  to  be  extracted. 

625*  indicates  that  the  cube  root  is  to  be  extracted. 

146-  indicates  that  the  square  root  of  the  cube  of  the  number  is  required. 

76.  The  ( .'. )  is  the  sign  of  Deduction.  It  is  read,  therefore,  hence,  or  con- 
sequently. 

77.  The  Interrogation  ( J )  is  the  sign  meaning  irhat,  or  how  many.  It  signi- 
fies that  the  answer  to  the  question  asked  is  to  be  found. 

78.  The  expressions  1°,  2°,  3°,  denote  first,  second,  third,  etc. 

79.  The  sign  of  the  comma  ( , )  is  the  sign  of  Xumeration.  It  is  used  to 
seiiarate  large  numbers  into  periods,  to  facilitate  the  reading  of  them. 


SIGNS  AND  ABBREVIATIONS. 


80.     The  following  are  the  principal  signs  and  abbreviations  in  general  use 
among  merchants  and  business  men,  and  are  used  in  this  work  : 


Abbreviatio)!,  conti-action  of  a  word  or  phrase  by 
omittiug  letters  or  substituting  certain  charac- 
ters. 

jier,  for,  or  for 


®,  at,  to 

each. 
%  or  acct.,  account. 
A.   D.,  the   year  of  our 

Lord. 
Agt.,  Agent. 

A.  M.,  Forenoon, 
amt.,  amount, 
ans.,  answer. 
Apr.,  April, 
ass't'd,  assorted. 
Aug.,  August. 
bal.,  balance, 
bbl.,  or  brl.,  barrel, 
bgs,,  bags. 

bdla.,  bundles, 
bis.,  bales. 

B.  L.,  Bill  of  Lading, 
bkts.,  V)askets. 

bot.  bought. 

B.  M.  the  gay  world. 

B.  Pay.,  Bills  Payable. 

B.  Rec,    Bills    Receiv- 
able. 

b   c,  buyer's  option, 
bque.,  barque, 
br. ,  brig, 
bu.,  bushel. 
bxs.,  boxes. 

c.  or  f ,  cents. 
Cap.,  Capital, 
chts.,  chests. 
cUs.,  checks,  casks. 
Co.,  Conipanv. 

C.  O.   v.,  Cash  on   De- 
li ver,y. 

C.  B.,  Cash  Book, 
com.,  commission, 
const.,  consignment. 
Cr.,  Creditor. 

cs.,  cases. 

cwt.,  hundredweight. 

d.  pence. 

D.  or   d.,   or  dol.,   or  |, 
dollar. 

do.,  or  "  the  same. 
dft.,  draft. 


Dec,  December. 

doz.,  dozen. 

Dr.,  Debtor. 

ds.,  days. 

dwt.  or  pwt.,  penny- 
weight. 

ea.,  each. 

E.  E.,  Errors  excepted. 

E.  &  O.  E.,  Errors  and 
omissions  excepted. 

emb'd.,  enil)roidered. 

Eng.,  Engli.sh. 

En  Ville,  or  E.  V.,  in 
the  town  or  city. 

ergo,  therefore. 

Et  als.,  and  others. 

Ex  Rel.,  on  the  relation 
or  information  of  the 
aggrieved  party. 

ex.,  example. 

exch.,  exchange. 

exp.,  expenses. 

f.,  farthing. 

fav.,  favor. 

Feb.,  February. 

fig'd,  figured. 

f.  o.  b.,  free  on  board. 

fol.,  folio. 

fr.,  franc. 

ft.,  feet  or  foot. 

fwd.  or  for'd,  forward. 

frt.,  freight. 

gal.,  gallon. 

gr.,  grain,  gross. 

hf ,  half. 

hf.  chts.,  half  chests. 

hhd.,  hogshead. 

in.,  inches. 

I.  H.,  Invoice  book. 

i.  c. ,  that  is. 

ins.,  insurance. 

inst.,  present  month. 

int.,  interest. 

iuv., invoice,  inventory 

I.  O.  U.,  I  owe  you. 

.Jan.,  January. 

16s.,  pounds. 


Led.,  Ledger. 

L.  F.,  Ledger  folio. 

M.,  Noon. 

m/a.,  mouths  after  date. 

Mar.,  March. 

nnlse.,  merchandise. 

M.S.,  manuscript. 

mo.,  month. 

No.,  number. 

Nov.,  November. 

N.  B.,  note  well. 

N.  P.,  Notary  Public. 

Oct.,  October. 

0.  I.  B.,  Outward  In- 
voice Book. 

O.  K.,  all  right. 

oz.,   ounce. 

p.,  page. 

pp.,  pages. 

pay't.,  payment. 

pd.,  paid. 

pkgs.,  packages. 

pes.  or  ps.,  pieces. 

pr.,  pair. 

per  or  ^,  by. 

lilts.,  plates. 

pwt.  or  dwt.,  penny- 
weight. 

P.  M.,  afternoon. 

P.  O.  Post  Office. 

P.  P.  C,  to  take  leave. 

Prox.,  Next  month. 

P.  S.  Postscript. 

pun.,  puncheon. 

Jits.,  pints. 

(jr.,  quarter. 

qts.,  quarts. 

R.  R.,  Railroad. 

rec'd,  received. 

rec't.,  receipt. 

R.  S.  V.  P.,  answer  if 
you  i^lease. 

s.  or  /  shilling. 


S.  B.  Sales  Book. 

Sept.,  Sejitember. 

Schr.,  Schooner. 

Sh.,  Ship. 

Shjit. ,  Sliipment. 

Sunds.,  Sundries. 

8.  o.,  seller's  option. 

Sr.,  Senior, 

Str.,   Steamer. 

Supt.,  Sujierinteudent. 

T.  O.,  Turn  Over. 

trcs.,  tierces. 

nit.,  last  month. 

U.  S.,  United  States. 

ves.,  vessel. 

W.  I.,  West  Indies. 

wt.,  weight. 

yd.,  yard. 

yr.,  year. 

£  or  L.,  pound. 

^01  I'Pi'  cent. 
,  number. 

-)-,  sign  of  addition. 

— ,  sign  of  subtraction. 

X,  sign  of  multiiilica- 
tiou. 

-i-,  sign  of  division. 

=,  sign  of  equality. 

1',  one  and  one-fourth. 

1^,  one  and  one-half. 

1',  one  and  three- 
fourths. 

Mk.,  Marks,  the  Ger- 
man monetary  unit. 

l/,  Clieik  nuirk  ;  cor- 
rect ;  approved. 

®  or  $,  Cil'rao,  used 
to  separate  the  mil- 
reis  from  the  reis  in 
Brazil  money. 


r  17  doz.,  4  of  which  are  ® 
2  and  7  ®  $15. 


17doz.  S,^„Sft,  $,'o 
$10  per  doz.,  6  ® 

8  doz.  J  '2)  5  /  J  ®  4/6,  =  2  doz.  No.  4  ®  5  shillings 
per  doz.,  and  6  doz.  No.  5  ®  4  shillings  six_ 
pence  per  dozen. 


Note.— The  origin  of  the  dollar  sign,  $,  has  been  variously  accounted  for.  It  is  thought  by 
some  to  be  an  abbreviation  of  the  letters  U.  S.,  the  initials  for  United  States.  But  it  wiis  in  u.se  long 
before  the  adoption  of  the  Federal  currency,  and  it  is  probably  a  nioditied  figure  8,  denoting  a  piece 
of  eight,  i.  e.,  eight  reals — an  old  Spanish  coin,  the  value  of  which  was  a  dollar. 

The  character  £, pound  sterling,  is  but  a  capital L  with  a  line  drawn  across  it,  and  represents 
the  initial  letter  of  the  Latin  word  Librw,  pound. 

(38) 


*  SIGNS    AND    ABBREVIATIONS,  39 

The  sign  /,  or  s.  is  believed  to  have  been  originally  a  capital  S  (written  thus,  f  ),  the  initial 
letter  for  the  word  shilling. 

The  sign  It.,  jiouuil  weight,  is  formed  from  the  first  and  thirdletters  of  the  Latin  -wordXiftj-CE, 
and  connected  with  a  horizontal  line. 

The  sign  ®  i.s  a  moditication  of  the  Latin  word  ad,  which  means  at  or  to. 

The  sign  d,  for  penny,  one-twelfth  of  an  English  shilling,  is  a  coin  corresponding  in  value  to 
the  7:f  jT  of  the  old  Saxon  pound,  which  was  the  same  in  value  as  the  Komau  denarius,  the  if^  of  the 
Libra.     Hence  used  to  denote  penny. 

The  character  °o  is  a  modification  of  -I-,  the  sign  of  division.  Thus  eight  per  cent,  may  be 
expressed  by  toitj  or  8  -^  100,  or  by  omitting  the  denominator  and  writing  it  thus,  8-H,  or  iu  rapid 
■writing  thus,  8  I .  or  as  we  now  write  it,  thus  8  %. 

The  character  %*,  is  another  form  of  p,  the  initial  letter  of  the  Latin  word  Per. 

The  abbreviation  O.  K.  is  used  by  business  men  and  telegraph  operators,  and  signiiies  "all 
right,"  or  "  inspected  and  approved." 

The  abbre\'iation  Fr.  is  a  contraction  of  the  word  France,  which  is  derived  from  the  German 
word  Franke,  a  name  of  the  German  tribe  who,  in  the  fifth  century,  overran  and  conquered  Gaul, 
and  established  the  Kingdom  of  France. 

S.  L.  &  C.  means  Shippers  Load  and  Count.  These  letters  are  used  on  Railroad  Bills  of 
Lading  and  restrict  the  responsibility  of  the  Kailroad  Company  regarding  the  correctness  of  the 
articles  shipped. 

C.I.  F.  or  Cif.  These  letters  are  used  by  buyers  when  making  offers  for  goods,  and  mean 
Cost,  IiiKurance  a\u\  Freiyht.  and  all  charges  to   the  wharf  of  the  buyer. 

Shippers'  Order  Notify.  AVheu  these  words  are  indorsed  on  a  Hill  of  Lading,  they  mean 
that  the  consignee  cannot  take  possession  of  the  goods  until  he  has  ]iaid  the  Draft  drawn  <m  him 
by  the  shipper  and  received  the  Bill  of  Lading  for  the  goods.  The  Draft  and  Hill  of  Lading  are 
sent  by  the  shiiiper  to  a  Hank  which  notifies  the  consignee  that  they  hold  a  Draft  ou  him,  which, 
by  paying  he  will  be  given  the  Bill  of  Lading  for  the  goods. 


ddition. 


81.  Addition — Increasuui — is  the  process  of  uniting  two  or  more  nxinibers 
of  tlie  same  name  or  kind,  into  one  equivalent  number. 

82.  The  number  obtained  by  this  process  is  called  the  sum  or  anioiiiit. 

83.  The  Sign  of  Addition  is  a  perpendicular  cross,  +,  called  i)lns ;  it 
means  more ;  thus,  7  +  9  is  read,  7  plus  9,  and  indicates  that  7  and  9  are  to  be 
added.  "WTien  used  after  a  number,  thus,  5  +,  which  is  read  5  plus,  it  means  5  and 
a  small  excess. 

84.  Tlie  Sign  of  E(£uality  is  =.  It  is  ve>A(\.  equals  or  is  equal  to,  and  denotes 
that  the  munbers  between  which  it  is  placed,  are  equal  to  each  other ;  thus,  7  +  9 
=  10.  means  that  7  and  9  added,  are  equal  to  16.  The  expressiou  is  read,  7  plus  9 
equals  16. 

85.  A  Numerical  Equation  is  an  equality  between  two  numerical  expres- 
sions, which,  though  differing  in  form  from  each  other,  are  equivalent.  Each  exjires- 
sion  is  called  a  term  of  the  equation.  Thus,  5  +  8  =  13  is  a  numerical  equation  in 
which  the  5  +  8  is  called  the  first  member  of  the  equation  and  13  the  second 
member,  and  both  are  called  the  terms  of  the  equation. 

86.  Principle  of  Addition.  Numbers  of  the  same  kind,  order,  or  character 
only,  can  be  added.  Thus  we  cannot  add  2  apples  and  3  oranges;  or  5  pounds  of 
sugar  and  6  boxes  of  peaches ;  or  0  units  and  5  hundreds ;  or  J  and  5,  etc.  We 
can  only  add  apples  to  apples,  oranges  to  oranges,  sugar  to  sugar,  peaches  to 
peaches,  units  to  units,  hundreds  to  hundreds,  halves  to  halves,  fourths  to  fourths, 
etc.  We  can  collect  together  things  of  different  kinds,  apples,  peaches,  oranges,  etc., 
but  by  collecting  them  together  we  do  not  increase  the  number  or  sum  of  either, 
and  hence  there  is  no  addition. 

Addition  is  the  basis  of  all  numerical  operations,  and  being  so  constantly 
iised  in  all  departments  of  business  life,  tlierefore  it  is  essentially  necessary  that  the 
accountant  and  business  man  should  be  both  accurate  and  rapid  in  the  various 
methods  of  adding  numbers.  By  due  attention  to  the  principles  and  combinations 
of  numbers,  as  herein  presented,  aided  by  practice,  the  learner  can,  in  a  few  weeks, 
acquire  a  degree  of  proficiency  that  will  place  him  in  the  front  rank  of  experts,  and 
doubly  capacitate  him  for  service  to  himself  or  to  others  for  whom  he  may  labor. 

To  assist  in  disciplining  the  mind  and  to  point  out  the  best  and  most  rapid 
systems  of  adding  columns,  we  present  the  following : 

(40) 


ADDITION    TABLES. 


41 


8V. 


ADDITION  TABLES. 


8 
9 


I'n.lolo   i^To.    X. 


1     1 

3 
4 


1 
1 

2.1 
2.3  ■ 

1.2 
4.3 

1.2.3 
0.4.3 

1.2.3 
G.5.4 

1.2.3.4 
7.0.5.4 

1.2.3.4 
8.7.G.5 

10 


Explanation. — lu  this  table,  T\e 
show  20  difi'ereut  combiuatioiis  of 
the  9  significant  figures,  to  jiro- 
dnce  results  from  1  to  9.  It  may 
be  said  tbat  three  I's  make  3, 
three  2"s  make  6,  etc.,  and  tbat 
they  are  regular  combinations ; 
but  we  see  by  the  table  that  two 
I's  are  2,  and  that  two  2's  are  4, 
etc.  Hence,  though  the  table 
does  not  contain  all  the  possible 
combinations,  it  does  contain  all  i  J"  (j;.; 
that  are  essential  and  of  value  in 
this  connection. 


1.2.3.4.5 
9.8.7.G.5 


Note.— In  learning  these  tables  and  in  handling  all  numbers,  all  intermediate  v-ords  and 
thoughts  thatoccur  between  the  numbers  to  be  combined,  should  be  omitted.  Thus,  instead  of  saying 
or  thinking  that  2  and  2  are  4,  3  and  5  are  8,  etc.,  say  or  think  4  ;  8 ;  etc. 

Ta-Tol©    3Xro.    2. 

Explanation.— In  this  table, 
we  show  the  25  ditierent  couibi- 
natious  of  the  9  significant  fig- 
ures, the  sum  of  which  equals 
(e«  or  more.  To  attain  rapidity 
in  aildiiig,  it  is  absolutely  neces- 
sary that  the  learner  should  be 
so  familiar  with  these  combina- 
tions that  he  can  instantly  see 
the  result  without  adding,  i.  e. 
he  7uust  know  the  result  by  the 
combiuatioii,  just  as  he  knows 
the  value  of  4  or  5,  by  the  com- 
bination of  lines  forming  the 
figure,  or  as  he  knows  the  juo- 
nnnciation  of  a  word  without 
spelling  it. 


11 


12 


13 


14 


15 


2.3.4.0 
9.8.7.(: 

3.4.5.( 
y.S.7.( 

4.5.G 
9.S.7 

5.(;.7 

'J.S.7 

G.7 
0.8 

7.S 


1< 


18 


The  rapid  increasing  and  decreasing  operations  in  the  science  of  iinmbers, 
depend  htrgely  upon  the  capacity  of  the  calcnh^tor  to  apprehend  instantly  and  to 
apply  accurately,  the  result  of  two  or  more  figures.  And  the  object  of  these  tables 
is  to  aid  the  learner  in  acquiring  the  desiied  rapidity. 


88. 


ADDITION  AND    SUBTRACTION  TABLE. 
Ta.t>le>    ^To.    3. 


1  Ik 

^=9 

1  & 

?=8 

1  & 

?  =  7 

1  &  ?=G 

1  &  ?=5 

1  & 

?=4 

1  &? 

=3 

1 

2  " 

"9 

2  " 

"  8 

1 1     " 

2  "     "  G 

2  "     '•  5 

2  " 

"  4 

2  " 

"  3 

3" 

"9 

3  " 

•'  8 

3  " 

"  T 

3  "     "  G 

3  "     "  5 

3  " 

u  4 

4" 

"9 

4  " 

"  8 

4  " 

li    T 

4  "     "  G 

4  "     "5 

5" 

"  9 

5  " 

"  8 

5  " 

fcb  7 

5  '<     '<  G 

6" 

"  9 

G  " 

"  8 

C  " 

a  T 

7" 

"  9 

7  " 

"  8 

8" 

"  9 

1  &  l.=2 


Explanation.— in  this  table,  we  present  the  36  combinations  of  the  significant  figures,  in  which 
the  difference  between  each  is  to  be  supplied  by  the  learner.  This  is  a  very  important  table  for 
rapid  work  in  subtraction,  by  the  addition  method,  and  should  receive  careful  attention. 


42 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


89. 


ADDITION  AND  SUBTRACTION  TABLE. 


1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

IG 

17 

18 

19 

20 

21 

22 

23 

24 

25 


& 


?=100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
''  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 


26  & 

27 

28 

29 

30 

31 

32 

33 

34 

35 

30 

37 

38 

39 

40 

41 

42 

43 

44 

45 

46 

47 

48 

49 

50 


?=100 

"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 


51  & 

52  " 
53" 

54  " 

55  " 

56  " 

57  " 

58  » 
59" 
GO  " 

61  " 

62  " 

63  " 

64  " 

65  " 

66  " 

67  " 

68  " 

69  " 

70  " 

71  " 

72  " 

73  " 

74  " 

75  " 


?  =  100 
"  100 
"  10(t 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 
"  100 


76  &  ?r 

=  100 

77  "  ' 

'  100 

78  "  ' 

'  100 

79  "  ' 

'  100 

80  "  ' 

'  100 

81  "  ' 

'  100 

82  "  ' 

'  100 

S3"  ' 

'  100 

84  "  ' 

'  100 

85  "  ' 

'  100 

86  "  ' 

'  100 

87  "  ' 

'  100 

88  "  ' 

'  100 

89  "  ' 

'  100 

90  "  ' 

'  100 

91  "  ' 

'  100 

92  "  ' 

'  100 

93  "  ' 

'  100 

94  "  ' 

'  100 

95  "  ' 

'  100 

96  "  ' 

'  100 

97  "  ' 

'  100 

98  "  ' 

'  100 

99  "  ' 

'  100 

Explanation. — We  present  this  table  to  akl  the  learner  in  instantly  seeing  the  difl'erence 
between  100  and  any  number  from  1  to  99.  It  is  of  special  valne  in  addition  and  subtraction,  and 
all  who  expect  to  become  rapid  Calculators  must  be  proficient  in  this  character  of  work. 

These  tables  constitute  the  alphabet  of  numbers,  and  reucler  obsolete  the  dis- 
gusting and  mind  weakening  practice  of  counting  lingers,  birds  on  limbs,  ducks  iu 
ponds,  apples  ou  trees,  or  rabbits  iu  the  yard,  etc.,  which  is  so  ofteu  seen  in  primary 
arithmetics. 


mm 


Brill  Exercise  No.  1. 


'Ia^ 


90.    With  one  look  at  each  of  the  following  groups  of  numbers,  name  the 
result  and  repeat  the  operation  until  you  can  call  the  amouuts  at  sight. 


6 

5 

4 

3 

o 

3 

1 

o 

4 

1 

G 

3 

7 

1 

8 

3 

3 

■1 

* 

5 

7 

4 

3 

3 

»> 

5 

*> 

G 

o 

** 

1 

8 

7 

3 

5 

G 

1 

4 

5 

G 

7 

3 

4 

1 

4 

2 

9 

3 

2 

7 

4 

3 

7 

4 

3 

o 

o 

9 

5 

G 

3 

5 

0 

o 

3 

i 

8 

- 

5 

9 

8 

9 

8 

9 

8 

9        8 

8 

7 

9 

5 

G 

7 

G 

5 

4 

8 

9 

8 

C 

G 

4 

4 

2         o 

9 

8 

7 

9 

7 

G 

S 

9 

o 

4 

3 

8 

7 

G 

8 

5        5 

G 

9 

4 

7 

5 

5 

7 

3 

<» 

1 

9 

3 

3 

4 

4 

7        5 

G 

1 

4 

« 

8 

4 

2 

3 

'> 

5 

3 

4 

5 

1 

G 

9 

8 

2 

3 

G 

5 

3 

8 

2 

4 

O 

5 

2 

i> 

3 

3 

2 

1 

5 

7 

4 

7 

6 

9 

8 

G 

7 

7 

9 

8 

7 

8 

'> 

4 

G 

8 

9 

3 

5 

4 

5 

G 

2 

7 

1 

8 

4 

G 

5 

4 

9 

7 

9 

7 

C 

5 

G 

3 

5 

4 

8 

2 

7 

7 

8 

5 

G 

8 

8 

5 

8 

7 

1 

8 

9 

5 

8 

9 

9 

8 

9 

8 

8 

5 

6 

9 

DRILL    EXERCISE    Xo.    2. 


1 

2 

3 

4 

5 

G 

7 

8 

9 

4 

7 

o 

9 

4 

1 

7 

8 

8 

G 

5 

8 

4 

G 

7 

5 

8 

3 

2 

9 

8 

G 

■) 

4 

G 

3 

7 

9 

1 

5 

3 

2 

8 

9 

7 

3 

7 

9 

1 

o 

3 

4 

2 

8 

7 

3 

2 

3 

7 

4 

4 

5 

G 

8 

4 

1 

3 

9 

3 

3 

3 

G 

7 

8 

9 

3 

i 

5 

0 

9 

8 

7 

G 

5 

4 

3 

7 

7 

8 

8 

9 

9 

5 

5 

4 

4 

3 

3 

2 

2 

1 

1 

1 

o 

3 

4 

5 

G 

7 

8 

9 

0 

3 

1 

8 

2 

4 

0 

5 

1 

4 

2 

3 

6 

0 

9 

6 

8 

2 

5 

2 

7 

5 

8 

(43) 


44  SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 

DRILL  EXERCISE   No.  2.— Continued. 


9 

0 

8 

7 

5 

3 

4 

o 

1 

8 

G 

4 

9 

4 

1 

7 

2 

8 

9 

1 

2 

3 

4 

5 

G 

7 

8 

!» 

0 

9 

1 

8 

*> 

7 

3 

7 

4 

5 

S 

1 

2 

7 

0 

4 

7 

8 

!) 

9 

8 

8 

3 

2 

5 

1 

9 

0 

1 

2 

7 

8 

1 

9 

O 

G 

7 

4 

5 

1 

2 

3 

4 

5 

G 

7 

8 

9 

7 

1 

8 

'> 

0 

3 

7 

4 

G 

5 

0 

9 

8 

7 

5 

G 

4 

3 

2 

1 

2 

3 

4 

5 

G 

7 

8 

9 

0 

8 

4 

7 

•> 

9 

1 

4 

8 

*> 

<) 

0 

9 

8 

8 

7 

7 

G 

G 

5 

5 

5 

4 

3 

3 

2 

1 

0 

G 

.") 

G 

9 

8 

5       4 

8 

2 

9 

3 

8 

4 

1 

5 

2 

6 

9 

5 

o 

8 

•> 

»> 

9 

9 

8       8 

7 

7 

G 

1 

0 

G 

8 

»> 

4 

9 

0 

3 

1 

0 

1 

•  ) 

3 

4 

5       G 

7 

8 

9 

0 

9 

8 

7 

G 

5 

4 

3 

2 

0 

1 

3 

4 

7 

5 

G       8 

o 

1 

9 

3 

0 

9 

G 

7 

4 

3 

2 

8 

1 

9 

1 

2 

3 

4 

5 

G 

7 

8 

9 

3 

8 

4 

9 

5 

2 

1 

G 

8 

!) 

8 

7 

G 

5 

3 

'> 

1 

7 

0 

4 

9 

1 

*> 

8 

9 

0 

3 

3 

5 

8 

9 

1 

0 

3 

8 

•> 

G 

4 

9 

0 

7 

8 

9 

1 

8 

4 

5 

4 

5 

4 

5 

4 

5 

4 

5 

4 

5 

4 

5 

4 

5 

4 

5 

9 

8 

7 

C 

5 

4 

3 

2 

1 

S 

5 

C 

4 

3 

8 

1 

6 

3 

1 

4 

2 

5 

9 

G 

0 

2 

5 

G 

8 

9 

1 

3 

9 

0 

2 

5 

9 

7 

3 

5 

7 

9 

1 

8 

3 

8 

G 

0 

4 

8 

2 

8 

G 

3 

9 

1 

7 

4 

1 

9 

2 

8 

3 

8 

4 

7 

5 

G 

0 

9 

1 

8 

3 

6 

2 

5 

8 

1 

1 

2 

G 

3 

5 

4 

8 

9 

1 

9 

2 

8 

3 

7 

4 

G 

5 

8 

9 

0 

G 

3 

9 

o 

8 

1 

0 

G 

2 

4 

5 

9 

6 

8 

2 

9 

0 

7 

8 

9 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

3 

8 

5 

9 

3 

t| 

1 

4 

9 

8 

1 

4 

0 

3 

6 

8 

9 

■  1 

6 

3 

8 

o 

G 

9 

3 

o 

8 

7 

8 

9 

7 

8 

3 

9 

3 

9 

8 

5 

o 

8 

2 

7 

7 

4 

5 

0 

7 

4 

8 

8 

9 

4 

7 

8 

9 

8 

8 

9 

3 

2 

1 

5 

7 

6 

5 

8 

3 

7 

9 

7 

5 

8 

4 

S 

7 

2 

7 

5 

') 

6 

5 

1 

2 

3 

4 

5 

G 

7 

8 

<) 

0 

1 

*> 

3 

4 

5 

6 

7 

8 

9 

DRILL    EXERCISE  No.  3. 

91.     1.     Write  all  the  combinations  of  two  figures  that  make  2, 3,4,  5,  G,  7, 
8  aiul  9. 

2.  Write  all  the  combinations  of  two  figures,  that  make  10,  11,  12,  13,  14,  15, 
IG,  17  and  18. 

Repeat  the  following  exercises  several  times,  naming  the  results  as  rapidly  as 
you  can,  and  continue  the  exercise  for  several  consecutive  days: 

3.  Commence  with  1  and  orally  add  thereto  2,  and  continue  to  add  2  to  the 
successively  occurring  sums,  until  you  produce  101.     Thus,  3,  5,  7,  9,  11,  13,  etc. 


DRILL     EXERCISE. 


45 


4. 

4,  7, 10, 

5. 

5,  9,  13, 

6. 
7. 
8. 

9. 
10. 

11. 
12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
30. 


DRILL  EXEEOISE  Xo.  3.— Continued. 

Coinmeuce  with  1  aiul  iu  like  mauuer  add  3  until  youx)i"oduce  100.  Tims, 
13,  etc. 

Coinmeuce  with  1  and  in  like  mauuer  add  4  until  you  produce  101.  Thus, 
17,  etc. 

Commence  vrith  1  and  in  like  manner  add  5  until  you  produce  101. 

Commence  with  1  and  in  like  manner  add  6  until  you  produce  CI. 

Commence  M'ith  1  and  iu  like  manner  add  7  until  you  produce  71. 

Commence  with  1  and  in  like  manner  add  8  until  you  pi'oduee  81. 

Commence  with  1  and  in  like  manner  add  9  until  yovi  produce  91. 

Orally  add  by  2's  until  you  produce  20. 


11 

3's 

a 

u 

4's 

it 

u 

5's 

u 

il 

O's 

(. 

u 

7's 

u 

u 

8's 

u 

u 

O's 

u 

;& 

lO's 

iL 

u 

ll's 

u 

11 

12's 

11 

(1 

13's 

11 

u 

ll's 

11 

li 

lo's 

11 

11 

30. 

11 

40. 

11 

50. 

11 

60. 

(1 

70. 

11 

SO. 

11 

90. 

11 

100. 

11 

110. 

11 

120. 

11 

130. 

11 

140. 

11 

150 

Commence  at  1  and  orally  add  3  and  5,  alternately,  until  you  produce  100. 
Commence  at  1  and  orally  add  4  and  7,  alternately,  until  you  produce  100. 
Add  by  2's,  3's  and  4's  alternately  from  0  to  99. 


92.     Add  the  following-  problems  orally 


5+C+S=? 

7+4+9= 

6+7+5= 

9+8+8=? 

4+7+9= 

8+9+6= 


14+  7+  3=? 
21+  0+  2=1 

7+12+  0=? 
2G+  9+12=? 

8+23+  5=? 
10+15+10=? 

What  is  the  sum  of  8  peaches  and  10  oranges  ? 

What  is  the  amount  of  four  O's  plus  three  O's  ? 

Add  twelve  thousand,  twelve  hundred  and  twelve. 

What  is  the  sum  of  twenty  blackbirds,  twenty  sick  birds  and  twenty-six  red  birds  ? 


22+  8+  1  =  ? 
12+  9+  7  =  ? 
4+  9+  8=? 
20+40+  3  =  ? 
22+14+  4  =  ? 
17+19+18=? 


24+1G+  5+  S=1 
7+  9+30+  5=? 
18+12+15+  C=? 
43+  7  +  14+  n  =  ? 
72+21+  9+14=? 
09+15+12+36=? 


46 


SOULES    I'lIILOSOPHIC    PRACTICAL    MATHEMATICS. 


WKITTEX    TKOBLEMS    IN    ADDITION. 


93.     Ada  tlie  folluwiiiy  iiuinbers,  0370,  504,  309,  485,  aud  5092: 

OPERATION. 

^  .  KxpJanalion. — In  all  nddition  prolileras,  we  first  write  the  mimbeM 

"5.3  so  tliat  units  iif  tlie  saiuu  onU^r  stand  in  the  same  column,  i.  e.  units  in  the 

r.i:  .^  units',    or   first  colunin;    tina  in   the   tens',   or  second  column;    hundreds 

in  the  linndreds',  or  third  colnnui,  and  so  ou  through  the  unmliers.  We 
tlien  liegin  at  the  units' (u- first  column  aud  add  the  columns  separately. 
In  adding  the  first  colunui,  we  commence  with  the  2  aud  5,  aud  name  only 
the  successive  results;  thus,  7,  16,  20,  26,  which  is  2  lens  and  6  units ;  the 
6  we  write  in  the  first  jilace,  or  column  of  units,  and  place  the  2  tens 
which  is  to  he  carried  to  the  column  of  tens,  directly  below  the  6  in  a 
small  figure.  Then  adding  the  2  tens  to  the  tens'  column,  we  say,  11,  19, 
25,  32 — which  is  3  hundreds  and  2  tens;  the 2  tens  we  write  in  the  column 
of  tens,  and  place  the  3  hundreds  which  is  to  he  carried  to  the  hundreds' 
column,  directly  under  it.  Then  adding  the  3  hundreds  to  the  hundreds' 
column,  we  say,  7,  10,  1.5,  18,  which  is  1  thousand  and  8  hundreds;  the  8 
hundreds  we  write  in  the  hundreds'  column,  and  the  carrying  figure,  1 
thousand,  directly  under.  Then  adding  the  1  thousand  to  "the  fourth,  or 
thousands'  column,  we  say,  6,  12,  which  is  1  ten  thousand  and  2  thousands,  aud  this  being  the  last  col- 
umn to  add,  we  write  the  figures  in  their  respective  columns  aud  produce  12826  as  the  sum  of  all 
the  numbers. 

Note. — When  adding,  set  the  result  in  pencil  figures,  being  careful  to  place  the  carrying  fig- 
ure or  figures  directly  beneath  the  answer  figure  of  each  column  added,  as  shown  in  the  preceding 
problem. 

Different  methods  of  using  the  carrying  figure,  elucidated  by  tlie  following 
examples : 


HWHtJ 

6370 
564 
309 

485 
5092 


Sum,    12,82  6 
Carriling  figures,  132 


FIRST — REGULAR  METHOD. 

$  5462.48 

$  5462.48 

848.19 

848.19 

761.78 

761.78 

81484.52 

81484.52 

7948.93 

7948.93 

852.57 

852.57 

$97358.47 

.37 

1432.33 

3.1 

25. 

33 

40 

13 

8 

SECOND   METHOD. 


or 


37 
31 
25 
33 
40 
13 
8 

$97358.47 


THIRD  METHOD. 

$  5462.48 

848.19 

761.78 

81484.52 

7948.93 

852.57 

8 

13 

40 

33 

25 

3.1 

.37 

$97358.47 

$97358.47 

Explanation. — By  the  first  or  Regular  Method,  which  is  preferable  for  accountants,  the  carry- 
ing figure  is  written  in  a  small  figure  under  the  resultant  figure  of  the  column  added. 

By  the  second  method,  we  conunence  ou  the  right  and  add  each  column  independently  of  the 
other  columns,  and  set  the  results  under  the  numbers  added,  or  to  the  right  of  them,  as  is  shown  in 
the  operation;  then  the.se  several  results  are  added,  and  produce  the  sum  or  amount. 

By  the  third  method,  we  commence  on  the  left  and  add  each  column  independently  of  the 
other  columns,  and  set  the  results  as  shown  in  the  operation  ;  then  we  add  these  several  results  and 
produce  the  sum  or  amount. 

It  will  be  observed  that  in  the  second  and  third  methods  the  units,  tens,  hundreds,  thousands, 
etc.,  are  written  in  their  respective  columns. 

Note. — In  the  second  aud  third  methods,  the  results  of  the  separate  columns  may  he  written 
to  the  right  or  left  of  the  numbers  added,  or  on  a  separate  piece  of  paper,  and  the  final  result 
thereof  written  beneath  the  numbers  added. 

Note  2.  The  second  method,  shown  above,  is  sometimes  called  the  Civil  Service  Method, 
because  it  is  used  by  many  of  the  Civil  Service  Clerks. 


"■  GENERAL    DIRECTIONS    FOR    ADDITION.  4/ 

GENERAL   DIRECTIONS    FOE    ADDITION. 

94.  From  the  foregoiug  elucidatious,  we  derive  tlie  following  general  direc- 
tions for  addition  : 

1.  Write  the  numbers  so  that  units  of  the  same  order  stand  in  the  same  column. 
See  explanation,  page  4G. 

2.  Begin  at  the  units\  or  first  column  on  the  right,  and  add  each  column  sepa- 
rately, irritiiig  the  unit  result  under  the  column  added,  and  carrying  the  tens,  if  any, 
to  the  tens^  column.  At  the  last  column,  write  the  full  sum  of  the  column.  See  explan- 
ation, page  4G. 

3.  To  add  horizontally,  the  numbers  are  not  icritten  in  columns  of  like  orders ; 
and  the  result,  or  sum,  is  icritten  to  the  right  of  the  numbers. 

PROOF    OF    ADDITION. 

95.  Tbe  best  proof  of  the  correctness  of  addition  is  for  the  calculator  to  be 
proficient  in  his  work,  and  then  re-add  the  columns  in  the  reverse  direction.  Cast- 
ing out  the  9's,  as  is  sometimes  done,  is  not  positive  proof  of  correctness,  as  the 
figures  may  be  transposed,  or  replaced  by  wrong  ones  having  the  same  sum,  or 
mistakes  may  be  made  that  balance  each  other,  and  the  work  may  appear  to  prove 
right  when  it  is  wrong;  though  it  will  never  prove  wrong  when  it  is  right;  hence 
many  accountants  in  verifying  their  calculations,  prefer  to  repeat,  or  go  over  the 
addition  in  a  reverse  direction. 

When  a  column  is  added  in  the  reverse  direction  and  the  result  disagrees 
with  the  first  addition,  then  omit  the  bottom  figure  of  the  column  when  adding  it  up 
and  the  top  figure  of  the  column  when  adding  it  doicn,  until  all  the  other  figures  of 
the  column  have  been  added,  and  then  add  the  omitted  figures.  By  this  means,  the 
combination  in  which  the  error  has  been  made  is  changed,  and  the  error,  which 
might  be  repeated  indefinitelj',  is  avoided. 

See  page  152  for  a  system  of  jiroof  of  addition,  subtraction,  multiplication, 
and  division,  and  also  of  the  proof  of  the  correctness  of  the  transfer  of  numbers. 

WEITTEN  PROBLEMS. 

96.  Add  the  following  numbers,  and  verify  the  work  by  reverse  addition : 

3i>9  Explanation. — Commence  at  the  top  of  the  right  hand  column  and  add,  naming 

fi'i7  results  only,  thus :  13,  20,  22,  30,  37,  42,  49  ;  then  add  the  4  tens  to  the  second  column 

goo  and  proceed  in  the  same  manner,  9,  10,  13,  22,  28,  32,  34,  etc. 

<68  In   adding,    the  thought  of   words  and  the   repetition   of   results   should   be 

K4.7 

Tt'  banished  from  the  mind.     Never  allow  yourself  to  spell  your  way  or  crawl  up  or 

4--0 

"07  down  a  column,  as  is  too  often  done  by  many  teachers  and  their  pupils.     Thus:  9 

and  4  are  13,  13  and  7  are  20,  20  and  2  are  22,  22  and  8  are  30,  30  and  7  are  37,  37  and 

6429  5  are  42,  42  and  7  are  49. 


48 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


WEITTEN  niOBLEMS.— Coiitiiuied. 


0) 

(3) 

W 

(5) 

(6) 

(7) 

(8) 

(9) 

47 

63 

124 

845 

4072 

6913 

540314 

687659 

72 

28 

690 

324 

1867 

2456 

123450 

789021 

12 

34 

658 

97 

504 

4379 

342189 

90876 

45 

98 

67 

8 

1090 

308 

67310 

427803 

68 

92 

623 

326 

68 

6547 

102938 

608091 

21 

32 

489 

72 

607 

7085 

125439 

623784 

33 

99 

109 

216 

92 

2123 

334455 

9S9790 

64 

55 

327 

123 

4173 

3458 

121212 

602548 

25 

C7 

432 

78 

8293 

6565 

304958 

553241 

35 

23 

399 

666 

8902 

4507 

12345G 

789021 

ADDITION    BY    GEOUPING    THE    FIGUEES. 

97.  This  is  a  very  rapid  system  of  adding  aad  consists  in  groupiug  two  or 
more  figures  in  a  column  and  naming  tlie  result  of  tlie  group,  in  the  same  manner 
as  we  group  and  name  the  result  of  several  letters  in  vords,  when  we  read.  The 
following  problems  will  illustrate  this  method: 


17 


-I 

37  I 
49! 


3  7^13 

6  5  (  00 

84  i" 

n|34 


61 
69  I 


49 


92  > 
30( 

(»  4  \ 

09  1 


Explanation. — Commence  at  tlie  top  of  the  first  column  and  add  thus  :  13,  22, 
34,  41,  49,  Gl.  Then  add  the  6  tens  or  carrying  figure  to  the  second  column  and 
coutiuue  as  before,  17,  31,  37,  49,  61,  69. 

Note. — To  attain   a  high  degree  of  skill  in  this  method,  the  student 
^'•ould  know  perfectly  the  alphabet  of  numbers,  as  given  on  page  41. 


DEILL  EXEECISE. 


98 

grouiied,  as 
in  practical 


In  the  second,  third  and  fourth  examples  fo 
indicated  by  the  braces,  as  they  would  uatur; 
addition : 


lUowing,  the  columns  are 
iilly  be  grouped  mentally, 


(■>) 

4  6  3 

18  9 

13  5 

2  4  5 

8  9  0 

9  8  7 


2  3 
0  8 
4  5 
2  3 
5 
7 
3 
3 


(7) 
3  0 


3  4 
0  6 
5  6 
2  4 
2  0 
7  9 

4  3 

5  2 
0  9 


HORIZONTAL    ADDITION'. 


49 


lu  adding  long  columns,  the  rapidity  may  be  often  increased  by  grouping  and 
adding  suck  numbers  as  make  10  or  i!0,  independently  of  other  groupings. 

Where  the  same  figure  occurs  more  than  once,  multiply  it  instead  of  adding. 
Thus,  if  the  figure  7  occurs  4:  times,  multiply  it  by  -1  and  use  the  product  instead  of 
adding. 

Where  figures  occur  in  regular  order  and  the  uumber  is  odd,  thus,  5,  G,  7,  or 
19,  IS,  17,  IG,  15,  then  as  many  times  the  middle  number  as  there  are  numbers  iu 
regular  order,  will  be  their  sum. 

Wheu  the  luimber  iu  regular  order  is  even,  thus,  2,  3,  4,  5,  or  17,  IG,  15,  14, 
13,  12,  then  oue-half  as  manj^  times  the  sum  of  the  extremes  as  there  are  numbers 
iu  regular  order,  will  be  their  sum. 

Accountants  generally  add  down  the  column  so  that  the  sum,  when  found,  is 
at  the  foot  of  the  column  where  it  is  to  be  written,  and  to  prove  the  result,  they 
re-add  the  column  upward. 


HORIZO:^fTAL  ADDITION. 

99.  This  method  of  addition  consists  in  adding  numbers  from  right  to  left,  or 
from  left  to  right,  wheu  they  are  written  in  horizontal  lines  or  when  they  are  uot 
■written  in  vertical  columns  of  like  orders. 

The  method  is  very  valuable  in  business  and  is  in  general  use  by  the  most 
efficient  clerks  and  accountants. 

Note   1. — Wheu  using  this  method  great  care    is  required  to  add,  only  figurks  of   like 

ORDER. 

Note  2. — The  method  of  adding  two  figures  of  the  numbers  at  a  time  may  be  used  -when 
adding  horizontally. 

Add  horizontally  the  following  numbers  : 

(1)  45G,  248,  3947,  1084,  2G81,  89020  =  9743G. 

Note. — When  adding  horizontally,  add  toward  the  side  at  which  the  total  result  is  to  be 
written.  Thus  in  the  above  we  commence  on  the  left  and  say  6.  14,  21,  25,  26;  then  (carrying  the  2 
to  the  5)  7,  11,  1.5,  23,  31,  33;  then  (carrying  the  3)  7,  9,  18,  24 ;  then  (carrying  the  2)  5,  6,  8,  17; 
then  (carrying  the  1)  9,  which  completes  the  result. 

Add  the  following  uumbers  horizontally : 

(2)  245,     3G2,  2581,  409,  G87,  5498,  G4,  7910G2  = 

(3)  79,     508,     309,     84,  50G,  1742,  18,  164307,  4235,  78215  = 

(4)  51G,  4027,     684,     97,     G8,  1493,  54,  352085,     781,     5638,  721409  = 


Add  the  following  uumbers  horizontally  and  vertically ; 


(5) 


4345 
2214 
3625 
5G25 
1212 
1198 
4621 
5731 


1542 

214 

2153 

2101 

3120 

28 

982 

4842 


5136 

136 

214 

9876 

2152 

73 

4215 

1563 


6362 
1043 
256 
2123 
1312 
2006 
2982 
2842 


3809 
2132 
3429 
1215 
2121 
7849 
1215 
1563 


3908 
1025 
836 
3724 
3002 
2150 
1098 
4910 


2621 

47 

215 

2152 

82 

6 

4635 

3125 


***** 
***** 
***** 
***** 
»*#*• 
***** 
•  ••»• 
•»•*• 


5o  soule's  niiLosoPHic  practical  mathematics. 

DKILL    EXEECISES. 
100.     What  is  the  sum  of  each  of  tlie  followiug  columus  of  uunibers? 


(1) 

(2) 

(3) 

(4) 

(5) 

(B) 

4304 

7S0 

890 

777 

9040 

98475 

291 

12C1 

706 

888 

1288 

8697 

G43 

537 

73 

999 

9907 

46789 

f|S 

309 

4009 

066 

6543 

57698 

1400 

0987 

8888 

645 

2018 

94576 

2099 

72SG 

5679 

h»S 

5281 

49567 

8877 

1896 

2834 

777 

6745 

84685 

101,  Add  the  following  numbers  horizoutally : 

(1)  248,     3936,       409,     1278,       97,         563,     9210,  45682  = 

(2)  335,     1468,         87,       911,  1809,     19068,         54,  17401  = 

(3)  72,  13615,  41848,     1905,         8,       9703,       691,  8625  = 

102.  Drill  daily  on  the  following  columus  of    numbers  with  and  without 
grouping,  until  you  acquire  great  rapidity: 


(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

818 

412 

582 

328 

809 

981 

864 

677 

595 

849 

390 

297 

578 

346 

523 

350 

363 

305 

249 

283 

970 

318 

757 

386 

605 

269 

629 

420 

463 

327 

276 

824 

420 

672 

848 

789 

145 

982 

830 

651 

752 

932 

731 

793 

945 

696 

174 

217 

221 

543 

843 

373 

542 

864 

397 

136 

144 

326 

232 

502 

865 

576 

853 

965 

684 

169 

176 

111 

151 

113 

129 

876 

684 

448 

976 

295 

767 

871 

387 

438 

768 

444 

743 

404 

666 

468 

644 

512 

516 

455 

904 

102 

915 

151 

217 

687 

747 

814 

247 

328 

972 

814 

686 

148 

879 

825 

156 

376 

331 

633 

114 

331 

637 

263 

516 

951 

106 

468 

281 

624 

346 

554 

917 

295 

259 

784 

872 

189 

828 

581 

545 

161 

650 

101 

896 

122 

694 

177 

986 

491 

622 

197 

411 

461 

864 

440 

788 

885 

817 

888 

749 

490 

237 

874 

565 

450 

866 

264 

918 

992 

717 

876 

349 

898 

150 

414 

944 

294 

289 

202 

222 

902 

489 

769 

514 

654 

922 

896 

259 

548 

234 

396 

698 

243 

446 

789 

116 

597 

381 

365 

166 

484 

228 

174 

576 

458 

911 

814 

329 

208 

365 

235 

433 

952 

489 

747 

866 

277 

678 

662 

272 

386 

949 

683 

394 

636 

179 

476 

640 

764 

729 

624 

687 

574 

407 

241 

129 

716 

821 

287 

955 

897 

762 

956 

812 

477 

659 

802 

457 

848 

177 

477 

849 

658 

798 

681 

778 

584 

587 

255 

12978 

Note. — By  prefixing  1  to  the  second  line  from  tbe  top,  anil  affixing  the  unit  figure  of  the  third 
line  to  the  second  line,  the  footing  or  sum  of  each  problem  is  shown. 


*  ADDITION    OF    FRACTIONAL    NUMBERS  5 1 

TO  ADD  THE  FRACTIONAL  NUMBERS,  i  OR  J. 

103.  lu  business,  the  iiieasuiemeiits  of  some  lines  of  goods  are  recorded  to 
the  accuracy  of  J  and  ^  of  the  unit  of  iiicasurenicnt,  and  such  parts  or  fractions  are 
written  in  small  Hyures  to  the  right  and  at  the  top  of  the  whole  number  thus,  41', 
39^,   40-  would    be   read,    in  cases  where  the   small   figures   represent  fourths, 

FORTY-ONE  AND  ONE-FOURTH,  THIRTY-NINE  AND  THREE-FOURTHS,  and  FORTY  AND 

TWO-FOURTHS.     lu  case  tlie  small  hgures  indicated  eicjhths,  the  numbers  would 
be  read  41  and  one-eighth,  39  and  thkee-eiohths,  and  40  and  two-eighths. 

Note. — This  work  is  presented  in  ndvance  of  fractions,  on  acconut  of  its  importance  in  liusi- 
ness.  It  is  believed  that  it  can  be  readily  nuderstood  by  the  practical  sense  of  students  who  have 
not  yet  studied  fractions. 

Add  the  following  numbers  vertically  and  horizontally: 

41',  39%  40=,41%  39',  38-',  38%  40%  41%  39^ 

operation  horizontally. 

41%     39%     40%     41%     39%     38%     38%     40%     41%     39^  =  401^. 

Explanation. — Commencing  at  the  top  of  the  vertical  column  or  at  the 

left  of  the  horizontal  line  and  adding  the  small  figures  indicating  fourths,  we 

obtain  23  fourths.     Then,  since  4  fourths  equal  one,  in  23  fourths  we  have  5 

and  3  fourths.     Then  adding  the  5  to  the  other  whole  numbers,  we  produce  401 

and  3  fourths. 

XoTE. — In  business,  where  there  are  many  bales,  bags,  gallons,  yards,  pounds,  etc.,  to  add, 
it  is  usual  to  set  the  numbers  to  be  added  in  columns  of  10,  as  shown  in  the  following  examples : 

What  is  the  total  weight  of  the  following  78  bales  of  cotton  ■? 
(Add  the  footings  of  the  columns  horizontally). 


OPERATION 

41^ 

39' 

40* 

413 

391 

38» 

38* 

402 

413 

393 

401  a 

(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

425 

465 

525 

546 

499 

500 

458 

519 

465 

519 

423 

424 

426 

427 

428 

425 

501 

502 

503 

501 

504 

506 

507 

508 

509 

510 

511 

512 

513 

514 

515 

516 

430 

431 

423 

433 

444 

445 

446 

447 

517 

518 

519 

520 

523 

524 

525 

526 

448 

449 

450 

451 

452 

453 

454 

455 

527 

546 

537 

536 

534 

.532 

529 

544 

528 

529 

530 

531 

532 

533 

534 

546 

545 

543 

540 

538 

458 

455 

iH^r-i*  i.*rlt  *#**  »»»#  #♦•»  #-^»  *»*»  »»»#_.*'•»*'»    lbs. 

Note. When  weighing  bales  of  cotton,  it  is  not  the  custom  to  consider  the  fractions  of  pounds. 

Find  the  total  weight  of  the  following  96  bags  of  coffee : 
(Add  the  footings  of  the  columns  horizontally). 


(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

160 

161 

162 

163 

164 

165 

163 

160 

162 

161 

161 

165 

164 

160 

163 

162 

161 

165 

163 

163 

165 

164 

163 

162 

161 

160 

161 

162 

161 

162 

164 

163 

162 

161 

160 

165 

164 

163 

161 

162 

163 

162 

161 

160 

165 

164 

163 

162 

162 

160 

162 

161 

160 

165 

164 

163 

162 

160 

160 

161 

165 

160 

161 

162 

163 

164 

165 

162 

162 

160 

161 

162 

163 

164 

165 

160 

164 

161 

165 

164 

163 

162 

161 

160 

165 

161 

160 

160 

161 

162 

163 

164 

165 

163 

160 

163 

_    ••••♦IbB. 


52 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Weekly  Report  of  Mail  Matter  Received  at  a  City  Post  Ofliice. 


Class  ov   Mail. 

Mox. 

TUES. 

Wed. 

Thuus. 

Fri. 

Sat. 

Total. 

29246 

1657 

5129 

1844 

950 

41695 

28561 

1823 

4214 

1297 

720 

42322 

29203 

1654 

3942 

1329 

604 

44275 

27693 

1236 

4870 

16.56 

898 

46889 

27367 

1342 

5986 

894 

471 

57318 

39985 

394 

4652 

773 

504 

40207 

***»*« 

«•.««# 

Postal  Ciirtls 

#**#*!*■ 

Book  Packets 

#  »f*»» 

Parcels 

Newspaiters 

*  ##  *  # 

Total 

*#»f  #  * 

Weekly  Report  of  a  Circulating  Library. 


Class  of  Books. 

Mox. 

TUES. 

Wed. 

TnuRS. 

Fri. 

Sat. 

Total. 

4761 
1322 
1215 
2424 
1316 
198 

3857 
1421 

1110 

1324 

1218 

408 

1935 
1202 
1313 
1387 
1400 
108 

4560 
2121 
1214 
1216 
878 
135 

2976 
1322 
1124 
1317 
420 
308 

4085 
1382 
1174 
1566 
817 
297 

***** 

***** 

***** 

***** 

***** 

***** 

Total         

*  J'  *  • 

».,. 

.... 

.«.. 

#■*#• 

**** 

***** 

ADDITION    OF    DOLLARS    AiSTD    OEXTS. 


104.  Dollar  and  Cent  Signs.— The  dollar  sign  is  $,  and  tbe  cent  sign  is  ^. 
When  the  dollar  sign  is  placed  before  numbers,  they  are  read  as  dollars.  Thus,  $45 
is  read  45  dollars.  When  the  cent  sign  is  placed  after  uninbers,  they  are  read  as 
cents.  Thus,  14/?  is  read  14  cents.  When  dollars  and  cents  are  written  together, 
the  cents  are  separated  from  the  dollars  by  a  decimal  point  (  .  )  and  the  sign  of 
cents  is  omitted.    Thus,  $10.45  is  read  10  dollars  and  45  cents. 

Since  there  are  100  cents  in  1  dollar,  cents  always  occupy  two  places  and  only 
two,  in  connection  with  dollars.  When  the  number  of  cents  is  less  than  10,  a  naxf/lit 
must  be  used  to  till  the  tens^  column,  or  the  first  place  at  the  right  of  the  point. 
Thus,  8  dollars  and  5  cents  are  written  $8.05. 

When  cents  only  are  written,  tbey  are  expressed  as  follows :  25  cents  or  25/ 
or  $.25. 

When  writing  numbers  representing  dollars  and  cents  for  the  purpose  of 
addition,  they  must  be  set  so  that  dollars  will  be  under  dollars,  and  cents  under 
cents,  in  the  regular  order  of  units,  tens,  hundreds,  etc.,  and  the  decimal  points  ( , ) 
that  separate  dollars  and  cents  must  be  in  a  vertical  line. 

The  dollar  sign  (  $  )  and  the  point  ( . )  should  never  be  omitted  when  writing 
dollars  and  cents,  except  when  writing  in  books,  or  on  paper,  which  contains  dollar 
and  cent  columns. 

105.  The  United  States  Monetary  Units  are  as  follows : 

10  mills  (m)  =  1  cent,  /.  10  dimes    =  1  dollar,  $. 

10  cents        =  1  dime,  d,  10  dollars  =  1  eagle,  E. 


106.    Add 


PROBLEMS. 

^3 

PROBLEMS. 

(1) 

(2) 

(3) 

W 

(5) 

(6) 

$321. 

$521.16 

$  9.45 

$    431, 

$194.15 

$684.22 

640.80 

83.25 

80. 

124. 

8.05 

708.10 

9.13 

19.30 

17. 

381. 

73.75 

276.14 

75.20 

S. 

.65 

569. 

6.13 

582.98 

100.05 

4.07 

0.10 

827. 

.95 

928.35 

$1146.18        $635.78       $113.20 


$2332 


$283.03         $3179.79 


7.  Add  $108,  $97.16,  $84.12,  $  .75,  $8,  $6.40,  25/,  $18.        Aus.  $322.68. 

8.  Add  $580.10,  $671.23,  $794.98,  $88,  45/,  5/,  $3.10.       Ans.  $2137.91. 

9.  Add  $999.99,  $888.88,  $777.77,  $066.66,  $555.55,  $444.44,  $333.33,  $222.22, 
$111.11,  and  1/.  Ans.  $4999.90. 

10.  Add    $987.65,     $876.54,     $765.43,     $054.32,     $543.21,    $123.45,    $234.56, 
$345.67,  $456.78,  $567.89,  $678.90,  and  $789.  Ans.  $7023.40. 

11.  James  has  $420;  Conrad  has  $130  more  than  James  ;  and  Henry  lias  as 
much  as  James  and  Conrad  tofretber.    What  sum  have  the  three  ?        Ans.  $1940. 

12.  A  lady  paid  $42.75  for  a  dress,  $80  dollars  for  a  shawl,  $21  for  a  bonnet, 
and  $8.75  for  a  pair  of  shoes.     What  was  the  total  cost  1  Ans.  $152.50. 

]  3.    Add  the  following  coluunis  vertically  and  horizontally,  and  verify  the  work 
by  adding  the  results : 

Totals. 


1468 
123 
371 
987 
818 
972 
729 
235 
328 
848 


$306 
789 
930 
270 
390 
434 
217 
490 
809 
789 


»  48 
437 
270 
103 
970 
545 
297 
684 
981 
169 


$516 
213 
418 
682 
276 
274 
318 
582 
346 
123 


18 

$233 

40 

78 

478 

09 

55 

407 

54 

76 

823 

47 

43 

752 

34 

90 

717 

30 

82 

249 

32 

57 

V57 

42 

24 

523 

45 

45 

234 

56 

$108 
890 
173 
527 
843 
222 
576 
543 
350 
456 


$410 
289 
157 
193 
865 
166 
554 
684 
386 
890 


$337 
869 
687 
184 
976 
527 
161 
560 
605 
256 


ADDITIOX  OF  SEVERAL  COLUMNS  IN  ONE  OPERATION. 


107.  In  many  cases,  the  addition  of  several  columns  at  one  operation  will 
greatly  expedite  the  accountants'  work ;  but  in  long  columns  of  solid  figures,  the 
addition  by  grouping,  as  above  explained,  is  by  far  the  most  rapid  and  practical. 

The  following  will  illustrate  the  primary  work  of  adding  several  columns  at 
once: 


Add: 

47 
18 
71 
62 
94 
26 

318 


Explanation.— Commence  at  the  top  and  proceed  by  adding  to  tbe  upper 
number  47,  first  the  tens  of  the  next  number  below,  then  the  units,  and  in  like 
manner  continue  till  all  are  added,  thus :  47  +  10  =  57  +  8  =  65  +  70  =  135  + 
1  =  136  +  60  =  196  +  2  =  198  +  90  =  288  +  4  =  292  +  20  =  312  +  6  =  318, 
the  sum  of  the  two  columns. 


54  soule's  philosophic  practical  mathematics.  * 

By  iiamiiiff  results  only,  which  in  practice  should  be  doue,  we  have  47,  57,  65, 
135,  13G,  196,  19S,  2SS,  292,  313,  318. 

2.        Add : 

^Ij"^;  Explanation. — Commeuciug  at  the  top, 

15S7  1     3691  2     5091 

^^^*^  =^     6796  «     7296 


we  have : 

^     5791 

4 

5796 

•^  7376 

8 

7383 

11593  »  11383         1°  11583         "   11593,  the  answer. 

These  operations  show  the  basis  of  adding  several  columns  siniultaueously,- 
and  though  the  method  is  too  laborious  to  be  of  material  advantage  in  long  columns, 
it  should  be  practiced  until  the  mind  can  easily  and  readily  retain  large  and  vary- 
ing numerical  results. 

In  practice,  we  shorten  the  work  very  much  by  combining  and  adding  whole 
amounts  at  once.  Thus,  in  the  second  problem,  by  combiuiug,  as  we  would  naturally 
do  iu  practice,  we  would  have  iu  the  two  upper  numbers  5796,  then  adding  the 
bundreds'  and  thousands'  figures  (15)  of  the  third  number,  we  have  7296;  then  add 
the  units'  and  tens'  figiu-es  of  the  third  number,  we  have  7383,  to  which  add  the 
fourth  number  at  once,  and  we  have  11593,  the  correct  result. 

To  show  clearly  how  to  contract,  and  to  practicalize  the  above  system,  we 
present  the  following  combinations,  which  should  be  carefully  studied. 

ADD  THE  FOLLOWING  NUMBERS  : 

(3)         (4)          (5)          (6)          (7)          (8)          (9)  (10)         (11)  (12)  (13)  (14) 

23        52        84        68  98        88        97  96        95  89  99  89 

16        36        95         72  59        46         89  99        88  87  98  98 

39        88       179       140       157       134       186       195       183       176       197       187 

Explanation. — In  tbo  tliird,  fourth  and  fifth  problems,  the  sum  of  the  units'  figures  being  less 
than  10,  the  whole  sum  is  instantly  seen. 

In  problems  G,  7,  8,  9,  10,  11,  12,  13  and  14,  we  instantly  see  that  the  sum  of  the  units'  figures  is  10, 
or  an  excess  of  10,  and  hence  we  know  that  the  snm  of  the  tens  is  to  be  increased  l)y  1,  and  without 
thinking  what  the  excess  is,  we  write  the  result  of  the  teus'  figures,  which,  according  to  the  combi- 
nations, shown  in  tables  1  and  2,  we  know  without  adding,  and  while  writing  this  result,  we  give  an 
instantaneous  thought  to  the  exact  sum  of  the  units'  figures. 

ADD   THE   FOLLOWING  NUMBERS  : 

(19)          (20)           (21)            (22)          (23)  (24)  (25)  (26) 

126        342        485        1627        178  526  395  2653 

75          92         304          211        201  308  296  1210 


(15) 

(16) 

(17) 

(18) 

74 

86 

88 

68 

98 

95 

97 

91 

172       181       185       159        201        434         789         1838        379        834        691        3863 

Explanation. — In  these  problems,  from  15  to  20  inclusive,  we  show  how  to  facilitate  and  expe- 
dite the  work  when  one  number  apiiroximates  100.  Thus,  iu  problem  13,  we  first  mentally  add  100 
to  the  74  which  makes  174,  and  then, -because  the  100  added  is  2  more  than  the  98,  which  should 
have  been  added,  we  mentally  deduct  2  from  174,  and  write  172,  the  correct  sum.  In  problems  16,  17, 
18,  19,  and  20,  we  mentally  add  to  each  upper  number,  100,  and  then  from  the  several  sums 
thus  produced,  wo  mentally  deduct,  respectively,  5,  3,  9,  25,  and  8.  Iu  problems  21  and  22,  we  first 
meutally  add,  respectively,  300  aiul  200,  and  thcu  to  the  respective  sums  thus  obtained,  we  mentally 
add  4  and  11.     The  other  numbers  are  added  iu  like  manner. 

Note. — Combinations  similar  to  the  foregoing  problems,  3  to  26,  must  be  practiced  until  the 
student  can  easily  and  rapidly  iierform  the  work,  otherwise,  proficiency  in  addition  and  multipli- 
cation cannot  be  attained.     _ 

Rapid  work  in  simultaneous  multiplication  requires  the  instantaneous  sum  of  two  pairs  of 
numbers,  and  hence,  again,  the  im])ortaiice  of  proficiency  in  this  work. 

Note. — Iu  using  numbers  mentally  or  orally,  the  calculator  shoubl  always  think  or  speak  the 
result  in  pairs,  thus,  2476  and  158,  should  be  read  mentally  or  orally,  24,  76,  and  1,  58. 


DRILL     PROBLEMS. 


55 


DRILL  PROBLEMS. 
108.     Add  the  followiug  coluinus  at  oue  operatiou,  and  drill  on  the  same,  at 
difl'ereut  sittings  or  lessons,  until  you  achieve  rapidity  and  accuracy  : 

87         UO         184        04        85        240        264        429         703        843         708 
96  9.5  72 


OS 


13 


103 


187 


367 


241 


427 


.524 


— 

04 

27 

92 

58 

42 

321 

284 

316 

475 

762 

27 

81 

30 

21 

88 

406 

343 

509 

666 

533 

18 

45 

47 

76 

98 

463 

220 

704 

988 

487 

74 

47 

362 

432 

567 

4807 

463 

65342 

123456 

21 

88 

400 

234 

405 

621 

247 

372>6 

906342 

36 

92 

29 

444 

219 

1282 

5002 

22406 

421362 

48 

81 

391 

843 

924 

1096 

1603 

1321 

187307 

19 

67 

132 

777 

633 

2761 

4120 

74350 

8899 

27 

41 

254 

96 

111 

7932 

2031 

64275 

364819 

86 

78 

345 

123 

456 

101 

501 

11111 

300405 

32 

72 

789 

202 

404 

8763 

1004 

24350 

871400 

59 

33 

876 

463 

99 

5832 

803 

60(i 

463220 

63 

12 

101 

508 

399 

1847 

160 

10000 

768 

58 

92 

999     387     555 
ADDITION  BY  OXE  OF  THE 

1234 

4512 
RTIES  ( 

56780 

102938 

PROPE 

)F  0 

109. 

1.  Wl 

lat  is  the 

•ium 

of  5 

ine 

s  of  numbe 

rs,  the  lii-st  being 

407?   A 

us.  2465' 

OPERATION:     467  1st  line. 

382  2d      " 

815  3d      " 

617  difference  between  2d  line  and  9. 
184  "  "        3d  line  and  9. 


2465  Ans. 

Explanation,— la  all  problems  of  this  character,  with  any  number  of  odd  lines,  the  answer 
may  be  produced  as  soon  as  the  first  line  of  numbers  is  given,  bi/  writing  tkejirst  line  minus  lite  num- 
ber of  9's  to  be  produced,  and  then  prefixing  the  number  of  9's  to  this  result.  Then,  wlieu  tlio  otber 
lines  are  set,  to  insure  the  result,  every  two  lines  must  make  9,  or',  in  otber  words,  one-baif  of  the 
other  lines  must  be  such  numbers  as  will,  when  added  to  the  remaining  half  alternately,  produce  9 
in  each  column  of  the  two  lines  added. 

In  this  problem,  we  subtract  2  from  the  first  liue  467,  and  write  465,  to  which  we  prefix  the  2 
and  thus  produce  the  answer  2465.  It  should  be  borne  in  mind  that  5  Hues  give  two  9's,  7  lines  three 
9's,  9  lines /oHc  9's,  etc. 

2.    What  is  the  sum  of  9  lines  of  numbers,  the  first  being  10644  ? 

OPERATION :      10644  1st  line. 

23456  2d      " 

13482  3d      " 

96780  4th     " 

25621  5th    " 

76543  difference  between  2d  line  and  9. 
86517  "  ■  "        3d      "   and  9. 

03219  "  "         4th    "    and  9. 

74378  "  "        5th    "    and  9. 


Ans.  410040, 


What  is  the  sum  of  7  lines  of  numbers,  the  first  being  86021 1      Ans.  386018. 


56 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


THE  VICENARY  SYSTEM  OF  ADDITION 

110.  The  Viceuary  System  combines  the  grouping  method  of  adding,  with 
a  method  of  recording  the  20's  ou  the  joints  of  the  fingers  as  fast  as  they  are  pro- 
duced by  the  operation  of  adding. 

Note. — This  system  was  eluc'ul.ateil  and  publisbed  by  the  author  iu  1873. 

All  expert  calculators  admit  that  the  most  rapid  system  of  adding  long  col- 
umns of  numbers  is  by  grouping  two,  three  or  more  numbers  and  mentally  naming 
the  result  of  the  group. 

But  such  a  .system  requires  a  constant  strain  on  the  mind  to  retain  the  large 
and  varying  results  and  to  make  the  several  additions  of  the  different  groups,  that 
with  many  it  is  wholly  impracticable.  To  avoid  this  mental  strain,  several  different 
methods  of  operation  have  been  suggested  by  difterent  calculators,  but  all  have  had 
some  demerit  that  has  rendered  them  more  or  less  objectionable.  With  a  view, 
therefore,  to  obviate  the  objections  and  difficulties  of  the  usual  methods  and  render 
the  work  simple,  comprehensible  and  practical,  we  present  a  new  system,  which, 
from  the  character  of  the  work,  we  name  the  Vlcenary  System.  The  leading  pecu- 
liarity of  this  system  consists  iu  recording  on  the  different  joints  of  the  fingers  and 
thumb  of  the  left  hand,  the  20's  as  soon  as  they  are  produced,  and  thus  freeing  the 
inind  from  all  effort  to  retain  large  results,  or  of  making  the  rather  difiBcult  addi- 
tions resulting  from  the  several  groupings. 


THE   VICENARY   SYSTEM  ELUCIDATED. 

111.  We  have  before  stated  that  the  leading 
characteristic  or  peculiarity  of  the  V^icenary  Sys- 
tem of  addition  consists  in  recording  the  20's  as 
soon  as  produced,  on  the  joints  of  the  fingers  of 
the  left  hand;  and  in  order  to  render  this  part  of 
the  operation  clear,  we  present  a  cut  of  a  hand 
with  the  different  joints  of  the  fingers  numbered, 
so  as  to  represent  the  values  that  we  record  on  them  in  adding.  The  manner  of 
making  the  record  of  the  20's  on  these  joints  is,  by  simply  placing  the  end  of  the 
thumb  on  the  finger,  or  the  end  of  the  first  finger  ou  the  thumb,  at  such  joint  as 
represents  the  value  to  be  recorded.  The  manner  of  making  the  record,  and  the 
value  of  the  several  joints,  should  be  well  understood  before  the  operation  of  adding 
is  commenced.  The  addition  tables,  ou  preceding  pages,  should  also  be  well  under- 
stood. 

In  making  the  addition,  we  first  mentally  group  enough  figures  to  produce  a 
result  not  less  than  10,  and  then  to  this  result  add  the  result  of  enough  other  fig- 
ures to  produce  a  second  result  not  less  than  20 ;  then  having  20  or  an  excess  of  20, 
we  record  the  20  on  the  hand,  by  placing  the  end  of  the  thumb  ou  the  first  joint  of 
the  first  finger  which  represents  the  value  of  20,  then  we  add  the  excess,  if  any,  to  the 
next  group,  and  continue  to  i>roduce  and  record  the  20's  until  the  column  is 
added.  If  it  is  desired,  cnouf/h  figures  to  produce  a  result  in  excess  of  30,  may  he 
grouped  at  once,  and  the  record  made  and  the  excess  carried  on  as  above  explained. 


■^■'^m^. 


*  THE    VICENARY    SYSTEM    OF    ADDITION.  5? 

112.     To  better  elucidate   the  viceuaiy   system,  we  present   tlie  following 
problems  and  explanations : 

(1)     Add  the  following  numbers  : 

82  ^_^-r— — ^  Fx])!a>intion. — Conimenciug  at  the  liottora  of 
gy  ^^-^^  S^^v'  *^'''  ""'•'^'  I'liliiiurii  ■^^t'  first  mentally  sec  13,  (7,  6), 
r-r.  ^Bk.  ^""^^^  \^_3--^  t(i  whiih  we  addit  ("i.  4),  whieli  makes  2:2;  we  then 
'"  f^^f  /J  ;\  M  ''•~^ — -  -^  leeoid  the  20  by  jdaeiiig  the  end  of  the  thumb  on 
89  m  pi.  '  /  \  g  '■  S  ^  S  3  the  first  joint  of  the  liist  tiiijjer.  Next,  we  see  11 
64  m\  ?■"■-»-  -•'■  /^  M  ■  X  ^  +  ^  '"'  ^>  ''"''  -^  '"  fxi'i'ss  of  21)1,  to  which  we  add  9 
38  \\l  1  "'""/  /  [  °  '  °-^  (3,  4,  2),  and  obtain  23;  then  wc  record  the  20  by 
Of.  jl  Iki't"''  /  /  ■  "  S  •■  °^  Jiassiug  the  end  of  the  thumb  to  the  end  of  th'o 
no  vl©^  ;  •'  >^o°  -  .TT'^  second  iiuKcr.  We  then  see  17  (6,  8,  and  3  in  ex- 
Si^                V      ^''^'M:.'-.,-...: ; -./   f  '  "  •    °-^           cess  of  20),   to  which   we  add    4  and   obtain   21  ; 

74  then  we  record  the  20  by  passing  the  end  of  tlio 

53  thumb  to  the  end  of  the  third  finger.     Next  we  see  16  (9,  6,  and  1  in  excess  of  20),  to 

,r  which  we  add  9,  {7,  2),  and  oiitaiu  25,  the  20  of  which  we  record  by  passing  the  end  of 

*)^  the  thumb  to  the  first  joint  of  the  fourth  finger,  and  the  5  we  write   in   the  units'  phico 

27  of  the  answer.     We  now  have  the  first  column  added,  and  by  inspecting  the  position  of 

84^  the  thumb  and  fingers  of  the  left  baud,  we  find  SO  recorded,  which,  with  the  5,  the  last 

(55  excess  of  20,  make  85,  the  sura  of  the  column.     Then,  to  add  the  second,  or  tens'  colunni, 

nn  we  first  mentally  see  16  (8  and  the  8  tens  from  the  units'  column)  plus  9  (3,  6)  =  25;  wo 

then  record  the  20  on  the  first  joint  of  the  first  finger,  as  above  directed.      Next  we  see  15 

8<  (8,  2,  and  the  5  in  excess  of  20),  plus  9  (4,  5),  =  24,  then  recording  the  20,  by  moving  tlio 

end  of  the  thumb  to  the  first  joint  of  the  second  finger,  we  next  see  20  (7,  9,  and  the  4  in 


104.5  excess),  which  we  record  by  passing  the  end  of  the  thumb  to  the  first  joint  of  the  third 

finger.  Then  we  see  12  (3,  3,  6),  plus  8  =  20,  this  we  reconl  liy  placing  the  euil  of  the 
thumb  on  the  first  joint  of  the  fourth  finger.  Then  we  see  16  (7,  9),  jilus  8  =  24  ;  the  20  we  reccud 
by  placing  the  end  of  the  first  finger  on  the  first  joint  of  the  thumb,  anil  as  the  addition  of  the  col- 
umn is  now  completed,  we  inspect  the  position  of  the  fingers  and  thumb  of  tlie  recording  hand  .nnd 
from  the  position  last  above  named,  find  the  record  to  Ije  100,  which,  with  the  4  last  excess,  makes 
104,  the  result  of  the  column.  This  104  we  write  in  the  total  result,  and  obtain  1045,  as  the  sum  of 
all  the  numbers. 

^)  (3)     Add  the  following  numbers : 

5  •  16 

2S 

(J  i 21 

g  \  Explanation. — Having  made  the  explanation  of  the  preceding  example  very 

7       7  )  24 

A  J  ^i  >  full,  we  will  therefore,  in  this,  omit  some  of  the  details.     To  make  the  operation 

^  ^  clearer,  we  have  linked  together,  with  brackets,  the  numbers  used  to  produce  tlio 

2  S       /•  various  intermediate  results.     Commencing  at  tiie  bottom  of  the  column,  we  pro- 

TueS 

8i  ceedthus:  11,  14,  25;  then  recording  the  20  and  adding  the  5  excess  to  the  next 

3  ^  11  )  21 

g  ^       '  "  group,  we  have  15,  7,  22 ;  then,  recording  the  20  and  adding  the  2  excess  we  havo 

K>       '  15,  13,  28;  then,  recording  the  20  and  adding  the  8  excess  to  the  next  group,  we 

^  (         -'^  have  23 ;  then,  recording  the  20  and  adding  the  3  excess  to  the  next  group,  we 

9  (  10  ^  28  ijave  18,  10,  28 ;  then,  recording  the  20  and  adding  the  8  excess  to  the  next  group, 

7  M8  >  we  have  21 ;  then,  recording  the  20  and  adding  the  1  excess  to  the  next  group,  we 

8  \ 

6  ) 23  have  10,  11,  21 ;  then  recording  the  20  and  adding  the  1  excess  to  the  next  group, 

9  ( 

8  J  13  )  ''8  ■^^  have  16,  11,  27 ;  then  recording  the  20  and  adding  the  7  excess  to  the  next  group, 

^  J  , .  (  we  have  17,  7,  24 ;  then  recording  the  20  and  adding  the  4  excess  to  the  next  group, 

9  i    ^    _  we  have  21 ;  then  recording  the  20  and  adding  the  1  exce.ss  to  the  next  group, 

8  n5  i  we  have  16.     This  completes  the  addition,  and  by  inspecting  the  recording  hand,  we 
2  S 
5  /  I'l  )  25  find  the  end  of  the  first  finger  on  the  second  joint  of  the  thumb,  which  shows  that 

9  (       / 

7  >  11  S  200  has  been  recorded,  which,  with  the  last  obtained  16  makes  216,  the  total  sum 


A 

216 


of  the  column. 


58 


SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


DRILL  PROBLEMS. 


Add,  separately,  tlie  following  columus  of  uumbers : 


1) 

3 

4 

II 
■'■} 
'\ 

8  S 


(3) 
13 


11 


uS 


(4) 


>13 


73 


73 


Explanation. — In  the  tliird  example,  -we  added  in  tlie  same  manner  as  in  the  preceding  example. 

In  the  fourth  and  fifth  examples,  in  stating  the  results  of  the  grouped  figures,  and  also  in  the 
result  in  excess  of  20,  we  have  set  only  the  excess  figure,  or  figures,  which,  in  practice,  are  the  only 
figures  that  should  lie  mentally  noticed.  In  the  fourth  exami)le,  we  grouped  difi'erently  from  the 
bracketing  of  the  third  example  in  order  to  show  that  no  particular  form  of  grouping  is  necessary, 
and  that  the  more  figures  grouiied,  the  more  rapidly  the  addition  is  made. 

In  the  fifth  example,  we  find /re  figures  alike,  and  hence,  by  multiplying,  we  instantly  see  45; 
•we  then  record  the  40,  and  continue  in  the  usual  manner,  until  we  come  to  three  figures  that  are 
alike ;  we  then  multiply  them,  and  to  the  product,  add  the  3  excess,  and  4,  the  last  figure  in  the 
column,  and  produce  11  in  excess  of  20. 


CONTRACTING  METHODS. 


113.    To  be  obsei'ved  when  adding  by  the  Vicenary  System : 

1.  In  mentaUy  naming  the  results  of  the  various  groups,  the  unit  figure  only 
should  be  named ;  thus,  in  the  addition  of  the  fourth  example,  instead  of  mentally 
naming  or  thinking  17,  11,  etc.,  name  or  think  only  7,  1,  etc.  Remember,  in  prac- 
tice, these  results  are  never  set  on  paper.     It  is  here  done  to  elucidate  the  work. 

2.  In  adding  the  two  results  of  the  grouped  figures  to  produce  an  excess  of 
20,  add  only  the  unit  figures,  and  only  mentally  name  the  excess  of  20. 

These  points  were  elucidated  in  adding  the  fourth  and  fifth  examples. 

3.  Whenever  the  same  figure  occurs  connectedly  several  times,  the  sum  should 
be  obtained  by  multiplying  instead  of  adding.  This  principle  was  elucidated  in 
example  5. 


*  DRILL    PROBLEMS.  Sg 

Add  the  following  columns  of  numbers  by  the  Yiceuaiy  JMetliod  until  you 
acquire  rapidity  and  accuracy : 


(6) 

(7) 

(8) 

(9) 

(10) 

(11) 

(12) 

(13) 

(14) 

(15) 

3 

C3 

24 

5 

5 

49 

39 

5 

7563 

19G3 

G 

25 

63 

o 

9 

83 

77 

3 

28325 

2345 

5 

23 

75 

4 

3 

27 

62 

7 

461523 

1215 

<2 

93 

82 

8 

0 

51 

35 

o 

6393 

1872 

*> 

83 

29 

2 

1 

43 

56 

8 

21783 

7312 

4 

72 

76 

O 

O 

02 

50 

9 

151672 

910 

1 

43 

84 

1 

1 

13 

46 

4 

3243 

2311 

7 

72 

57 

3 

7 

38 

34 

8 

72 

617 

9 

11 

38 

5 

0 

55 

40 

5 

311 

99 

3 

63 

92 

') 

7 

28 

19 

9 

3263 

313 

2 

15 

14 

3 

1 

33 

58 

3 

70015 

4632 

0 

63 

68 

2 

1 

24 

49 

1 

80C3 

516 

3 

00 

74 

8 

8 

81 

68 

o 

1000 

3313 

8 

92 

99 

9 

6 

91 

62 

6 

50792 

88 

0 

31 

76 

8 

7 

88 

93 

6  . 

107331 

200 

5 

41 

57 

9 

8 

92 

82 

6 

2441 

3915 

r- 
t 

57 

81 

2 

9 

02 

55 

3 

3457 

617 

f) 

23 

92 

2 

4 

48 

23 

2 

07423 

3129 

s 

35 

25 

3 

4 

65 

2L 

5 

87035 

28 

9 

21 

63 

3 

1 

08 

30 

5 

297521 

319 

5 

00 

48 

6 

1 

62 

74 

8 

SOO(tO 

4615 

3 

51 

78 

6 

2 

64 

28 

5 

6751 

313 

9 

24 

97 

8 

2 

87 

84 

5 

324 

3239 

2 

33 

62 

4 

1 

48 

25 

6 

233 

272 

9 

49 

54 

5 

4 

55 

62 

2 

49 

99 

3 

75 

38 

9 

8 

82 

49 

3 

275 

333 

7 

17 

91 

1 

8 

78 

89 

3 

3617 

8917 

o 

35 

56 

9 

I 

57 

98 

3 

81535 

222 

7 

23 

72 

8 

2 

20 

28 

6 

7123 

G717 

C 

87 

85 

0 

1 

31 

33 

7 

GO  73  7 

516 

7 

15 

39 

1 

1» 

42 

40 

S 

3(115 

817 

S 

23 

24 

t 

6 

53 

87 

9 

40123 

999 

2 

95 

76 

7 

5 

84 

62 

9 

891395 

1322 

C 

64 

58 

9 

1 

43 

49 

2 

20G 

400 

7 

53 

49 

1 

9 

90 

77 

5 

1482 

620 

5 

08 

62 

3 

3 

18 

83 

0 

5014 

189 

8 

72 

14 

8 

6 

oo 

92 

4 

382 

4002 

9 

19 

17 

1 

4 

37 

11 

4 

1571 

841 

4 

68 

27 

4 

1 

4(t 

13 

7 

0321 76 

264 

2 

47 

19 

o 

7 

50 

14 

6 

35489 

974 

MISCELLANEOUS   PROBLEMS  IX  ADDITION. 

114.  1.  Add  seven  million  four  thousand  uiuety-six,  and  three  hundred 
eighty-seveu  thousand  five  hundred  sixty-two.  Aiis.  7391658. 

2.  Find  the  sum  of  488S7G5.  92238,  1600084,  SS88S88,  99099099999, 
4100000S0S707  and  222222333333444444.  Aus.   222226533349723125. 


6o 


SOULE  S    PHILOSOI'IHC    PRACTICAL    MATHEMATICS. 


3.  A  inercbaut  bouglit  at  one  time  708  barrels  tlour  for  $4032  ;  at  another, 
273  barrels  for  $1774.50;  and  at  another,  410  barrels  for  $2921.25.  How  many  bar- 
rels did  he  buy  and  what  was  the  total  cost  ?  Aus.  1451  bbls.     $8727.75  cost. 

4.  From  Xew  Orleans  to  MacDonoughville  is  1  mile ;  thence  to  Algiers,  1 ;  thenee 
to  Old  Spanish  Fort  St.  Leon,  10;  thence  to  Poverty  Point,  18;  thence  to  Point 
Celeste,  7;  thence  to  Poiute-a-la-Hache,  3 ;  thence  to  Sixty  Mile  Point,  15  ;  thence  to 
Quarantine,  9;  thence  to  Bolivar  Point,  3;  thence  to  Forts  St.  Phlip  and  Jackson,  2; 
thence  to  The  Jump,  10;  thence  to  Head  of  Passes,  11 ;  thence  to  Pilot  Town,  10; 
thence  to  Port  Eads,  1.     How  many  miles  from  New  Orleans  to  Port  Eads? 

Ans.  107  miles. 

5.  From  New  Orleans  to  Algiers  Depot  is  1  mile;  thence  to  Gretna,  3;  thence  to 
Jefferson,  9;  thence  to  St.  Charles,  0;  thence  to  Boutte,  6;  thence  to  Bayou  des 
Alemendes,  8;  thence  to  Racelaud,  8  ;  thence  to  Ewing's,  6  ;  thence  to  Lafoui'che,G; 
theuce  to  Terrebonne,  3 ;  thence  to  ChucaUoula,  G  ;  thence  to  Tigerville,  5 ;  thence  to 
L'Ourse,  4 ;  thence  to  Bayou  Boeuf,  3  ;  thence  to  Ramos,  3  ,  thence  to  Morgan  City, 
4  ;  theuce  to  Galveston,  240.     How  many  miles  to  Galveston  ?        Ans.  321  miles. 

6.  During  the  fiscal  year  ending  September  1,  1893,  the  sugar  production' 
in  the  ditferent  sugar  producing  parishes  of  Louisiana,  was  as  follows: 


Parishes  In  Pounds. 

Ascension 39,754,703 

Asaumiitiou 49,939,541 

Avoyelles 2,574,891 

East  J5atou  Rouge. .  4,324,21:? 

Iberia 18,247,813 

Iberville 38,755,.524 

Jefferson 6,661,494 

Lafayette 323,438 


Pahishes.  In  Pounds. 

Lafourche 39,875,463 

Orleaus 2,626,000 

Plaquemines 18,943,010 

Pointe  Coupee 7,0,58,314 

Rapides 4,567,404 

St.  Beruar.l 2,666,908 

St.  Charles 13,693,265 

St.  James 30,554,329 


Parishes.  I 

St.  John  the  B.aptist 

St.  Landry 

St.  Martin 

St.  Mary 

Terrebonne 

Vermilion 

West  Baton  Ronge. . 
Other  small  iiarishus 


N  Pounds. 
21,196,526 

4,282,245 

7,024,291 
79,641,726 
68,456,522 

3,452,639 
17,424,725 

2,640,000 


How  many  pounds  were  produced  in  the  State  during  the  fiscal  year. 

Ans.  484,084,984  pounds. 

7.     During  the  fi.scal  year  ending  September  1,   1893,  the  different  countries 
of  the  world  produced  the  following  quantities  of  sugar : 

OOXJ3XrTHIE!S. 


Cane  Sugar. 

Cuba 

Porto  Kico 

Trinidad 

Barbados 

Jamaica 

Autigiiii  and  St.  Kitta, 

Martiniiiue 

Guadeloupe 

Lesser  Antilles 

Deraerara 

Reunion 

Mauritius 


In  Tojre. 

850,000 
70,000 
.50.000 
70,000 
28,000 
30,000 
30,000 
50,000 
27,000 

110,000 
35,000 
75,000 


Cane  Sugar. 

Java 

Briti.sh  India. . . 

Brazil 

Natal 


Fiji 

Phillipine  Lslands. . . 
Manila,  Cebu,  Hoilo. 

United  States 

Pern 

Egypt 

Saudwich.  Islaada  . . . 


In  Tons. 
430,000 

50,000 

200,000 

8,000 

23,000 
225,000 
250,000 
246,094 

40,000 

65,000 
125,000 


Bket  Sugar.  In  Tons. 

Germany 1,225,000 

Austria-Hungary 780,000 

France 590,000 

Russia 460,000 

Belgium 180,000 

Holland  70,000 

United    States,    short 

tons 13,284 

Other  countries 97,000 


How  many  tons  each,  of  cane  and  beet  sugar,  were  produced  in  the  world 
during  the  fiscal  year,  and  liow  many  tons  of  both  kinds  ? 

Ans.  3,087,094  tons  of  cane  sugar, 

3,415,284  tons  of  beet  sugar , 

and  6,502,378  tons  of  both. 


*  MISCELLANEOUS     PROBLEMS    IN    ADDITION.  6 1 

8.  From  IS"ew  Orleans  to  tlie  mouth  of  Eecl  Eiver,  is  210  miles ;  tlieuce  to 
Black  Eiver,  40 ;  thence  to  Alexandria,  110 ;  thence  to  Grand  Ecore,  120 ;  thence 
to  Grand  Bayou,  9u ;  thence  to  New  Hope,  GO ;  thence  to  Waterloo,  30  ;  thence  to 
Shreveport,  35.     How  many  miles  to  Shreveport  by  river  ?  Ans.  700  miles. 

9.  From  Xew  Orleans  to  Carrolltou,  is  7  miles;  thence  to  Doualdsonville, 
71 ;  thence  to  Plaquemines,  3U  ;  thence  to  Baton  Eouge,  20 ;  thence  to  Port  Hudson, 
23 ;  thence  to  Bayou  Sara,  12 ;  thence  to  mouth  'of  Eed  Eiver,  -40 ;  thence  to  Nat- 
chez, 72 ;  thence  to  Eodney,  45 ;  thence  to  Grand  Gulf,  18 ;  thence  to  Vicksburg, 
61;  thence  to  the  Louisiana  Line,  97 ;  thence  to  Helena,  230;  thence  to  Columbus, 
329;  thence  to  Cairo,  20;  thence  to  Cape  Girardeau,  50;  thence  to  St.  Louis,  151. 
How  many  miles  to  St.  Louis  by  river.  Aus.  1278  miles. 

10.     According  to  the  most  reliable  information,  the  following  figures  show  the 
area,  in  square  miles,  and  the  population  of  the  difl'erent  countries  of  the  world : 

Square  Miles.  Population.  Square  Milea.  Population. 

KortU  America 9,349,741  88,005,69.5      Africa 11,514,985  168,497,091 

South  America   6,887,794  33,565,882  Australa.si a   ami   Poly- 
Europe 3,942,.530  360,.580,788         iiesia 3,4.56,103          5,083,968 

Asia,  including  Malaysia  16,956,284  823,155,251  South  Polar  Regions  .. .         2.53,678         

What  is  the  entire  area  and  the  whole  population  of  the  world  ? 

Ans.  52,301,115  square  miles  area,  1,479,488,075  j)opulation. 

11.  Bought  at  one  time  343  yards  of  calico  and  132  yards  of  silk  ;  at  another, 
104  yards  of  calico  and  24  yards  of  silk;  at  another,  90  yards  of  calico  and  48  yards 
of  silk.     How  many  yards  of  each  kind  did  I  buy?       Ans.  calico,  543  ;  silk,  204. 

12.  Newman,  Soule,  and  Lynch,  form  a  copartnership;  Newman  invests 
$5400,  Sovde,  $5000,  and  Lynch,  $1000  more  than  both  Newmau  and  Soide.  What 
is  the  capital  of  the  firm  ? 

13.  The  cotton  crop  of  the  Southern  States,  from  1885  to  Sept.  1, 1893,  was 

as  follows : 

1885-6.  6,550,215  bales New  Orleans  received  1,764,013  bales. 

1886-7.  6,513.263  "  1,762,798  " 

1887-8.  7,017,707  "  1,780,375  " 

1888-9.  6,935,082  "  1,697,376  " 

.      1889-0.  7,311,322  "  1,973,.571  " 

1890-1.  8,6.52,597  "  2,077,744  " 

1891-2.  9,035,379  "  2,503,251  " 

1892-3.  6,700,365  "  1.602,079  " 

How  many  bales  were  produced  in  the  eight  years,  and  how  many  of  them 
did  New  Orleans  receive '?        Ans.  58,715,930  bales.     N.  O.  rec'd  15,101,207  bales. 

14.  The  Mayor  of  the  City  of  New  Orleans,  receives  a  yearly  salary  of 
$3500;  the  treasurer,  $3500;  the  Commissioner  of  Public  Works,  $3500;  the 
Comptroller,  $3500 ;  the  Commissioner  of  Police  and  of  Public  Buildings,  $3000 ; 
the  City  Attorney,  $3500;  the  City  Surveyor,  $2500;  the  City  Superintendent  of 
Public  Schools,  i'JOOO;  the  City  Superintendent  of  Fire  Alarm  Telegraph,  $1800. 
What  are  the  total  salaries  of  these  ofiflcers  1  Ans.  $27800. 

15.  The  Governor  of  Louisiana,  receives  a  salary  of  $4000  per  annum ;  the 


62 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Lieutenant  Governor,  $8  per  day  during  the  GO  days'  session  of  the  Legislature ; 
the  Secretary  of  State  receives  $1800  per  annum  ;  the  Auditor  of  Accounts,  $13500  ; 
tlie  State  Treasurer,  $2000 ;  the  Attorney  General,  $3000 ;  the  five  Justices  of  the 
Supreme  Court,  $5000  each  ;  the  two  Judges  of  Criminal  Court  in  N.  O.,  $4000 
each ;  the  two  Judges  of  the  Court  of  Appeals  in  N.  O.,  $4000  each ;  the  five 
Judges  of  the  Civil  District  Court,  Parish  of  Orleans,  $4000  each ;  the  State  Super- 
intendent of  Ediication,  $2000.  What  is  the  total  salary  of  all  these  officers,  includ- 
ing the  per  diem  of  the  Lieutenant  Governor  ?  Ans.  $76780. 

16.  A  father  gave  to  his  son  seven  thousand  eiglit  hundred  dollars  ;  to  his 
daughter,  nineteen  hundred  and  fifty  dollars  more  than  he  gave  to  his  son  ;  and  to 
his  wife,  three  thousand  five  hundred  more  than  he  gave  to  both  the  son  and  the 
daughter.     What  sum  did  he  give  away  ?  Ans.  $38600. 

17.  Tlie  length  of  the  Mississippi  River  is  4200  miles;  of  the  Nile,  4000; 
Amazon,  3750;  Yenisei,  3400;  Obi,  3000;  Yang-tse-Kiang,  3320;  Niger,  3000; 
Lena,  2700 ;  Amoor,  2650 ;  Volga,  2000 ;  Ganges,  1600 ;  Brahmapootra,  2300 ;  La 
Plata,  2300;  Mackenzie,  2300 ;  St.  Lawrence,  2000;  Saskatchewan,  1900;  Orinoco, 
1550;  Columbia,  1020;  Colorado,  600;  Yukon,  1600;  Bed  River,  1500.  What  is  the 
combined  length  of  all?  Ans.  50690. 

18.  The  following  statement  compiled  from  the  New  Orleans  Cotton 
Exchange  Report  of  September  1st,  1893,  shows  the  consumptiou  of  cotton  by  the 
Southern  States,  from  Sept.  1,  1892,  to  Sept.  1,  1893 : 


STATES. 


Alabama 

Arkansas 

Georgia 

Keutucky 

Louisian.a 

Mississijijii.  ... 

Missouri 

Nortli  Carolina. 
South  Carolina 

Tennessee 

Texas 

Virginia 


No.  of  Mills 

ju 
Operation. 


1 

57 
6 
4 
7 
1 
122 

53 

26 
7 

12 


Looms 

ia 

Operation. 


2,875 

108 

11,243 

688 

1,518 

1,812 

240 

11,108 

13,537 

2,621 

1,362 

3, 109 


Spindles 

in 
Operation. 


1,59,912 

3,000 

511,336 

50,543 

52,800 

53,692 

10,500 

524,770 

525,457 

128,215 

44,436 

106,486 


Bales 
Consumed. 


Pounds 
Consume*!. 


50,222 

765 

184,403 

17,004 

14,408 

15,955 

1,931 

181,996 

203,533 

33,443 

12,  ,396 

27,792 


24,449,891 

382,500 

86,931,608 

8,339,968 

6,862,896 

7,479,433 

942,115 

82,913,899 

94,013,304 

16,239,159 

6,478,066 

12,961,202 


Average 
Weight 
Per  Bale. 


486.83 
500.00 
471.45 
490.47 
476.32 
468.78 
487.89 
4.55.50 
461.90 
485.58 
522.59 
466.36 


How  many  each,  of  mills,  looms  and  spindles  were  in  operation,  and  how 
many  each,  of  bales  and  pounds  of  cotton  were  consumed  during  the  year  by  the 
twelve  states?  Ans.  318  mills,  50,221  looms,  2,171,147  spindles, 

743,848  bales,  347,994,041  pounds. 

19.  A  man  made  a  will  and  bequeathed  $4000  to  each  of  three  sons ;  and  to 
each  of  two  daughters,  $500  more  than  to  each  son ;  and  to  his  wife,  $1000  more  than 
to  a  son  and  a  daughter  together ;  and  the  remainder  of  his  estate,  which  was  $1000 
more  than  he  bequeathed  to  his  wife  and  children,  he  left  to  benevolent  institutions. 
Wliat  was  the  sum  received  by  the  wife  and  by  each  daughter ;  what  was  the 
sum  given  to  benevolent  institutions,  and  what  was  the  amount  of  the  whole  estate  ? 

Ans.   Wife,  $9500  ;  each  daughter,  $4500 ;  Benevolent 
Institutions,  $31500 ;  whole  estate  $62000. 


SY;ivof sjs  F^oR  RBvmjv, 


117.     Define  the  following  words  and  phrases : 

Signs  and  Abbreviations  in  Busi- 
ness. 

Addition. 

Sum,  or  Amount. 

ITumerical  Equation. 

Principle  of  Addition. 
87,  88,  and  89.    The  Alphabet  of  IJ'um- 
bers. 

Proof  of  Addition. 

Horizontal  Addition. 

Adding  Fractional  Xumbers,  ^or  ^. 

Adding  Dollars  and  Cents. 

U.  S.  Monetary  Units. 

Addition   of  Several   Columns  in- 

Oiie  Operation. 

Addition    by  one    of  the  Proper- 
ties of  9. 


64. 

Signs  and  Symbols. 

What 

are  they 

SO. 

used  for  ? 

65. 

Sign  of  Addition. 

81. 

66. 

Of  Subtraction. 

82. 

67. 

Of  Multiplication. 

85. 

68. 

Of  Division. 

86. 

69. 

Of  Equality. 

87, 

70. 

The  Parenthesis  and  Vinculum. 

71. 

Decimal  Point. 

9G. 

72. 

Sign  of  Katio. 

99. 

73. 

Sign  of  Proportion. 

103. 

74. 

Sign  of  Involution. 

104. 

75. 

Sign  of  Evolution. 

105. 

76. 

Sign  of  Deduction. 

107. 

77. 

luterrogatiou  Point. 

78. 

The  Signs  for  First, 

Second, 

etc. 

109. 

79. 

Sign  of  the  Comma. 

(63) 


ulbtraction. 


(DECREASING) 


118.  SuLtractioii  is  the  process  of  finding  the  difference  between  two  num- 
bers of  tbe  same  kind. 

119.  The  result  obtained  by  subtraction  is  called  the  Difference,  or 
Remaiiuler. 

130.  The  greater  number  is  called  the  Minuend,  which  means  a  number  to 
be  decreased. 

121.  The  lesser  number  is  called  the  Subtraliend,  which  means  the  number 
to  be  subtracted. 

122.  Tlie  Sign  of  Subtraction  is  a  horizontal  line  — .  It  is  read  minus, 
and  means  less. 

Wbeu  this  sign  is  jjlaced  between  two  Tiumbers,  it  indicates  that  the  number 
after  it  is  to  be  subtracted  from  the  number  before  it.  Thus,  S  —  3  is  read,  8  minus 
3  =  5,  or  8  less  3  =  6. 

123.  The  Principle  governing  all  problems  in  subtraction  is,  that  like  num- 
bers and  units  of  the  same  order  only,  can  be  subtracted,  the  one  from  the  other. 

121.  To  Prove  the  operation  of  subtractioTi,  add  the  remainder  to  the  sub- 
subtrahend  ;  if  the  sum  is  equal  to  the  minuend,  the  work  is  correct. 

125.  Subtraction  is  the  leverse  of  addition,  and  by  it  we  find  what  number 
added  to  the  lesser  of  two  numbers  will  produce  the  greater. 


126. 


SUBTliACTIO:N^  TABLE. 


2  —  2  =  0 

3  —  3  =  0 

4  —  4  =  0 

5  —  5  =  0 

6-6  =  0 

7  —  7  =  0 

8  —  8  =  0 

9  —  9  =  0 

3  —  2  =  1 

4  —  3  =  1 

5  —  4  =  1 

6  —  5  =  1 

7  —  6  =  1 

8  —  7=1 

9  —  8  =  1 

10  —  9  =  1 

4  —  2  =  2 

5  —  3  =  2 

0  —  4  =  2 

7  —  5  =  2 

8  —  6  =  2 

9  —  7  =  2 

10  —  8  =  2 

11  —  9  =  2 

5  —  2  =  3 

(3  —  3  =  3 

7  —  4  =  3 

8  —  5  =  3 

9  —  6  =  3 

10  —  7  =  3 

11  —  8  =  3 

12  —  9  =  3 

6  —  2  =  4 

7  —  3  =  4 

8  —  4  =  4 

9  —  5  =  4 

10  —  0  =  4 

11  —  7  =  4 

12  —  8  =  4 

13-9  =  4 

7  —  2  =  5 

8  —  3  =  5 

9  —  4  =  5 

10  —  5  =  5 

11  — 6  =  5 

12  —  7  =  5 

13  —  8  =  5 

14  —  9  =  5 

8  — 2==G 

0-3  =  0 

10  — 4  =  6 

11  —  5  =  6 

12  —  6  =  6 

13  —  7  =  6 

14  — 8  =  6 

15  —  9  =  6 

9  —  2  =  7 

10  —  3  =  7 

11  — 4  =  7 

12  —  5  =  7 

13  —  6  =  7 

14  —  7  =  7 

15  —  8  =  7 

16  —  9  =  7 

10  —  2  =  8 

11  —  3  =  8 

12  — 4  =  8 

13  —  5  =  8 

14  —  6  =  8 

15  —  7  =  8 

16  —  8  =  8 

17  —  9  =  8 

11  — 2  =  9 

12  —  3  =  9 

13  —  4  =  9 

14  —  5  =  9 

15  —  6  =  9 

16  —  7  =  9 

17  —  8  =  9 

18  —  9  =  9 

127. 


OllAL  EXERCISES. 


1.  Commence  at  50  and  orally  couut  to  0  by  continually  subtracting  1,  thus: 
49,  48,  47,  4G,  45,  etc. 

2.  Commence  at  50  and  orally  count  to  0  by  continually  subtracting  2,  thus: 
48,  40,  44,  42,  etc. 

3.  Commence  at  51  and  orally  count  to  0  by  successively  subtracting  3,  thus: 
47,  44,  41,  38,  etc. 

(64) 


■*  ORAL    EXERCISES.  65 

4.  lu   like  manner,  commence  at  50  and  subtract  respectively  4,  5,  G,  7,  8, 
i),  10,  until  you  produce  1,  tlius:  4(5,  41,  35,  28,  etc. 

5.  CouiMicnce  at  50  and  .subtract  alternately  2  and  5  until  you  produce  1, 
thus:  48,43,41,30,  etc. 

C.     Commence  at  50  and  subtract  alternately  8  and  3  until  you  produce  G, 
thus:  42,  .30,  31,  etc. 

7.  GO  _2  +  12  +  G  —  2— 2  —  3  +  5  —  8+11  —  G  +  0—  3  equals  1 

8.  75  +  5—  20  —  20  —  5  +  8  +  7  —  9  +  G  —  5  +  4  —  3  +  2  —  45  =1 

128.     To  subtract  one  numher  from  another,  ichen  any  Jigxire  of  the  subtrahend 
is  less  than  the  corresponding  fifjure  of  the  minuend. 
1.     From  897  subtract  041. 

OPERATION. 

807         ,      041  F.Ji>lai>aiiov. — First   ■write  tlie   nnmliprs  witli  the  lesser  iinder  or 

(jj^i       or      yjj-  oivc  tile  greater,  so  that   uuits  of  the  same  order  stand  iu  the  same 

column.       Then   comnieuoe   with  tbe  nnits'  figure  and  snbtraet  each 
^~^  ]~~  order  separately,  thus:   1  from  7  leaves  6;  4  from  9  leaves  5;  6  from  8 

250  250  lea\es  2.     By  this  work  \ve  obtain  the  difference,  or  remainder,  25(3. 

Subtract  the  foUowiug  : 

278  843  384         •  978  425  9870  6203 

499  521  762  655  679  3450  7021 


129.  Demonstration — to  prove  tbat  tbe  difference  between  two  uumbers  is 
tbe  same  as  tbe  diflereoce  between  tbe  two  uumbers  wbeu  equally  increased. 

OPERATION. 

6  6  +  3  =  9  5  5  +  2  =  7  23  23  +  10  =  33 

4  4  +  3  =  7  2  2  +  2  =  4  11  11  +  10  =  21 

—  The  difference  is   —  —  xhe  difference  is    —  —  The  difference  is       — 

2  the  same.             -^  ^  ■      the  same.             3  12  the  same.             12 

Explanation. — Here  we  see  that  the  difference  between  6  and  4  is  2,  and  that  the  difference 
between  6  and  4  equally  increased  by  3,  is  also  2.  The  o]ierations  with  5  and  2,  and  23  and  II  show 
similar  results.  Hence  the  law  lliat  the  difference  between  two  numbers  is  the  same  as  the  difference 
between  the  two  numbers  tchtn  eqiialli/  increased. 

The  application  of  tbis  numerical  law  is  sbown  iu  tbe  following  problem,  and 
it  governs  all  operations  in  subtraction,  wbere  tbe  subtrabeud  ligure  of  any  order 
exceeds  tbe  minuend  ligure  of  tbe  same  order. 

130.  To  subtract  one  number  from  another,  when  any  figure  of  the  subtrahend 
is  greater  than  the  corresponding  figure  of  the  minuend. 

1.    From  4173  subtract  234G. 

FIRST   OPERATION. 


Minuend    -    -     -    4173  Subtrabeud     -     -     -    2346 

Subtrabeud    -     -     2346       "'       Minuend     -     -     -     -     4173 


Difference       -    -    1827  1827 

Explanation. — Having  written  the  numbers  with  the  lesser  under  or  over  the  greater,  so  that 
units  of  the  same  order  stand  in  the  same  column,  we  commence  at  the  right  hand  to  perform  the 
operation. 

We  first  observe  that  6  units  cannot  be  taken  from  3  units.    We  therefore,  according  to  the 


66  soule's  philosophic  practical  mathematics.  -• 

foregoing  nnmerioal  law,  mentally  increase  the  3  units  liy  10  units,  making  13  units  ;  from  tbis,  -we 
subtract  the  (i  units  and  set  tlic  r('nKiin<lor,  7  units,  in  the  line  of  (Uttereuce.  Then,  as  we  added  10 
vnilK  til  till'  minuiMiil,  we  now.  tn  roiuiirusate  tlu'rel'iir,  uieutally  ad<l  1  ten,  the  e(|uivalent  of  10  units 
to  the  tens'  lijjure  of  the  subtrahend,  and  say  5  from  7  leaves  2,  which  we  write  lu  the  line  of  dif- 
ference. 

We  next  observe  that  3 /iMorfrerfs  cannot  be  taken  from  1  hnndird  ;  and,  therefore,  for  reasons 
above  given,  we  mentally  add  10  hundreds  to  the  1  hundred,  uuikiug  11  liundreds,  and  then  say  3  from 
11  leaves  8.  Then  having  added  10  hundreds  to  the  hundreds'  tignre  of  the  minuend,  we  now  men- 
tally add  1  thonsiind,  the  equivalent  of  the  10  hundreds,  to  the  thousands'  (ignre  of  the  subtrahend, 
and  say  3  from  4  leaves  1,  This  completes  the  operation  and  gives  1827  as  the  diii'ereuce  between 
the  two  numbers. 

131.        THE  BORROWING  METHOD  IMPROPER. 

The  foregoing  is  the  only  r.ational  and  true  inetbod  of  subtraction,  and  it 
should  be  universally  substituted  for  the  absurd  and  unniatlieniatical  "  borroiring^^ 
inetbod,  \vliich  is  given  by  nearly  all  the  authors  of  arithmetics  now  before  the 
public.    In  the  language  of  Shakespeare :  "  Neither  a  borrower  iior  a  lender  be." 

Subtract  the  following: 

7843  5813  6213  98761  1234  140728  34007 

5907  427  7302  5496  87601  60893  65100 


132.  TO  SUBTRACT  BY  ADDITION. 

1.  From  4173  subtract  2346. 

OPERATION. 
4173  2346  Explanation. — We  will  here  perform  the  operation  by  addition, 

''346        ^^'        4173  which  consists  simply  in  adding  to  the  subtrahend  such  a  number  as 

"  will  make  it  equal  to  the   minuend.     Thus,    commencing  with   the 

~  units'  figure  of  the  subtrahend,  or  smaller  number,  we  say,  6  and  7 

1827  1827  make  13;  and  write  the  7  in  the  units'  place  of  the  difference;  then 

carrying  1,  we  say  5  and  2  make  7,  and  write  the  2  in  the  tens'  column  of  the  difference;  then  we 
eay,  3  and  8  make  11,  and  write  the  8  iu  the  third  column,  or  hundreds'  place  of  the  difference;  then 
carrying  1,  we  say  3  and  1  make  4,  and  write  the  1  iu  the /our(/i  place  of  the  diti'erence.  This  com- 
pletes the  operation. 

2.  From  73245  subtract  1228. 

OPERATION   BY   SUBTRACTION. 

73245 
,  Q.^Q  Explanation. — Here  we  say  8  from  15  leaves  7 ;  3  from  4 

''"  leaves  1 ;  2  from  2  leaves  0 ;    1  from  3  leaves  2 ;    0  from  7 


72017 

OPERATION   BY   ADDITION. 


leaves  7. 


joog  Explanation. — Here  we  say  8  and  7  make  15;  3   and  1 

.  make  4  ;  2  and  0  make  2 ;  1  and  2  make  3 ;  0  and  7  make  7. 

72017 
3.     From  56802  subtract  50531. 


OPERATION   BY   SUBT7IACTION. 

56802 
50531 


Explanation. — Here  we  say  1  from  2,  I ;  3  from  10,  7 ;  6 
from  8,  2 ;   0  from  6,  6 ;  5  from  5,  0 ;  which  being  the  last 


/joTi  figure  on  the  left,  has  no  value,  and  hence  is  not  written. 


OPERATION   BY   ADDITION. 

56802 
50531 


Explanation. — Here  we  say  1  and  1  =  2;    3  and  7  =  10; 
6  and  2  =  8 ;   0  and  6  =  6  ;  5  and  0=5.     The  naught  is  not 


QO'ji  written  for  the  reason  given  in  the  first  solution. 


GENERAL    DIRECTIONS    FOR  SUBTRACTION.  6/ 

Subtract  the  following  by  both  methods  : 

428  284  2087  4802  7S091  5647832 

537  ia2  5301  1309  G0742  4909271 


GENERAL  DIRECTIONS  FOR  SUBTRACTIOX. 

133.  From  the  foregoing  elucidations,  we  derive  the  following  general 
directions  for  subtraction : 

1.  Write  the  numbers  so  that  units  of  the  same  order  stand  in  the  same  column. 

2.  Begin  at  the  units^  figure  and  take  successively  each  figure  of  the  subtrahend 
from  the  figure  of  the  corresponding  order  of  the  minuend. 

3.  When  a  figure  of  the  subtrahend  is  greater  than  the  figure  of  the  same  order 
in  the  minuend,  add  10  to  the  minuend  figure,  perform  the  subtraction,  and  then  add  1 
— the  equivalent  of  the  10— to  the  next  subtrahend  figure. 

4.  To  subtract  by  the  addition  process,  proceed  as  elucidated  in  problem  1 
above. 

PROBLEMS. 

13-1:.  Write  the  following  groups  of  numbers  as  they  are  here  written,  and 
subtract  the  lesser  from  the  greater  of  each  group : 

(1)  (2)  (3)  (4)  (5)  (6)  (7) 

467  1807  3842  607  3001  6879  12004 

342  4251  1291  8013  1009  9640  17862 


Find  the  diflerence  between  the  following  numbers  ; 


8.  624  and  507.  Ans.  117. 

9.  1148  and  5162.  Ans.  4014. 

10.  7013  and  904.  Ans.  6109. 

11.  237  and  894.  Ans.  657. 


12.  754  and  621.  Ans.  133. 

13.  41074089  and  1875429. 

Ans.  39198660. 

14.  9876543210  and  1234567890 

Ans.  8641975320. 


TO  SUBTRACT  DOLLARS  AND  CENTS. 
135.     What  is  the  difference  between  $483  and  $51.65  ?  Ans.  $431.35. 

OPERATION. 

$483.00  Explanation. — In   all   problems  of  this  kind,  we  first  write  the 

51.6j  numbers  iu  the  same  manner  as  when  adding  dollars  and  cents,  with 

dollars  nnder  dollars,  and  cents  under  cents,  so  that  units  of  the  same 

$431.oO  order  stand  in  the  same  column,  and  the  points  in  a  vertical  line. 

When  there  are  no  cents  iu  the  minuend,  we  fill  the  place  of  cents  with 
naughts. 

The  operation  of  subtraction  is  performed  with  dollars  and  cents,  the  same  as 
with  other  numbers. 

What  is  the  difference  between    the    numbers   in   each  of   the    following 


groups : 

(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

$16.25 

$8.00 

$.75 

$41.04 

$10.50 

$1.93 

$408.76 

9.38 

3.75 

.50 

6.61 

4.78 

.47 

291.85 

68  soule's  philosophic  practical  mathematics.  * 

(8)  •         (9)          (10)  (U)  (12)  (13) 

$081.85  $127.0.5     $248.00  $49.11  $8529.00  $500784290.15 

9().;58  l().5..-.0       181.15  9.89  2798.17  91842531.83 


(14)         (15)  (16)  (17)  (18)  (19) 

$570.00     $87.45     $  482.08     $14.00     $279107.10     $4704000.24 
132.85      19.;58      2240.10      08.25       91020.48      7150831.50 


BUSINESS  COXTRACTIONS   IN    SUBTRACTING   DOLLARS    AND  CENTS. 

136.     What  is  the  "Change"  (diftereucc)  to  be  giveu  to  the  purchasers  in 

the  following  ti'ansactious  f 

Note. — Make  the  subtractions  mentally,   using  pen  or  pencil,  only  to  write  the  diflereuce. 
Kepeat  the  work  uutil  you  acij^ulre  rapidity  and  accuracy. 


Amount  of 

Bills 

Amount  of 

Bills 

A 

nount  of 

Bills 

Purchase. 

Presented. 

Purchase. 

Presented. 

P 

irchase. 

Presented. 

$   1.45 

$   3.00 

Aus. 

$ 

$  2.75 

$   5.00 

Aus 

$ 

$   4.05 

$   10.00  Ans 

.05 

2.00 

'• 

$ 

1.03 

5.00 

$ 

43.40 

00.00       " 

4.27 

10.00 

$ 

23.42 

50.0(1 

$ 

72.20 

100.00       " 

1.85 

5.00 

$ 

3.85 

10.00 

$ 

1.15 

5.00       " 

12.20 

15.00 

$ 

38.25 

50.00 

$ 

07.75 

100.00       " 

l(i.03 

20.00 

$ 

14.10 

20.00 

$ 

25.10 

40.00       " 

7.18 

10.00 

$ 

(i.38 

1(1.00 

$ 

Explanation. — In  all  cases  like  tlie  above,  first  add  (mentally)  to  the  amount  of  purcliase 
enounjh  cents  to  make  even  dollars;  then  make  the  (mental)  subtraction,  and  to  the  difference  add 
(mentallv)  as  manv  cents  as  vou  added  to  the  purchase.  Thus,  in  the  first  example,  55c.  added  to 
the  11.45,  makes  $2.00  ;  and  $2.00  from  ,§3.00,  leaves  Sl-00 ;  and  55c.  added  to  $1.00,  makes  $1.55  chanffe. 

Witli  a  little  practice,  these  mentiil  additions  and  subtractions  maybe  performed  instantly 
and  without  any  mental  labor.  See  page  42,  for  table  to  qualify  to  see  instantly  the  difference 
between  any  number  and  100. 

Note. — In  business  practice,  the  book-keepers,  salesmen  and  "change"  clerks,  generally 
count  out  to  the  purchaser,  when  the  amount  of  purchase  embraces  dollars  and  cents,  sullicicnt 
change  to  make  witli  the  purchase  even  dollars,  and  then  count  even  dollars  to  the  amount  of 
money  presented  by  the  purchaser.  Thus,  if  tlie  purchase  were  $2.85  and  the  money  presented  was 
S5.00,  the  party  making  change  would  count  out  15  eeuts,  thus  making  $3.00,  and  then  count  out 
$2.00  more,  thus  making,  with  the  purchase,  i$5.00. 

HORIZONTAL  SUBTRACTION. 

137.  Accountants  and  business  men  frequently  find  it  convenient  to  sub- 
tract horizontally,  i.  e.,  when  the  minuend  and  subtrahend  are  on  the  same  line. 
Tlie  process  or  operation  in  this  case,  is  the  same  as  explained  in  the  General  Direc- 
tions, except  that  neither  number  is  written  under  the  other. 

Find  the  difference,  by  horizontal  subtraction,  between  the  following 
numbers : 

(1)  (2)  (3)  (4) 

84  —  52  =  32.  143  —  02  =  81.  721  —  95  =  020.  02  —  18  =  "? 

(5)  (6)  (7)  (8) 

87  —  39=?  043-214  =  '?  $480.48  —  $91.35  =  ?     $327.15  — $108.08  =  ? 

(9)  (10)  (11) 

$386.14  —  $109.87  =  1      $480.23  and  $291.45  =  ?    $2470.40  and  $0871.55  =.  1 

(12)  (13) 

$84936.14  antl  $57170.28  =  1  $4238.04  and  $50789.34  =  ? 


*  SL'BTRACTION    BY    THK    MKTHOn    OF    rOMMF.NCING    OX    THE    LEFT  69 

SCnBTRACTIOX  BY  THE  METHOD  OF  C0MMEXC1:NG  OX  THE  LEFT. 

13S.  Ill  tlie  operations  of  addition  and  STibtraction,  we  generally  commence 
to  add  and  to  subtract  ou  the  riglit,  but  we  have  sliowu  in  adding  several  columns 
at  once,  tliat  we  may  commence  to  add  on  tlie  left;,  and  in  like  manner,  we  may 
commence  on  tlie  left  in  the  operations  of  subtraction,  and  perform  tbe  work  towards 
tbe  right  with  the  same  rapidity  and  ease  as  in  the  ordinary  way  of  work. 

It  is  often  convenient,  and  sometimes  necessary,  to  know  almost  instantly  the 
difference  between  the  hundredth,  thousandth  or  millionth  figures  independent  of 
the  difference  of  the  lesser  orders  of  figures,  and  by  this  new  method  of  work,  this 
important  result  is  obtaiued. 

The  following  example  and  tlie  suliition  uf  the  same  will  render  the  operati(m 
clear : 

From  4381*7  subtract  174113. 

OPERATION.  Explanation. — We  commence  on  the  left  at  tlio  1,  and  liecanse  the  a<lioininK 

438''7  snhtrahend  tijj;Mie  (7)  is  greater  than  the  niinueml  tignio  of  the  sauio  onk-r,  wo 

^-.~_  add  1  to  the  lirst  sulitiahend  tiguve  and  say  2  from  4  leaves  2,  then  l)ecause  wo 

'  '"^-''^  added    1    to   the   snlitrahend    tignre  of   the   fifth  older,  we  now  add  its  equiva- 

lent,  10,  to  the  luinueml  lignre  of  the  next  lower  order,  and  say  7  from  13  leaves  6; 

li(Jo32  then  ])assiug  to  the  next  lower  order,  for  the  reason  that  the  subtrahend  figure 

of  the  next  lower  order  is  greater  than  the  minuend  figure  of  the  same  order,  we 

add  1  to  the  4  and  say  5  from  »  leaves  3;  then,  because  we  added  1  to  the  subtrahend  figure  of 

the  third  order,  we  now  add  its  eijuivalent.  If),  to  the  minuend  figure  of  the  second  order,  and  say 

9  from  12  leaves  3;  then  5  from  7  leaves  2,  ami  thus  we  have  the  correct  remainder. 

"Working  by  this  method,  what  is  the  difference  between  3S9210  and  1907GS  I 

Ans.  19844L'. 
2487  and  5GS;  4784  and  4819;  9021  and  G87G  ;  534G72  and  270S19. 

SUBTRACTIOX  BY  THE  COMPLEMENT  OF  10. 

139.  In  many  lines  of  business,  banking  especially,  and  in  the  operation  of 
closing  accounts,  it  is  often  convenient,  and  a  saving  of  time,  to  be  able  to  ^\Tite  the 
difference  between  the  sums  of  several  unadded  debit  and  credit  numbers,  without 
first  finding  the  amount  of  the  iesi)ective  debit  and  credit  numbers  of  which  the 
difference  or  balance  is  desired. 

1.     What  is  the  difference  or  balance  of  the  following  account? 

Ans.  6790, 

DE.      CR.  Explanation. — Here  we  add  the  units'  fignres  of  the  eraaller,  or  the 

credit  side,  and  obtain  22;  this  we  subtract  from  the  next  higher  number 
of  tens  (30)  and  obtain  S,  which  we  add  to  the  units'  figures  of  the  debit 
side  and  ]>rodnce  40.  The  0  we  write  as  the  first  figure  of  the  ditfer- 
ence.  Here  we  observe  that  there  were  three  tens  iu  the  credit  side  and 
/one  in  the  debit,  and  hence  we  have  1  ten  to  orfrf  to  the  tens  of  the  debit,  or  to 
subtract  from  the  tens  of  the  credit.  We  will  subtract  it  from  the  credit. 
We  now  add  the  second  column  and  obtain,  minus  the  1  ten,  28  ;  this  taken 
from  30,  leaves  2,  which,  added  to  the  second  column  of  the  debit,  gives  2D; 
ice.  ^jjg  9  ^^g  write  as  the  .second  figure  of  the  balance.  Here  we  observe  that 
there  is  one  more  ten  iu  the  credit  side  than  iu  the  debit.  This  1  ten  we  add 
to  the  third  column  of  the  credit  and  obtain  25,  and  tlii.s  taken  from  30 
leaves  5,  which  we  add  to  the  third  column  of  the  debit  and  produce  27 ;  the  7  is  written  as  tho  third 
figure  of  the  balance.  We  now  observe  that  there  is  an  excess  of  one  ten  in  the  credit,  which  \7e 
add  to  the /0HC//1  column  of  credit  auil  obtain  9,  this  taken  from  10  leaves  1,  which  is  added  to  the 
fourth  cidumn  of  debit  and  produces  16  ;  the  (>is  written  as  the  fourth  figure  of  tho  <liii'erenco  ;  aud 
as  the  tens  are  now  equal  iu  the  debit  and  credit  numbers,  the  diflerence  or  balance  is  complete. 


8472 

3876 

21G8 

4589 

4607 

764 

596 

1483 

1659 

6790  B 

70 


SOULE  S    PHILOSOnilC    PRACTICAL    MATHEMATICS. 


1)48 
793 
4:37 


4351 


2.    What  is  the  balance  of  the  following  account  ? 

DR.  CR.  DiKECTioNS.     First.    7  +  3 +8=  18  ;  20  —  IS  =2  ;  2  + 1  =  3,  tlie  first 

^^_^_ figiiro  of  tLe  balance  —  2  tens  excess  on  tbe  <lel>it. 

tSiroiid.  2+2  +  9  +  4  =  17;  20  —  17  =  3.  3  +  5  =  8,  the  second  fig- 
ure of  the  bnlinice  —  2  t<'Ms  excess  on  the  debit. 

Tliird.  2  +  4  +  7  +  y  =  22.  30  — 22=  8  ;  8  + 3=  11.  Tliis  Hives  1  as 
the  third  li.!j;iii-e  of  the  balance,  and  as  there  were  3  tens  in  the  debit  and 
1  ten  in  the  credit,  tlieni  are  two  tens  excess  in  the  debit,  which,  sub- 
tracted from  tlie  fourth  figure  of  the  credit,  gives  2  as  thi:  funrlli  and  final 
figure  of  the  balance. 


Balance,  2183 

3.  A   (lepo.sitor   has  a  credit  balance   of  $720(1.     He   draws   the   following 
checks:  $527,  $1318,  $98,  and  $1042.     What  is  the  credit  balance  ?         Ans.  $3021. 

Directions.  First.  Add  horizontally  the  iinits'  figures  of  the  checks  =  2.5;  30  —  25=5. 
Then 5  +  6=11.     (2  tens  excess  on  the  checks). 

Second.  Add,  horizontally  tbe  tens'  figures  of  the  checks  and  the  2  tens  excess,  =  18.  20  —  18 
=  2.  Then  2  +  0=2,  the  second  figure  of  the  balance.  (2  tens  excess  in  the  check  numbers).  lu 
the  same  manner,  proceed  with  the  remaining  orders. 

4.  What  is  the  debit  balance  of  the  following  account  ? 


DR. 


CR. 


1375 
8692 


350 


DiHECTloxs.     First.     10  —  6=4;   4  +  2  +  5=11. 
Second.     10  —  5  =  5;  5  +  9  +  7  =  21. 
Third.     (3  —  1  =  2)  10  —  2=8;  8  +  3  +  6=17. 
Fourth.     8  +  1  =  9. 


9711  Balance. 

6.     Find  the  difference  between  the  folU)wing  numbers  by  the  same  process. 
38071  and  934008. 
Explanation.    9  and  8  =  17 ;  3  and  0  =  9 ;  9  and  0  =  9  ;  1  and  4  =  5;  6  and  3  =  9;  9  —  1  =  8. 
6.     Find  the  balance  of  the  following  accounts  : 

DR.  CR.  DR.      CR.  DR.  CR. 


$  482.45 
1243.91 

502.87 


$345.36 

92.85 


571 

899 
223 


0410 


4123.14 


5421.45 

342.37 

1091.52 


7.     Find  the  balance  of  the  following  accounts : 

Kohlmnn,  Levy  &  Hiller. 


1891 
Jan. 
Feb. 

March 
April 


To  Merchandise. 


4 

184 

50 

10 

75 

65 

46 

212 

13 

54 

478 

14 

1891 
Jan. 
Feb. 
April 
May 


By  Cash.. 

It         It 

Balance. 


225 
65 


00 
55 
25 


Bolton,  Bicks  &  Schumnehet: 


14 

1468 

15 

46 

527 

28 

74 

193 

16 

98 

2809 

75 

116 

794 

88 

1894 
July 
Aug. 

Oct. 


I.  B.. 
C.  B. 

J 

C.  B. 


31 

2476 

23 

3823 

.35 

1068 

67 

483 

Note.  — If  it  is  desired  to  balance  and  foot  the  account,  as  in  the  work  of  closing  Ledgers, 
first  add  the  greater  side  of  the  account,  and  set  tbe  amount.  Then  add  the  lesser  side  and  lind  the 
difi"erence  by  the  method  shown  in  iirobleni  2,  above. 

This  method  of  subtraction  should  be  practiced  by  all  students  who  aspire  to  be  accountants, 
until  they  acquire  accuracy  and  rapidity.  Some  bankers  -will  not  employ  book-keepers  who  cannot 
subtract  "in  this  manner. 


*  MISCELLANEOUS    PROBLEMS    IN    SUBTRACTION.  7 1 

8.  A  depositor  has  a  credit  balance  of  $4208.10,  lie  deposits  $084.50  and 
draws  checks  for  tlie  following  amounts:  $528.00,8476.40,  $78.00,  $287.44.  What 
is  his  credit  balance  '?  Ans.     $3522.70. 

9.  My  cash  balance  in  bank  was  $1840.15;  I  drew  out  $920.00,  then  depos- 
ited in  currency,  $1200.00.  and  in  checks,  $1719.25;  I  then  drew  out  $408.50.  How 
much  is  my  i)reseut  cash  balance  iu  bank  ?  Ans.  $3436.90. 

MISCELLANEOUS    PROBLEMS    IX    SUBTEACTIO]^. 

140.  1.  Bought  a  lot  of  flour  for  .$2225,  and  sold  the  same  for  $2800, 
What  was  the  gain  ?  Ans.  $575. 

2.  It  is  700  miles  to  Shreveport  and  321  miles  to  Galveston.  How  much 
farther  is  it  to  Shreveport  than  to  Galveston  !  Ans.  379  miles. 

3.  The  ant  has  tifty  eyes,  and  the  dragon  fly  12000.  How  many  more  has 
the  dragon  fly  than  the  ant?  Ans.  11950  eyes. 

4.  The  Old  Testament,  King  .Tames'  edition,  contains  39  Books,  929  chapters, 
23214  verses,  592439  words,  and  2738100  letters.  The  ifew  Testament  contains  27 
Books,  260  chapters,  7950  verses,  182253  words,  and  933380  letters.  How  many 
more  of  each.  Books,  chapters,  verses,  words  and  letters,  does  the  Old  Testament 
contain  than  the  New  ? 

Ans.  12  Books,  669  chapters,  15264  verses,  410186  words,  and  1804720  letters. 

5.  Mercury,  the  smallest  planet  of  the  Solar  system,  is  30.000,000  miles  dis- 
tant from  the  Sun.  Neptune,  the  most  distant,  but  not  the  largest  planet,  is 
2,746,000,000  miles  distant  from  the  Sun.  How  much  farther  from  the  Sun  is  Nep- 
tune than  Mercury  '?  Ans.  2,710,000,000  miles. 

6.  Physicists  have  determined  that  to  produce  the  color,  dark  red,  395,000,- 
000,000,000  ethereal  waves  strike  the  eye  per  second ;  and  to  produce  violet, 
760,000,000,000,000  ethereal  waves  strike  the  eye  per  second.  How  many  more 
waves  per  second  are  required  to  produce  violet  than  dark  red  ? 

Ans.  365,000,000,000,000. 

7.  Sound  travels  through  tlie  air  at  the  rate  of  1118  feet  per  second,  and  a 
bullet  tired  from  a  rifle,  travels  1750  feet  per  second.  How  much  faster  does  the 
ball  travel  than  sound  ?  Ans.   632  feet  per  second. 

8.  Physiologists  have  determined,  with  the  aid  of  the  microscope,  that  the 
lungs  of  a  man  contain  not  less  than  600,000,000  air  cells ;  they  have  also  deter- 
mined that  a  single  drop  of  human  blood  contains  more  than  4,000,000,000  of 
corpuscles.  How  many  more  corpuscles  in  one  drop  of  blood  than  air  cells  in  the 
lungs  ?  Ans.  3,400,000,000. 

9.  General  George  Washington  was  born  iu  1732,  and  died  in  1799  ;  General 
Robt.  E.  Lee  was  born  iu  1807,  and  died  in  1870.  How  much  older  was  General 
Washington  than  General  Lee,  when  he  died?  Ans.  4  years. 

10.  What  is  the  difference  between  23222  and  11  thousand  11  hundred  11  ? 

Ans.  11111. 


72  SOULES    J'lIILOSOPHlC    PRACTICAL    MATHEMATICS.  ♦ 

11.  What  number  iiiiist  W  added  to  0,S741  to  make  a  million  1 

Alls.  n312r)9. 

12.  Huljbanl  lias  $.")(!()  wliicli  is  $!.")()  more  than  T,  and  I  have  87.">  more  than 
KeitFer.     How  mneh  has  Keiffer,  and  how  much  have  I '!        Ans.  Keiti'er  has  $275. 

I  have  $350. 

13.  There  are  two  parties  who  owe  me  $8000,  and  one  of  tliem  owes  $1250. 
The  other  wishes  to  pay  me  |1700  ou  account.     IIow  much  will  he  theu  owe  ? 

A  us.  $2050. 

11.  A  speculator  bought  a  lot  of  apples  for  $215,  and  sold  them  at  such  a 
price,  that  if  he  had  gotten  $22.50  more,  he  would  have  gained  as  much  as  they 
cost  him.     How  much  did  he  sell  them  for  ?  Ans.  $407.50. 

15.  I  gave  a  $20  bill  for  a  purchase  of  $8.15.  How  much  change  should  I 
receive!  Ans.  $11.85 

IG.  The  common  house  fly  makes  3,30  beats  per  second  with  his  wings,  and 
the  honey  bee  makes  190  beats  per  second.  How  many  more  beats  will  the  fly  make 
in  one  minute  than  the  bee  1  Ans.  8400. 

Note. — There  are  sixty  seconds  in  a  minnte. 

17.  From  New  Orleans  to  Yicksburg  is  401  miles,  and  to  Natchez,  277  miles. 
How  far  is  it  from  Natchez  to  Vicksburg  ?  Ans.  124  miles. 

18.  Wliat  is  the  difference  between  one  million,  seventeen  thousand  seven, 
and  one  thousand,  sixteen  hundred  sixteen  "?  Ans.  1014391. 

19.  The  sum  of  two  numbers  is  1463,  and  one  of  the  numbers  is  028.  What 
is  the  other  ?  Ans.  835. 

20.  The  velocity  of  our  earth  on  its  yearly  voyage  through  space,  around 
the  sun,  is  99733  feet  per  second ;  the  velocity  of  a  12-pound  cannon  ball,  tired 
from  a  gun,  with  an  average  charge  of  powder,  is  1734  feet  per  second.  How  many 
feet  farther  does  the  earth  travel,  in  each  second,  than  a  cannon  ball  ? 

Ans.  97999  feet,  or  18  miles  and  2959  feet. 

21.  What  number  is  that,  to  which,  if  17821  be  added,  the  sum  will  be 
37907?  Ans.  20086. 

22.  At  an  election,  the  defeated  candidate  received  23742  votes ;  but  had  he 
received  5112  votes  more  from  new  voters,  he  would  have  been  elected  by  1000 
majority.    How  many  votes  did  the  elected  candidate  receive  ?  Ans.  27854. 

23.  How  many  years  have  elapsed  since  the  birth  of  the  following  named 
persons:  Zoroaster,  according  to  Aristotle,  was  boru  B.  C.  5429;  Abraham,  B.  C 
2000;  Menes,  B.  C.  2000 ;  Moses,  B.  C.  1570;  Sohmion,  B.  0.  1033;  Homer,  B.  0, 
1000 ;  Lycurgus,  B.  C.  850 ;  Thales,  B.  C.  G40 ;  Solon,  B.  C.  038 ;  Pythagoras,  B.  C 
GOO ;  Confucius,  B.  C.  551 ;  Buddha,  or  Guataraa,  B.  C.  500 ;  Sophocles,  B.  0.  495 
Socrates,  B.  C.  470 ;  HjiJpocrates,  B.  C.  4G0  ;  Blato,  B.  C.  429  ;  Aristotle,  B.  C.  384: 
Demosthenes,  B.  C.  382;  Praxiteles,  B.  C.  3G0;  Alexander,  B.  C.  356  ;  Euclid,  B.  C, 
323 ;  Cicero,  B.  C.  106  ;  Seneca,  B.  C.  5 ;  Plutarch,  A.  D.  50 ;  Justinian,  A.  D.  483 ; 
Bacon,  A.  D.  1561. 

24.  A  father  divided   his   ijlantatiou,  consisting  of  4500  acres,  among  his 


"  iMISCELLANEUUS     PROBLEMS     I.\    SUBTRACTION.  73 

five  sous  :  Albert,  Edward,  William,  Frank,  and  Robert.  To  Albert,  lie  gave  800 
acres;  to  Edward,  be  gave  loO  acres  more  than  lie  gave  Albert;  to  William,  be  gave 
100  acres  less  tban  be  gave  Edward  ;  to  Frank,  be  gave  as  luucb  as  be  gave 
Edward;  and  the  remainder  be  gave  to  Kobert.  How  many  acres  did  Robert 
receive  ?  A  us.  930  acres. 

25.  The  Equatorial  diameter  of  the  eartb  is  792.5.05  miles,  and  tlie  Rolar 
diameter  is  7899.17  miles.  How  miicli  greater  is  tbe  Equatorial  diameter  tban  tlie 
Polar?  Ans.  20.48  miles. 

20.     Find  tbe  difl'ereuce  between  1001734  and  .ALMDCCXLIV.  Ans.  10. 

27.  A  carriage,  a  borse,  4  mules,  and  25  sbeep,  are  wortb  $1423.  Tbe  car- 
riage is  wortb  $430,  tbe  borse,  $130,  mules,  8173  a  piece.  Wbat  are  tbe  sbeep 
wortb  ?  Ans.  $125. 

28.  Tbe  casli  on  band  in  tbe  moruing  was  $278.50.  Received  during  tbe 
day  from  sales,  §482.10,  and  from  otber  sources,  $073.  Tbere  is  on  baud  at  tbe 
close  of  tbe  day,  $330.50.     How  much  was  paid  out  duriug  tbe  day  ? 

Aus.  $1103.10. 

29.  The  cash  balance  in  bank,  in  tbe  morning,  was  $2800.13.  Deposited 
duriug  the  day,  $472.  Drew  checks  iu  favor  of  A.  $95,  of  B.  $10(3,  and  of  (J.  $400. 
There  remains  on  band  undeposited,  $150.  What  is  the  balance  in  bank  and  on 
haud,  at  tbe  close  of  tbe  day's  business  '  Ans.  $2881.15. 

30.  December  1,  1893,  a  mercbaut  had  cash  in  bis  safe,  $2340,  and  on  deposit 
in  bank,  $0800.  During  tbe  month  be  received  $4520,  and  paid  out  $5350.  What 
is  the  balance  in  safe  and  bank  1  Ans.  $8310. 

31.  January  1,  1894,  a  retailer  <leposited  $35  change  in  his  cash  drawer,  in 
the  moruiug,  as  per  ticket:  paid  out  during  tbe  day,  $320;  received  from  J.  Jones, 
duriug  tbe  day,  $410;  from  S.  Smith,  $80.  At  uigbt  there  is  $328  in  the  drawer. 
What  bave  the  cash  sales  been  !  Aus.  $123. 

32.  There  is  $22.40  change  in  the  drawer  in  the  morning,  as  iier  ticket: 
duriug  the  day,  cash  was  received  from  A.  $00,  from  B.  $40.50,  from  C  $121,  and  from 
bills  receivable,  $000.  There  was  paid  out  for  expenses,  $34.50,  and  to  D.  $28. 
There  was  $990.13  on  hand  in  the  drawer  at  uight.     What  were  the  cash  sales? 

Ans.  $214.73. 

33.  A  merchant  purchased  mercbaudise  to  the  ainouut  of  $03,421.10,  and 
sold  mercbaudise  to  the  amount  of  $42,412.90.  He  then  took  au  account  of  stock 
and  fouud  that  he  had  $31,018.40  on  baud.     How  much  did  be  gain  ? 

Ans.  $8010.20. 


74 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


SViVOf»SJS    JPOR    JRBVIBW, 


141.     Define  tbe  following  words  and  jibrases: 


lis.     Subtraction. 

119.  Difference,  or  Remainder. 

120.  Minuend. 

121.  Subtrabend. 

122.  Sign  of  Subtraction. 

123.  Principle  of  Subtraction. 

124.  To  Prove  Subtraction. 

129.  Wliat  is  tbe  Numerical  Law  re- 
garding tbe  Difference  between 
Two  Numbers? 

131.  Tbe  Borrowing  Method  Impro- 
j)er. 


132.  To  Subtract  by  Addition. 

133.  General  Directions  for  Subtrac- 

tion. 

135.  To  Subtract  Dollars  and  Cents. 

136.  Business    Contractions  in   Sub- 

tracting Dollars  and  Cents. 

137.  To  Subtract  Horizontally. 

138.  Subtraction   by  tbe  Metbod  of 

Commencing  on  tbe  Left. 

139.  Subtraction  by  tbe  Complement 

of  10, 


.-Mt' 


ultiplicatfon. 


142.  Multiplication  is  tlio  process  of  increasing  one  of  two  numbers  as 
many  times  as  there  are  imitrj  in  the  other.  Or,  differently  explained,  it  is  the  pro- 
cess of  finding  the  product  of  two  numbers. 

143.  The  number  to  be  multiplied  is  called  the  Multiplicand. 

144.  The  number  which  shows  how  many  times  the  multiplicand  is  to  be 
increased,  or  repeated,  is  called  the  Multiplier. 

145.  The  result  obtained  by  nmltiplying  is  called  the  Product. 

146.  The  multiplicand  and  multiplier  are  called  Factors.  The  meaning  of 
the  word  factor  is  maker  or  producer. 

147.  The  Sign  of  Multiplication  is  an  oblique  cross,  x.  It  shows  that  the 
numbers,  between  which  it  is  placed,  are  to  be  multiplied  together,  and  is  read 
multiplied  by,  or  times.     Thus,  8  x  3,  is  read  8  multiplied  by  3,  or  3  times  8. 

Changing  the  order  of  the  factors  does  not  change  the  product  or  result. 
Thus,  8x3  may  be  read  3  times  8,  or  8  times  3. 

148.  To  Prove  the  operations  of  multiplication,  repeat  the  work,  or  multiply 
the  multiplier  by  the  multiplicand.  If  the  result  is  the  same  as  the  first,  the  work 
is  i^robably  correct. 

Note. — See  page  145  for  proof  of  niulfiplicatiou  l)y  casting  out  9's  and  ll's. 

PEINCIPLES    OF  MULTirLICATION. 

149.  1.  In  all  multiplication  ojierations,  the  product  is  the  same,  in  name  or 
Tcind,  as  the  multiplicand.  2.  In  all  cases,  the  multiplier  nuist  be  regarded  as  an 
abstract  number.  Two  denominate  numbers  cannot  be  multii^lied  together  as  denom- 
inate numbers.  Thus,  we  cannot  multiply  5  apples  by  5  apples,  10  cents  by  10 
cents.  VJ,  yards  by  8  pounds,  or  G  boxes  by  $2.  All  such  questions  are  absm-d  and 
insolvable. 

To  illustrate  this  principle,  we  present  tlie  following  problem : 

1.     What  will  S  pounds  of  rice  cost  at  11  cents  a  pound  ? 

OPERATION.  Exploiintion. — Here  we  have  lie,  the  price  of  1  pourid,  to  be  increased 

1    -ti  ^  8  times,  and  our  reasoning  is,  since  1  pound  costs  lie,  8  pounds  will  cost  8 

Multiplicand,  11/  times  as  much,  or  8  times  lie,  which  is  88c.     To  use  the  8  as  a  denominate 

Multiplier,  8  number  auil  say  8  pounds   times  lie.,  would  show  a  deliciency  of  head 

seuse'on  the  part  of  the  calculator.     It  is  true  tliat  in  this  problem  the 

Prndn  't  8H^         multiplicand  and  multiplier  are  both  denominate  numbers;  but  since  the 

'  ^  multi))lier  shows  the  number  of  times  the  imiltiplicaud  is  to  be  increased 

or  taken,  it  must  always  be  considered  and  iiaed  as  an  abstract  number.  Tlius,  in  this  example,  we 
do  not  multiply  lie.  by  8  pounds,  but  we  take  or  increase  lie.  S  limes,  for  the  reason  that  8  pounds 
■will  cost  8  times  as  much  as  1  pound. 

To  our  mind,  nothing  in  connection  with  the  science  of  numbers  is  more  absurd  or  meaning- 
less than  to  speak  of  jnultiplying  lie.  by  8  pounds,  or  $2  by  5  gallons,  or  50c.  by  50c.  To  propose 
to  multiply  4  oranges  by  5  peaches  would  be  equally  good  sense.  We  know  what  is  meant  by  tak- 
ing or  increasing  He.  or  4  oranges  a  certain  number  of  times ;  but  we  cannot  understand  or  com- 
prehend the  taking  of  any  thing  so  many  cent  times,  gallon  tiuies,  or  peach  times. 

(75) 


76 


SOULES    I'lIILOSOPHIC    TRACTICAL    MATHEMATICS. 


150. 


MUI.TIPLICATION    TABLE. 


1  1 

'> 

3 

4 

5 

6 

7 

8 

9 

lOJJ  11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

24 

25 

2 

4 

6 

8 

10 

12 

14 

16 

18 

20 

22 

24 

26 

28 

30 

32 

34 

36 

38 

40 

42 

44 

46 

48 

50 

3 

6 

i 

12 

15 

18 

21 

24 

27 

30 

33 

36 

39 

42 

45 

48 

51 

54 

57 

60 

63 

66 

69 

72 

75 

4 

8 

12 

16 

20 

24 

28 

32 

36 

40 

44 

48 

52 

56 

60 

64 

68 

72 

76 

80 

84 

88 

92 

96 

100 

5 

10 

13 

21 

25 

30 

35 

40 

45 

50 

55 

60 

65 

70 

75 

80 

85 

90 

95 

100 

105 

110 

115 

120 

125 

6 

12 

18 

24 

30 

36 

42 

48 

54 

60 

66 

72 

78 

84 

90 

96 

102 

108 

114 

120 

126 

132 

138 

144 

150 

7 

14 

21 

28 

35 

42 

40 

56 

63 

70 

77 

84 

91 

98 

105 

112 

119 

126 

133 

140 

147 

154 

161 

168 

175 

8 

16 

24 

32 

40 

48 

56 

64 

72 

80 

88 

96 

104 

112 

120 

128 

136 

144 

152 

160 

168 

176 

184 

192 

200 

9 

18 

27 

30 

45 

54 

63 

72 

81 

90 

99 

108 

117 

126 

135 

144 

153 

162 

171 

180 

189 

198 

207 

216 

225 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

110 

120 

130 

140 

150 

160 

170 

180 

190 

200 

210 

220 

230 

240 

250 

" 

22 

33 

44 

55 

66 

77 

88 

99 

110 

121 

132 

143 

154 

165 

176 

187 

108 

209 

220 

231 

242 

253 

264 

275 

12 

24 

36 

48 

60 

72 

84 

96 

108 

120 

132 

144 

156 

168 

180 

192 

204 

216 

228 

240  252 

264 

276 

288 

300 

13 

26 

39 

52 

65 

78 

91 

104 

117 

130 

143 

156 

169 

182 

195 

208 

221 

234 

247 

260273 

286 

299 

312 

325 

,14 

28 

42 

56 

70 

84 

98 

112 

126 

140 

154 

168 

182 

196 

210 

224 

238 

252 

266 

280'294 

308 

322 

336 

350 

15 

30 

45 

60 

75 

90 

105 

120 

135 

150 

165 

180 

195 

210 

225 

240 

255 

270 

285 

300  315 

330 

345 

360 

375 

16 

32 

48 

64 

80 

96 

112 

128 

144 

160 

176 

192 

208 

224 

240 

256 

272 

288 

304 

32o'336 

352 

368 

384 

400 

17 

34 

51 

68 

85 

102 

119 

136 

153 

170 

187 

204 

221 

238 

255 

272 

289 

306 

323 

340  357 

374 

391 

408 

425 

18 

36 

54 

72 

90 

108 

126 

144 

162 

180 

198 

216 

234 

252 

270 

288 

306 

324 

342 

360378 

396 

414 

432 

450 

,19 

38 

57 

76 

95 

114 

133 

152 

171 

100 

209 

228 

247 

266 

285 

304 

323 

342 

361 

380  399 

418 

437 

456 

475 

20 

40 

60  80 

100 

120 

140 

160 

180 

200 

220 
231 

240 
252 

260 
273 

280 
294 

300 
315 

320 
336 

340 
357 

360 
378 

380 
399 

400  420 

420  441 

440 
462 

460 
483 

480 
504 

500 
525 

21 

42 

()3 

84 

To5 

126 

147 

168 

Isi) 

210 

22 

44 

66 

88 

no 

132 

154 

176 

198 

220 

242 

264 

286 

308 

330 

352 

374 

396 

418 

440  462 

484 

506 

528 

550 

23  46 

69 

92 

115 

138 

161 

184 

207 

230 

253 

276 

299 

322 

345 

368 

391 

414 

437 

460  483 

506 

529 

552 

575 

24 

48 

72 

96 

120 

144 

168 

192 

216 

240 

264 

288 

312 

336 

360 

384 

408 

432 

456 

480  504 '528 

552 

576 

600 

,25 

50 

75 

100 

125 

150  175 

200 

225 

250 

275 

300 

325 

350 

375 

400 

425 

450 

475 

-,00'525'550 

575 

600 

625 

We  here  present  the  Multiplication  table  of  rytliaj,^oras,  which  extends  to  25 
times  25.  If  the  coiiiinercial  student  excuses  himself  trom  learning  the  whole 
table,  he  should  at  least  learu  from  1  to  20  times  10,  or  that  part  bounded  by  the 
heavy  lines. 

This  portion  of  the  table  is  far  more  valuable  than  the  12  times  12  table  which 
is  given  in  school  arithmetics,  and  should  be  thoroughly  learned. 

Wbeu  lecarning  this  table,  or  when  using  it  in  the  solution  of  problems,  be 
careful  not  to  use  any  intermediate  words  orally  or  menfciilly,  but  simply  note  the 
figures  and  pronounce  the  product  at  once;  thus,  do  not  say  or  think  7  times  3  are 
21 ;  14  times  8  are  112;  17  times  9  are  153,  etc.;  merely  look  at  the  numbers  to  be 
multiplied  and  say,  or  think,  21,  112,  153,  etc.  In  reading,  we  do  not  stop  to  spell 
orally  or  mentally  the  words  that  compose  the  sentences  ;  from  the  combination  of 
the  letters  we  see  what  the  words  are,  without  looking  specially  at  each  Individual 
letter ;  and  to  read  or  operate  with  rapidity  in  the  combination  of  numbers,  we 
must  omit  all  superfluous  talk  or  thought. 


'ib 


*  EXAMPLES,    WITH    PHILOSOPHIC    SOLUTION'S.  "J^ 

THE    IMrOETA]SrCE    OF    MULTIPLICATION. 

151.  Next  to  a  tlioroiigli  knowledge  of  addition  stands  multiplication,  as  an 
important  acquisition  to  be  possessed  by  business  aspirants.  All  wlio  are  experts 
in  these  two  subjects  of  practical  arithmetic,  combined  with  a  neat  and  rapid  hand- 
writing, hold  passports  to  well  paying  positions.  No  young  man  or  lady  who  is 
ambitious  to  achieve  success  and  honor  in  the  fields  of  business  can  afford  to  be  non- 
efficient  in  these  topics.  Such  proficiency  is  worth  more  than  unnuuibered  testi- 
monials of  competency  from  honored  merchants,  noted  bankeis,  railroad  kings, 
political  leaders,  or  i>opular  governors. 

THE  THILOSOPniC  OR  LOGICAL  SYSTEM  OF  MLT^TIPLICATION. 

152.  At  this  point  of  our  work,  we  introduce,  and  throughout  the  book 
shall  continue  to  use  the  logic  of  numbers — the  Philosophic  System  of  solving 
problems.     (See  page  26  Article  23). 

By  tliis  system,  the  reasoning  faculties  of  the  mind  are  brought  into  action, 
invigorated,  strengthened,  and  capacitated  to  see  fine  distinctions,  to  consider  con- 
ditions, to  investigate  facts,  to  reason  logically,  and  to  deduce  correct  conclusions, 
from  not  only  the  i)remises  and  conditions  of  problems,  but  upon  all  matters  and 
questions  that  the  changing  a'ffairs  of  this  world's  life  may  present  for  consider- 
ation, 

EXAMPLES,    WITH   PHILOSOPHIC    SOLUTIONS. 

153.  1.     One  orange  costs  5  cents,     \\1iat  will  G  oranges  cost  ? 

Ans.  30  cents. 

SOLUTION     STATEMENT. 

5/ 

6  Heason. — One  orange  costs  5  cents.     Since  1  oianije  costs  5  cents, 

—  6  oranges  will  cost  6  times  as  much,  ■nliicli  is  30  cents. 
30/    Ans. 

2.  If  one  hat  costs  $2,  what  will  4  hats  cost  ?  Ans.  $8. 

SOLUTION   STATEJLENT. 

$2 

4  iieason.— One  hat  costs  $2.     Since  1  hat  costs  $2,  4  hats  ^vill  cost 

—  4  times  as  much,  Tvhich  is  $8. 
$8     Ans. 

3.  At  8  cents  per  yard,  what  will  5  yards  cost  1  Ans.  40/. 

SOLUTION   STATEMENT. 

8/ 

5  Reason. —One  yard  costs  8  cents.     Since  1  yard  costs  8  cents,  5 

—  yards  will  cost  5  times  as  much,  which  is  40  cents, 
40/     Ans. 

.  (Continued. ) 


78  soule's  philosophic  practical  mathematics.  * 

THE  reason,  why,  AND  WHEREFORE,  CONTINUED. 

Question. — How  do  you  know,  that  if  1  yunl  costs  8  cents,  5  yards  Avill  cost 
5  times  as  much  ? 

Answer. — By  the  exercise  of  my  judgment — by  the  use  of  the  reasoning 
faculties  of  the  mind. 

Question. — What  do  you  mean,  in  this  connection,  by  judgment? 

Answer. — The  conclusion  arrived  at  by  the  operations  of  the  mind,  after  duly 
considering  the  premise,  the  facts,  and  the  conditions  of  the  problem. 

Question. — What  do  you  mean  by  premise  or  premises? 

Answer. — The  proposition,  declai-ation,  truth,  or  fact,  which  is  asserted  as 
the  basis  or  i)redicate  of  a  question.  In  this  problem,  the  i)remise  is,  one  yard  costs 
8  cents. 

Question. — Why  will  5  yards  cost  5  times  as  mnch  as  1  yard  * 

Answer. — Because  5  is  live  times  as  much  as  1., 

Question. — W^hat  kind  of  reasoning  is  the  foregoing? 

Answer. — Analogical  and  axiomatical.  Analogical,  because  there  is  analogy, 
relationship  or  likeness  existing  between  the  cost  of  1  yard  and  the  cost  of  5  yards. 
Axiomatical,  because  the  premise  and  question  considered,  the  conclusion  is  self- 
evident. 

Question. — What  is  reason  ? 

Answer. — The  faculty  or  power  of  the  human  mind,  by  which  truth  is  dis- 
tinguished from  falsehood,  right  from  wrong,  and  by  which  correct  conclusions  are 
reached  by  considering  the  logical  relationship  which  exists  between  the  premises, 
the  facts,  and  the  conditions  of  particidar  statements  and  questions. 

4.  What  will  4  books  cost  at  40  cents  each  ? 

SOLUTION   STATEMENT. 

40^  Questioifs. — 1.     How  'do    yon 

4  Heason— One   book   costs  40c.      l«'i"W   that  4  books   will  cost  4 

„  11.    1,1        t       I      times   as  much    as   1   book?     2. 

Since  one  Ijook  costs  40  cents,   4      -.,..     .   .  i      •     i  a» 

'  What  do  you  mean  byjiidgmentf 

$1.G0      AnS.  boulis  will  cost  4  times  as  much.      3,   Why  will  4  books'cost  4  times 

as  much  as  1  book  ? 
Note. — The  answers  given  to  like  questions  in  the  preceding  problems,  are  the  proper  answers 
to  these  aud  similar  questions. 

5.  At  18  cents  per  dozen,  what  will  5  dozen  cost  ? 

SOLUTION   STATEMENT. 

18/  Questions. — 1.    How  do   you 

5  Reason. -One  dozen  costs  18c.      1-no"'  that  5  dozen  will  cost  5 
_.          ,    ,  1     io         J.      -      times  as   much  as   1   dozen?     2. 

_  Since   1  dozen   costs  18  cents,  .-,      ^y^^^,  ^^.^^^  .  ,5^^^,^  cost  5  times  as 

90/      Ans.  dozen  will  cost  5  times  as  much,      much  as  1  dozeu  f     3.     What  do 


you  mean  by  judgment? 


6.    If  1  hat  costs  $4,  what  will  6  hats  cost? 

SOLUTION    STATEMENT. 


^  4  Questions. — 1.    How    do    you 

6  iJm8(.n-Onehatcost8$4.  Smce  ^^^^  tj^^y  will  ?    2.    Why  will 

1  ^-"^  ''°^*^  *^'  6  l^^its  "ill  <;"«*  6  ^^^y ,    3    ^^,^^^^  ig  judgment  in 

$24  Ans                   times  as  much.  this  connection  ? 


EXAMPLES,    WITH    PHILOSOPHIC    SOLUTIONS. 


79 


7.     Flour  is  $7  per  barrel,  what  \yill  20  barrels  cost? 

SOLUTION   STATEMENT. 


20 


Season. — One  barrel  costs  $7. 
Since  1  barrel  costs  $7, 20  barrels 
will  cost  20  times  as  much. 


$140    Ans. 
Bought  12  pounds  of  sugar  at  7/  per  pound. 


■whole  ? 

SOLUTION    STATEMENT. 

7/ 
12, 


84/   Ans. 


Reason. — One  pound  costs  7c. 
Since  1  pound  costs  7  cents,  12 
pounds  will  cost  12  times  as 
much. 


9. 


At  $7  per  cord,  what  will  123  cords  cost  ? 

Explanation. — In  the  .second 
statement,  for  convenience,  the 
multiplicand  is  used  as  the  mul- 
tiplier. 

Reason — One  cord  costs  |7. 
Since  1  cord  of  wood  costs  $7, 
123  cords  will  cost  123  times  as 
much. 


SOLUTION    STATEirENTS 

1st.  2d. 

$     7  123 

123  7 


Questions. — 1.  How  do  you 
know  this  ?  2.  ^Vhy  will  they  ? 
3.  What  do  you  understand  by 
j  udgment  in  this  connection  f 

What  was  the  cost  of  the 


Questions. — 1.     How  do  you 
know  this  ?    2.  Why  will  it  f 


Questions. — 1.     How   do  you 
know  this?     2.  Why  will  itf 


$861  Ans.    $861   Ans. 

10.    If  an  employ^  receives  $4  per  day  for  services  and  he  works  22  days, 


how  much  money  has  he  earned  ? 

SOLUTION     STATEMENT 

22 
4 


$88    Ans. 


Reason. — One  d-ay's  service  is 
worth  $4.  Since  1  day's  service 
is  wortli  $4,  22  days'  services  are 
worth22  times  asmuch  or  since 
he  receives  $4  for  1  day's  work, 
for  22  (lays'  work  he  will  receive 
22  times  as  much. 


Questions.— 1.  How  do  you 
know  this?  2.  Why  will  he  f 
3.  What  do  you  mean  by  judg- 
ment ? 


11.    There  are  60  minutes  in  an  hour, 
day  of  24  hours  ? 


How  many  minutes  are  there  in  a 
Ans.  1440  minutes. 


SOLUTION    STATEMENT. 
60 
24 


1440     Ans. 


Reason. — In  1  hour  are  60  min- 
utes. Since  there  are  60  minutes 
in  1  hour,  in  24  hours  there  are 
24  times  as  many. 


Questions. — 1.  How  do  you 
know  this  f  What  do  you  mean 
by  judgment? 


The  reasoning  of  these  problems  and  the  answers  to  the  questions  following 
the  reasoning,  should  be  repeated  until  the  mind  is  fiilly  capacitated  to  solve,  in 
like  manner,  all  similar  problems. 

DRILL  PROBLEMS. 


154.     Solve  the  following  problems  in  like  manner  as  the  foregoing,  and 

write  the  reason  for  each : 

Ans.  $2.40. 
Ans.  75/. 


1.  What  will  6  books  cost,  at  40/  each  ? 

2.  At  15 «'  per  dozen,  what  will  5  dozen  cost? 

3.  If  1  box  costs  $2,  what  will  24  boxes  cost  ? 


Ans. 


8o  soule's  philosophic  practical  mathematics.  * 

4.  At  $6  per  cord,  what  will  41  cords  of  wood  cost  ?  Ans.  $24G. 

5.  Paid  $4  per  barrel  for  potatoes  and  bought  3G  barrels.     What  did  they 
cost?  Ans.  $144. 

6.  What  will  9  yards  cost  at  10/ per  yard  ?  Ans.  $1.44. 

7.  Flour  is  worth  $6  per  barrel.     What  are  25  barrels  worth  ?    Ans.  $150. 

8.  12  inches  uiake  a  foot.    IIow  many  inches  in  15  feet  ?    Ans.  ISO  inches. 

9.  4  quarts  make  a  gallon.     Uow  many  quarts   in   a  barrel  that  holds  41 
gallons  ?  Ans.  1(54  quarts. 

10.  If  you  buy  17  boxes  of  peaches  ®  $2  per  box,  what  will  they  cost? 

Alls.  $34. 

11.  If  you  buy  9  pencils  at  5  cents   each,  and  hand  to  the  seller  50/,  how 
much  change  should  you  receive  ?  Ans.  5/. 

12.  A  merchant  bought  43  barrels  of  apples  at  $4  iier  barrel,  and  paid  $120 
on  accouut.    How  much  does  he  still  owe  ?  Aus.  $52. 

TO  JIULTIPLY  ABSTRACT  NUMBERS  AND  GIVE  REASONS  THEREFOR. 

155.  1.     Multiply  7  by  6. 

OPERATION. 

7  Explanation  and  Heason. — Plato  tells  us  that  one  is  the  basis  of  all 

g  things  ;  but  whether  this  statement  be  true  or  false,  1  (one)  is  certainly 

the  basis  of  all  numbers.     Mnltijilieatiou  is  the  process  of  rejieatiufj  one 
number  as  many  times  as  there  are  units — ones — in  another.     Considering 
42   Ans.       these  facts,  we  tirst  multiply  the  7  by  1,  and  in  the  product  obtain  a  pre- 
mise for  our  argument.     Thus,  asiomatically  1  time  7  is  7.     Since  1  time  7 
is  7,  fi  times  7  is  6  times  as  many,  which  is  42.     This  is  the  long  looked  for  reason  for  the  multij)!!- 
cation  of  abstract  numbers. 

Multiply  the  following  uumbers,  and  write  the  reason  : 
0x4;  8x5;  17x12;  23x10;  234x157;  341x250. 

TO  MULTIPLY,  WHEN  THE   MULTIPLIER    CONSISTS    OF    ONLY 

ONE    FIGURE. 

156.  1.     'WTiat  is  the  product  of  947  imilti])li('(l  by  0  ? 

OPERATION.  Explanation. — In  all  jiniblems  of  this  kind,  write. the  multiplier 

under  the  units'   tigure  of  the    multiplicand;   then    commence   with 

•  the  units'  figure,  and  say,  6  times  7  are  42,  which  is  4  tens  and  2  units ; 

i      .  the  2  units  write  in  the  units'  place  of  the  product  and  retain  in  the 

1  is  mind  the  4  tens  to  add  to  the  column  of  tens  ;  next  say  G  tinu'S  4  ten8 

£.-.?  are  24  tens  plus  the  4  tens  retained  in  the  mind,  are  28  tens,  which  is  2 

Mnltiiilifind  '14  7  hundreds  and  8  tens ;  the  8  tens  write  in  the  ft-iis' column  of  the  jprodnct, 

-..    ,.•    ,•    ',  '     /I  and  retain  in  the  mind  the  2 /iHHrfref?.s  to  add  to  theeolumn  of /iHiirfrerf*. 

Multiplier  o  Then  say  6  times  9  are  54  hundreds,  plus  2  hundreds  are  56  hundreds, 

which  is  5  thousand  and  6  hundreds,  which  write  respectively  in  the 

Product  5082  t/iuHsnHrfs' and /noidreds' columns  of  the  product.     This  completes  the 

operation  and  gives  a  j)roduct  of  5682. 
In  practice,  instead  of  saying  6  times  7  are  42,  6  times  4  are  24,  etc.,  we  should  only  name 
the  result  of  the  combination.     Thus,  42,  24,  etc.     In  handling  figures,  we  should  aJ ways  pronounce 
the  result  of  the  romhinations  without  naming  the  figures  that  make  the  result,  Juat  08  we  pronounce  word* 
without  spelling  or  naming  the  letters  that  make  the  words. 


MULTIPLICATION    ELUCIDATED.  8 1 

Perform  the  following  multiplicatious : 


Multiplicand, 
Multiplier, 

(2) 

7 

3SU1 

(3) 
983 
8 

7864 

(4) 
2769 

5 

(5) 

76895 
9 

(6) 
81453 
6 

Product, 

13845 

692055 

4SS718 

7.  At  $85  each,  what  will  7  wagons  cost  ?  Ans.  $595. 

8.  What  will  8  lots  of  ground  cost,  at  $1875  each  ?  Ans.  $15000. 

9.  At  $6  per  barrel,  what  will  be  the  cost  of  245  barrels  of  flour  ? 

SOLUTION   STATEMENT.  Tfenson.— One  Ijarrel  of  flour  costs  S6.     Since  1  barrel  of  flour 

Multi])lier,  245  costs  |6,  245  ban-els  will  cost  245  times  as  much.     The  S6  is  the 

Multil)Iicaud       $      6  '^^''^'  multiijlicaml,  but  iu  the  operation  we  use  it  as  the  multiplier. 

^       '        '  This  is  ilone  for  convenience,  in  all  problems  ■where  the  multipli- 

'^  canil  is  less  than  the  multiplier.     The  result  is  the  same  which- 

$1470  Ans.    •  ever  factor  is  used  as  a  multiplier. 

10.  What  will  42  dozen  boxes  cost,  at  $9  per  dozen  %  Ans.  $378. 

11.  At  $7  a  piece,  what  will  48  chairs  cost?  Ans.  $336. 

12.  A  clerk  receives  $75  per  month.     If  he  spends  $40  per  month,  how  nuich 
cau  he  save  in  one  year,  or  12  months?  Ans.  $420. 

13.  Multiply  1,  2,  3,  4,  5,  6,  7,  0,  8,  9,  and  10  together.  Ans.  0. 

14.  Bought  44000  pounds  of  cotton  at  8/  per  pound  and  sold  it  at  9/  per 
pound.     What  was  the  gain  %  Ans.  $440. 

15.  What  is  the  diflereuce  in  the  cost  of  150  sheep  at  $4  a  head,  and  80  head 
of  cattle  at  $12  a  head.  Ans.  $300. 


TO  MULTIPLY,  WHE2f  THE   MULTIPLIER   CONSISTS   OF   MOPvE   THAN" 

ONE   riGUKE. 

157.     1.     What  is  the  product  of  397  nuiltiplied  by  653  % 

OPERATION. 


^1. 


Multiplicand,         39  7 
Multiplier,  653 


First  partial  product  by  3  units,  1191  =  3  times  the  multiplicand. 

Second     "  "         by  5  tens,  1985     =50     '■         "  " 

Third       ''  "        by  6  huud's,     2382       =  GOO  "         "  " 


Total  product,  259241  =  653"        "  " 

Explanation. — In  all  problems  of  this  kind,  write  the  multiplier  tuidertbe  multiplicand,  so 
that  units  of  the  same  order  stand  iu  the  same  column,  and  then  multiply  by  one  figure  at  a  time. 


82 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


First  iiniltiply  l)y  tlio  units'  figuro,  tlicn  hy  the  tens',  hundreds',  and  so  on,  in  rcgnlar  order,  tlirongh 
the  mult  i]dicr  ;  then  add  the  several  i)artial  jiroduets  toij;ether  and  thus  obtain  the  required  product. 

In  this  ]inildein,  first  multiply  by  :>,  the  units'  fifjiire,  in  tin',  same  manner  as  explained  in  the 
first  probh'm  where  there  was  but  one  tifjure  in  the  mnltijilier  and  olitain  llfll  as  the  first  jiartial 
prndnet.  Write  this  below  the  multiplier  so  that  units  of  the  same  order  stand  iu  the  same 
column. 

Next  multiply  by  the  5  lens ;  say  .'>  times  7  are  35,  'which  is  3  hundreds  and  5  tens  :  write  the  5 
tens  iu  the  tens'  eolunni  directly  below  the  multiplying  figure,  and  reserve  iu  the  miud  the  3  hundreds 
to  add  to  the  hundreds'  column.  Then  say  5  times  9  are  45  +  3  hundreds,  which  were  reserved,  mak- 
ing 48  hundreds  which  is  4  thousands  aixl  8  hundreds  ;  write  the  8  hundreds  iu  the  column  of  hundreds, 
and  reserve  the  4  thousands  to  adil  t<i  the  tliousands'  column  ;  then  say  5  times  3  are  15  plus  4  thous- 
ands reserved,  are  19  thousands  which  is  1  ten  thousand  and  9  thousands,  which  write  in  their  respective 
columns. 

Then,  in  like  manner,  multiply  by  the  6  hundreds  in  the  multiplier,  being  careful  to  write  the 
first  figure  obtained  (2)  in  the  /lunrfrerfs' column,  directly  under  the  6  of  the  multiplier,  and  the 
other  figures  in  their  respective  columns,  thousands,  ten  thousands,  and  hundred  thousands  ;  then  add 
thepariial  products  together,  which  gives  259241  as  the  whole  product  of  397  multiplied  by  653. 

NoTK. — In  practice,  remember  to  name  or  thiuk  only  the  results  of  the  numerical  combina- 
tions, when  adding  or  multiplying. 

EXAMPLES. 

2.    Multiply  3426  by  457. 

OPERATION. 


Multiplicand, 
Multiplier, 


s  S  =  o  S  s-a 
aSbHKHP 

3426 
457 


First  partial  product  by  7  nnits,        23982  =  7  times  the  multiplicand. 
Second    "  "        hy  5  tens,         17130     =50     "       "  " 

Third      "  "        by  4 /m red's,  13  704       =400"       "  " 


Whole  product. 


1565082  =  457 


Multiply  28433 
by  4172 


Multiply  989769 
by  248193 


5.    Multiply  21794 
by  2365 


Multiply  the  following  numbers : 


5368 
3355 

340868  Aus. 


12.  1683  by  328 

13.  2346  by  127 


Explanation. — In  all  problems  Tyhere  there  are  naughts  in 
the  multiplier,  multiply  by  the  significant  figures  only,  for  the 
reason  that  the  product  of  any  number  by  0  is  0 


483  by  569             8.       581  by  76 

10. 

671  by    508 

924  by  237            9.     1847  by  84 

11. 

8765  by  2046 

Operation  of  the 
10th  i)roblem. 

671 

508                            p^„,„ 

»/if.-/>« 

MULTIPLICATION. 


83 


TO    MULTIPLY,    WHEX    EITHER    FACTOR    IS    1    WITH    KAUGHTS 
ANiS^EXED,  AS   10,   100,  1000,  Etc. 


EXAMPLES. 


158.     1.    Multiply  465  by  100. 


OPERATION. 


4C500 


sucli  rmmber  into  ten: 


Explanation. — In  .ill  problems  of  this  kind,  Tve  annex  as  many  nanghts  to 
tbe  multiplicand  as  there  are  naughts  in  the  multiplier.  In  this  problem,  we 
annex  two  O's  and  produce  46500,  the  anssver.  The  basis  or  reason  for  this  is 
in  accordance  with  the  Principles  of  Notation,  (Arts.  44  and  4.5)  where  it  was 
shown  that  annexing  one  naught  to  a  number  removes  or  changes  the  unit  of 
the  tens  into  hundreds,  the  hundreds  iuto  thousands,  and  so  on,  and  hence  mul- 


tiplies by  10,  100,  1000,  etc.,  for  the  higher  powers  of  ten. 


2.  Multiply  500  by  10. 

3.  MiUtiply  1000  by  100. 


Aus.        5000 
Aus.     100000 


4.  Multiply  1850  by  1000. 

5.  Multiply  7020  by  10000. 


Aus. 
Aus, 


1850000 
70200000 


TO  MULTIPLY,  WHEN  EITHER  THE  MULTIPLICAND  OR  MULTIPLIER, 
OR    BOTH,    HAVE    NAUGHTS    OX    THE    RIGHT. 


159.     Multiply  463  by  200. 


OPERATION. 

463  463 


or 


200 


92600  Aus.    92600  Aus. 


Explanation. — In  all  problems  of  this  kind,  we  first  multiply 
together  the  signitjcant  lignres  and  to  the  product  annex  as  many 
nanghts  as  there  are  on  tlie  right  of  the  multiplier  or  multiplicand, 
or  of  both.  In  writing  the  numbers  to  be  multiplied,  place  the 
right  hand  significant  figures  of  the  multiplicand  aud  multiplier  in 
the  same  column,  omitting  the  nanghts  or  extending  them  to  the 
right.  In  this  problem,  we  first  use  the  multiplier  2  hundred  as  2  units  ;  hence  the  first  partial  pro- 
duct, 926,  was  100  times  too  small.  We  then,  by  annexing  the  two  naughts,  multiplied  it  by  100, 
and  olitained  92600  as  the  correct  product.  For  further  explanation  and  reason,  see  problem  1, 
Art.  158. 


94 

47 


2.    Multiply  940  by  4700. 

OPERATION. 

940 
4700 


658 
376 


or 


658 
376 


4418000  Aus.         4418000  Aus. 


Explanation. — 94  multiplied  by  47  equals  4418  as  the 
first  partial  product ;  then,  since  the  94  was  used  as  94 
units  instead  of  94  tens,  we  annex  one  naught  to  4418 
making  44180;  then,  since  the  47  was  used  as  47  units 
instead  of  47  hundreds,  we  annex  two  naughts  to  44180, 
making  4418000,  the  correct  answer.  See  Problem  1.  Ait. 
158. 


Multiply  3400  by  26. 

OPERATION. 

3400 
26 

204 
68 


Multiply  5020  by  420. 

OPERATION. 

5020 
420 


1004 
2008 


88400  Aus., 


2108400  Aus. 


4.  Multiply  82000  by  483. 

OPERATION. 

82000  483 

483  ^^  82000 


246 
656 

328 


966 
3864 


39606000  Ans. 


39606000  Aus. 


5. 
6. 

7. 


Multiply  842  by  600. 
Multiply  1208  by  1020. 
Multiply  9900  by  707. 


8.  Multiply  23500  by  12030. 

9.  Multiply  1000  by  6208. 
10.    Multiply  81000  by  90200. 


84  soule's  rHiLosoriiic  practical  mathematics.  * 

TO    MULTIPLY   LY  THE    FACTORS  OF  A  NUMBER. 

100.  Factors  of  a  number  are  such  numbers  as  will,  wlieii  niultiplieil 
togetlier,  produce  the  number.  Thus,  C  and  G  are  the  factors  of  30 ;  7  and  S  are 
the  factor-    of  50. 

PRIiSrCIPLE. 

The  product  of  any  number  of  factors  will  be  the  same,  iu  whatever  order 
they  may  be  multiplied. 

1.     Multiply  2^35  by  42. 

OPERATION. 

"     '^  Explanation. — Iu  all  problems  of  tliis  kind,  separate  the  multiplier 

iuto  two  or  uiore  factors;  tbeu  multiply  tbe  multiplicaud  by   one  of 

17045  t^^  factors,  tbe  resultiug  product  by  auotber  factor,  and  so  ou,  uutil 

0  all  tbe  factors  have  beeu  nsed.     The  last  product  will  bo  the  correct 

product. 


102270  Ans. 

Use  the  factors  and  multi]ily  the  following  numbers :  2.  4S0  by  301. 
3.  1750  by  125.  4.  3281  by  128.  5.  781  by  03.  G.  3140  by  30.  7.  588  by  81. 
8.     473  by  432.     9.     5107  by  330. 


TO    irULTrPLY^,    WHEN   THE   MULTIPLTCA^TD   OR    MULTIPLIER   CON- 
TAINS   DOLLARS    AND    CENTS. 

IGl.     1.    ^Multiply  $342.15  by  0. 

OPEKATION. 

$342.15  Explanation — In  all  problems   of   this    kind,    multii)ly  in 

(]  tbe  rcj^ular  mauuer,  then  prefix  the  dollar  sign  ($)  aud  i)laco 

the  point  (  .  )  two  places  from  the  right.     The  answer  is  then 

Product,        $2052.90  i"  dollars  and  cents. 

2.  "Wliat  will  1C82  pounds  of  sugar  cost,  at  9/  per  pound  1     Ans.  $151.38. 

3.  A   nierchaut's  monthly  expenses  are  $1342.75.      What  are  they  for  12 
months  1  Ans  $10113.(10. 

4.  It  costs  a  family  $2.30  a  day  for  marketing.     What  will  be  the  expense 
for  30  days?  Ans.  $09.00. 

5.  WTiat  -will  37  boxes  of  oranges  cost,  at  $3.75  per  box  ?        Ans.  $138.75. 
0.     At  10  cents  per  pound,  what  is  the  value  of  23780  pounds  of  cotton  ? 

'  Ans.  $3804.80. 

7.  If  it  costs  $17500  to  construct  one  mile  of  railroad,  what  will  be  the  cost 
to  build  304  miles  1  Ans.  $0370000. 

8.  What  will  875  tons  of  railroad  iron  cost  at  $55  per  ton  1      Aus.  $48125. 


*  MISCELLANEOUS    PROBLEMS    IN    MULTIPLICATION.  85 

GEXEEAL    DIRECTIONS    FOR    MULTIPLICATION. 

162.  From  the  foregoing  elucidations,  we  derive  tlie  following  general 
directions  for  multiplication : 

1.  Write  the  muUipUer  tinder  the  multiplicand,  so  that  units  of  the  same  order 
stand  in  the  same  column,  and  draw  a  line  beneath. 

2.  When  the  multiplier  consists  of  one  figure,  begin  at  the  units'  figure  and 
multiply  each  figure  of  the  multijyUcand  by  the  multiplier.  Write  in  the  product  line 
the  units  of  each  result,  and  add  the  tens,  if  any,  to  the  next  result. 

3.  When  the  multiplier  consists  of  two  or  more  figures,  begin  at  the  units^  figure 
and  multiply  successively,  each  figure  of  the  multiplicand  by  each  figure  of  the  multiplier, 
placing  the  right  hand  figure  of  each  partial  product  under  that  figure  of  the  multiplier 
which  produced  it. 

4.  Draw  a  line  beneath  the  several  partial  products  and  add  them  together  ;  the 
sum  irill  be  the  required  product.  If  there  are  any  decimals  in  the  factors,  point  off 
as  many  figures  from  the  right  of  the  product  as  there  are  places  of  decimals  in  the 
multiplicand  and  multiplier. 

Proof. — 1st.     Carefully  review  tlie  work.    2d.    Multiply  the  multiplier  by 
the  multiplicand ;  if  the  results  are  the  same,  the  work  is  jirobably  correct. 
Note. — See  proof  of  multiplication  by  casting  out  the  9's  and  It's,  page  154. 

MISCELLANEOUS   PROBLEMS   IN  MULTIPLICATION. 

1G3.     1.     What  is  the  value  of  the  following  numerical  expressions  : 

(43— 6)  + (S  X  2).  Ans.  53. 

2.  What  is  the  product  of  8  +  7,  nuiltiplied  by  204  —  101?         Ans.  1545. 

3.  What  is  the  product  of  IG  +  18  —  10  by  12  x2?  Ans.  576. 

4.  What  is  the  diflerence  between  50  —  (5  x  4)  and  25  +  4  —  8  ?      Ans.  9. 


5.  Multiply  240  —  50  +  22  by  14x16—  112  —  65.  Ans.  29736. 

6.  Multiply  16  thousand  16  hundred  16,  by  11  thousand  11  hundred  forty 
and  11.  Ans.  214052016. 

7.  Multiply  together  the  following  numbers  : 

9x8x7x6x5x4x3x0x2x1.  Ans.  0. 

8.  What  will  6  dozen  dozen  boxes  cost,  at  one  half  a  dozen  dozen  cents 
per  box  1  '  Ans.  $622.08. 

9.  The  pulse  of  a  healthy  middle-aged  person  beats  72  times  per  minute, 
how  many  times  does  it  beat  in  1440  minutes  ?  '^Ans.  103680. 

10.  Multiply  one  million  twenty-six,  by  nineteen  thousand  s^even  hundred 
ten.  Ans.  19710512460. 

11.  One  cubic  foot  contains  1728  cubic  inches.     How  many  cubic  inches 
in  324  cubic  feet  ?  Ans.  559872. 


86  soule's  philosophic  practical  mathematics.  * 

12.  One  square  foot  contains  144  square  inches.  How  many  square  inclies 
in  95  square  feet?  Ans.  13G80. 

13.  One  gallon  contains  231  cubic  inches.  How  many  cubic  inches  in  a  cis- 
tern that  holds  3500  gallons  ?  Ans.  808500. 

14.  One  bushel  contains  2150.43  cubic  inches.  How  many  cubic  inches  in 
20  bushels  1  Ans.  43008.40. 

15.  One  mile  contains  5280  feet.    How  many  feet  in  25  miles  ? 

Ans.  132000. 

16.  The  human  heart  beats  4200  times  an  hour.  How  maiiy  times  does  it 
beat  in  10  years,  there  being  24  hours  in  one  day,  and  allowing  365  days  in  each 
year?  Ans.  367920000. 

17.  Sound  travels  1118  feet  per  second.  How  far  will  it  travel  in  10  minutes, 
there  being  60  seconds  in  a  minute  ?  Aas.  670800  feet. 

18.  A  railroad  train  runs  25  miles  an  hour.  How  far  will  it  go  in  3  days, 
allowing  3  hours  for  lost  time  in  stoppages  1  Ans.  1725. 

19.  Light  travels  192500  miles -per  second.  How  many  miles  will  it  travel 
in  1  day,  there  being  24  hours  in  a  day,  60  minutes  in  an  hour,  and  60  seconds  in  a 
minute?  Ans.  16632000000. 

OPERATION   INDICATED. 

24  hours,  X  60  minutes,  x  60  seconds,  x  192500  miles  per  second,  =  the  answer. 

20.  If  a  person  respires  17  times  in  a  minute,  how  many  times  will  he  breathe 
in  a  day?  Ans.  24480. 

21.  At  $17  per  ounce,  what  is  the  value  of  9  pounds  of  gold,  there  being  12 
ounces  in  a  pound,  Troy  or  Mint  weight?  Ans.  $1836. 

22.  What  will  27893  pounds  of  tobacco  cost,  at  56  cents  per  pound  ? 

Ans.  $15620.08. 

23.  What  will  1870  acres  of  land  cost,  at  $18  per  acre  ?  Ans.  $33660. 

24.  The  Senate  and  the  House  of  Kepresentatives  of  the  State  of  Louisiana, 
consist  of  137  members,  who  receive  $4  per  day.  The  regular  session  continues 
60  days.     What  is  the  yearly  expense  for  the  salaries  of  the  State's  law  makers  ? 

Ans.  $32880. 

25.  How  many  hours,  minutes,  and  seconds  in  the  month  of  July  ? 
Note. — See  Problem  19. 

Ans.  744  hrs.     44640  min.     2678400  sec. 

26.  What  will  1463  ounces  of  silver  cost,  at  87/  per  ounce  ? 

Ans.  $1272.81. 

27.  A  contractor  has  865  men  employed  at  $1.50  per  day.  What  are  the 
weekly  wages  of  all,  for  6  days'  labor?  Ans.  $7785. 

28.  What  will  it  cost  to  build  37428  cubic  yards  of  levee,  at  23  cents  i>er 
cubic  yard  ?  Ans.  $8608.44. 

29.  A  steamboat  arrives  with  3840  bales  of  cotton,  1320  sacks  of  cotton  seed, 
and  580  barrels  of  molasses.     Her  freight  charges  are  $2  per  bale  for  cotton,  25/  per 


*  MISCELLANEOUS    PROBLEMS    IN    MULTIPLICATION.  8/ 

sack  for  cotton  seed,  ami  5(»/'  i)er  barrel  for  molasses.     What  is  the  amount  of  her 
freight  bill  ?  Aiis.  88300. 

30.  A  draymau  charges  75  cents  a  load,  and  he  has  hauled  63  loads.  How 
much  is  due  him  ?  Aus.  $47.25. 

31.  What  will  it  cost  to  slate  the  roof  of  a  house  coutaiuiug  52  squares,  at 
$13.25  per  square  ?  Ans.  $GS9. 

32.  The  walks  around  a  dwelling  contain  129  square  yards.  What  will  it 
cost  to  flag  thein  with  German  flags,  at  S3.10  per  square  yard?  Ans.  $399.90. 

33.  What  will  it  cost  to  pave  a  street  coutaiuiug  20000  square  yards,  with 
stoue,  at  §4.75  per  square  yard  !  Ans.  $95000. 

34.  Bought  2180  barrels  of  coal  at  48/  per  barrel.     What  was  the  cost  ? 

Ans.  $1040.40. 

35.  IMultiply  5  billion  IC,  by  5  million  1  thousand. 

Ans.  2500500008001GOOO. 

3G.  A  hogshead  of  sugar  contains  1085  pounds.  Dow  many  pounds  iu  107 
hogsheads  of  equal  weight  ?  Aus.  11G095. 

37.  A  i>lanter  produced  G8  bales  of  cotton.  If  the  average  weight  of  the 
bales  was  400  jtouuds,  and  the  cotton  sold  for  13  cents  per  pouud,  how  uuicli  money 
would  it  bring  ?  Ans.  $4060.40. 

oS.  What  will  3  cases,  coutaiuiug  2  dozen  pairs  each,  of  shoes  cost,  ©  $2.90 
per  pair  ?  Ans.  $208.80. 

39.  If  it  costs  $1.50  a  day  to  support  one  person,  what  will  it  cost  to  sup- 
port a  family  of  13,  for  one  year  or  305  days  '?  Aus.  $7117.50. 

40.  A  merchant  sold  three  dozen  dozen  ladies'  hose  at  one-quarter  of  a  dozen 
dozen  ceuts  a  pair.     How  much  did  he  receive  for  them  1  Ans.  $155.52. 

41.  The  pressure  of  the  atmosphere  is  15  pounds  on  every  square  inch  of 
surface.  The  exterior  surface  of  a  man  of  average  size  is  about  2500  square  inches. 
How  many  jrouuds  weight  does  he  sustain  ?  Ans.  37500  pounds. 

42.  How  many  dollars  are  375  $10  gold  pieces  worth  ?  Ans.  $3750. 

43.  What  is  the  value  of  2146  dimes  ?  Ans.  $214.60. 

44.  What  is  the  value  of  1010  quarter  dollars?  Ans.  $252.50. 

45.  What  is  the  value  of  728  nickles  ?  Ans.  $36.40. 

46.  What  is  the  value  of  1612  half  dollars  1  Ans.  $806.00. 

47.  4875  is  the  thirteenth  part  of  a  number.    What  is  the  number  ? 

Aus.  63375. 

48.  The  sun  is  1384500  times  as  large  as  the  earth ;  the  earth  is  45  times  as 
large  as  the  moou.     How  many  times  is  the  sun  as  large  as  the  moon  % 

Ans.  62302500. 

49.  A  man's  receipts  are  $1800  a  year,  and  his  disbursements  are  $1125  a 
year.    How  much  are  his  net  receipts  iu  three  years  ?  Ans.  $2025. 

50.  The  iiopulatiou  of  a  city  is  250,000.  If  the  average  expense  of  each 
inhabitant  is  $1  per  day,  what  is  the  total  yearly  exi^euse  for  an  ordinary  year  ? 

Ans.  $91,250,000. 

51.  A  merchant  sold  52  barrels  of  molasses,  each  containing  41  gallons,  at 
38/  iier  gallon.     How  much  did  he  receive  for  it  1!  Ans.  $810.16. 


88  soule's  thilosophic  practical  mathematics.  * 

i>'2.  Light  travels  192500  miles  a  second,  :ui<l  it  requires  100000  years  to  travel 
to  us  from  some  of  tlie  fixed  stars  that  are  seen  with  the  telescope.  Allowing  365 
days,  5  hours,  48  minutes,  nnd  40  seconds  to  a  year,  and  I'cmembering  that  there 
are  24  hours  in  a  day,  (iO  minutes  in  an  hour,  and  00  seconds  in  a  minute.  Low  far 
distant  are  such  stars  1  Ans.  607470SS3250000000  miles. 

SIMULTANEOUS,  OR  CROSS  MULTIPLICATION. 

164.  This  system  of  nuiltiiilicatiou  is  of  iuestimable  value.  It  is,  so  to 
speak,  the  accountants'  aiul  calculators'  magic  wand,  by  which  they  may  produce 
results  in  multiplicatiou  operations  with  almost  lightning  rapidity. 

No  one  can  be  proficient  in  the  handling  of  uumbers  without  a  thorough 
knowledge  of  this  system  of  work  in  addition  to  the  several  other  contracted 
methods,  which  follow  this  work. 

The  operations  of  this  system  are  based  upou  the  foUowiug  ijriuciples : 

1.  Units  X  units  i)roduce  units. 

2.  Tens  X  units  "         tens.    ) 

3.  Units  X  tens  "         tens.    ) 

4.  Ilnndredsx  units  i)roduce  hundreds.  ) 

5.  Tens  x  tens  "       hundreds,  y 
C.  Units  X  hundreds        "        hundreds. 

7.  Thousandsx  units       "        thousands. 

8.  Hundredsxtens  "        thousands 

9.  Tensx  hundreds  "        thousands 

10.  Units  X  thousands       "        thousands.  J 

11.  Ten  thousandsx  units  produce  ten  thousauds."^ 

12.  Thousandsx  tens  "        ten  thousands.  | 

13.  Hundreds  X  hundreds        "         ten  thousands.  )► 

14.  Tensx  thousands  "         ten  thousands.  I 

15.  Unitsx  ten  thousaiuls        "        ten  thousands.  J 

16.  Hundred  thousands  x  units  produce  hundred  thousands.^ 

17.  Ten  thousandsx  tens  "         hundred  thousands. 

18.  Thousandsx  hundreds  "         hundred  thousands. 

19.  Hundreds X  thousands  "        hundred  thousands. 

20.  Tensx  ten  thousaiuls  "         hundred  thousands. 

21.  Unitsx  hundred  thousands  "         hundred  thousands. 

and  so  on  for  hkjher  numbers,^ 

PROBLEMS. 

1.     Multiply  54  by  37. 

OPEKATION.  ExpJanalioii.    First  iniiltiply  together  in  the  ordinary  manner,  the  nnits'  fig- 

rjj.  nres,  thus,  7  times  4  =  28,  and  write  the  8  in  the  first  plate  of  the  prodnct,  and 

.,„  carry  2.    Next  multiply  the  tens' ii<;nre  of  the  multiplicand  by  the  units'  tigure 

"*  of  tlie  multiplier,  thus,  7  times5(+2  tocany)=37,  which  retain  in  the  mind  and 

V  adil  thereto  the  product  of  the  units'  figure  of  the  multiplicand  liy  the  tens, 

ligure  of  the  nniltii)licr,  thus,  3  times  4  =  12,  +  37  =  49.     Write  the  9  in  the 

1998  tens'  place  of  the  ])roduct  and  retain  the  4  hundreds  in  the  mind  to  lie  added  to 

the  column  of  hundreds.  The  product  of  the  multiplicand  hy  the  units'  figure 
of  the  multiplier,  and  also  of  the  units'  figure  of  the  multiplicand  by  the  ten.s'  figure  of  the  multi- 
plier is  now  ])roduced  in  the  two  figures  (98)  of  the  final  proiluct,  and  hence  we  have  no  further  use 
for  the  units'  figure  of  either  factor,  therefore  check  and  pass  the  unit  column  and  proceed  to  mul- 
tiply the  tens'  figures  of  the  two  factors,  thus,  3  times  5  are  1.5,  plus  the  4  retained  in  the  mind,  makes 
19,  which  write  to  the  left  of  the  two  product  figures  first  obtained,  and  complete  the  product. 


Is.    I 

Is.  r 


SIMULTANEOUS    OR    CROSS    MULTIPLICATION. 


89 


To  elucidate  tbe  opeiatiou  by  figures  only,  we  present  the  followicg; 
Explanation. — 


OPERATION. 

54 

37 

V 


Carryinjr 

ti  Lilies. 


X  -t  = 


7  X  5  +  2  +  .-5  X  4  = 


1998 
Multiply  tbe  following  numbers : 


3x^  +  4 


Product 
tigures. 


8  units. 

9  tens. 
9  liunds 


(2) 

(3) 

(4) 

(3) 

(6) 

(T) 

(8) 

(9) 

35 

62 

87 

7G 

93 

89 

93 

86 

46 

23 

42 

58 

64 

97 

78 

92 

10.     Multiply  5417  by  62. 


Ans.  335854. 


Explanation. — In  tbis,  as  in  the  above  problems,  first  miiltiiily  the  units 
figures  and  set  the  result  in  the  first  place  of  tbe  product.  Thus,  2  times  7 
are  14;  set  tbe  4  and  carrying  the  1  proceed  as  follows:  2  times  1  are  2  plus 
1  makes  3,  to  which  we  add  6  times  7  are  42,  making  45;  set  the  5  and  carry 
4.  Then  2  times  4  are  8  plus  4  makes  12,  to  which  we  add  6  times  1  are  6, 
_  .^_  making  18;  set  the  8  and  carry  the  1.     Then  2  times  5  are  10  jilus  1  makes  11, 

o35bo4  to  which  we  add  6  times  4  are  24  making  35  ;  set  the  5  and  carry  the  3.     Hav- 

ing now  as  many  figures  in  the  partial  product  produced  as  wo  have  in  the  multi]dicaud,  for  the 
reason  given  in  tbe  first  problem  we  check  and  pass  tbe  units'  figure  of  the  uniltiplier,  and  with  the 
tens'  figure  multiply  6  times  5  are  30  plus  the  3  to  carry  makes  33,  which  completes  the  product. 


OPEKATION. 
5417 
G2 

V 


11.     Multiply  62549  by  53. 

Explanation. — 

OPERATION.  ^ 

62549 
53 

V 


Carrying  and 
prmhu-t  figures. 

X  9  =  27  or  tlius: 


U 


12,  14,  45,  59; 


3x4  +  2  +  5_ 
3  X 


3315097 


X  9  =  59 

5  +  5  +  5^^  =  40 

3x2  +  4  +  5^xT>  =  35 
3x6  +  3  +  ox  2  =  31 
5x6+3     =33 


15,  20,  20,  40  : 


Multiply  tbe  following  numbers : 

(12)                        (13)                       (14)  (15) 

13502                3243                14907  897 

43                    27                      64  48 


6,  10,  25,  35 ; 

18,  21,  10,  31 ; 

• 

(cbeck  tbe  3). 

30,  33. 

(16) 

(17) 

52061 

2981453 

83 

39 

18.     Multiply  7524  by  346. 


OPERATION. 

7524 
346 

VV 


Explanation. — In  the  elucidation  of  this  problem,  tbe  principles 
used  beiug  the  same  as  in  the  above  examples,  we  will  therefore  con- 
dense the  explanation  of  the  work. 

6  times  4  =  24,  set  the  4  and  carry  the  2. 

6  times  2  =  12 +  2  =  14  +4  times  4  =  30,  write  tbe  0  and  carry 
tbe  3. 

6  times  5  =  30 +  3  =  33 +  4  times  2  =  41  +  3  times  4  =  53,  write 
the  3  and  carry  tbe  5. 
6  times  7  —  42 +5  =  47 +  4  times  5  =  67  +  3  times  2  =  73,  write  the  3  and  carry  tbe  7. 

Having  now  as  many  figures  in  tbe  partial  and  final  product  as  we  h.ave  figures  in  the  multi- 
plicand, we  observe  that  the  product  of  the  multiplicand  by  tbe  units'  figure  of  tbe  multiplier  is 
already  produced  in  the  final  product,  and  hence  we  cheek  and  pass  the  units'  figure  and  commence 
with  the  tens'  figure  of  the  multiplier ;  and  as  the  product  of  the  first  three  figures  of  the  multiplicand 


2603304 


90  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

has  been  taken  by  tho  tens'  figure  of  the  multiplier,  we  commence  with  the  fourth  or  last  tigure  of 
the  unilti]iliiaiiil!     4  times  7  =  :i8  +  7  =  35  +  3  times  5  =  50,  set  the  0  ami  cany  tlic  5. 

Now,  as  tlie  product  of  the  multiplicand  by  the  teus'  tigure  of  the  multiplier  is  produced  in 
the  final  product,  we  check  and  pass  the  same  and  commence  with  the  third  or  hundreds'  figure  of 
the  nuiltiplier ;  and  as  the  produi't  of  the  first  three  figures  of  the  multiplicand  by  the  third  or 
hundreds'  figure  of  tlie  multiplier  has  already  been  produced,  we  commence  with  the  fourth  or 
thousands'  figure  of  the  multiplicand. 

3  times  7  =  21  -|-  5  =  26,  which  we  set  and  complete  the  jiroduct. 

19.    Multiply  74018  by  45G02. 

OPERATION. 

4  0002  Explanation. product  tigmes. 

WW  2  X  8  =  10  or  thus: 


3375368830 


2x1  +  1  + I>X8=    3  10;2,  3;48; 

2x  0  +  Ox  1  +  0  X  8  =  48  8,12,0,18,40,58; 

2  x  4  +  4  +  0x0  +  0x1  +  5x8  =  58  14,  19,  5,  24,  32,  50 ; 

2x7+  5  +  0  X  4  +  0x0  +  5x1  +  4x8  =  50  (Check  2)  5,  24,  29  ; 

0x7  +  5  +  0  X  4  +  5x0  +  4x  1  =  33  33  ;  (check  0)  42,  45, 

0x7  +  3  +  5  X  4  +  4x0  =  Go  20,  05 ;  (check  6)  35, 

5x7  +  0  +  4  X  4  =  57  41,  10,  57  ;  (check  5) 

4^r7  +  5  =  33  28,  33. 

The  multiplication  sign  X,  in  the  above  elucidation,  should  be  read  "times." 
Note  1. — It  will  be  observed  that  the  general  principles  governing  this  system  of  multiplication 
have  been  strictly  used  in  the  operations  and  explanations.  Thus,  in  the  last  problem,  units  are  multi- 
plied by  units,  (8  by  2);  tens  by  units,  (1  by  2) ;  units  by  tens,  (8  by  0) ;  hundreds  by  units,  (0  by  2); 
teus  by  tens,  { 1  by  0)  ;  units  by  hundreds,  (8  by  6) ;  thousands  liy  units,  (4  by  2) ;  hundreds  by  tens, 
(0  by  6) ;  tens  by  hundreds,  (1  by  6) ;  units  by  thousands,  {8  by  5) ;  ten  tliousands  by  units,  (7'by  2) ; 
thousands  by  tens,  (4  by  0) ;  hundreds  by  hundreds,  (0  by  (i) ;  tens  by  thousands,  (lby5);  units 
by  ten, thousands,  (8  by  4) ;  ten  thousands  by  tens,  (7by  0) ;  thousands  by  hundreds,  (4  by  6) ;  hun- 
dreds by  thousands,  (0  by  5)  ;  tens  by  ten  thousands,  (1  by  4) ;  ten  thousands  by  hundreds,  (7  by  6); 
thousands  by  thousands,  (4  by  .5);  hundreds  by  ten  thousands,  (0  by  4);  ten  thousands  by  thous- 
ands, (7  by  5) ;  thousauds  by  ten  thousands,  (4  by  4) ;  ten  thousands  by  ten  thousands,  (7  by  4). 

NoTK.  2. — By  these  lengthy  elucidations,  the  work  seems  more  tedious  than  it  really  is  in 
practice.  In  performing  the  operation  practically,  we  name  results  only,  and  thereby  very  much 
lessen  the  labor 


DRILL  PROBLEMS,  IX  SIMULTANEOUS  MULTIPLICATION. 

165.  Sttidents  should  drill  on  the  followiii;;  problems,  at  different  times 
until  they  can  call  the  intermediate  results  as  rajtidly  as  they  can  talk. 

A  few  Aveeks  of  practice,  an  hour  or  two  per  day,  will  capacitate  the  learner 
to  see  mentally  the  intermediate  results  faster  than  he  can  talk  or  \\Tite  them. 

1.     Multiply  1234  by  37. 

OPERATION. 

^^'l^  Intermediate  results.— ^2S;  21,23,  12,  35;  14,  17,  9,  26; 

V  7,  9,  0,  15 ;  (check  7)  3,  4. 

45658 


SIMULTANEOUS    OR    CROSS    MULTIPLICATION.  9I 

2.  Multiply  438  by  395. 

OPERATION.  Intermediate  results.— iO ;   15,   19,  72,  91 ;   20,  29,  27, 

•4  3  8  56,  24,  80  ;  (check  5)  30,  44, 

39  5  9,53;  (check  9)  12,  17. 

or  thus:       40  ;  91 ;  80 ;  (check  5). 

17  3010  53;  (check  9)  17. 

3.  Multiply  3G02  by  5184. 

OPERATION.  Intermediate  results.— S;  10;  24,  25,  2,  27  ;  12,  14, 
3602  48,  62,  10,  72;  (check  4)  24, 

518  4  31,  6,  37  ;  (check  8)  3,  6,  30,  36  ; 

VVV  (check  1)  15,  18, 

18672768  or  thus :  8;  16;  27;   72; 

(check  4)  37  ;  (check  8)  36  ; 
(check  1)  18. 

4.  Multiply  90472  by  8706. 

OPERATION.  Intermediate  results. — 12  ;  42,  43;  24,  28,  14,  42. 

o  0  a  7  o  4'-^'  '^^  53,  16,  69  ;  54,  60, 

oi^^P  28,  88,  56,  144;  (check  6). 

'/y,  14,  32,  46  ;  (check  0)  63,  67  ; 

(check  7  W2,  78; 


787649232  ■  ortlius:  12;  43;  42;  69; 

144;  (check  6)  46; 
(check  0)  67  ;  (check  7)  78. 

5.    Multiply  503102  by  430240. 

OPERATION.  Intermediate  results. — 8;  4;  4;  12,2,14,6,20; 

503102  2,0,8,8,10;  20,21,3,24; 

4  3  0  2  4  0  (flieck  4)  10,  12,  9,  21,  4,  25  ; 

^^^^  (check  2)  2,  12,  14;  (check  0) 

15,  16;  (check  3)  20,  21; 


216454604  480  theu  auuex  the  0. 

or  thus:         8  ;  4  ;  4;  20  ;  16  ;  24  ;  (check  4; 
25;  (check  2)  14;  (check  0) 
16 ;  (check  3)  21 ;  theu  aunex 
the  0. 

NoTE.-7-Wben  there  are  naughts  on  the  right  of  either  or  both  of  the  factors,  treat  them  a£ 
explained  in  Problem  1,  Article  159. 

0.    IMultiply  4081700  by  53920. 

OPERATION. 

40  817  0  0  Explanation.- 14 ;  66  ;  52  ;  115  ; 
5  3  9  2  0  48  ;  (check  2)  80  ;  (check  9)  20 ; 
^^^ (check  3)  26  ;  annex  3  naughts. 


220085264000 


92  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

7.     Multiply  240700  by  G8100. 
Multiply  by  the  following  problems  : 


(8) 

(9) 

(10) 

(11) 

(12) 

(13) 

321 

2334 

4354 

270018 

554422 

9080706 

34 

130 

809 

30402 

12345 

1122033 

10914 

(14) 

(15) 

(16) 

(IT) 

(18) 

(19) 

45() 

4120 

5841 

10190 

2730 

88997 

123 

54 

325 

2305 

2107 

11238 

50088 

(20) 

(21) 

(22) 

(23) 

(24) 

(25) 

G208 

894 

2382 

6709 

4087 

19205 

340 

143 

751 

284 

2614 

41003 

2110720 

(26) 

(27) 

(28) 

(29) 

(30) 

(31) 

14901 

789 

999 

666 

5566 

5070608 

782 

567 

888 

777 

4488 

107024 

11652582 

32. 

Multiply  222333444555  by 

219543620324 

OPERATION 

22233;, 

.444555 

219543620324 

48811889336710025135820  Ans. 


A  DIFFEEEXT  METHOD  OF  SIMDX,TANEOUS   MULTIPLICATIOK 


166.     Multiply  673  by  452. 


Aus.    304196. 


OPERATION. 

673 
254 

304196 


JExplanaiion. —  2x3=    6 

2x7  +  3x5  =  29 
2x6  +  2  carried  +  7x5  +  3x4  =  61 


5x6  +  6  carried  +  4  x  7  =  64 
4x6  +  6  carried  =  30 


It  will  be  observed  that  the  problem  is  to  repeat  673,  452  times ;  this  has  been 
done  in  the  operation,  though  by  reason  of  reversing  the  multiplier  it  appears  at 
first  sight  that  the  multii>licand  has  been  repeated  only  254  times.    By  following 


*  SIMULTANEOUS    OR    CROSS    MULTIPLICATION.  93 

the  (liftereut  steps  of  the  solution  with  care,  it  is  clear  that  the  673  has  t>een  repeated 
2  times  +  50  +  400  times,  as  the  problem  required. 

This  method  of  simultaueoiis  multi]>licatioii  is  preferred  by  some  calculators 
to  the  method  first  presented.  We  however  much  prefer  the  first  method,  for  the 
following  reasons : 

1.  There  are  no  less  figures  to  make  and  no  less  thiukiug  to  do  in  this 
method  thau  there  is  iu  the  first  method. 

2.  By  reversiug  the  multiplier  errors  are  much  more  liable  to  occur,  and. 
increased  work  is  required  iu  rewriting  the  multiplier. 

3.  When  making  extensions  in  bills  or  invoices,  the  leversing  of  the  juulti- 
plier  would  add  mental  labor  to  the  operation,  lead  to  confusion,  and  increase  the 
liability  to  make  errors. 

We  present  this  method  more  to  show  its  deficiencies,  when  compared  with 
the  regular  simultaneous  multiplication  first  presented,  thau  for  any  merit  it  pos- 
sesses. 

2.     Multiply  G758  by  927.  Ans.  62646GG. 

OPEKATION    BY    THE    ABOVE   METHOD. 


6758 

Explanation. — 

7x8=    56 

729 

7x5  +  5  +  2x8=     56 

G2646G6 

7x7  +  5  +  2x5  +  9x«  =  136 

7x6  +  13  +  2x7  +  9x5           =1U 

2x6+ 11 +  9x7=                            86 

9X6  +  8=                                              62 

TO  MULTIPLY  WHEN  THE  MULTIPLIER  CONSISTS  OF  TWO  FIGURES. 
167.     1.    Multiply  5234  by  23. 

*^^'^^.t'"°^'  Explanation.— Yivnt,     4  X  23  =  92  ;    set  the  2  .iiul  cnrry  9. 

^-•^^  .Secuuil.     3  X  23  =  69  +  9  =  78 ;  set  tbe  8  and  carry  7. 

23  Third.       2  x  23  =  46  +  7  =  53  ;  set  the  3  nud  carry  5. 


Fuiirtb.     5  X  23  =  115  +  5  =  120. 


120382 
2.     Multiply  G845  by  34. 

OPERATION.  Explanation.— 

6845  5X34  =  170. 

34  4  X  34  =  136  +  17  =  153. 
8  X  34  =  272  +  15  =  287. 


232730  6  X  34  =  204  +  28  =  232. 

Multiply  the  following  problems  in  the  same  manner : 


(3) 

928 

42 

(4) 
1837 
26 

(5) 
40281 
54 

(6) 
63024 
63 

(7) 

725 

37 

(8) 
309 
86 

(9) 
8523 
47 


94 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


CONTRACTIONS   WHEN    THE    MULTirLIERS    ARE    CONVENIENT    ALI- 
QUOTS  OF  10,  100,  OR  1000,  OR  SOME  MULTIPLE   TUEREOF. 


Js'oTE. — Those  who  are  uufauiiliar  with  division  and  with  fractious  may  pass  sncb  iiroblems 
as  involve  this  knowledge. 


168. 


TABLE    OF    ALIQUQTS. 


To  multiply  by 


u 
u 
a 
li 
a 
li 
II 
li 
li 
11 
a 
II 
li 
li 

li 
II 


1,7  multiply  by 
li 

If 

H 

14f 
16§ 

im 

25 

31i 

33^ 

37| 

62^ 

66| 


10  and  divide  tbe  product  by  9 


75 

83J 

87J 

112J 

125 

133i 

16C| 

250 

333J 

375 

625 

833J 

875 


u 
u 
a 
li 
II 
II 
ii 
II 
II 
(1 
a 
II 
II 
a 
a 
u 
II 
a 
II 
ii 
II 
li 


10 

10 

10 

10 

10 

100 

100 

100 

100 

100 

300 

100 

500 

100 

300 

500 


u 


II 
il 
II 
II 
II 
II 
II 
II 
II 
li 
II 
II 
II 
il 
u 


8 
7 
0 
4 
3 

IG 

12 
8 
7 
6 

16 
4 

16 
3 
8 


100  and  subtract  J  of  product ;  or 

200  and  divide  by  3. 

100  and  subtract  \  of  product. 

500  and  divide  by  6,  or  x  1000  and  -rl2. 

700  and  divide  by  8,  or 

100  and  deduct  J  of  product. 

100  and  add  J  of  product. 
1000  and  divide  by  8. 

100  and  add  J  of  product. 
1000  and  divide  by  6 
1000  "  4 

1000  "  3 

3000  "  8 

5000  "  8 

5000  "  6 

7000  "  8 


Note. — Learn  this  table  iierfectly. 
Tbe  reasou  for  tlie  above  methods  of  work  will   appear  plain  by  a  careful 
examination  of  the  following  problems : 

PROBLEMS. 
169.     1.    Multiply  289274  by  2^. 

OPERATION. 

4)  289274°  Explanation. — In  answering  the  conditions  of  this  simple  problem, 

which  is  to  repeat  the  mnltii)licaiid2i  times,  we  first  observe  that  2^  is  i  of 

r-cr>-i  cr    A     «         ■^*''  '""  *^''*  ^  times  2i  make  10,  and  hence  to  facilitate  the  work,  wo  tirst 
r^OiOc)  AnS.       niultiply  by  10,  which  is  done  by  annexing  one  naught.     This  gives  ns  a 
product  of  2892740  whicb  is  four  times  too  great;  for  the  reason  that  10  is  4  times  2J.     To  produce 
the  correct  result,  therefore,  we  divide  by  4,  which  gives  ns  723185. 

Note. — In  practice,  we  would  simply  divide  by  4  and  carry  the  division  one  place. 


CONTRACTIONS    IN    MULTIPLICATION    BY    ALIQUOTS.  qS 

2.     Multiply  79802  by  12J. 

OPERATION, 

8)    79802""  Eiplanalion. — The  conditions  of  this  prolilem  require  that  the  multi- 

Iilioiuul  be  taken  or  repeateil  12A  times  ;  but  iu  perfoniiing  the  operation, 


0(K'>~~    A      ■         "  '^  ''"**  ol)ser\  e  that  V2\  is  ^  of  100,  and  hence  to  save  time  and  figures,  we 
JJo_  (a  Allb.        fiist  take  or  repeat  the  uiultiplicand  100  times,  as  indicated  by  the  small 

naughts,  which  gives  a  product  as  many  times  too  great  as  100  is  times  as  great  as  \2i,  which  is 

8  times.     Therefore,  we  divide  by  8  to  produce  the  correct  product. 

Note. — In  ])ractice,  the  annexing  of  naughts  should  bo  made  mentally,  or,  we  would  sim2>l>j  ^ 

S  and  extend  the  operation  two  places. 

3.  Multiply  937104  by  25. 

OPEEATION. 

4)    937104"°  Explanation. — We  first  multiply  by  100,  which  is  done  by  annexing 

two  naughts  to  the  multiplicand,  and  then  divide  by  4. 

OQioTfin    A  "^"^  reason  for  this  work  is  the  same  as  that  giveu  in  the  preceding 

-Oiw/UUU  AUS.  example. 

Note. — In  practice,  divide  by  4  and  extend  the  operation  two  places. 

4.  Multiply  9420  by  6G§. 

OPERATION. 

3)    9420"°  £jj>;(jHO(ioH.— We  first  multiply  by  100,  and  then   deduct  one-third  of 

314'*00  *'^^  product  from  itself;  the  remainder   is  the  correct  result  or  product. 

■^  The  reason  for  this  is,  that  as  66f  is  only  f  of  100,  when  we  multiply  by  100, 

we  produce  a  product  J  too  great,  and  hence  by  deducting  J  of  the  product, 

628400  we  have  in  the  remainder  the  correct  result. 

Multiiily  tbe  following  numbers  by  the  above  methods : 


(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

(11) 

32 

36 

96 

108 

40 

124 

238 

6i 

8i 
300 

1C§ 
1000 

62J 

37i 
1500 

125 

112.J 

200 

10500 

15500 

26775 

(12) 

(13) 

(14) 

(15) 

(16) 

(17) 

(18) 

817 

640 

264 

640 

800 

696 

1215 

375 

625 

875 

75 

18? 

833J 

133i 

Multiply  321,  192,  248,  576,  643,  and  764, 

by  2^,  3i,  8i,  12^,  14f,  161,  18^  25,  33J,  37^,  62J,  75,  S7J,  112J,  125, 
375,  and  625. 


TO  MULTIPLY  BY  7J,  17J,  22J,  27^,  32.J,  15,  35,  45,  AND  55. 

170.  1.  To  multiply  by  7§,  multiply  by  10,  and  deduct  J  of  the  product. 
The  reason  of  this  is  7i  being  |  of  10,  nuiltiplying  by  10  gives  a  product  \  too  large, 
therefore,  by  deducting  ^  of  the  iH'oduct  by  10,  wo  obtain  the  correct  result. 

2.  To  multiply  by  17J,  multiply  by  20,  and  deduct  J  of  the  product.  The 
reason  is  17J  being  1  of  20,  multiplying  by  20,  gives  a  product  J  too  large. 
Hence,  by  deducting  J  of  the  product  by  20,  we  produce  the  correct  resuLt. 

3.  To  multiply  by  22J,  multiply  by  20  and  add  J  of  the  product  to  itself. 


/' 


96  soule's  philosophic  practical  mathematics.  * 

Cousideriiij;'  tlie  reasons  given  in  the  two  i)receding  problems,  this  operation  is  clear, 
from  the  fai-t  that  22i  is  ^  more  than  20. 

4.  To  ninltiply  by  27i,  nuiltiply  by  .'iO,  and  deduct  xa  of  the  product.  This 
oi)eration  is  evident  because  27^  is  \\  of  30. 

5.  To  multiply  by  32J  nuiltiply  by  30  and  add  -^.,  of  the  product. 

Note  1. — 32|  is  i\r  more  than  30. 

Note  2. — There  are  (tro  24's  in  5 ;  ihree  2.V's  in  7i  ;  four  2J's  in  10 ;  six  2J's  in  15  ;  e'ujht  1\  in  20  ; 
nine  2i's  iu  22i ;  eleven  2i's  iu  2l\,  and  thiriee)i  2"i's  in  32^. 

C.  To  multiply  by  15,  multiply  by  10  and  add  i  of  the  product  to  itself,  or 
multiply  by  30  and  divide  by  2. 

7.  To  nuiltiply  by  35,  45  or  55,  multiply  respectively  by  70,  90,  and  110,  and 
divide  the  product  by  2  ;  or  halve  the  nudtiplicand  and  double  the  iiuiltiplier. 

Note. — Tliis  jirinciplo  may  lie  applied  to  6i5,  75,  85,  and  95,  by  parties  who  know  the  nnilti- 
catiou  table  to  20  times  10. 

Multiply  the  followino-  numbers: 

(11                       (2)  (3)  (4)  (5) 

247                    214  5S6  637  4826 

15                      35  45  55  15 


7410  7490  26370  70070 


3705  35035 


PRACTICAL  OPERATIOlSrS  WITH  ALIQUOTS  AND  OTHER  NUMBERS. 

Note. — Students  who  are  not  familiar  with  division  and  fractions,  may  omit  such  prol)lems  as 
involve  this  knowledge  in  their  solution. 

171.     1.     AVhat  will  402  pounds  cost  at  25/. 

Directions.     Divide  by  4,  carrying  the  division  two  places  for  cents. 

2.  Wliat  will  840  yards  cost  at  12^C  per  yard  ? 

Directions.     Divide  by  8,  carrying  the  division  two  places  fin-  cents. 

3.  What  will  503  bushels  cost  at  $1.25  per  bushel  1 

Directions.  To  the  number  of  bu.shels  add  ^  of  same,  carrying  the  divis- 
ion two  i>laces  for  cents.     Or,  divide  by  8,  carrying  the  division  three  places. 

4.  What  cost  1267  gallons  at  $1.62.i  ! 

Directions.  To  the  number  of  gallons  add  i  and  J  of  same,  carrying  the 
division  two  places  tor  cents. 

5.  What  co.st  4102  bushels  at  $1.37 J  ? 

Directions.  To  the  luimber  of  bushels  add  ^  and  J^  of  same,  carrying  the 
division  two  places  for  cents. 

6.  What  cost  65  turkeys  at  $12.33=^  per  dozen  ? 

Directions.  Multiply  the  number  of  turkeys  by  12,  add  to  the  product 
J  the  iihmber  of  turkeys  and  divide  the  sum  by  12,  carrying  the  division  two  places 
for  cents. 


Jm 


'^  PRACTICAL    OPERATIONS    WITH    ALIQUOTS.  AND    OTHER    NUMBERS.  97 

7.  What  cost  1183  pounds  at  $1.16  ? 

Directions.     Add  to  the  price  J  and  -j^  of  itself.     Or,  add  to  the  price  J 
of  itself  phis  J  of  the  ^. 

8.  Wliat  cost  37.5  boxes  at  $2.35 1 

Directions.    Multiply  the  iirice  by  3  aud  divide  the  product  by  8,  carry- 
ing the  division  three  places. 

9.  What  cost  7()  yards  at  $1.75  ? 

Directions.     Multiply  the  number  of  yards  by  2  and  deduct  J  of  product. 
Note.     SI. 75  is  |  of  |2. 

10.  What  cost  1128  pounds  at  95  c-  ? 

Directions.     From  the  number  of  pounds  deduct  .^5,  carrying  the  divis- 
ion two  places  for  cents. 

11.  What  cost  50  barrels  at  $7.15  ? 

Directions.     Divide  the  price  by  2,  carrying  the  division  2  places  for  cents. 

12.  What  cost  84.J  gallons  at  75c'  ? 

Directions.     From  the  gallons,  with  the  fraction  expressed  decimally, 
deduct  J  of  the  same. 

13.  What  cost  2317  pounds  corn  at  37.J/'  per  bushel  ? 

Directions.     Multiply  the  pounds  by  3  and  divide  the  product  by  8,  7, 
and  8,  carrying  the  division  two  places. 

14.  What  cost  87  dozen  at  $1.20  jier  dozen  ? 

Directions.    Add  i  of  the  number  of  dozens  to  the  same,  carrying  the 
division  two  places  for  cents. 

15.  What  cost  025  pounds  at  $1.G8  per  pound  ? 

Directions.     Multiply  the  price  per  pound  by  5  aud  divide  the  product 
by  8,  carrying  the  division  three  places. 
NoTK.     625  is  I  of  1000. 

16.  What  cost  83^  dozen  at  $6.70  per  dozen  ? 

Directions.     ^Multiply  the  price  by  5  aud  divide  the  product  by  6,  carry- 
ing the  division  two  places. 
Note.    83i  is  5  of  100. 

17.  WTiat  cost  91  yards  at  7i/  per  yard  ? 

Directions.     Annex  0  to  the  yards  and  deduct  ^  of  the  product  from  itself. 

18.  What  cost  95  yards  at  17i/  jier  yard  ? 

Directions.     Multiply  the  number  of  yards   by  20  aud  deduct  ^  of  the 
product  from  itself. 

Note.     17|  is  ^  of  20. 

19.  What  cost  187A  gallons  at  $3.42  per  gallon  ? 

Directions.     Multiply  the  price  by  200  and  deduct  tV  of  the  product  from 
itself     Or,  remove  the  decimal  point  two  jjlaces  to  the  right  of  the  price,  and  add  f 
of  the  amount  to  itself.     Or,  shorter  still,  multiply  the  price  by  3  and  divide  the 
product  by  10,  carrying  the  division  three  places. 
Note.    187i  is  ,3,  of  1000. 

20.  What  cost  250  barrels  at  $8.15  per  barrel. 

Directions.    Divide  the  j)rice  by  4,  carrying  the  division  three  iilaces. 


gS  soule's  philosophic  practical  mathematics.  * 

21.  What  cost  139  pounds  at  $1.1G§  per  pound  1 

Directions.  To  the  number  of  pounds  add  ^  of  the  same,  carrying  the 
division  two  jilaces  for  cents. 

22.  What  cost  6871  yards  at  2i/  per  yard  1 

Directions.  Divide  the  number  of  yards  by  4,  carrying  the  division  one 
place  and  point  oft"  two. 

23.  What  cost  433  yards  at  $1.1G§  ? 

Directions.  To  the  yards,  with  the  fraction  expressed  decimally,  add  -^ 
of  the  result.  • 

24.  What  cost  87J  pounds  at  $2.03  ? 

Directions.  From  the  price  per  pound  omit  decimal  point,  deduct  J  of 
same,  carrying  the  division  two  places.  Or,  from  the  price  deduct  J  and  express 
the  fractional  remainder  decimally.     Thus,  265  —  33|  =  231^  =  .$231.87.^. 

25.     What  cost  235  pounds  at  $1.80  ? 

Directions.    Multiply  the  pounds  by  2  and  deduct  xo  of  the  product,  or  I  of 
the  multiplicand  from  the  product. 

20.     What  cost  196  pounds    at  $1.75? 

Directions.  Multiply  the  pounds  by  2  and  deduct  ^  of  the  product  from 
itself.     Or,  to  the  number  of  pounds  add  J  and  ^  of  the  same. 

27.  What  cost  164  yards  at  55/  ? 

Directions.  To  J  the  number  of  yards  extended  two  places,  add  -jV  of 
the  J. 

28.  What  cost  50  pounds  at  $1.29  ? 

Directions.  Divide  the  price  by  2,  carrying  the  division  two  places  for 
cents. 

29.  What  cost  58  pounds  at  97;'  ? 

Directions.  Annex  2  naughts  to  the  58  and  deduct  3  times  58  from  the 
product. 

30.  What  cost  8803  pounds  of  wheat  at  $1.13  per  bushel  1 

Directions.    Multiply  the  pounds  by  2  and  from  the  product  deduct  jV  of 

the  number  of  pounds.     Or  thus,  — ' 

Note.     $1.15  —  60  iiouuds  per  busbel  =  IHc-  per  pound. 

31.  What  cost  6420  pounds  corn  at  49/  per  bushel  ? 

Directions.     From  the  number  of  jtounds  deduct  J  and  i>oint  off  two  fig- 

r.     ^.1  6420  X  49 

ures.     Or  thus,  ^ 

Note,     f^  =  Jc.  per  pounil. 

32.  What  cost  582  gallons  at  $1.31  J  1 

Directions.  To  the  number  of  gallons  add  J  and  jir  (J  of  the  i)  of  the 
gallons. 

33.  What  cost  465  yards  at  $2.935c  ? 

Directions.  Multiply  the  number  of  yards  by  3  and  from  the  product 
deduct  tV  of  the  number  of  yards. 


♦  PRACTICAL    OPERATIONS    WITH    ALIQUOTS    AND    OTHER    NUMBERS.  99 

3i.     What  cost  3840  pouiuls  of  liay  at  $183  per  ton  ? 

Directions.     — — ^jqqq or  point  off  three  figures  in  tlie  pounds  and  mul- 

tiplj^  the  poiiiuls  by  J  the  price,  or  the  price  by  i  the  pounds. 

NoTK. — In  this  case,  when  the  prire  is  in  dnllars  there  will  be  three  decimals  in  the  answer. 
When  the  price  contains  cents,  there  will  he  live  decimals  in  the  answer. 

35.  What  cost  14C75  pounds  of  oats  at  48;^  per  bushel  ? 

Directions.      — ^^75 or  multiply  the  pounds  by  3,  divide  the  product 

by  2  and  i)oiiit  off  two  places. 

36.  What  cost  2S7C0  pounds  of  barley  at  52/  per  bushel  1 

Directions.     ^ — ^-— — —  or  to  the  pounds  add  -^  of  the  same  and  point  off 
two  places. 

37.  What  cost  34  bushels  and  24  pounds  of  corn  meal  at  75/  per  bushel  ? 

Directions.    To  |  of  the  bushels  add  tr  of  75, 

thus,  34. 

8.50  or  thus,  34  at  75  =  $25.50 

$25.50    =  cost  of  34  bushels. 

.375=    "    "    i  "        or22Ibs.  24x75=    .41 

35  =    "    "    2  lbs.  44 


$25.91    practically. 


$25.91  practically. 


38.    What  cost  475  lbs.  oats  at  36/  per  bushel  ? 

Directions.     Since  there  are  33  lbs,  in  a  bushel,  mentally  sej)aratie  the 
price  36/  into  32/+ 4/.    Point  off  two  places  in  475  and  then  add  J  of  it  to  itself. 


Thus:     4.75 

.591 


$5.34f 
39.    What  cost  1663  pounds  of  beans  at  $1.30  per  bushel  ? 
Directions.     ^'^'^^^>ji.30  ^^  ^^g^  4.  C2  =  26H  bus. 

26  bus.  $1.30  =  $33.80  $26.      =  26  bus.  at  $1.00 

311hs.  .65  7.80  =  26       "  .30 

15i  "    say  .33  or  thus,         .65  =  31  lbs. 

4i"      "  .  9  .33  =  15J" 

.  9  =    4i  " 


$34.87  practically. 


$34.87  practically. 
40.    What  cost  2781  pounds  of  small  hominy  at  64  cents  per  bushel  ? 

Directions.  ^  Or,  since  there  are  50  pounds  in  a  bushel  and  since 
the  price  is  equal  to  50  +  |  +  -^  of  ^  of  50,  therefore  to  the  number  of  pounds,  add 
i  and  IT  of  i  of  the  pounds,  and  point  off  two  places. 

Thus:        i)  27.81 


A)    5.562 
2.2248 


$35.5968 


lOO  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

41.    ATliat  cost  473  bushels  and  55  pounds  of  sweet  potatoes  at  $1.40  per 
bushel  ? 

Note.     60  pounds  make  a  bushel  of  sweet  potatoes  hi  Now  Orleaus. 

Directions.     473  x  $1.40  =  $CG2.20  to  which  add  U  of  $1.40  which  is  $1.28 
iU  or  {^  of  $1.40)  =  $1.40-13  (jJ^-  of  140)  =  $1.28  +  $002.20  =  $063.48.     Aiis. 

or  thus :  To  the  bushels  add  f  of  themselves,  and  to  this  result 
add  H  of  $1.40  thus : 

473. 
94.  GO 
94.  GO 

1.28 


J;G63.48 

42.  What  cost  53  tons,  1670  pounds  of  hay  at  $16.25  per  ton  ? 

43.  What  cost  25  sacks,  3750  pounds  corn  at  63/  per  bushel  ? 
Directions.    Add  to  the  weight  J  of  itself. 

Note.     63  is  i  more  than  56,  the  numher  of  jiouuds  in  a  bushel  of  corn. 

44.  What  cost  25  sacks,  4375  pounds  of  bran  at  87^/  per  cwt.  1 

Directions.     Subtract  from  the  weight  ^  of  itself. 

87i- 
NoTE.     87Jc.  for  100  pounds  is  i7|H~=  |c.  per  pound  :=  |c.  less  than  Ic.  per  pound. 

45.  What  cost  30  quarter  bales,  3214  pounds  hay  at  $15  per  ton  ? 
Directions.    Subtract  from  the  weight  J  of  itself. 

NoTK.     $15  per  ton  is  |Jo3  =  Jc.  per  pound. 

46.  What  cost  176  sacks,  35220  pounds  oats  at  37i,c'  per  bushel  ? 
Directions.     Add  to  the  pounds  ^  of  itself,  ^  of  the  J  and  ^  of  the  J. 

Note.     Since  32  pounds  make  a  bushel,  37J  is  i,  J  of  the  J,  and  i  of  the  J  more  than  32. 

47.  What  cost  35220  pounds,  llOOf  §  bushels  oats  at  38/  per  bushel  ? 
Directions.    Add  to  the  weight  ^  of  itself,  and  J  of  the  J.    Or,  add  to 

the  weight  (38-32)  =  6/  for  each  bushel,  and  4/  (32/)  for  20  pounds  ®  6('  per  bushel. 

The  above  problems  should  be  carefully  noted  by  the  student,  as  they  contain 
many  points  that  Avill  aid  him  in  becoming  a  rapid  calculator. 

In  all  problems  of  multiplication,  a  quick,  critical  view  should  be  given  to  the 
factors,  the  price  and  the  quantity,  to  see  what  advantages  may  be  taken  or  what 
short  combinations  may  be  made  with  them  as  aliquots,  or  as  approximations  to  hun- 
dreds or  easy  fractions  of  hundreds,  or  in  consequence  of  their  family  properties 
.as  tens,  complements,  supplements,  etc.  By  so  doing,  the  mind  is  made  acute  and 
great  facility  is  acquired  in  handling  numbers.  The  table  of  aliquots  should  be 
thoroughly  understood. 

There  is  almost  no  end  to  the  various  combinations  that  may  be  used  in  the 
'Operations  of  midtiplication.  rarticnlar  attention  is  invited  to  the  methods  follow- 
ing, wherein  numbers  are  classed  into  families  and  where  each  family  possesses  some 


*  CONTRACTIONS   IN    MULTIPLICATJON.  lOI 

iiidividuality  of  relation  to  the  principles  of  mimbers  by  which  the  operations  of 
multiplication  may  be  contracted. 

It  is  often  convenient  to  consider  the  price  as  the  quantity,  and  the  quantity  as 
the  price,  before  making  the  mnJtiplication,  as  indicated  in  the  directions  for  work- 
ing several  of  the  above  problems. 

For  further  work  of  this  character,  see  the  miscellaneous  problems  following 
Denominate  Numbers. 


TO  MULTIPLY  WHEN  THE  MULTIPLIER  IS  A  MIXED  NUMBER  WHOSE 

FRACTIONAL  PART  IS  ONE  FRACTIONAL  UNIT  LESS  THAN 

A  WHOLE   NUMBER. 

172.     1.    Multiply  243  by  llj. 

OPERATION. 

243  £jr^)/«Ha/io)i.— Multiply    by    12,    and    deduct    from    tlie    product 

II4  i  of   the  nuiltiplieand.      The  reasou    for  this  is   clear,   for  wheu  we 

multiplied  by  12  we  repeated  the  multijilicaud  J  time  more  than  the 
problem  requires,  and  hence  the  deduction  of  one-fourth  thereof  gives 
the  correct  product. 


291G 
60f 


2855^ 


PROBLEMS. 


Multiply  the  following  problems  in  like  manner: 

(2)  (3)  (4)  (5) 

358  149  87  246 

^  5|  95  i-t* 


(6) 

(7) 

(8) 

892 

6^5 

486 

2^5 

68i 

37H 

TO  MULTIPLY  BY  ANY  NUMBER  BETWEEN  EIGHTY-EIGHT  AND  ONE 
HUNDRED,  OR  NINE  HUNDRED  AND  EIGHTY-EIGHT  AND 

ONE  THOUSAND. 

173.     1.     Multiply  21G  by  98. 

OPERATION.  Explanation — /»  aJl  prohlems  of  tins  l-incl,  mullipln  h)/ 100  and 

llien  deduct  an  manii  limes  the  multiplicand  ax  the  multiplier  is  less 

210°"  (/,„,(    100.     In   this  iirobleui,  we   lirst  uuiltijdy  the  216  by   100, 

432  Avbieh  is  done  by  annesin<;  two  naughts ;  this  gives  us  a  product 

2  times  216  too  great,  for  the  reason  that  100  is  two  more  than 

01  ICQ       A  ^^-     Hence,  to   produce  the  correct' result,    we  deduct  2  times 

-1  IDS     Ans.  216  which  is  432  from  the  product  by  100,  and  in  the  remainder 

we  have  the  correct  product, 

2.     Multiply  4268  by  997. 

OPERATION.  Explanation.— In  all  prohlems  of  this  lind.  multiply  i if  1000 

and  then  deduct  as  mamj  iimes  the  multiplicand  as  the  multiplier  is 

4268°  "o  less  than  1000.     In  this  problem,  we  first  multiply  the  4268  by 

12804  1000,  which  is  done  l>v  annexing  three  naughts  ;  this  gives  us  a 

product  3  times  4268  too  great,  for  the  reason  that  1000  is  three 

^o--i  oc      A  "'ore  than  997.     Hence,  to  produce  the  correct  result,  we  deduct 

4-001  JO     Ans.  3  ^jjjjgj.  4268,  which  is  12804,  from  the  product  by  1000,  and  in  the 

remainder  we  have  the  correct  product. 


3. 

342  by  94 

C. 

528  by  93 

9. 

4. 

684  "  88 

7. 

763  "  91 

10. 

5. 

507  "  91 

8. 

842  "  99 

11. 

I02  soule's  philosophic  practical  mathematics.  ^ 

Multiply  tlie  following  numbers  in  the  same  manner : 

3284  by  995  12.     800  by    89 

4009  "    989  13.     498  "     92 

3864  "    996  14.  6407  "   988 


TO  MULTIPLY  BY  ANY  NUMBEli  BETWEEN  ONE  HUNDRED  AND  ONE 

HUNDEED  AND  THIRTEEN,  AND  ONE  THOUSAND  AND  ONE 

THOUSAND  AND  THIRTEEN. 

174.     1.    Multiply  5239  by  107. 

OPERATION.  Explanation. — In  all  problems  of  this  kind,  first  multiply  Ijy 

503900  100,  then  adfl  to  the  product  thus  obtained,  as  many  times  the 

ofj^iyo  multiplicand  as  the  multiplier  is  greater  than  100. 

The  reason  for  this  is  obvious,  for  the  multiplicand  is  to  be 
increased  107  times,  and  by  annexin;;  two  O's  we  increase  it  100 
560573      Aus.  times,  to  which  product  we  add  the  iiroduct  of  the  multiplicand 

by  the  7,  aud  thus  repeat  the  multiiilicaud  107  times. 

2.     Multiply  2871  by  1005. 

OPERATION.  Explanaiion. — In  all  problems  of  this  kind,  first  multiply  by 

oii"i  000  1000,  then  add  to  the  product  thus  obtained,  aa  many  times  the 

-8'1   _  multiplicand  as  the  multiplier  is  greater  than  1000. 


14355 


288.5355    Ans.         the  work. 


Note  1. — In  the   foregoing  problems,  small   naughts   have 
lieen  annexed,  in  order  to  show  more  clearly  the  ojierations  of 


Note  2.— The  operations  of  the  two  preceding  methods  may  be  extended  to  the  extent  of  the 
students'  knowledge  of  the  multiplication  table. 

Multiply  the  following  numbers  in  the  same  manner : 

3.     2564  by  103  6.       521  by  1001  9.     48071  by  1007 


4. 

809  "  106 

7. 

1685  "  1006 

10. 

2910  "  111 

5. 

4712  "  112 

8. 

2364  "  1012 

11. 

17030  "  1008 

TO  MULTIPLY  BY  ANY  NUMBER  BETWEEN  TWELVE   AND   TWENTY. 


?5.  1 

]\Iultiply  564  by  13. 

OPERATION. 

1692 

664 

or   564  or  ,564x13 

13 

13      1692 

Explanation. — In  all  problems  of  this  kind,  multiply 
by  the  unit  figure  of  the  multiplier  aud  write  the  product 
over  or  under  the  multiplicand,  one  place  to  the  right,  and 

7r<w  ^ooi)  •i<l<l  same  to  the  multiplicand. 

By  this  means,  we  save  multiplying  by  the  tens'  figure 

rr-iot)  of  the  multiplier. 

Note. — This  method  of  arranging  the  figures  may  also  be  used  to  advantage  when  the  multi- 
plier is  any  number  similar  to  the  following:  106  or  1004.  In  such  cases,  the  iiroducts  would  be 
placed  two  places  to  the  right  for  hundreds,  aud  three,  for  thousands. 


CONTRACTIONS    IN    MULTIPLICATION. 


IQ- 


Multiply  the  following'  numbers  as  above ; 


2. 

681  by  14 

5. 

12G9  by  IG 

8. 

728  by   13 

11. 

286  by   19 

3. 

754  "  17 

c. 

5028  "   LS 

9. 

1407  "   15 

12. 

1G75  "   17 

4. 

425  "  100 

7. 

1232  "  108 

10. 

341G  "  1004 

13. 

40G2  "  1005 

TO    MULTIPLY    BY  ANY  NUMBER  OF  TWO  FIGURES   WHICU    ENDS 

IX  ONE. 


17G.     1.    Multiply  452  by  41. 


OPERATION. 


1808 
452 
41 

18532 


or 


452 
41 


or      452x41 

1808 


1808 


18532 


18532 


Explanation. — In  all  problems  of  this  kind,  multiply 
Ly  the  tens'  figure  of  the  multiplier  and  write  the  product 
over  or  iiudcr  the  multiplicand  one  place  to  the  left  and 
add  same  to  the  multiplicand. 

By  this  means,  we  save  multiplying  hy  the  units'  figure 
of  the  multiplier. 


Note  1.— This  method  of  aiTanging  the  figures  may  also  ho  used  to  advantage  when  the  multi- 
plier is  auy  number  similar  to  the  following  :  tiOl  nr  4001.  In  such  cases,  the  products  would  be 
placed  two  places  to  the  left  for  hundreds,  and  three,  for  thousands. 

Note  2. — In  all  these  contracted  methods,  the  ojieratiou  should  be  repeated  several  times  on 
each  problem. 

PROBLEMS. 


jMultiply  tbe  following  numbers  as  above: 

2.  048  by    31  5.     4235  by    51  8.     2890  by      61 

3.  872  "      41  G.     2806  "      81  9.     5075  "        91 

4.  425  "    501  7.     1242  "    601  10.     6322  "   4001 


11.  9027  by      71 

12.  23G4  "      111 

13.  4327  "    7001 


TO    MULTIPLY    BY    THE    FACTORS    OF    THE    MULTIPLIER. 


177.    Multiply  8421  by  64. 


OPERATION. 
8421 
64 


67368 


538944  Ans. 


Explanation. — Instead  of  multiplying  by  64,  we  use  the  factors  8  and 
8;  first  multiplying  hy  8  we  produce  a  product  of  673(38,  which  wo  multi- 
ply hy  8,  and  thus  produce  the  correct  result.  By  this  system  of  work,  we 
save  one  line  of  figures  and  the  addition  of  two  lines. 

See  page  84  for  further  explanations. 

PROBLEMS. 


Multiply  tbe  following  numbers  as  above : 


2. 

2407  by  32 

4. 

796  by  63 

6.  527  by  48 

8. 

2840  by  56 

3. 

682  "  45 

5. 

1428  "  36 

7.  290  "  72 

9. 

7065  "  49 

104 


SOULE  S   PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


TO  MULTIPLY  ANY  NUMBER  BY  ELEVEN  OE  ONE  HUNDRED  AND 

ELEVEN. 

178.     1.    Multiply  02  by  11, 

OPERATION. 


G82  Ans. 


Explanation. — In  iiiultiplyinK  nny  ii«ml)t'r  liy  11,  it  is  evident  that 
■n-e  have  but  to  add  tlie  imiiiljei-  to  its  decuple,  i.  e.  to  teu  times  the 
number,  heuco  in  this  and  all  operations  of  two  lijfures  we  have  but 
to  add  the  two  figures  together  and  place  the  sum  between  them. 
When  the  sum  of  the  two  ligures  is  in  excess  of  9,  carry  one  to  the 

left  hand  figure.     Thus  multiply  87  by  11 ;  8  and  7  added  make  15;  place  the  5  between  the  figures 

ami  add  the  1  to  the  8  which  gives  a  product  of  957. 

2.    Multiply  3456  by  11. 


OPERATION. 

3456 

11 


38016 


Explanation. — In  this  problem,  according  to  the  principle  stated 
in  the  first  problem,  we  add  the  number  to  its  decuple,  thus  0  +  G  = 
6;  6  +  5  =  11;  5  +  4  +  1  to  carry  =  10 ;  4  +  3  +  1  to  carry  =  8 J 
then  bring  down  the  3,  and  we  have  the  correct  product. 


3.    Multiply  3156  by  111. 

OPERATION.  Explanation. — In  this  problem,  we  are  required  to  repeat  3456 

100  times  plus  10  times,  plus  1  time.     Hence,  if  we  add  3456  to  10 

315o  times  3456  and  100  times  3456  we  shall  have  the  correct  result.   This 

111  we  do  by  bringing  down  the  first  figure  6,  and  adding  as  follows: 

6  plus  5  =  11;  6  plus  1  to  carry,  plus  5  plus  4  =  16;   5  plus  1  to 

SS'i61  fi  carry,  phis  4  plus  3  =  13 ;  4  plus  1  to  carry,  plus  3  =  8;  and  then 

o  ooio  bring  down  the  3.     By  performing  these  operations  in  the  usual 

manner,  the  work  will  ajjpear  jilaiu. 

Multiply  the  followiug  numbers  a.s  above : 

463  by  11  6.    245  by  111  8.     823  by  111  10.     2407  by    111 

386  "    11  7.     741   "     11  9.     568  "     11  11.  14586   "    1111 


TO  MULTIPLY  BY  ANY  NUMBER  ONE  PART  OF  WHICH  IS  A  FACTOR 

OF    ANOTHER    PART. 

179.     1.    Multiply  3246  by  328. 


OPERATION. 

3246 

328 


25968 
1038720 

1064688 

9 


Explanation. — The  conditions  of  this  problem  are,  that  we  take  or 
repeat  the  multiplicand  328  times.  In  performing  the  operation,  we 
first  multiply  by  8,  which  repeats  it  8  times;  we  yet  have  it  to  repeat 
320  times,  and  as  320  is  efiual  to  8  40  times,  therefore,  40  times  the  pro- 
duct by  8  (25968)  is  the  s.ame  as  320  times  the  multiplicand ;  hence,  to 
shorten  the  operation,  we  add  to  the  product  l)y  8,  40  times  itself. 
The  sum  of  these  two  products  is  the  correct  product. 


Multiply  7251  by  618. 


OPERATION. 
7251 


618 


4350600 
130518 

4481118 


Explanation. — In  this  problem,  we  6rst  multiply  by  the  6  hundreds 
which  repeats  the  7251  600  times,  and  leaves  it  unrepeated  18  times. 
To  repeat  it  18  times  more,  we  observe  that  18  is  equal  to  6,  3  times, 
and  having  already  repeated  the  7251  600  times,  we  therefore  repeat 
the  one  hundredth  part  of  the  product  by  the  six  hundreds  3  times, 
which  is  equal  to  18  times  7251,  and  add  the  same  to  the  product  by  6 
hundreds.     The  result  is  the  correct  product. 


CONTRACTIONS    IN    MULTIPLICATION. 


io5 


3.     Multiply  530183  by  27945. 


FIRST    OPERATION. 

530183 
27945 


■482504700 

14470941000 

24128235 


14983033935 

SECOND    OPERATION. 
Omitting  the  naughts. 

530183 
27945 


4825047 
14470941 
24128235 

14983633935 


Eiplanation. — In  this  problem,  we  observe  that  the  first  two  figures 
43,  are  equal  to  the  hundreds'  figure  5  times,  aud  that  the  thousands' 
and  ten  thousauds'  figures  27,  are  equal  to  the  hundreds'  figure  3  times. 
■\Ve  therefore  multiply  first  by  the  hundreds'  figure  9,  which  repeats  the 
multiplicand  900  times.  We  have  yet  to  repeat  it  27045  times,  and 
(using  the  27000  first),  since  27000  is  equal  to  9  3000  times,  wo  therefore 
repeat  the  product  by  9  3000  times,  instead  of  the  multiplicand  27000 
times,  which  gives  us  14476941000;  and  since  4.5  is  equal  to  9  five  times, 
we  therefore  repeat  the  one  hundredth  part  of  the  product  by  the  nine 
humlreds  5  times,  instead  of  repeating  the  multiplicand  io  times,  which 
gives  us  24128235.  AVe  then  add  the  several  products  together,  the  sum 
of  which  is  the  true  product. 


TO  MULTIPLY  BY  AXY  NUMBER   OF    TWO   FIGURES   EXDIXO    WITH 

NINE. 


180.     1.    Multiply  503  by  39 

OPERATION. 

503 


503 
39 


22520 


or 


2520 


21957  Ans. 


X  40  Explanation. — In  all  problems  of  this  kind,  in- 

crease the  multiplier  by  1  and  from  the  product  sub- 
tract the  multiplicand.     In  this  problem  we  multi- 
ply by  40,  which  gives  a  product  of  22.520,  from 
21957  Ans.  which  we  subtract  563  aud  obtain  21957,  the  answer. 

PROBLEMS. 

Multiply  the  following  numbers  as  above. 

2.  720  by  29  4.     804  by  49  6.     627  by  59  8.     9081  by  69 

3.  683  "   79  5.     927    "   89  7.  1074   "  19  9.     2345   "  99 


TO  MULTIPLY  ANY  TWO   NUMBERS    OF  TWO   OR  THREE    FIGURES 

EACH,  WHEN  THE   TENS   AND   HUNDREDS   ARE    ALIKE 

AND  THE  SUM  OF  THE  UNITS   IS  TEN, 


181.     L    Multiply  86  by  84. 


OPERATION. 
86 

84 


7224 


Explanation. — In  all  problems  of  this  kind,  first  multiply  the  units' 
figures  and  write  the  whole  result,  then  add  one  to  the  multiplier  of 
the  tens,  and  multijily  the  other  tens,  or  tens  and  hundreds,  by  i  t,  and 
prefix  the  result  to  the  product  of  the  units'  figures.  In  this  problem, 
we  first  multiply  4  times  6  are  24,  which  we  set.  Then  add  1  to  the  8 
of  the  tens'  column,  making  it  9;  wo  multiply  9  times  8  are  72,  which 
is  set  to  the  left  of  the  24  and  produces  the  correct  product. 


io6 


SOULE  S    PIIILOSpPHIC    PRACTICAL    MATHEMATICS. 


2.     Multiply  132  by  138. 

OPERATION.  Explanation. — We  first  multiply  8  times  2  are  16,  wliicli  coustitutea 

132  the  units'  ami  tens'  figures  of  the  iiroduet.     We  then  atUl  one  to  13, 

■10Q  making  it  14,  and  iuulti)ily  14  times  13  are  182,  which  completes  the 

■    product.     Or,  after  adding  the  1  to  the  13,  we  may  multiply  14  times  3 

are  43  ;  set  the  2  and  multiiily  14  times  1  i)lus  4  to  carry,  are  18. 

18210  NoTK  1. — -When  the  multiplication  of  the  units'  flgures  does  not 

give  a  product  of  two  figures,  the  tens'  ))laie  must  be  tilled  with  a  0. 

Note  2. — The  reason  for  this  work  is  that  since  the  sum  of  the  units'  figures  is  10,  and  the  tens' 

figures  are  alike,  it  is  plain  that  the  sinu  of  the  products  of  the  tens  liy  the  units  and  the  units  by 

the  tens,  is  equal  to  10  times  the  tens'  figure,  and  since  its  first  figure  occupying  the  second  place 

is  naught  it  does  not  change  the  original  product  of  the  units,  and  since  the  second  or  carrying 

figure  of  the  units  by  the  tens  is  always  the  same  as  the  tens  of  the  factors,  we  therefore  add  1  to 

either  factor  and  multiply  by  the  other  for  the  final  product. 


Multiply  tlie  foUowiiiij,'  immbers  as  above: 


(3) 
38 

(4) 
01 

(5) 
223 

(6) 

149 

(7) 
320 

(8) 
417 

(9) 
311 

(10) 
208 

32 

09 

227 

141 

324 

413 

319 

202 

1216 


4209 


50021 


31009 


105024 


172221 


(U) 

(12) 

(13) 

(14) 

(15) 

(1(5) 

(17) 

(18) 

(10) 

29 

47 

81 

91 

120 

200 

104 

102 

113 

21 

43 

89 

99 

124 

204 

100 

108 

117 

TO  MULTIPLY  TWO  NUMBERS  OF   TWO  FIGURES  EACH  WHEN  THE 
TEKS'  FIGURES  ADD  TEI^  AND  THE  UNITS'  FIGURES  ARE  ALIKE. 

182.     1.     Multiply  47  by  07. 


OPERATION. 

47 
07 


3140  Aus. 


Explanation. — In  all  problems  of  this  kind,  first  multiply  the  units' 
figures  and  set  the  whole  result  in  the  product  line,  then  multiply  the 
tens'  figures  and  add  to  their  ])roduct  one  of  the  units'  figures,  and 
jirefis  the  result  to  tlie  ]>roduct  of  tlie  units'  figures.  In  jiroblem  1,  we 
lirst  multiply  7  times  7  ^  40  ;  then  6  tiuu'S  4  =  24,  to  which  we  add  7 
and  obtain  31.     Problems  2,  3,  4,  and  5,  were  similarlv  worked. 


Note — ^^Vhen  the  ])roduct  of  the  units'  figures  is  but  one  figure,  the  tens'  place  in  the  product 
must  be  filled  with  a  0. 


Multiply  tlie  followiug  numbers  as  above : 


(2) 

(3) 

(t) 

(•J) 

(0) 

(7) 

(8) 

(9) 

(10) 

(11) 

84 

73 

90 

41 

78 

24 

98 

40 

42 

77 

24 

33 

16 

01 

38 

84 

18 

66 

62 

37 

2016 


2409 


1536 


2501 


CONTRACTIONS    IN    MULTIPLICATION. 


107 


TO    MULTIPLY    TWO    NUMBERS    WHEN    THE    UNITS'    FIGUEES    ARE 

FIVES  AND   THE   SUM   OF    THE   TENS'  FIGUEES   IS   MOEE 

OK  LESS  THAN  TEN. 


183. 

65 

85 

5525 


IMiiltiply  05  X  85. 


OPERATION. 


5x5  =  25.     8t. X  Ot.=  48h.+  J  of 
(S+G)  =  7h.=  551i.  or  5525. 

2.     Multiply  45  x  35. 


Explanation.     First,  5  x  5  =  25  units. 

Second,  8  X  6  ^  48  liundreils,  to  wliicU  add 
A  of  tlie  sum  of  tlie  tens'  figures,  8  -f-  I'  -i-  2  =  7, 
:=  55  hundreds. 


45 
35 


1575 


OPERATION. 

5  X  5  =  25.      3t.  X  4t.  = 
of  (3  +  4)3*11.+  I2I1. 
the  25  =  1575. 


Explanation.     First,     5  x  5  =  25  units. 
Second,     3  X  4  =  12  hundreds,  to  which  add 
I2I1.  X  h      *  "f  tlie  sum  of  the  tens'  figures,  3  +  4  -2  =  3 
=  1550  +       hnnilreds   50  units,  =  1550  +  the    25  units   first 
obtained  =  1575  product. 


45 
35 


1575 


SECOND   OPERATION. 


3x4  =  I2I1.+  iof  (3  +  4)  =  3h., 
50  uuits  =  1550  +  5x5,=  1575. 


Explanation. — In  this  solution,  since  the  snm 
of  the  tens'  figures  is  odd,  we  first  find  the  pro- 
duct of  the  tens'  figures  to  wliicli  wo  .add  the  pro- 
duct of  the  units'  figures  and  thus  jiroduce  the 
final  product.  This  method  may  he  used  in  all 
cases,  if  desired. 


PROBLEMS. 

Multiply  tlie  following  problems  as  above : 


(3) 

(i) 

(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

(11) 

85 

75 

45 

105 

165 

115 

175 

225 

185 

25 

65 

95 

85 

95 

115 

15 

35 

145 

TO  MULTIPLY  TWO  NUMBERS  OF   TWO    OR   THREE   FIGURES   EACH, 
WHEN    THE  HUNDREDS   AND    TENS,  OR   THE    UNITS'  OR 
TENS'    FIGURES    ONLY,    ARE    ALIKE. 


184.     Thus,  multiply  the  following  numbers : 


(1) 
54 

34 


1836  Alls. 


(2) 
87 
82 


(3) 
124 
123 


7134 


15252 


(4) 
116 
146 


16936  Aus. 


Explanation. — In  all  problems  of  this  kind,  first  multiply  the  unit  figures  and  write  the  unit 
result  in  the  product  line,  then  add  the  colunni  of  units  or  tens,  that  is  not  alike,  or  both,  where 
three  figures  are  used,  and  with  the  sum  multij)ly  one  of  the  numbers  in  the  column  where  thejf  are 


io8 


SOaLKS    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


alike,  iuiil  :ul(l  to  it.  tlui  cMrrviiif;  riijiiro;  tlipn  luiiltiply  tlio  fignros  in  tbo  tens  or  tons  ami  liuudreds, 
autl  adil  the  cairvin^  ligmv;  tUo  result  will  be  the  correct  jirodiict. 

In  jirobleiii  1,  we  iir.st  multiply  4  times  <  =  IG;  the  6  \vo  set  and  carry  1  ;  then  3  jdiis  5  =  8, 
and  8  timis  4  =  313  jilus  1  to  carry  make  33;  the  3  we  set  and  carry  3;  then  3  times  5  ^  15  pins  3 
to  carry  make  18. 

Ill  problem  2,  we  first  mnUiply  2  times  7  =  14  ;  then  7  plus  2  =  9  and  9  times  8  =72  plus  1  to 
carry  =  73;  then  8  times  8=  G4  plus  7  to  carry  =71.     Problems  3  and  4  are  similarly  worked. 

PROBLEMS. 

Multiply  the.  followiuf;-  imuibers  as  above. : 


(5) 

(6) 

(7) 

(8) 

(9) 

(10) 

73 

65 

106 

45 

123 

89 

23 

64 

107 

85 

113 

85 

(11)  (12)  (13) 

154  93  146 

154  97  106 


TO    MULTIPLY    BY    THE    COMPLEMENTS    AXD    SUPPLEMENTS    OF 

NUMBEES. 

[See  page  26  for  a  definition  of  Complements  and  Supplements  of  Numbers]. 

185.  The  following  operations  involving  the  complements  and  supplements  of 
iinmber.s  are  deduced  from  Algebraic  principles,  and  while  these  principles  are  easily 
applied  to  Arithmetical  Computations,  the  reasons  therefor  cannot  be  fully  com- 
prehended without  a  knowledge  of  Algebra,  and  hence  are  not  given. 


TO    MULTIPLY     TWO    NUMBERS,    BOTH    OF    WHICH    ARE   CONVEN- 
IENT   NUMBERS    UNDER    ONE    HUNDRED,    ONE  THOUSAND,  Etc. 

186.     1.    Multiply  97  by  94. 


OPERATION. 


97 

94 
9118 


(3  Complement  of  97) 
(6  Complement  of  94) 


Explanation. — In  all  problems  of  this  kind,  1st  multiply 

I  complements  of  tbo  two  numbers  and  write  the  result  in 

'    "'  '"■      Subtract   the   complement   of  either 


the  complc 

the     product     line.         2.        ou'JM.H-i.     mv     «„uiai|jiciuciiu     yfL    ciiinTi 

number  from  the  other  number  and  prefix  the  remainder  to 
the  product  of  the  complements.  Or,  add  the  multiplicand 
and  multiplier,  rejecting  1  at  the  left  baud  of  the  sum. 

KoTK.  1. — When  tbe  product  of  the  complements  does  not  contain  as  many  figures  as  therd 
are  naughts  in  tbe  order,  100,  1000,  etc.,  to  which  the  numbers  belong,  prefix  naughts  to  supply  the 
deficiency.     See  Problems  2  and  3. 

NoTK  2. — When  tbe  number  of  figures  in  the  product  of  the  complements  exceeds  the  naughts 
in  the  order,  100  and  1000,  etc.,  to  which  tbe  numbers  belong,  add  the  excess,  or  left  band  figure 
to  the  result  obtained  by  subtracting  the  complement  of  either  number  from  the  other  number. 
See  Problems  4  and  5. 


CONTRACTIONS    IN    MULTIPLICATION. 


109 


2.     Multiply  98  by  97 

OPEKATION. 

9S   (I'Comp.) 
97   (3Comp.) 


950G  Aus. 


Explanation.  Ist.  3  times 
2  =  0,  which  we  write  in  the 
product  line  and  prefix  a 
uaiight  as  directed  in  Note  1. 

2d.  3  from  98  =  95,  which 
is  prefixed  to  the  06  aud  com- 
pletes the  product. 


4.     Jlultiply  89  by  88. 


OPERATION. 
89  (UComp.) 
88   (12Comp.) 


7832  Aus. 


Explanation.  1st.  12  times 
11  =132,  32  of  which  is  writ- 
ten in  tlio  product  liue,  as  di- 
rected in  Note  2. 

2(1.  12  from  89  =  77  -)- 1  = 
78,  which  is  prefixed  to  the 
32  aud  completes  the  product. 


3.     Multiply  991  by  989. 

OPERATION. 

991    (  9Comp.  of  991.) 
989   (11  Comp.  of  989.) 


Explanation.  1st. 
11  times  9  =  99, 
which  is  written 
in  the  product  liue 
aud  a  uauj;ht  pre- 
lixed  as  directed  iu 
Note  1. 
2d.     9  from  989  =  980,  which  is  prefixed  to 


980099  Ans. 


the  099  aud  completes  the  product. 

5.     Multiply  978  by  940. 

OPERATION. 
978   {-'2  Comp.) 
940   (60  Comp.) 


919320  Aus 


Explanation.  1st.  60 
times  22  =  1320,  320  of 
whicli  is  written  in  the 
])roduct  lino  as  directed  in 
Note  2. 
2d.  60  from  978  =918-|- 
1=919,  which  is  prefixed  to  the  320 
and  completes  the  jiroduct. 


PROBLEMS. 


Multiply  tbe  followiug  numbers  as  above  ; 


(6) 

(7) 

(8) 

(9) 

(10) 

(11) 

(12) 

(13) 

(M) 

(ir.) 

94 

92 

88 

82 

99 

73 

97 

74 

Si, 

99 

93 

93 

88 

91 

96 

98 

62 

85 

91 

99 

(16) 

(17) 

(18) 

(19) 

(20) 

(2I> 

(22) 

(23) 

996 

988 

991 

969 

950 

940 

965 

999 

994 

997 

985 

985 

988 

930 

987 

909 

990024 


985036 


970135 


TO  MULTIPLY  TWO  NUMBERS,  AVHOSE  COMBINATION  IS  SUCH   THAT   THE   PRODUCT 

OF  THE  COMPLEMENT  OF  THE  UNITS'  FIGURE  OF  THE  MULTIPLIER,  BY  THE 

TENS'  FIGURE  OF  THE   MULTIPLICAND  IS   THE  SAME  AS  THE  PRODUCT 

OF    THE    TENS'   FIGURE   OF    THE   MULTIPLIER    BY   THE  UNITS' 

FIGURE    OF    THE    MULTIPLICAND. 


187.     1.     Multiply  48  by  34. 


Explanation. — In  this  prohlem,  the  complement  of  the  unit  figure  of 
the  multiplier  is  6,  and  6  times  the  tens'  figure  of  the  multiplicand  is  24 ; 
which  is  the  same  as  the  product  of  the  tens'  figure  of  the  multiplier,  and 
the  unit  figure  of  the  multiplicand.  In  all  cases  of  this  kind,  we  multiply 
the  units'  tigures  aud  write  the  result  iu  the  product  liue.  Then  we  add  1 
to  the  tens'  figure  of  the  multiplier,  and  with  the  sum  multiply  the 
tens'  figure,  or  tens  and  hundreds  of  the  multiplicand  and  prefix  the  result 
to  the  product  of  the  units'  figures.     In  this  prohlem,  we  say  4  times  8  are  32  ;  4  times  4  are  16. 

Note. — Any  multiple  of  11,  may  he  multiplied  as  ahove  when  the  figures  of  the  multiplier 
add  10. 


OPEKATION. 
48 
34 

1632   Aus. 


IIO  SOULES    PIIILOSOrillC    PRACTICAL    MATHEMATICS. 

PROBLEMS. 
2.     Multiply  tbo  followinj,^  muiibors  as  above  : 


(2) 

(3) 

W 

(5) 

(0) 

(■) 

m 

(9) 

(10) 

(") 

(12) 

(13 

63 

32 

88 

146 

128 

42 

96 

147 

96 

84 

44 

55 

29 

94 

73 

77 

94 

86 

3S 

124 

66 

67 

73 

82 

TO     MULTIPLY    TWO     iNTUMBEES,   BOTH    OP   WniCH    AEE    COISTVEN- 
lENT  miMBERS  UNDER  ANY  NUMBER  OF  HUNDREDS. 

188.    1.    Multiply  497  by  294. 

OPERATION.  Explanation.  —  In  this  problem,  we  first    multijjly 

>    i«-T  o  ^        1  ^    r  n-,      the      complements     of    the      multiplicaud      and      the 

(500)  497  3  Complement  of  97.     „„iltiplier  and  set  the  result  18,  in  the  product  line. 
(300)  294  6  Complement  of  94.      (Where  this  result  is  but  one  figure,  the  tens'  place  is 

filled  with  a  0.     When  the  result  is  more  than  two  figures, 

the  hundreds'  figure  is  added  to  the  hundreds'  column  of 

[   „  the  product).     2.     We  mentally  add  the  complement  fig- 

1461118  nres  to  their  respective  numbers,  and  thus  produce  500 

'  and  300  as  shown  in  the  parentheses  on  the  left.     3.     We 

multiply  the  5  of  the  500,  by  the  complement  of  the  multiplier,  and  the  3  of  the  300,  by  the  com- 
plement of  the  multiplicaud  and  add  the  products,  which  gives  39,  as  shown  by  the  number  within 
the  brace  on  the  left  of  the  operation.  4.  We  subtract  the  39  from  100  and  thus  obtain 
61  which  is  prefixed  to  the  18  in  the  iiroduct  line.  5.  We  multiply  5  and  3,  the  significant  figures 
of  500  and  300,  and  tlius  obtain  15,  from  which  we  subtract  1,  the  number  of  hundreds  used  when 
the  39  was  sul)tractcd,  and  prefix  the  remainder  14,  to  tlie  6118,  which  completes  the  product.  All 
similar  numbers  of  hundreds  and  thousands  may  be  multiplied  in  a  similar  manner. 

Note  1. — The  figures  shown  within  the  brace  and  parentheses  should  be  ixientally  made. 
Note  2. — When  the  sum  of  the  products  of  the  increased  hundreds'  figures  by  the  comple- 
ments, in  that  part  of  the  operation  where  the  39  was  produced,  exceeds  100,  tlien  subtract  the  sum 
from  200,  and  from  the  product  of  the  increased  significant  figures  of  the  multiplicand  and   multi- 
plier, deduct  2. 

PROBLEMS. 

2.     Multiply  595  by  393.  3.  Multiply  698  by  497. 

OPERATION.  Explanation  hy  figures.                              operation. 

(6)     595     (5)  7  times  5  =  35.                                            (7)     698     (2) 

(4)     393     (7)  7  times  6  +  5  times  4  =  62.                      (5)     497     (3) 


233855  Alls 

4  times  6 

23. 

346906 

Aus. 

Mult 

iply  the  following  numbers  as  above : 

ii) 

(5)       (6) 

(7) 

(8) 

(9) 

(10) 

(11) 

(9) 

891 

(9) 

291     892 

991 

394 

791 

392 

488 

(6) 

592 

(8) 

295     596 

892 

592 

495 

788 

692 

(12) 

(13) 

(14) 

(15) 

(16) 

(17) 

(18) 

(19) 

493 

692 

790 

392 

488 

698 

691 

894 

395 

597 

691 

593 

489 

496 

394 

389 

CONTRACTIONS    IN    MULTIPLICATION. 


Ill 


TO    MULTIPLY    TWO    LUMBERS,    BOTH    OF     WHICri     ARE     COIfVEN- 
lENT  XUMBERS  ABOVE  ANY  NUMBER  OF  HUNDREDS. 


1 89.     1.    Multiply  704  by  305. 


OPERATION. 
704 

305 


214720 


NOTK   1. 


Explanniion. — lu  this  problem,  we  first  multiply  tlie supplement  figures 
4  and  5,  iiud  set  tbe  result  I'O  iu  tbe  ])ru(Iuet  line.  2.  \Ve  multiply  tbo 
buiiilfeds'  ligures  of  eaiU  luimber  Ijy  tbe  supplement  figure  of  tbe  otber, 
;iu(l  lubl  tbe  two  ]>ro(luets  thus,  5  times  7  =  35  ;  4  tiiues  3  =  12  ;  3.")  and  12 
=  47;  tbis  47  we  ]irefis  to  the  20  in  tlio  i>rodnetline.  3.  Wemultiply  tbo 
bundreds'  figures  and  obtain  21,  whieb  we  write  in  tbe  product  and  com- 
plete tbe  operation.  All  similar  nuniliers  of  bundreds  and  tbousands  may 
be  similarly  worked. 


two 
exceeds 


— Wben  tbe  product  of  tbe  units'  or  supplement  figures  do  not  give  two  figures,  fill 
iwitbaO.     See  Problem  2.     Wben  this  product  exceeds  2  figures,  add  tbe   bundreds' 

...  , —  juudre<ls'  column  of  tbe  product.     See  Problem  3. 

Note  2. — Wben  tbe  sum  of  the  products  of  tbe  hundreds'  figures  by  tbe  units'  figures  is  not 


fiorures,  till  the  thousands'  place   in  the  ]tro(liiet  with  a  0.     8eo  Problem  4.     When  this  siiin 
eds  2  figures,  add  the  tea  thousands  to  the  ten  thousands'  column  of  the  product.     See  Problem  5. 


(2) 

301 

202 

60802 


(3) 
312 

209 

65208 


(4) 

501 
301 

150801 


PROBLEMS. 

(5; 

912 
811 


739632 


(6) 

813 

206 


(7) 

612 

509 


(8) 

1206 

911 


(9) 

406 

507 


(10) 
612 
109 


(11) 

308 

702 


(12) 
810 
305 


(13) 
511 
406 


(U) 
203 
202 


(15) 

1412 
1211 


TO     MULTIPLY     TWO     NUMBERS,  BOTH   OF    WHICH   ARE    CONVEN- 
IENT   NUMBERS    OVER    ONE    HUNDRED,    ONE 
THOUSAND,  Etc. 


190.     1.    Multiply  108  by  104. 


OPERATION. 


;:08  (8  Supplement.) 
104  (4  Supplement.) 

11232  Ans. 


Explanaiion — In  all  problems  of  this  kind,  first  write  tbe  product 
of  tbe  supplement  figures  iu  tbe  product  line.  2.  Cancel,  mentally, 
one  of  tbe  left  baud  figures  iu  one  of  tbe  numbers,  add  the  reuuiin- 
ing  figures  and  the  carrying  figure,  if  any,  and  prefix  tbe  sura  to  tbe 
product  of  tbe  supplement  figures.  Thus,  4  x8  =32;  tbeu  cancel 
1  in  the  108  and  adil  8  to  104  =  112,  which  is  prefixed  to  tbe  32  and 
comiiletes  tbe  product. 

Or,  first  cancel  the  left  hand  figure  iu  one  of  tbe  numbers,  then  add 
the  remaining  figures  and  annex;  thereto  the  product  of  tbe  supplements  of  the  two  numbers. 

Note  1. — In  case  of  bundreds,  wben  the  product  of  tbe  supplement  numbers  does  not  contain 
two  figures,  fill  tbe  tens'  place  with  a  0. 

In  case  of  tbousands,  wben  the  product  or  tbe  supplement  numbers  does  not  contain  three 
figures,  fill  tbe  hundreds'  place  or  the  tens'  and  hundreds'  places  with  O's. 

Note  2. — In  case  of  bundreds,  wben  the  product  of  tbe  supplement  numbers  contains  more 
than  two  figures,  add  the  hundreds  to  the  buuilreds'  column,  when  adding  the  multiplicand  and 
multiplier. 

In  case  of  thousands  when  the  product  of  tbe  supplement  numbers  contains  more  than  three 
figures,  add  tbe  thousands  to  the  thousands'  column  or  i>lace,  wben  adding  tbe  multiplicand  and 
multiplier. 


I  12 


soule's  philosophic  practical  mathemXtics. 


2.     Multiply  1014  by  1009. 

OPERATION. 


Explanation. — Here  we  first  nraltiply  tlie  supplement  numbers  and  write 
tbe  result  iu  the  product  Hue.  2.  Cancel,  mentally,  the  left  hand  figure 
in  one  of  tlie  numbers,  add  the  remainiuf;  figures,  and  the  carrying  fig- 
ure, if  any,  and  prefix  the  sum  to  the  product  by  the  supplement  figures. 
Thus,  in  this  prnldem,  'J  X  14  =  12U;  then  cancel  the  left  hand  1  in  the 
1014,  add  It  to  1009,  and  prefix  the  sum,  1023,  to  the  product  by  the  sup- 
plement numbers. 

Note  1. ^Vllen  the  product  of  the  supplement  numbers  does  not  contain  three  figures,  fill  the 

hundreds'  place,  or  the  hundreds'  and  tens',  with  Os.     See  Problem  4. 

Note  2.  AVlien  the  pro<luct  of  the  supplement  numbers  contains  more  than  three  figures,  add 
the  thousands  to  the  thous;iuds'  column.  For  the  higher  numbers,  as  10000,  etc.,  the  same  principle 
maintains.     See  Problem  5. 


/014 

loot) 

1023126  Alls. 


(9) 
103 
108 


PROBLEMS. 

ply  the  following- 1 

umber.s  as  above : 

(3) 

103 

102 

(i) 
1004 
1001 

(5) 
1036 
1040 

(6) 

116 

111 

12876 

(7) 
121 
104 

(8) 
1342 
1015 

10506 

1005004 

1077440 

12584 

1362130 

(10) 
114 
112 

(11) 
109 
112 

(12) 

1008 

1007 

(13) 
1024 
1016 

(14) 

1135 
1010 

(1,-;) 

1037 

1040 


TO  MULTIPLY  TWO  NUMBERS,  OXE  OF  WHICH  IS  A  COXVEXIEXT 

NUMBER  OVER,  AND  THE  OTHER  A  CONVENIENT  NUMBER 

UNDER  ONE  HUNDRED,  ONE  THOUSAND,  Etc. 


191.     1.     Multiply  105  by  93. 


OPERATION. 
105      (5  Supplement.) 
93      n  Complement.) 

9705 


Explanation. — In  all  proldems  of  this  kind,  first  multiply  the 
complement  and  supplement  numbers  and  subtract  the  product 
from  100,  or  1000,  according  as  the  numbers  multipliedbelong  to  the 
order  of  hundreds  or  thousands,  and  write  the  remainder  in  the  pro- 
duct lilu',  as  the  fixst  two  figures  of  the  product.  2.  Cancel  the 
left  hand  1  in  the  multiplicand,  add  the  remaining  figures  of  the 
same,  from  which  sum  subtract  1,  or  the  number  of  hundreds  used 
in  subtractiug  the  ])roduct  of  the  complement  and  supplement  fig- 
ures, and  prefix  the  rem.ainder  to  the  product  first  obtained.  Thus,  in  this  problem,  we  say,  7x5  = 
35  ;  35  from  100  =  65,  which  is  written  in  the  ])roduct  line.  Then  cancel,  mentally,  the  left  hand  1, 
we  add  93  and  5  =  98,  less  1,  for  the  hunilred  used  in  subtractiug  the  35  =  97,  which  is  prefixed  to 
the  65;  and  completes  the  product. 

Note  1.  AVben  the  remainder,  resulting  from  the  subtraction  of  the  products  of  the  comple- 
ment and  supplement  numbers  does  not  contain  two  figures,  then  fill  the  tens'  place  with  a  0.  See 
Problem  2. 

Note  2. — AVhen  the  product  by  the  complement  and  supplement  figures  exceeds  100,  or  1000, 
then  su1>tract  the  same  from  200  or  300,  or  2000  or  3000,  according  as  the  numbers  multiplied  are  of 
the  hundreds'  or  thousands'  orders,  and  then,  from  the  sum  of  the  a<lded  figures  of  the  multiplicand 
and  multiplier,  subtract  as  many  as  you  used  hundreds  or  thousands.     See  Problems  2  and  3. 


CONTRACTIONS    IN    MULTIPLICATION. 


113 


Multiply  tlie  following  miinbers  as  above  ; 


(2) 
112 
89 

9908 


Mental  Operatiox.— 11x12  =  132  ; 
200  —  132  =  68.  Tbeii9  +  2=11;  11  —  2 
=  9.  Then  8  + 1  =  9.  Left  liaiul  1  is 
c.Tiiceled. 


116 
97 

11252 

(12) 

108 

95 


(3) 

1012 
988 

999856 

(13) 

1012 

990 


(6) 
133 

85 

11305 

(U) 
1104 

988 


(3) 
1223 
994 

1215062 


Mextai,  Operatiox. — 6  x  223  = 
1338 ;  2000  — 1338  =  002.  Then  4+3 
=  7  —  2  =  5.       Theu  9  +  2  =  11. 

Then  9  +  2  +  1  to  curry  =  12. 

Left  liniul  figure  of  uiultiplicarnl 
is  cauceleil. 


(7) 

1023 

997 

1019931 


(15) 

121 

96 


(8) 
125 
96 


(9) 

123 

92 


(10) 

1013 

996 


(U) 

1123 
988 


12000       11316    1008918     1109524 


(16) 
111 

88 


(17) 

109 

89 


(18) 
1018 
940 


(19) 
1201 
996 


(20) 
1302 
975 


A  PECULIAR  PROPERTY  OF    THE    NUMBER   NINE. 

192.  If  vre  multiply  the  9  figures  iu  their  order  :  1,  2,  3,  4,  5,  6,  7,  8,  9,  by  9,  or 
auy  multiple  of  9,  not  exceeding  9  times  9,  the  product  will  be  in  like  figures, 
except  the  tens'  place,  which  will  be  a  0.  The  significant  figure  of  the  product  will 
be  the  quotient  of  the  multiplier  divided  by  9.  Thus,  to  multiply  by  9,  will  give  a 
product  of  Is ;  18,  which  is  two  times  9,  will  give  a  product  of  2s,  and  so  on  with 
the  other  nniltiples  of  9,  uji  to  81. 

OPERATIONS. 
(1)  (2)  (3)  (4) 

123450789  123456789  123456789  123456789 

9  27  45  72 


1111111101 


3333333303 


5555555505 


8888888808 


If  we  omit  the  8  iu  the  multiplicand  the  product  figures  will  all  be  the  same. 

There  are  many  peculiar  properties  belonging  to  the  figure  nine,  by  reason  of 
its  being  1  less  than  the  radix  of  our  system  of  notation,  but  being  mostly  of  no 
practical  value,  we  cannot  give  them  space.  There  is,  however,  one  property  of  the 
nine,  that  may  often  be  used  with  advantage  by  accountants,  in  the  detection  of 
errors  in  posting  or  transferring  accounts,  and  we  will  state  it.    It  is  as  follows : 

That  the  difference  between  any  number,  and  the  figures  composing  the  num- 
ber reversed  iu  any  maunei",  is  a  multiple  of  9.  Thus,  the  difference  between  6871, 
and  any  transposition  or  reversing  possible  to  be  made  with  the  same  figures,  will 
lie  a  multiple  of  9. 

EXAMPLES. 

6871        6871        6871        6871       6871       6871 
7861        1786        0718        6781       8671       8167 


9)  990 


9)  5085 


110 


9)  153 


17 


9)  90 


9)  1800 


9)  1296 


10 


200 


144 


Note. — See  Expert  Accounting,  in  Scull's  New  Science  and  Practice  of  Accounts,  for  an 
extended  application  of  tlie  Properties  of  9's  and  ll's,  to  tbe  finding  of  Errors  in  Accounts.  Also,  see 
the  Properties  of  9's  and  ll's  following  Division  in  this  book. 


114 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


ALGEBEAIC  FORMULAE  FROM  WHICH   ARE   DEDUCED    THE   OPERA- 
TIONS OF  MULTIPLYING  COMPLEMENT  AND  SUPPLEMENT 

NUMBERS. 


193.        (See  Problem,  page  105). 

8G  =  90  —  i 
84  =  !)()  —  0 


7224      <J0  X  00  — 4  X  00 

—  C  X  90  +  4X  0 

81()(»  — 10  X  00  +  24  = 
8100  —  900  +  24  =  7224 

(See  Problem,  page  107). 

65  =  60  +  5 

85  =  80  +  5 


5525       60  X  80  +  5  X  SO 

+  5x00  +  5x5 

4800  +  400  +  300  +  25  =  5525 


(See  Problem,  page  109). 

48  =  50  —  2 
34  =  50  — 16 


1632       50x50—2x50 

—  16  X  50  +  2  X  16 

2500  —  18  X  50  +  32  = 
2500  —  900  +  32  =  1032 

(See  Problem,  page  111). 

704  =  700  +  4 
305  =300  +  5 


214720 


700  X  300  +  4  X  300 

+  5  X  700  +  4  X  5 

21000()+12()(»+35(»0+20=214720 


(See  Problem,  page  112). 

105  =  1(»()  +  5 
93=  1(M»  — 7 


9705 


1()(»  X  100  +  5  X  100 

_7  X  100  —  5  X 

1()()()()_:>  X  100  —  35  = 
10000  —  200  —  35  =  9765 


47 

67 


(See  Problem,  page  106). 

50—3 
50  +  17 


3149 


97: 

94 


50  X  50  —  3  X  50 

+  17  X  50  — 3  X  17 

2500+  14  X  50  —  51  = 
2500  +  700  —  51  =  3149 

(See  Problem,  page  108). 

100  —  3 
:  100-6 


9118       100  X  100  —  3  X  100 

—  0  X  100  +  3  X  0 

10000  —  9  X  100  +  18  = 
10000  —  900+  18  =  9118 

(See  Problem,  page  110). 

497    =500—3 
294    =300  —  6 


140118 


500  X  300  — 3  X  300 

—  0x500+3x6 

150000—900—3000+18=140118 


(See  Problem,  page  111). 

108  =  100  +  8 
104  =  100+4 


11232 


100  X  100  +  8  X  100 

+  4  x  100  +  8  X  4 

10000  +  800  +  400  +  32  =  11233 


32  =  40  —  S 
43  =  40  +  3 


137G 


40  X  40  —  8  X  40 

+  3x  40  —  8  X  3 

1600  —  5  X  40  —  24  = 
1600  —  200—24  =  1376 


CONTRACTIONS    IN    MULTIPLICATION. 


Il5 


MISGELLA^TEOUS  DRILL  PROBLEMS  Ilf  CONTRACTED  METHODS. 

194.  The  student  who  aspires  to  rapidity,  accuracy,  and  superior  merit  as  a 
business  calculator,  should  jiractice  a  few  minutes  on  one  of  the  following  exercises, 
each  day,  before  entering  upon  the  regular  work  of  arithmetic. 

This  exercise  is  the  secret  road  that  leads  to  honor  and  fame  as  au  exx)ert 
calculator. 

Note. — Students  who  do  not  understand  division  and  fractions,  may  omit  sucli  problems  as 
require  such  knowledge. 


Add  atone 
Operation. 

36 
42 

EXERCISE    No.    1. 

MULTIPLY  SIMULTANEOUSLY. 

Add  by  the 
Ticenary  System 

58 
34 

57 
28 
81 
79 
14 
68 

34 

21 

46 
26 

62 
34 

58          2416          124 
47              23          237 

342 
276 

3204 
2063 

68 
17 
45 
78 
12 
34 

MULTIPLY  BY  ALIQUOTS. 

94 

86 
23 

246 

63 
25 

432 
33i 

56          344            1368 
12^          62i            125 

4026 

375 

72 
183 

56 
90 
13 
51 

COMPLEMENTS. 

SUPPLEMENTS. 

42 
67 

346 
481 
209 
526 

93 
92 

98 
97 

91 

88 

991 

988 

108          104 
105          102 

162 
107 

1012 
1011 

19 

78 
55 
89 

140 
37 

COMPLEMENTS  AND  SUPPLEMENTS. 

76 
90 

68 
34 

109 
93 

126 
92 

146 

77 

1012          497          594 
988          294          395 

806 
307 

714 

512 

2161 

TENS'  FAMILY. 

79 
23 
31 

1024 

3248 

721 

1809 

44 
46 

61 
69 

124 

126 

323          87           73        84 
327          27          33        24 

ADD  HORIZONTALLY. 

109 
101 

68 
48 

84 
45 
54 
55 
66 
77 

4708, 

290 

728, 

49,        864, 

-1403. 

89 

ii6 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Add  at  one 

195 

Operutluu. 

327 

114 

208 

83            68 

75 

79            97 

320 

—                      

612 

184 

719 

406 

1462  at  12i/. 
544  at  18f  C. 

116  at  25;-. 

2362 

87                85 
89                91 

1427 

5134 

409 

1292 


3G021 
12003 
10400 
5026 
10820 
72035 


DRILL    EXERCISE    No.    2. 

SIMULTANEOUS    MULTIPLICATION. 


89 
76 


298 
89 


689 
98 


548 

287 


768 

678 


104 
102 


196. 


90 
94 


7004 
904 


ALIQUOTS. 

480  at  37i/. 
326  at  16|^. 
183  at  33JA 

COMPLEMENTS. 


78 
88 


SUPPLEMENTS. 


123 
116 


1014 
1009 


82 
89 


1126 
1012 


99 
99 


988 
9S9 


987 
876 


4067 
352 


654  at  87J^. 
5(52  at  SS^c'. 
275  at  $1.25 


985 
975 


976 
974 


COMPLEMENTS  &  SUPPLEMENTS. 


137 

88 


1045 
986 


DRILL  Exercise  no.  3. 

MISCELLANEOUS  CONTRACTING  METHODS. 


78 
38 


593 
491 


412 
418 


42 
29 


68 
36 


246 

127 


6843 
729 


58924 
36945 


3456 
111 


432 

17.-^ 


268 

75 


2695 
lOOS 


148 
95 


463 
162i 


7321 
996 


Subtract  by  additiou : 


4867 


248 

1261 

527 


3842 

903 

1125 


163 

49 

1407 


460 

588 
123 


6892 
4087 
9364 


65 

85 


5525 


OPERATION. 

5  X  5  =  25.     St. X  6t.=  48h.+ 
|of  (8+6)=  7h.=  551i.  or 
5525. 


45 
35 


1575 


OPEBATION. 

5  X  5  =  25.  3t.  X  4t.=  12b.+ 
i  of  (3  +  4)  =  3ib.+  12h.= 
1550 +tbe  25  =  1575. 


75    95 
55    25 


63 

32 

88 

146 

42 

29 

94 

73 

77 

S(i 

CONTRACTIONS   IN    MULTIPLICATION. 


117 


197. 


DRILL    EXERCISE    No.    4. 


1. 

o 

3. 
4. 

5. 


62  X  25 

87  X  12J 

140  X  ISa 

305  X  33^ 

740  X  G2J 


G. 

97^ 

X 

125 

7. 

65 

X 

22J 

8. 

184 

X 

187J 

9. 

212 

X 

75 

0. 

514 

X 

83i 

11.  328  X  375 

12.  261  X  750 

13.  414  X     35 

14.  807  X       SJ 

15.  694  X     16§ 


198. 


DRILL    EXERCISE    No.    5. 


KEUGASS  &  THOMPSON", 


New  Orleans,  Apr,  1, 1894. 
Bot.  of  DELERNO  &  GRANER. 


523i  Ib.s.  Butter       .        .        .        . 
136|    "     La.  Tecans 

56  doz.  C.  Egg;s       .         .         .         . 
249  boxes  Lobsters 
140  iais  Pickles        .        .        .        . 

48 "lbs.  Bice 

83  busb.  W.  Corn 

64      "      W.  Oats  -         -        -         - 
92  bags  Salt 

84  lbs.  B.  Tea         .        .        .        . 
108  gals.  11.  Whiskey       - 

143  doz.  Pine  Apples 

119  lbs.  Y.  H.  Tea  -         -         - 

324  boxes  Slirinip    -         .         .         - 

240  gals.  Fire  Proof  Coal  Oil 

164  lbs.  Lard  .        .        .        . 

28G    "    Green  Peas 

320    "    N.  Y.  C.  Cheese 

112  gals.  La.  Molasses     - 

124      "     B.  Whiskey       - 

72  bbls.  La.  Oranges 

61      »     N.  Potatoes       - 

16J  lbs.  C.  Soap      -         -         -         . 
9|    "    Persian  Dates  - 
1  bbl.  50/  Sugar,  241-24-217  lbs. 

43  lbs.  Lard  Oil      .         .         .         - 

45    "    Elgin  C.  Butter 

34  bush.  W.  Oats 
425  bbls.  Flour        .... 


® 

u 


u 
u 
u 
n 

it 
u 
u 
u 


25  / 
12^/ 

183/ 

37|/ 
62.i/ 

87  / 

44  / 

93  / 

97  / 

$1.06 

1.15 

96  / 

35  f; 

15  / 

7^/ 

2^/ 

22i/ 

$1.25 

3.75 

2.50 

6J  / 
5i  / 
58  / 
35  / 

48  / 

$8.75 


5620 


70 


KoTE  1. — All  extensions  of  the  above  bill  should  be  made  mentally. 

Note  2.— See  pages  96  to  101  for  other  Practical  Drill  Problems.     Also,  see  pages  53,  54,  57, 
58,  68,  69,  70,  90,  91,  and  95,  for  Eapid  Methods  of  Handling  Numbers. 


ii8 


SOULE  S    PHILOSOrmC    PRACTICAL    MATHEMATICS. 


199. 


MULTIPLIOATION  BY    SQUARING  NUMBERS. 

TABLE  OF  SQUAKES  AND  CUBES. 


Numb. 

Square. 

Cube. 

Numb. 

Square. 

Cube. 

Numb. 

Square. 

Cube. 

Numb. 

Square. 

Cube. 

1 

1 

1 

26 

676 

17576 

51 

2601 

132651 

76 

5776 

438976 

2 

4 

8 

27 

729 

19683 

52 

2704 

140608 

77 

5929 

456.533 

3 

9 

27 

28 

784 

21952 

53 

2809 

148877 

78 

6084 

4745.52 

4 

16 

64 

29 

841 

24389 

54 

2916 

157464 

79 

6241 

493039 

5 

25 

125 

30 

900 

27000 

55 

3025 

166375 

80 

6400 

512000 

6 

36 

216 

31 

961 

29791 

56 

3136 

1 7.5616 

81 

6.561 

531441 

7 

49 

343 

32 

1024 

32768 

57 

3249 

185193 

82 

6724 

551368 

8 

64 

512 

33 

1089 

35937 

58 

3364 

195112 

83 

6889 

571787 

9 

81 

729 

34 

1156 

39304 

fj9 

3481 

205379 

84 

7056 

592704 

10 

100 

1000 

35 

1225 

42875 

61) 

3.)00 

21601)0 

85 

7225 

614125 

11 

121 

1331 

36 

1296 

46656 

61 

3721 

226981 

86 

7396 

636056 

12 

144 

1728 

37 

1369 

5u653 

62 

3844 

238328 

87 

7569 

658.503 

13 

1G9 

2197 

38 

1444 

54872 

63 

3969 

250047 

88 

7744 

681472 

14 

196 

2744 

39 

1.521 

59319 

64 

4096 

262144 

89 

7921 

704969 

15 

225 

3375 

40 

1600 

64000 

65 

4225 

274625 

90 

8100 

729000 

16 

256 

4096 

41 

1681 

68921 

66 

4356 

287496 

91 

8281 

753571 

17 

289 

4913 

42 

1764 

74088 

67 

4489 

300763 

92 

8464 

778688 

18 

324 

5832 

43 

1849 

79507 

68 

4624 

314432 

93 

8649 

804357 

19 

361 

6859 

44 

1936 

85184 

69 

4761 

328509 

94 

8836 

830584 

20 

400 

8000 

45 

2025 

91125 

70 

4900 

343000 

95 

9025 

857375 

21 

441 

9261 

46 

2116 

97336 

71 

5041 

357911 

96 

9216 

884736 

22 

484 

10648 

47 

2209 

103823 

72 

5184 

373248 

97 

9409 

912673 

23 

529 

12167 

48 

2304 

110592 

73 

5329 

389017 

98 

9604 

941192 

24 

576 

13824 

49 

2401 

117649 

74 

5476 

405224 

99 

9801 

970299 

25 

625 

15625 

50 

2500 

125000 

75 

5625 

421875 

100 

10000 

1000000 

SQUARING  NUIVIBERS. 


200.  The  usual  methods  of  squaring  numbers  on  the  basis  of  complements  and 
supplements,  are  attended  with  so  many  variations  tliat  they  are  of  very  little 
practical  value,  especially  to  those  who  understand  tlie  sy.stem  of  simultaneous 
multiplication,  and  hence,  we  shall  present  but  little  work  of  this  kind. 


TO  SQUARE  NUMBERS  BETWEEN  TWENTY-FIVE  AND  FIFTY. 


201.     1.     What  is  the  square  of  26  ? 


OPERATION. 

26  —  25=  1 
25—  1=  24 
24=  =  576 

576+l»«=676   Aus. 


Explanation. — In  all  problems  of  this  kind,  nient.illy  subtract 
25  from  the  number  to  bi;  squared,  and  the  remainder  froni  25;  then 
square  tbis  second  remainder  and  add  to  its  hundreds'  ligurethe  tirst 
remainder. 

To  be  proficient  in  this  work,  the  squares  of  numbers  as  high  as 
25,  must  be  thoroughly  learned  from  the  table. 


I 


*  CONTRACTIONS    IN    MULTIPLICATION.  II9 

TO  SQUAEE  XUMBEHS  BETWEE]^  FIFTY  AXD  SEVENTY-FFV^E. 
202.     2.    AVliat  is  the  square  of  G3  ? 

OPERATION. 

g3 2o  =    38  Explanation. — In  all  proWems  of  tliis  kind,  mentally  suljtract, 

63  —  50  =     13  first  25,  and  then  50  from  tlie  given  munlier ;  then  square  the  second 

1"^"  =  I'jJ  remainder  and  add  to  its  hundreds'  figure  the  first  remainder. 

109+38'"'=  3969  Aiis. 


TO    SQUAEE  NTOMBERS   BETWEEIST   SEVEjSTTY-FFS^E    AKD   ONE   HUIT- 

DEED. 

203.     What  is  the  square  of  89  ? 

OPERATION.  Explanation. — In  all  problems  of  this  kind,   subtract 

89  —  75  =  14  75  from  the  mnnber,  and  the   remainder  from   25;  then 

25  — 14  =  11  square  the  second  remainder  and  add  to  its  hundreds'  fig- 

11"               121  ure  the  difference  between  the  number  and  the  second 

121+(S9<"'— 11'"')=  7921  Ans.           remainder. 

PROBLEMS. 
What  is  the  square  of  28,  42,  51,  07,  78,  and  93  ? 


TO  MULTIPLY  TWO  NUMBEES  OX  THE  PEINCIPLE  THAT  THE  SQUAEE 

OF  THE  MEAN  OF  TWO  NUMBEES  JNIINUS  THE   SQUAEE 

OF  ONE-HALF  OF  THE  DIFFEEENOE,  IS  EQUAL 

TO  THE  PEODUCT  OF  THE  TWO  NUMBEES. 

201.    Multiply  15  by  19. 

OPERATION. 

15  +  19  =  34  4-  2  =  17=  =  289  —  2"-=  (4)  285    Ans. 

Explanation. — From  the  square  of  the  mean,  subtract  the  square  of  half  their  difference. 

1.    Multiply  OS  by  72. 

OPERATION. 

(70  —  2)  X  (70  +  2)  =  70^  —  2=  =  4900  —  4  =  4896     Ans. 


I20  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

2.  MiUtiply  81  by  79. 

OPERATION. 

80=  —  1=  =  0400  —  1  =  6399     Ans. 

3.  Multiply  92  by  108. 

OPERATION. 

100=  —8=  =  10000  —  01  =  9936    Ans. 
Multiply  the  following  numbers  as  above: 

4.  28  by  31  6.     67  by  69  8.     57  by  63 

5.  31   "  32  7.     73    "  85  9.     74   "  86 


TO  MULTIPLY  TWO  NUMBERS,  THE  DIFFEEEISTCE  OF  WHICH  IS  ONE, 
THREE,  FIVE,  OR  ANT  SMALL  NUMBER. 

205.     1.    Multiply  21  by  22. 

OPERATION. 

21=  =  411  +  21  =  462     Ans. 

Explanation. — In  all  problems  of  this  kind,  square  the  lesser  number  and  add  to  the  product  as 
many  times  the  lesser  number  as  it  is  less  than  the  greater.  The  reason  is  that  the  lesser  number 
should  be  repeated  as  many  times  as  there  are  units  in  the  greater  number;  but  in  squaring  the  lesser 
number,  you  fail  to  repeat  it  as  many  times  as  there  are  units  difference  between  the  numbers. 

2.  Multiply  87  by  89. 

OPERATION. 

87=  =  7569  +  (2  X  87)  =  174  =  7743     Ans. 

3.  Multiply  236  by  241. 

OPERATION. 

236=  =  55696  +  (5  x  236)  =  1180  =  56876    Ans. 

4.  Multiply  96  by  92.     Ans.  8832. 

5.  Multiply  354  by  361.     Ans.  127794. 

6.  Multiply  32  by  43. 


SYNOPSIS  FOR  HEVIEW  OF  MULTIPLICATION. 


named 


Note.— The  numbers  before  the  words  and  phrases  refer  to  the  articles  treating  the  subjects 


206.     Define  the  following  words  and  phrases  : 


142. 
143, 
144. 
145. 
146. 
147. 
148. 
149. 
150. 
151. 
152. 

153. 

154. 
155. 

156. 

157. 


158. 

159. 

160. 
161. 


Multiplication. 

Multiplicand. 

Multiplier. 

Product. 

Factors. 

The  Sign  of  Midtiplication. 

Proof  of  JIultiplicatioii. 

Principles  of  ilultiplication. 

Midtiplication  Table. 

The  Importance  of  Multiplication. 

The  Philosophic  or  Logical  System 
of  INIvdtiplication. 

Examples  ■with  Philosophic  Solu- 
tions. 

Drill  Problems. 

To  Multiply  Abstract  Numbers 
and  give  reasons  therefor. 

To  Multiply,  when  the  Multiplier 
consists  of  only  one  figure. 

To  Midtiply,  when  the  ^Multiplier 
consists  of  more  than  one  fig- 
ure. 

To  ilidtiply,  when  either  factor  is 
1  with  naughts  annexed,  as  10, 
100, 1000,  etc. 

To  ^Multiply,  when  either  the  Mid- 
tiplicand  or  Mxdtiplier,  or  both, 
have  naughts  on  the  right. 

To  Multiply  by  the  Factors  of  a 
Xumber. 

To  Multiply,  when  the  Multipli- 
cand or  Multiplier  contains 
Dollars  and  Cents. 


102.  General  Directions  for  Multiplica- 
tion. 

163.  ^liscellaiieous  Problems  in  Multi- 
l)lication. 

104.  Sinuiltaneous  or  Cross  ]SIxdtiplica- 

tion. 

105.  Drill  Problems    in   Simultaneous 

JIultiplication. 

166.  A  Diftereiit   :\Iethod   of  Simulta- 

neous ^Midtiplication. 

167.  To  ^Multiply  when  the  Multiplier 

consists  of  Two  Figures. 

168.  Table  of  Aliquots. 

169.  Problems. 

170.  To  Multiply  by  7J,  17.J,  22J,  27J, 

32J,  15,  35,  45  and  55. 

171.  Practical    Operations    Mith     Ali- 

quots and  other  numbers. 

172.  To  Multiply,  when  the  Multiplier 

is  a  mixed  number  whose  frac- 
tional part  is  one  fractional 
unit,  less  than  a  whole  number. 

173-93.  Miscellaneous  Contractions  in 
Multiplication. 

193.  Algebraic  FormidiB,  for  Comple- 
ments and  Supplements, 

194-08.  Drill  Problems  in  Contracted 
Methods  in  Addition  and  Multi- 
plication. 

199.    Table  of  Squares  and  Cubes. 

200-5.  Squaring  Xumbers. 


a21) 


HlVISION. 


(  DECREASING). 


207.  Division  is  the  process  of  flndiiig  how  many  times  one  number  is  equal 
to  another,  which  is  used  as  a  unit  of  measure.  Or,  differently  defined,  it  is  the 
l^rocess  of  finding  oue  of  the  factors  of  a  given  product,  when  the  other  factor 
is  known. 

208.  From  the  first  and  the  better  definition,  we  see  tliat  division  is  a  process 
of  measuring  some  numbers  by  other  numbers.  And  that  it  is  not  the  process  of 
finding  "liow  many  times  one  number  is  contained  in  another,"  as  is  taught  by 
nearly  all  the  authors  of  Arithmetics.  One  number  cannot  go  into  another,  how- 
ever small  the  oue,  or  large  the  other;  and  the  question  of  division  of  numbers 
does  not  warrant  a  definition  so  unmathematical,  indefinite  and  illogical. 

THE    LOGIC    OF    DIVISION. 

209.  The  following  questions  and  operations  will  elucidate  the  true  meaning 
of  division  and  its  philosophy: 

1.  You  have  $6,  aud  I  have  $2.  How  many  times  is  your  money  equal  to 
mine? 

Is  it  not  clear,  by  the  terms  of  the  question,  that  your  sum  of  money  is  to  be 
comi>ared  with,  and  measured  by,  my  $2 1  And  the  contracted  thought  to  do  this 
is,  $6  is  equal  to  $2,  3  times. 

The  full  logical  thought,  recognizing  1,  or  unity,  as  the  basis  of  all  numerical 
computations,  would  be  as  follows :  f  G  is  equal  to  $1,  6  times.  Since  $G  is  equal  to 
$1,  C  times,  it  is  equal  to  $2,  one-half  the  number  of  times,  which  is  3. 

2.  Again — you  have  8  yards  of  cloth,  and  I  have  4  yards.  How  many  times 
is  your  quantity  of  cloth  equal  to  mine? 

Here  it  is  plain  that  my  4  yards  is  made,  by  the  terms  of  the  question,  the 
unit  of  measure ;  and  your  8  yards  is  the  thing  to  be  measured.  And  the  contracted 
thought  to  do  the  work  would  be,  8  yards  is  equal  to  4  yards,  2  times. 

The  full  logical  reason  is  as  follows :  8  yards  are  equal  to  1  yard  8  times. 
Since  8  yards  are  equal  to  1  yard,  8  times,  tliey  are  equal  to  4  yards  instead  of  1, 
t]ie  fourth  part  of  the  number  of  times,  which  is  2. 

3.  Divide  10  by  5. 

In  this  problem,  the  real  question  is,  how  many  times  is  10  equal  to  5,  not  hoio 
many  times  5  ca7i  go  into  10,  or  how  many  times  is  5  contained  in  10. 

5  is  the  unit  of  measure,  and  10  is  tlie  number  to  be  measured  or  divided. 
The  contracted  reasouiug  is,  10  is  equal  to  5,  2  times;  not  5  is  contained  or 

(122) 


*  DIVISION.  123 

goes  into  10,  2  times.  The  full  logical  reason  is  as  follows :  10  is  equal  to  1, 10  times. 
Since  10  is  equal  to  1,  10  times,  it  is  equal  to  5,  instead  of  1,  the  fifth  part  of  the 
number  of  times,  which  is  2. 

This  process  of  reasoning  is  applicable  to  every  possible  question  of  division, 
and  it  is  the  only  logical  reasoning  that  can  be  given  for  abstract  numbers.  Yet 
strange  to  say,  it  has  escaped  the  attention  or  the  approbation  of  all  other  authors 
of  Arithmetics.  "With  due  modesty,  we  claim  some  merit  for  its  first  introduction, 
and  earnestly  commend  it  to  the  thoughtful  consideration  of  authors,  of  students, 
and  of  the  jiublic. 

210.  The  Dividend  is  the  number  to  be  measured,  or  the  number  to  be  divided. 

211.  The  Divisor  is  the  number  used  as  the  unit  of  measure  or  the  number  by 
which  to  divide. 

212.  The  Quotient  is  the  result  of  the  division,  and  shows  how  many  times  the 
dividend  is  equal  to  the  divisor. 

213.  The  Remainder  is  the  number  left  after  dividing  dividends  which  are 
not  nudtiples  of  the  divisor,  or  which  are  not  equal  to  the  divisor  an  exact  number 
of  times.     It  must  always  be  less  than  the  divisor. 

214.  The  Sign  of  Division  is  a  horizontal  line  with  a  point  above  and  below, 
thus  -^-.  It  is  read  dirided  lnj,  or  is  equal  to  ;  and  it  indicates  that  the  number 
before  it,  is  to  be  divided  by  the  number  after  it:  thus  25  -^  5,  is  read  25  divided  by 
5,  or  25  is  equal  to  5. 

215.  Division  is  also  indicated  by  a  horizontal  line,  a  vertical  line,  or  a  curved 
line,  when  placed  between  the  dividend  and  the  divisor.  Thus,  the  following  state- 
ments : 

36  9     30  9)  30  3C(^ 

9 
are  all  read  36  divided  by  9,  or  how  many  times  is  36  equal  to  9,  or  30  is  equal  to  9 
how  many  times  ? 

PEINCIPLES   OF  DIVISION. 

216.  1.  Wlien  the  divisor  and  dividend  are  both  denominate  or  both  abstract 
numbers,  the  quotient  will  be  an  abstract  number. 

2.  When  the  divisor  is  an  abstract  number  and  the  dividend  a  denominate 
number,  the  quotient   will  be  a  denominate  number. 

3.  "WHien  there  is  a  remainder,  it  is  a  iiart  of  the  dividend,  and  is  therefore 
the  same  in  name  or  kind. 

4.  Multiplying  the  dividend  or  dividing  the  divisor,  multiplies  the  quotient. 

5.  Dividing  the  dividend  or  multiplying  the  divisor,  divides  the  quotient. 

6.  Multiplying  or  dividing  both  the  divisor  and  dividend  by  the  same  num- 
ber, does  not  change  the  quotient. 

217.  Proof. — Multiply  the  quotient  by  the  divisor,  and  to  the  product  add  the 
remainder,  if  any.  If  the  result  is  equal  to  the  dividend,  the  worli  is  correct.  See 
Proof  of  Division  by  the  Properties  of  9's  and  ll's,  at  the  close  of  this  subject. 


124 


soule's  philosophic  practical  mathematics. 


218.     The  (iperatioii  of  division  may  be  perforined,  either  by  addition  or  sub- 
traction.    Thus,  in  the  foUowiug  iiroblem  : 

How  many  times  is  43  equal  to  12 !       Ans.  3  times  and  7  remainder. 

OPERATION   BY    ADDITION.  OPERATION    BY   SUBTRACTION. 


i3  is  equal  to  12,  1  time. 
43  "       "      "  12,  2  times. 


43  " 


24 

"  12,  3  times. 

30 


and  7  remainder. 

43 


43 
43  is  equal  to  12,  1  time. 

31 

43  "       "      "  12,  2  times. 

19 

43  "       "      "  12,  3  times. 

and  7  remainder. 


OEAL  EXERCISES. 


219.     1.     How  many  times  is  0  equal  to  1  ?  or  0  -^  1  =  ?  no  times. 
2.  "  "  1         "        0  ?   or  1  H-  0  =  ! 

Ans.  An  infinite  number  of  times. 


3.  How  many  times  is  I  equal  to  1  ?  or    1-^1=  ? 

4  "  "  2  "  l?or    2— l=f 

5  "  "  3  "  l?or    3—1=? 

6  "  "  4  "  2?or    .1—2=? 

7.  "  "  8  "  2?  or    8—2=? 

8.  "  "  9  "  3?  or   9—3=? 
9  "  "  12  "  4  ?  or  12— 4=  ? 

10.  "  "  20  "  5  ?  or  20— 5=  ? 

11.  "  "  24  "  6  ?  or  24— 6=  ? 


12.  How  many  times  is  35  equal  to  7?  or  35—  7=  f 

13.  "  '      "  56  "  ft?or56—  8= » 

14.  "  "  63  "  9?  or  63—  9=  f 

15.  "  "  72  "  9?  or  72—  9=? 

16.  "  "  80  "  10?  or  80— 10=? 

17.  "  "  88  "  11?  or  88— 11=? 

18.  "  "  96  "  12  for  96^12=? 


20. 

21. 


14 


19.  \!^=1    4)42=?    9)45=?    77—7=?    12     84=  ? 

How  many  times  is  3G  equal  to  3,  to  4,  to  6,  to  9,  to  12,  to  3C  ? 
How  many  times  is  42  equal  to  2,  to  G,  to  7, 


to  42? 


21 


28 
4 


G4 


—  =  ?     24(^     '^-(^      7)  5G  =  ?     9 


FRACTIONAL    NUMBERS. 


220.  When  we  divide  a  unit  or  a  number  of  units  of  any  kind  into  equal  parts, 
these  parts  are  sometimes  called  fractions.  The  name  of  the  equal  parts  varies 
according  to  the  number  of  parts  into  Avhich  the  thing  or  number  is  divided. 

When  the  unit  or  number  is  divided  into  two  equal  parts,  1  of  the  parts  is 
called  one-half,  ami  is  written  thus,  J.  If  divided  into  4  equal  parts,  1  of  the  parts 
is  called  one-fourth,  and  is  written  thus,  ^ ;  3  of  the  parts  are  called  three-fourths, 
and  are  written  thus,  f. 

In  like  manner,  we  ohta,in  fifths,  sixths,  sevenths,  eighths,  ticelfths,  sixteenths, 
twenty-firsts,  etc. 


DIVISION. 


125 


In  -writing  fractional  numbers  in  figures,  we  place  the  number  which  shows  the 
name  of  the  parts  below  a  horizontal  Hue  as  a  divisor,  and  the  number  which  shows 
how  many  jiarts  are  taken,  or  used,  above  the  line  as  a  dividend. 

The  following  examjiles  will  fully  elucidate  this  work  : 


Two-thirds,  §. 
Three-fourths,  a. 


Five-eighths,  |. 
Seven-ninths,  | 


Seven-twelfths,  i^. 
Xine-tenths,  ^%-. 
Fifteen-sixteenths,  ||.  Eleven-eightieths,  ^. 

How  do  you  find  i,  J,  ^,  i,  etc.,  of  any  number  1 
How  do  you  find  §,  |,  |,  J,  etc.,  of  any  number  ? 
What  is  ^  of  4  f        What  is  i  of  15 1         What  is  ^  of    9  ?      What  is  f  of  15  ? 

"     a     i  "  G?  "      "  -V  "    18?  "       "  ?  "    12?  "     "    -r  "    G3  ? 


i  "  8? 


1  "   28  ? 


J  "    10? 


1  (i 


8U1 


221. 


ORAL  AX.D  WEITTEX  PROBLEMS. 


1.  Five  pounds  cost  40/. 
What  was  the  cost  of  1 
pound  ? 


SOLUTION   STATEMENT. 

10 


8/  Ans. 


Meason — Five  j)ouiuls  cost 
•10c.  Since  5  jiouiids  cost  40 
cents,  1  jiouud  will  cost  the  5th 
part,  which  is  8  ceuts. 


REASONS,  WHY,  AND  WHEEEFORE,  CONTINUED. 


Question  1.  How  do  you  know  that  if  5  pounds  cost  40  cents,  1  pound  will 
cost  the  5th  part  ? 

Answer.  By  the  exercise  of  my  judgment — by  the  use  of  the  reasoning 
faculties  of  my  mind. 

Question  2.     What  do  you  mean  in  this  connection  by  judgment  ? 

Answer.  The  conclusion  arrived  at  by  the  operations  of  the  mind,  after  hav- 
ing duly  considered  the  i)remise,  the  facts,  and  the  conditions  of  the  problem. 

Question  3.     What  do  you  mean  by  the  pi'emise  or  premises  ? 

Answer.  The  proposition,  declaration,  truth  or  fact  which  is  stated,  or 
asserted,  as  a  basis,  or  predicate,  of  a  question.  In  this  problem  the  premise  is, 
5  pounds  cost  40  cents. 

Question  4.    Why  will  1  pound  cost  one-fifth  part  as  much  as  5  pounds  ? 

Answer.    Because  1  is  the  fifth  jiart  of  five. 

Question  5.    What  kind  of  reasoning  is  the  foregoing  ? 

Answer.     (See  answer  to  the  same  question  on  page  78). 

Question  6.    What  do  you  mean  by  reason  ? 

Answer.     (See  answer  to  the  same  question  ou  page  78). 


I  26 


SOULE  S    PHILOSOrHIC    PRACTICAL    MATHEMATICS. 


■   2.    If  12  yards  cost   GO/. 
What  will  1  yard  cost  ? 


SOLUTION    STATEMENT. 

CO 


12 


~>c  Alls. 


Reason. — Twelve  yards  cost 
60c.  Since  12  yards  cost  60c., 
1  yard  wiU  cost  the  12th  part, 
"which  is  5  ceuts. 


Questions.    1.   Ho\y  do  you  know  this!    2.    Why  will  it?    3.   What  do  you 
mean  by  judgment? 


3.     Xine  barrels  of  tlour  cost  $72.     What  was  the  cost  of  1  barrel  ? 


Aus. 


Note. — Work  aud  reason  as  in  above  problems. 


4.  Paid  $18  for  6  days' 
labor.  What  was  the  rate 
per  day  ? 


SOLUTION   STATEMENT. 

$ 

18 
6 


$3  Ans. 
Questions.     1.  How  do  you  know  this  ?     2.  Why  will  it  ? 


Reason.- — Sis  days'  labor  cost 
$18.  Since  6  days'  labor  cost 
$18,  1  day's  labor  will  cost  the 
6th  part,  which  is  |;3. 


5.  The  freight  on  17  bales 
of  cotton  was  $51.  What 
■was  the  freight  per  bale  1 


SOLUTION   STATEMENT. 

$ 


17 


Ans. 


Reason. — The  freight  on  17 
bales  was  $51.  Since  the  freight 
on  17  bales]  was  $51,  on  1  bale 
it  would  be  the  17th  part, 
which  is  $3. 


Questions.     1.   How  do  you  know  this  ?    2.  Why  will  it  ?    3.  What  do  you 
mean  by  judgment  ? 


G.  If  you  divide  2  dozen 
oranges  equally  among  8 
persons,  how  many  oranges 
will  you  give  each  ? 


SOLUTION    STATEMENT. 
O. 

24: 


Reason. — Eight  persons  are  to 
receive  24  oranges.     Since  8  per- 
sons are  to  receive  24  oranges, 
_  1  person  will  receive   the  8th 

3  oran""es  Ans.        part,  -which  is  3  oranges. 


Questions.     1.     How  do  you  know  this  ?     2.     Why  ? 


7.  A  railroad  train  runs 
3G0  miles  in  12  hours. 
What  is  the  speed  per  hour  ? 


SOLUTION    STATEMENT. 

M. 

3G0 


12 


30  miles  Ans. 


Reason. — In  12  hours  360  miles 
are  run.  Since  360  miles  are  run 
in  12  hours,  in  one  hour  the  12th 
part  of  the  distance  would  be 
run,  which  is  30  miles.  Or,  since 
12  hours'  running  gives  360 
miles,  1  hour's  ruuuing  will  give 
the  12th  part. 


Questions.    1.    How  do  you  know  this  ?    2.     Why  ? 


DIVISION. 


127 


S.  Eice  is  7  cents  jicr 
iwuutl.  How  many  pounds 
can  you  buy  for  35  cents  ! 


1st  SOLUTION  state:ment. 
m. 
1 
35 

5  lbs.  Alls. 

2d  SOLUTION   STATEMENT. 


35 
5  lbs.  Ans. 


jRcason. — Seveu  cen(s  Imy  1 
pouuil  of  rice.  Siuce  7  cents 
will  buy  1  pound,  1  cent  will 
Imy  tlie  7th  part,  and  35  cents 
will  buy  35  times  as  much.  Or, 
siuce  7  cents  buy  1  pound,  for  35 
cents  you  can  buy  as  many 
pounds  as  35  cents  are  equal  to 
7  cents. 


Questions. 
by  judgment  ? 


1.    How  do  you  know  this  ?    2.    Why  ?    3.    What  do  you  mean 


Note — Work  and  reason  the  following  problems  in  the  same  manner. 

9.  At  9  cents  per  pound,  how  many  pounds  can  be  bought  for  45  cents  ? 

Ans.  5  pounds. 

10.  Flour  is  worth  $S  i)er  barrel.    How  many  barrels  can  be  purchased  for  $50 1 

Ans.  7  barrels. 

11.  For  $.95,  how  many  papers  can  you  buy  at  5  cents  a  paper  ?      Ans.  19. 

12.  At  $3  a  piece,  how  many  chairs  can  be  bought  for  $36  ?  Ans.  12  chairs. 

13.  If  the  printer  charges  $2.50  to  set  one  page  of  this  book,  how  many  pages 
can  be  set  for  $75  ?  Ans.  30  pages. 

14.  The  dividend  is  42  and  the  divisor,  7.     ^Vhat  is  the  quotient  ?      Ans.  6. 

15.  The  divisor  is  8,  the  quotient  3,  aud  the  remainder  2.     What  was  the 
dividend?  Ans.  26. 


TO  DIVIDE  WHEX  THE  DIVISOR  DOES  NOT  EXCEED  TWELVE. 


2''9 


Divide  3648  by  5. 


OPERATION. 

Divisor  5)  3648  dividend. 


Quotient  729,  and  3  rem. 


Explanation. — In  all  problems  of  this  kind,  we  write 
the  numbers  as  shown  in  the  operation,  and  then  begin  on 
the  left  of  the  dividend  to  divide.  We  begin  on  the  left 
in  order  to  carry  tlie  remainder,  if  any,  of  the  higher  order 
of  units  to  the  next  lower  order.  In  this  problem,  we  first 
take  the  3  thousands,  and  by  inspection,  wo  see  it  is  not 
equal  to  5;  we  therefore  unite  It  with  the  6  hundreds,  making  3G  hundreds,  which,  by  trial  multipli- 
cation and  subtraction,  mentally  jierformed,  we  find  is  equal  to  5,  7  hundreds  times  and  1  remainder ; 
the  7  we  write  in  the  hundreds'  column  of  the  qiiotienft  line,  directly  under  the  6,  the  last  figure  of 
the  dividend  used.  Then,  to  the  1  remainder,  we  mentally  annex  the  4  tens,  making  14  tens,  as  the 
second  partial  dividend,  and  which,  by  mental  multipMcation  and  subtracTion,  we  find  it  equal  to  5, 
2  (ens  times  aud  4  remainder ;  the  2  we  write  in  the  ten-i'  column  of  the  quotient  line,  and  to  the  4  we 
mentally  annex  the  units'  figure  of  the  dividend,  makiiig  48  units  as  the  third  aud  last  partial  divi- 
dend ;  this  we  find,  by  mental  multiitlication  aud  subti  Vction,  to  bo  equal  to  5,  9  times  and  3  remain- 
der. 

This  remainder  is  usually  expressed  fr.actionally  b    writing  it  over  the  divisor;  thus  f  which 
expresses  the  part  of  a  unit  of  times  that  the  remainder   «  equal  to  the  divisor. 


128  soule's  riiiLosopiiic  practical  mathematics.  ♦ 

SHORT  DIVISION. 

223.     Operations  in  division  according  to  the  foregoing  method,  are  called  short 
(liriKion,  l)u(ans(^  the  multiplication  and  subtraction  work,  in  tinding  the  remainder 
of  the  i)artial  dividends,  were  mentally  performed. 
'J.     How  many  times  is  84:6  equal  to  0  ? 

OPERATION.  Explanation. — In  the  preceding  prolilem,  we  gave  a  fnll  and 

T^'    ■      V   r\  '^ir  iliTi<1f>iw1  explicit  explanation  tor  each  stci)  of  tlie  operation.     In  practice 

iJl\  isoi    o;  o-±o  iii\  lueiiu.  ^^^^^^,,^  ^^j.  ^1^^^  (.xplanation  therein  Kiven  is  omitted,  anil  the  work 

)iert'ornie(l  thus:  Conmiencin.!;  with  tlie  lett  hand  lifjure  we  say  8 
Quotient   1-11  is  equal  to  G,  1  time  and  2  remainder;  21  is  equal  to  (i,  4  times; 

6  is  equal  to  6,  1  time. 

PROBLEM.S. 

Work  the  following  indicated  divisions  : 

(3)  (I)  (.">)  (fi)         (7)  (8) 

7)  S47     8)  l-'327      y)  1085     11)  2380     2.344     7  |  03020 

121        1540J        120|        216ia     3 

(9)  (10)  (11)  (12)  (13)  (14) 

8)  1471  4)  11899  9)  81018  12)  10S24        21345-^8        9)  7G451 


Divide  the  following: 

15.     9872  by  4  17.     1028G  by  G  10.       IGOl  by  9  21.       10008  by  9 

IG.     1483    '"'   7  18.     48710   ''   7  20.     41070   "    8  22.     199999   "  8 

23.  What  is  J  of  $  528  ?  25.     What  are  a  of  $  448  ? 

24.  "  are  f  of  $1005?  26.  "      "   5  of  $6448  ? 

27.  How  many  apples  can  be  bought  for  $2.25,  at  5  cents  a  piece  1 

Ans.  45  apples. 

28.  At  15  cents  per  pound,  how  many  pounds  can  you  buy  for  $3.15  ? 

Ans.  21  pounds. 

29.  Paid  $90  for  10  volumes  of  Chambers'  Cyclopedia.     What  was  the  price 
of  one  volume  f  Ans.  $9. 

30.  If  8  men  are  to  receive  $5791  in  equal  parts,  what  will  be  each  man's 
share?  Ans.  $723§ 

31.  The  dividend  is  03,  and  the  quotient  is  9.     What  is  the  divisor  ? 

Ans.    7. 

32.  The  quotient  is  36,  and  the  divisor  is  6.     What  is  the  dividend  ? 

Ans.  216. 

33.  The  dividend  is  72,  and  the  divisor  is  4.    What  is  the  quotient  ? 

Ans.  18. 

34.  The  quotient  is  15,  the  divisor  is  3,  and  the  remainder  is  2.     What  is  the 
dividend  ?  Aus.  47. 

35.  The  divisor  is  24,  the  quotient  is  6.     What  is  the  dividend?     Ans.  144. 

36.  How.  many  pounds  of  cottou,  at  11  cents  a  pound,  will  be  required  to  pay 
for  33  pounds  of  sugar  'S  8  cents  per  pound  1  Ans.  24. 


*  DIVISIOX,  129 

TO    DIVIDE    WHEX    THE    DIVISOR    EXCEEDS    TWELVE. 

224.     1.    Divide  7387  by  36. 

OPERATION.  Explanation.— ^Ve  first  Tvrite  the  numliers   as  sliown  in 

Divisnr   Divirlpixl    Oiinf-ipiit  ^^^  "Peration,  and  commence  to  divide  as  explained  in  the 

-^'^  ISOr,  UiyaciUl,  (Quotient.  g^st  written  example.     Bnt  as  the  divisor  is  tc.o  large  to  be 

30)  7387  (205-j%  conveniently  used  mentally,  we  therefore  write  the  operation 

72  of  mnltiplying  the  divisor  by  the  quotient  tigures,  and  sub- 

tracting  the  successive  products   from   the   several   partial 

^„_  dividends.       In   performing  the    division,    we    first    see    by 

■*-"'  comparison,  that  7  ihniisiinds  are  not  e(iMal  to  36,  and  hence 

180  tlicre  will  bo  no  thousands  in  tlie  quotient,     We  then  annex 

to  the  7  thousands  the  3  hiiudi-eds,  maliing  73  hundreds  as  the 

7   TPiiniiubir  ^^^^  partial  dividend:  this  is  equal  to  3l3,  2  times,  and  a  re- 

it-iiuuiiuti.  maiuder;    we  write  the  2   in  the  hundreds'  column  of  the 

quotient,  multiply  the  divisor  by  it,  write  the  product  under, 
and  subtract  the  same  from  the  73  hundreds  of  the  dividend.  Tliis  work  gives  us  1  hundred  remain- 
der, to  which  we  annex  tlie  8  tens,  making  18  tens  as  the  second  partial  dividend  ;  tliis  ])artial  dividend 
not  being  e(|ual  to  36,  we  write  0  (no  tens)  in  tlie  tens'  column  of  the  quotient,  and  annex  to  the  18 
tens  the  7  units,  making  187  units  as  the  third  and  last  partial  dividend.  This  is  equal  to  36,  5  times, 
and  a  remainder;  we  write  the  5  in  the  quotient,  multiply  and  subtract  as  we  dill  with  the  first 
obtained  figure  of  the  quotient,  and  thus  produce  7  remainder,  which  we  write  over  the  divisor  as 
explained  in  short  division. 

LONG  DmSIOK 

Operations  in  division,  according  to  the  above  method,  are  called  long  division, 
for  the  reason  tliat  the  multiplication  and  subtraction  work  in  finding  the  remain- 
ders of  the  partial  dividends  is  wi'itteu.     See  Contractions  in  Division. 

* 

GENERAL    DIRECTIONS    FOR    DIVISION. 


225.  From  the  foregoing-  elucidations,  we  derive  the  following  general  directions 
for  the  operations  of  division.  For  the  jirocess  of  reasoning  given  in  connection  with 
the  operations  or  solution  statements,  we  refer  to  pages  122  to  129. 

1.  Draic  a  vertical  or  curved  line,  and  tvrite  the  dividend  on  the  right,  and  Ute 
divisor  on  the  left. 

2.  Take  the  least  number  of  the  left  hand  Jiff  ures  of  the  dividend  that  are  equal 
to  or  greater  than  the  divisor,  find  how  many  times  the  same  is  equal  to  the  divisor,  and 
write  the  rcsitlt  in  the  quotient  line. 

3.  Multiply  the  divisor  by  this  quotient  figure,  subtract  the  product  from  the 
jaartial  dividend  used,  and  to  the  remainder  annex  the  succeeding  figure  of  the  dividend 
and  divide  as  before. 

4.  Proceed  in  lihe  manner  until  all  the  figures  of  the  dividend  have  been  used. 

5.  When  the  partial  dividend  is  not  equal  to  the  divisor,  ivrite  a  naught  in  the 
quotient,  annex  the  succeeding  figure  of  the  dividend  to  the  partial  dividend,  and  pro- 
ceed as  before. 

6.  If  there  is  a  remainder  after  the  last  division,  tcrite  it  in  the  quotient,  draw 
41  line  below.it,  and  tcrite  the  divisor  underneath,  as  apart  of  the  quotient. 


130  soule's  philosophic  practical  mathematics.  * 

Proof.    Multiply  the  quotient  by  the  divisor  and  to  the  product  add  the 
remainder,  if  any.    If  the  result  is  equal  to  the  dividend,  the  work  is  correct. 

Note  1. — Tho  product  referred  to  in  No.  3,  must  never  be  greater  than  the  partial  dividend 
from  which  it  is  to  bo  subtracted  ;  if  it  is  larger,  the  quotient  tigure  is  too  large,  and  must  bo 
diminished. 

Note  2. — The  remainder,  after  each  .subtraction  referred  to  in  No.  3,  miist  always  bo  less  than 
the  divisor  ;  if  it  is  not,  the  last  quotient  figure  is  too  small  and  must  bo  increased. 

Note  3. — The  order  of  each  quotient  figure  is  the  same  as  the  lowest  order  in  the  partial 
dividends  from  which  it  was  obtained. 

PROBLEM. 

How  many  times  is  6GS04  equal  to  53  ?  Ans.  1260| 

OPERATION. 

Divisor,  Dividend,  Quotient,  Proof. 

53)   GG804   (IL'GOgA  12C0  quotient. 

53  53  divisor. 


m. 


138  3780 

106  G300 


320 


24  remainder. 


318  GG804  dividend. 


2-1:  remainder. 

PROBLEMS. 

Divide  the  following  numbers  : 

1      784  by  82.  Ans.  9|f.  2.     91070  by  87C1.  Ans.  10| 

3'     107911  by  396.  Ans.  4.     7167901  l)y  11267.  Ans. 

5!     37021  by  2002.  Ans.  G.     8888888  by  332311.        Ans 

7.  Divide  $6805  equally  among  5  men.    What  will  be  the  share  of  each  ? 

An.s.  $1361. 

8.  What  is  the  sixty-fourth  part  of  $44800  ?  Ans.  $700. 

9.  145  men  picked  1305000  pounds  of  cotton.     Suppose  each  picked  an  equal 
quantity,  how  nuich  did  each  man  pick  !  Ans.  9000  jHrnnds. 

10.  A  father  gave  liis  7  sons  a  Christmas  present  of  $353.50  to  be  shared 
equally  among  them.     What  was  each  one's  share  ?  Ans.  $50.50. 

TO  DIVIDE  WHEN  THEEE   AEE   NAUGHTS   ON  THE   EIGHT  OF  THE 

DIVISOE. 

226.     1.    Divide  2843  by  200.  •  Ans.  14..tV 

OPERATION.  Explanation. — Since  by  our  scale  of  numbers  they  increase  from 

2  I  00)28  I  43  r'lglit  to  left  in  a  ten-fold  ratio,  and  decrease  from  left  to  right  in  a  cor- 

I  I  responding  maimer,  it  is  clear  that  the  removal  of  any  order  of  figures 

from   left  to  right  diminishes   its  value   ten  times  for  each  xil'ice  of 

14  and  43  rem.  removal.  And  as  ]ireviously  shown  on  page  83,  that  the  annexing  of 
naughts  mutliplics  numbers,  by  removing  them  to  places  of  higher 
value  so  in  like  manner,  cutting  figures  off  from  the  right  of  a  number  removes  the  remaining  orders 
to  the  riciht  and  hence  decreases  them  ten-fold  for  every  figure  cut  off.  Hence  to  cut  off  one  figure  is 
dividing  by  10;  to  cut  off  two  figures  divides  by  100;  to  cut  off  three  figures  divides  by  1000;  and 
so  on. 

Considering  these  principles,  in  all  cases  of  this  kind  we  cut  off  the  naughts  from  the  right 
of  the  divisor  and  the  same  number  of  figures  from  the  right  of  the  dividend;  and  then  divide 
the  remaining  figures  of  the  dividend  by  the  remaining  figures  of  the  divisor.  When  there  is  a. 
remainder,  annex  the  figures  cut  off',  and  we  obtain  the  true  remainder. 


DIVISION.  131 


2.     Divide  87931  by  1000.  Ans.  871^0 

OPERATION. 

1 1  000)87  I  931 

Quotient  87  aud  931  remainder. 
3.    Divide  178  by  10.  4.    Divide  6581  by  300. 

OPERATION.  OPERATION. 

1 1 0)17  18  31 00)65  I  81 


Quotient  17,  aud  8  remainder.  Quotient  21,  and  281  remainder. 

Aus.  17/j.  Ans.  21iU. 

5.     Divide  71468071  by  311000. 

OPERATION. 

341 1 000)71468 1  071(209iH^-J  Aus. 
682 


0^00 
3069 

199 

6. 

Divide  8897600  by  8100. 

7. 

Divide  1000000  by  10000. 

8. 

Divide  99999  by  9000. 

9. 

Divide  33440  by  270. 

10. 

Divide  140817  by  6800. 

Ans.  1098i|. 
Ans.  100. 

Ana      11    »»» 

•^u&.   -LJ-aooo' 

Ans.  123fff. 
Ans.  20A|ix 


TO   DIVIDE    BY   THE   FACTOES   OF  A  NUMBEE. 


227.      1.     Divide  936  by  24.  Explanation. — In  all  proljlems -where  tlie  divisor  is  a  com- 

OPERATION.  posite  number,   we  may  divide   by   the   factors    and  thus 

<  Nogg  shorten  the  operation,     In  this  example  the  factors  are  4  and 

6,  and  we  first  divide  by  4  which  gives  a  quotient  6  times 

g\034  too  large,  for  the  reason  that  4  is  but  ^  of  24,  the  true  divisor. 

We  therefore  divide  this  quotient  by  6  and  obtain  the  tru& 

39  quotieut. 

2.  Divide  588  by  28.    The  factors  are  4  and  7.  Ans.  21. 

3.  Divide  6976  by  32.    The  factors  are  4  and  8.  Ans.  218. 

4.  Divide  2583  by  63.     The  factors  are  7  and  9.  Aus.  41. 

5.  Divide  10206  by  81.    The  factors  are  9  aud  9.  Ans.  126. 

6.  Divide  11984  by  56.    The  factors  are  8  aud  7.  Aus.  214. 


132 


SOULE  S    raiLOSOPHIC    PRACTICAL    MATHEMATICS. 


PKOBLEMS    INl^OLVIXG    THE    ENGLISH    MONEY    OF    ACCOUNT. 


228.     1.     What  will  138i0  poiiiuls  of  cotton  cost,  at  8  i)euce  a  pouixl. 

Alls.  £4(51.  Gs, 


Sd. 


ExpUinnlion. — In  fliis  prolilrni,  tlio  prico  is  };ivi'ii  in  one 
of  tlie  snl)(li\'isi<>ns  of  tlic  I']n;;'li.sh  iii(nu't;iry  nnit,  and  Ltuice 
\\(i  must  know  what,  tlnit  nuit  anil  its  snIiiUvisions  are,  Ijefore 
Avo  oau  Kolvo  t]i(i  ]irolilcni.  Tlie  Enj;lisli  monetary  unit  is 
the  jioniid  sli'i-liii;/,  wliieh  is  divided  info  20  aliilliiins  ;  caeli 
sliillinf;  is  divided  into  12  peiinicn,  ami  each  penny  into  4 
fiiHIiiiifin.  With  this  knowledge  of  English  money,  we  can 
work  all  prol)lenis  of  tlio  above  cliaraeter.  In  this  example, 
we  first  iiinlti]ily  the  iirice  of  one  jioiind  by  the  iinmber  of 
ponnds,  and  thus  ])rodneo  the  valne  of  the  whole  in  pence. 
Then  to  reduce  the  j«'»ee  to  s/n7/i»f/s,  wiMlivide  them  liy  12, 
and  obtain  9226  s/n7/(»r/s  and  a  renniinder  of  ^,  which  being 
a  part  of  the  dividend,  is  therefore  Sd.  Then,  to  reduce  the  shilHiifls  to  jiohihIx,  ^v(^  divide  them  by 
20,  and  obtain  4(il  poiitids  and  a  remainder  of  6,  which  being  a  part  of  the  secoud  dividend,  is  there- 
fore 6s.  In  the  English  monetary  system  the  following  ald)reviations  .are  used:  £,.  represents 
pounds ;  s.  represents  shillings  ;  d.  represents  pence,  and  f.  represents  farthings. 


OPEKATION. 

i;]840 
Sd. 

12)  llOTL'Od. 

20)  922(5     Sd. 

£401       Gs. 


2.     What  is  the  value  of  483  yards 
of  cloth,  at  16  shillings  per  yard  1 

Aus.  £380.  8s. 


3.    What  will  241  boxes  cheese  cost. 


at  £3  per  box  ? 


Ans.  £723. 


OPERATION. 

483 
16 


OPERATION. 

241 
3 


20)  772^  shillings. 
£  3SG  Ss. 


£  723 


4.  Sold  486  yards  of  calico  at  5  pence  a  yard.    What  did  it  amount  to? 

Ans.  £10  2s.  6d. 

5.  Bought  38495  ijoniids  of  good  middling  cotton  at  7  pence  jier  i^ouud. 
How  much  did  it  cost  ?  Ans.  £1122  los.  5d. 

6.  What  is  the  value  of  850  barrels  of  flour  at  34  shillings  a  barrel  ? 

Ans.  £1445. 

7.  How  much  Avill  1812  tons  of  iron  cost,  at  £52  4s.  per  ton  ? 

Ans.  £94586  8s. 

8.  Bought  38421  pounds  of  cotton  at  9  pence  per  pound.     What  did  it  cost  ? 

Aus.  £1440  15s.  9d, 


DIVISION. 


133 


TO  FIND  THE   TRUE   REMAINDERS  WHEN   DIVIDING   BY  THE   FAC- 
TORS   OF   A   NUMBER. 


229.     Divide  1607  by  72,  using  the  factors  3, 4,  and  6,  and  find  tlie  true  remainder. 


Ans.  22  quotient,  and  23  remainder. 


FIRST 
3)  1007 


OPERATION. 


Explanation. — In  this  problem,  using  the  factorsS,  4,  and  6,  we 
obtain  for  lemaimlers  2,  3,  and  1. 

The  first  remainder  2,  is  clearly  units  of  the  given  dividend, 
and  hence  a  part  of  the  trne  remainder. 

The  second  remainder  3,  hiting  fourths  of  the  second  dividend 
535,  which  are  reciprocal  thirds  of  the  given  dividend,  it  is  hence 
f  of  the  reciprocal  of  i^  of  -J  =  9,  of  the  given  dividend  and  tfue 
remainder. 

The  tliird  remainder  1,  being  sixths  of  the  third  dividend,  133, 
which  are  reciprocal  twelfths  of  the  given  dividend,  it  is 
hence  ^  of  the  reciprocal  of  ^  of  i-  of  j^  =  12,  of  the  given  divi- 
dend and  trne  remainder.  Therefore  2,  the  first  remainder,  pins  9,  the  nnit  value  of  the  second 
remainder  plus  12,  the  unit  value  of  the  third  remainder  =  23,  the  true  remainder. 


2,  1  ist  remainder. 


3,  2d  remainder. 


22     1,  3d  remainder. 


SECOND    OPERATION. 


1607 


535  =  2 


§of  1. 


133  =  3  =  f  + 


XI 


XI  ^  4.  XX 


_iw    i    12     -f-     12     12    T-"    7  2> 

or  23  the  true  remainder. 


THIRD   OPERATION. 

1607 


4  I  535  =  2,  1st  remainder 

0  I  133  =  3,  2d  remainder. 

22  =  1,  3d  remainder. 

First  remainder      ------        2 

Plus  2d  remainder  3, xfhe  preceding  divisor  3=9 
Plus  3d  remainder  1,  xall  the  preceding  divisor  s 
4  and  3  = 12 


which  added,  gives  the  trne  remainder 


23 


Eplanation. — The  second  operation  involves  fractions,  and  hence  not  regularly  in  order  at  this 
place.  In  this  operation  each  subsequent  remainder  is  reduced  to  the  fractional  unit  of  the  preced- 
ing fraction. 

By  the  third  operation,  we  see  that  the  true  remainder  may  be  obtained  without  considering 
the  reciprocal  relationship  of  the  quotients  and  divisors. 

The  general  method  shown  in  the  third  o])eration  is  as  follows  r  Add  to  the  first  remainder,  the 
product  of  the  other  remainders  by  all  the  divisors  i)receding  the  one  which  produced  it. 


2.     Divide  7851  by  64,  using  the  fac- 
tors 8  and  8. 

Ans.  122  quotient,  43  remainder. 


8)7851 


OPERATION. 


ExpJanation. — 
Here  the  1st  re- 
mainder is  3,  to 
which  we  add 
the  product  of 
the  2d  |remain- 
der  5, multiplied 
by  the  preced- 
ing divisor  8,  equals  40,  making  43,  the  true  re- 
mainder. 


8)981—3,  1st  remainder. 
122 — 5,  2d  remainder. 


3.     Divide  17803  by  90,  using  the  fac- 
tors 2,  3,  4,  and  4. 

OPERATION. 


Ans.  185||. 


3)17803 


Explanation. 


3j8901 — 1      1st  remainder 


4)2907 — 0     2d  remainder,  0x2  = 


4)741 — 3     3d  remainder,  3x3x2  = 

185 — 1      4th  remainder  1x4x3x2  = 
True  remainder, 


1 
0 

18 

24 
43 


134  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

4.  Divide  27865  by  the  factors  of  81.     Aiis.  34:4^^  or  344^  and  1  remainder. 

6.  Divide  101041  by  the  factors  of  84.  Aiis.  1202|f. 

6.  Divide  899  by  the  factors  of  108.  Ans.  8-1^^. 

7.  If  $4091  are  divided  equally  among  35  men,  Avhat  will  each  one  receive  ? 

Ans.  $134gV. 


THE  TAEENTHESIS  AND  VII^CULUM. 

230.  The  parenthesis,  (  ),  or  vinculum,  — ,  denotes  that  the  numbers  within  the 
parenthesis  or  below  the  vinculum  are  to  be  considered  as  one  quantity.  Thus,  as 
Bhowu   on  page  30,  12  —  (4+3)  =  5,  or  12  —  4  +  3  =  5. 

In  all  operations  involving  the  signs  of  +,  — ,  x,  and  -4-  the  following  prin- 
ciples govern  : 

1.  The  signs  +,  and  —  affect  only  the  number  or  expression  which  immedi- 
ately follows  either  of  them. 

2.  The  signs  x  and  -^  indicate  an  operation  to  be  iierformed  with  the 
numbers  or  expressions  between  which  either  may  be  placed,  and  suc^h  operation 
must  be  jierformed  before  that  of  any  +  or  —  which  may  immediately  precede 
or  follow. 

3.  When  the  parenthesis  or  vinculum  is  iised,  the  operations  indicated  tliereby 
must  be  performed  or  simplified  into  one  expression  before  connecting  them  with 
any  other  numbers  in  the  problem.  This  is  done  for  the  reason  that  all  numbers 
included  within  the  parenthesis  or  the  vinculum  are  considered  as  one  number. 

4.  In  the  operations  of  all  problems  involving  these  principles,  commence 

"with  the  (  ),  or ,  if  any,  then  with  the  expressions  indicated  by  the  x  and  -^ 

signs,  and  lastly  with  the  numbers  connected  with  the  signs  of  +  and  — ,  which 
should  be  considered  from  left  to  right. 

5.  In  cases  where  there  are  double  parentheses,  the  operation  indicated  by 
the  interior  parenthesis  must  be  performed  first. 

PROBLEMS. 

1.  Simplify  the  following  expression  : 

16  +  24  +  8  -^  4  —  3  x  5.  Ans.  27. 

2.  What  is  the  value  of  the  following  numerical  expression  ? 


(740  +  63  —  200)  ~  450  —  325  +  76  x  {Go  —  35  +  10)  ? 

OPERATION. 

740  +  63  =  803  —  200  =  603  ;  then  450  —  325  =  125  +  76  =  201 ;  then  603 
-i-  201  =  3 ;  then  65  —  35  =  30  +  10  =  40 ;  then  3  x  40  =  120  Answer. 

3.     Simplify  the  following  expression  : 

(740  +  63  —  200)  +  (450  —  325  +  76  )  x  (65  _  35  +  10)  ? 


*  DIVISION.  135 

OPERATION. 

740  +  63  =  803  —  200  =  603 ;  then  450  —  325  =  125  +  76  =  201 ;  then  603 
+  201  =  804;  tlieu  60  —  35  =  30  +  10  =  40;  then  804  x  40  =  32160  Ans. 

4.     Simplify  the  following  expression : 


(740  +  03  —  200)  +  450  —  325  +  70  x  (05  —  35  +  10). 

OPERATION. 

450  —  325  =  125  +  70  =  201 ;  then  65  —  35  =  30  +  10  =  40 ;    then  201    x 
40  =  8040;  then  740  +  63  =  803  —  200  =  603 ;  then  603  +  8040  =  8643  Ans. 

5.    What  is  the  value  of  the  following  expression  ? 


(420  —  00)  —  576  —  490  +  312  —  10  +  10  x  5  4-  8  +  6  —  4.        Ans.  584. 


6.  What  is  the  value  of  9500  +  6200  —  9000  X  3400  +  2100  —  5490  -^  (0500 
—  6400  +  200) '?  Ans.  15400. 

7.  What  is  the  value  of  2765  —  1624  x  (320  +  96)  4-  (3645  —  3376). 

Ans.  1764iAf. 

8.  What  is  the  value  of  47  +  1501  4-  7  +  142  x  20  +  522  -4-  87  +  207  4- 
23  +  (342  —  150)  X  62.  Ans.  15029. 


231.  MISCELLANEOUS  PROBLEMS  IN  DIVISION. 

1.  One  of  the  factors  of  10800  is  225.     What  is  the  other  ?  Ans.  48. 

2.  What  number  multiplied  by  137,  will  give  959137  for  the  product  f 

Ans.  7001. 

3.  Multiplying  372  by  an  unknown  number  gives  44640.  What  is  the  num- 
ber? Ans.  120. 

4.  What  is  the  quotient  of  9126  divided  by  9  ?  Ans.  1014. 

5.  Divide  four  million  eight  thousand  sixteen  by  MMDCXLIV. 

Ans.  1515fff|. 

6.  The  great  church  of  St.  Peter,  in  Rome,  will  accommodate  57000  peojile. 
The  largest  church  in  New  Orleans  will  accommodate  2500.  How  many  times  is  the 
capacity  of  the  church  of  St.  Peter  equal  to  that  of  the  largest  church  in  New 
Orleans  ?  Ans.  22|f . 

7.  What  number  is  that  to  which,  if  sixteen  be  added,  the  sum  multiplied 
by  8,  and  13  subtracted  from  the  product,  the  remainder  will  be  339  ?        Ans.  28. 

OPERATION. 

339  +  13  =  352  -4-  8  =  44  —  10  =  28  Ans. 
The  student  will  write  tlie  full  reasoning. 

8.  There  is  a  number  from  which,  if  you  subtract  55,  and  divide  the  remain- 
der by  12,  your  quotient  will  be  30.     What  is  that  number  ?  Ans.  487. 


136  soule's  philosophic  practical  mathematics.  * 

operation. 
3G  X  V2  =  -132  +  55  =  487  Ans. 

Tho  studout  will  -n-rito  tho  full  reasoniug. 

9.  A  merchant  owes  a  debt  of  $1875,  which  lie  agreed  to  pay  by  weekly 
iustallments  of  $25.  He  has  made  55  payments.  How  mauy  more  payments  has  he 
to  make  ?  Ans.  20. 

10.  The  velocity  of  the  earth  on  its  yearly  voyage  around  the  sun  is  99733 
feet  per  second.  The  velocity  of  a  cannon  ball  fired  from  a  gun  with  an  average 
charge  of  powder  is  about  1750  feet  a  second.  How  many  times  as  fast  as  the  veloc- 
ity of  a  cannon  ball,  is  the  velocity  of  our  earth  ?  A71S.  5G^ff. 

11.  About  one-sixth  of  a  man's  weight  is  blood.  How  many  pounds  of  blood 
in  a  man  whose  weight  is  1G8  pounds  ?  Ans.  28  pounds. 

12.  A  merchant  bought  350  barrels  of  flour  at  $6  per  barrel,  and  sold  it  at 
$7.50  per  barrel.  The  gain  he  gave,  in  equal  parts,  to  four  worthy  boys,  to  aid 
them  in  obtaining  an  education.  What  was  the  cost  and  the  selling  price  of  the 
flour,  and  how  much  money  did  each  boy  receive  ?  Ans.  $2100  cost, 

$2625  selling  price,  $131.25  each  boy  received. 

13.  The  air  which  surrounds  our  earth,  and  of  which  we  each  inhale  about 
600  gallons  every  hour,  is  composed  of  four  parts  of  nitrogen,  and  one  part  of  oxy- 
gen. How  many  gallons  of  each  are  there  in  a  room  22  feet  long,  21  feet  wide,  and 
10  feet  high,  and  which  contains  34560  gallons  of  air  ? 

Ans.  6912  oxygen,  27648  nitrogen. 

14.  An  acre  contains  160  square  rods.  How  many  acres  in  a  plantation  con- 
taining 123200  square  rods  1  Ans.  770  acres. 

15.  A  boy  sold  50  oranges  at  5/  each,  and  thereby  gained  $1.50.  At  what 
rate  did  he  buy  the  oranges  !  Ans.  2/  a  piece. 

16.  How  many  times  136  will  produce  1768  ?  Ans.  13. 

17.  Divide  the  product  of  750  and  875,  by  their  difference.  Ans.  5250. 

18.  The  diameter  of  the  earth  at  the  equator  is  7925  miles.  How  long  woidd 
it  take  a  locomotive  to  travel  that  distance,  at  the  rate  of  25  miles  an  hour  ? 

Ans.  317  hours  =  13  days,  5  hours. 

19.  It  is  estimated  that,  by  reason  of  intemperance,  the  United  States  loses 
annually  $98400000.  llow  many  school  houses  costing  $5000  each,  and  how  many 
libraries  costing  $3000  could  be  established  with  this  amount  of  money  1 

Ans.  12300  of  each. 

20.  The  first  Atlantic  telegraph  cable,  as  originally  made,  cost  $1258250. 
Ten  miles  of  deep  sea  cable  were  made  at  a  cost  of  $1450  per  mile,  and  25  miles  of 
shore  ends  were  made  at  a  cost  $1250  per  mile.  The  remainder  cost  $485  per  mile. 
How  many  miles  of  cable  were  made  'I  Ans.  2535  miles. 

21.  The  circumference  of  our  earth  at  the  equator  is  24899  miles,  and  the 
mean  diameter  of  the  earth  is  7912  miles.  How  many  times  is  tlie  circumference  as 
great  as  the  mean  diameter  1  Ans.  3|if  |  times. 

22.  A  grocer  wishes  to  put  3335  pounds  of  sugar  in  three  kinds  of  boxes, 
containing  respectively,  20,  50,  and  75  pounds,  and  use  the  same  number  of  boxes 
of  each  kind  or  size.     How  many  boxes  wUl  he  require  ?        Ans.  23  of  each  size. 


*  DIVISION.  137 

23.  The  capacity  of  steam  engines  is  measured  by  horse  lioicer  ;  and  1  horse 
power  is  a  force  that  will  raise  3300U  pounds,  one  foot  iu  one  minute.  IIow  much 
horse  power  has  a  steam  engine  that  possesses  a  capacity  of  1188000  pounds  ? 

Ans.  3G. 

21.  The  average  weight  of  a  man  is  150  pounds,  and  about  ^  of  this  weight 
is  blood.  Allowing  that  the  heart  throws  out  two  ounces  of  blood  at  each  pulsatiou, 
that  it  beats  72  times  a  minute,  and  that  IG  ounces  make  a  pound,  how  long  will  it 
take  the  heart  to  circulate  all  the  blood  in  the  body  ?  Ans.  2||f  minutes. 

25.  Ten  freedmen  agreed  to  pick  20000  pounds  of  cotton  and  receive  for 
their  labor  i  of  the  cotton  iiicked.  After  they  had  picked  7000  pounds,  four  freed- 
men quit,  leaving  the  other  six  to  finish  the  work.  How  much  cotton  is  each  entitled 
to  when  the  work  is  finished  ?  Ans.  110  pounds  each  for  those  who  left,  and 

573|  each  for  those  who  remained. 

2G.  The  Old  and  ^ew  Testaments  contain  1189  chapters  and  31164  verses. 
If  a  person  were  to  read  12  chapters  every  Sunday,  how  many  Sundays  would  it 
require  to  read  the  contents  of  the  whole  Bible  ?  If  he  were  to  read  85  verses  a 
day,  how  many  days  would  it  require  to  read  the  contents  of  the  whole  book  ? 

Ans.  99  -iV  Sundays ;  366  |i  days. 

27.  A  merchant  bought  800  gallons  of  molasses  at  65/,  and  sold  ^  of  it  at 
72/  a  gallon.  From  the  ijrofit  he  bought  his  children  a  set  of  Cutter's  Anatomical 
and  Physiological  Charts,  and  had  $8.20  left.    What  did  the  charts  cost  ? 

Ans.  $10.80. 

28.  A  man  produces,  by  breathing,  at  least  six  gallons  of  carbonic  acid  gas 
every  minute ;  a  single  burning  gas  jet,  ten  gallons ;  an  ordinary  stove,  sixty  gal- 
lons. How  many  gallons  of  carbonic  acid  gas  will  two  heated  stoves,  fifty  burning 
gas  jets,  and  an  audience  of  1000  people,  produce  in  thi-ee  hours,  and  how  many 
times  would  the  quantity  fill  a  room  100  feet  long,  50  feet  wide,  and  30  feet  high  ? 

Ans.  1191600  gallons.  1 -215^0-0-0-0%  times. 

29.  A  merchant  pays  $180  for  12  dozen  shirts.  At  what  price  per  shirt  must 
he  sell  them  to  gain  $72  ?  Ans.  $1.75. 

30.  A  tailor  paid  $117.60  for  a  piece  of  cloth,  the  price  was  $2.80  a  yard. 
How  many  yards  did  it  contain  ?  Ans,  42. 

31.  Paid  $4569  for  paving  1523  square  yards.  What  was  the  price  per  square 
yard?  Ans.  $3.00. 


138 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


CONTEACTIONS  IN  DITISION. 
233.     1.    Divide  9746521  by  634. 


OPEEATION. 

!S34)974G521(15373-a,\  Ans. 


3406 


2305 


4633 
1941 
39 


Explanation, — The  operation  of  this  problem  is  abridged 
by  performing  mentally,  the  subtraction  of  the  product  of 
the  divisor  by  the  quotient  figure,  instead  of  placing  it 
under  the  dividend  and  then  subtracting,  as  is  ordinarily 
done. 


Or,  to  arrange  tlie  figures  differently,  which  we  much  prefer,  as  follows : 


3241 

4369 

0634 

634)9746521 


15373^  Ans. 


lu  this  problem,  the  subtraction  is  performed  mentally, 
as  in  the  example  above,  but  the  remainders  are  placed  in  a 
vertical  position  over  the  dividend,  so  as  to  avoid  repeating 
the  figures  of  the  dividend. 


TO  DIVIDE  BY  THS  PACTOES  OF  THE  DIVISOE. 


283.     Divide  8316  by  36. 


OPERATION. 

36  )  8316 


1386  quotient  by  6 


231  Ans. 


Explanation. — Here  tve  first  divide  by  6,  which  gives 
us  a  quotient  6  times  too  large,  because  6  is  but  \  of  36, 
the  true  divisor ;  and  hence,  to  obtain  the  correct  result, 
we  divide  the  first  quotient  by  6. 


■    DIVISION. 

CONTRACTIONS. 


139 


2.34:.    Peculiar  to  problems  when  the  divisor  is  an  aliquot  part  of  10, 100,  or  1000, 
or  some  convenient  multiple  of  10,  100,  or  1000.     Thus,  to  divide  by 


1^ 
14 
If 
1§ 
2J 
3i 
H 

12J 
14| 
1G§ 
18f 
22^ 
25 
31i 
33J 
37J 
62i 
66| 
75 
83^ 
87J 
88a 
125 
133i 
166| 
225 
250 
333J 
375 
625 
833J 
875 
15 
35 
45 


midtiply  the  dividend  by         9  and  divide  the  jiroduct  by 


u 
a 

u 


a 
u 
u 
u 
a 
u 
u 
u 
a 
u 
a 
u 
ik 
u 
u 
II 
il 
II 
u 
a 
u 
u 
il 
u 
a 
a 


u 

u 

a 


u 

u 
u 
il 
li 
u 

u 

a 
u 
a 
u 
(( 
u 
a 

u 
a 
a 
u 
a 
u 


u 

8 

4; 

7 

u 

6 

u 

4 

u 

3 

u 

16 

u 

12 

u 

8 

u 

7 

u 

6 

u 

IG 

i.' 

4 

u 

4 

u 

16 

a 

3 

il. 

8 

a 

8 

u 

3 

a 

4 

ii 

6 

u 

8 

u 

9 

u 

8 

ii 

3 

ii 

6 

li 

4 

li 

.4 

a 

3 

11 

8 

a 

8 

il 

6 

il 

8 

(( 

2 

u 

2 

u 

0 

u 
if 

a 
(. 
a 
II 
II 
II 
II 
a 
il 
a 
II 
il 
il 
II 
u 
il 
II 
a 
II 
li 
a 
a 
II 
il 
li 
a 
il 
li 
il 
u 
il 


il 
a 
li 
a 
li 
il 
il 
il 
i. 
il 
li 
a 
li 
II 


a 

II 
li 
li 
II 
li 
a 
II 
li 
il 
il 
li 
il 
II 
II 
II 


;by 

10 

11 

10 

11 

10 

11 

10 

li 

10 

a 

10 

il 

100 

11 

100 

il 

100 

il 

100 

a 

100 

a 

300 

il 

90 

11 

100 

a 

500 

a 

100 

u 

300 

11 

500 

il 

200 

u 

300 

li 

500 

il 

700 

il 

800 

li 

1000 

il 

400 

11 

1000 

il 

900 

11 

1000 

a 

1000 

il 

3000 

11 

5000 

il 

5000 

11 

7000 

a 

30 

il 

70 

il 

90 

The  reasons  for  these  contractions  are  based  upon  the  fact  that  in  all  division 
operations  the  residt  or  quotient  is  not  changed  by  midtiplying  the  divisor  and 
the  dividend  by  the  same  number. 


140  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

To  elucidate  this  work,  we  present  the  following  problems  : 
3.     Divide  72i2  by  2J. 


OPERATION. 

4 


10 


7242  Explanation. — l?y  inspection,  Tve  observe  that  4  times  2}  are  10  ;  we 

J.  thereloro,  to  save  time  and  labor,  lirst  multiply  it  liy  4,  to  jirodnce  a 

new  and  luoro  convenient  divisor;  and  in  or(lor  not  to  effect  a  cliange 

_^       ,""  or  error,  in  the  qnotient,  wo  also  multiply  the  dividend  by  4  for  a  new 

wbJO.o  Alls.  dividend,  and  then  divide  in  the  regular  mauuer  which,  in  this  case,  is 

done  by  simply  poiutiug  off  one  figure. 

4.  Divide  179801  by  33 J. 

OPERATION'.  Explanation. — In  this  problem,  -we  also  see  by  inspection  that  3 

ooi\   -1  "n*.;/!!  times  the  divisor  makes  100,  hence  for  the  reasons  given  above,  we  nnil- 

OOj;   -1  i  .lOOi  ^j^^jy  ^j^^  dividend  by  3,  and  divide  the  product  by  100.     In  i)ractice,  we 

—  do  iiot  set  the  figure  (3  ju  this  example)  by  which  we  multiply  the 

5394.03  Alls.  dividend. 

5.  Divide  12358  by  37 J. 

OPERATION. 

3<2)  1-ouS  Explanation. — We  here  observe,  by  inspection,  that  the  divisor  37^^ 

~^  is  I  of  a  hundred,  and  hence,  if  we  multiply  it  by  8,  we  will  have  for 

"PP)  Jo^pi  ;i  1,,5-iyr  divisor  300 ;  therefore,  for  reasons  given  above,  we  multiply  the 

divisor  and  the  dividend  by  8,  and  then  divide  in  the  usual  manner. 
329if^  Alls. 

6.  Divide  97450  by  75. 

OPERATION. 

75) 97450 

Explanation. — By   inspection,  we  here  observe  that  4   times  the 

300)  389Spp  divisor  makes  300  ;  we  therefore  multiply  the  divisor  and  dividend  by  4 

and  with  their  products  proceed  to  divide. 

1299J   Alls. 

7.  Divide  4307491  by  125. 

OPERATION. 
125)  4307401  Explanation. — In  this  problem,  we  observe  that 
,                            8  times  the  divisor  makes  1000;  we  therefore  mnlti- 


1000)  34459p^^  =  34459  |if  Aus.  ply  the  dividend  by  8  and  divide  the  product  by  1000. 


PECULIAR  COISTTRACTIOXS. 

235.  Ill  tlie  preceding  contractions  the  work  was  based  upon  the  principle  that 
multiplying  the  divisor  and  dividend  by  the  same  number  does  not  effect  a  change 
in  the  quotient. 

In  the  following  contractions  the  work  is  based  upon  the  principle  that,  to  add 
to,  or  subtract  from  both  divisor  and  quotient,  the  same  aliquot  parts,  will  effect  no 
change  in  the  result. 


M 


DIVISION. 


141 


By  tbe  application  of  these  principles,  we  can  often  increase  or  diminish 
either  divisor  or  dividend,  or  both,  by  aliquot  parts,  and  thus  obtain  a  more  con- 
venient divisor. 

8.     Divide  426  by  7  J. 


OPERATION. 

7i)  42.G 
142 


56.8  =  501  Ans. 


Explanation. — lu  this  problem,  we  first  (Tivide  by  10,  and  then 
add  i  of  the  quotieut  to  itself.  Tliis  we  do  because  by  inspection, 
we  observe  that  7+  is  J  of  10,  or,  that  it  is  i  of  itself  less  than  10; 
hence  it  follows  that  if  we  divide  by  10,  our  quotient  will  be  J  of 
itself  too  small,  and  to  obtain  the  correct  quotient  we  must  add  ^ 
of  the  quotieut  by  10  to  itself. 


Note. — To  add  to,  or  deduct  from,  the  dividend  before  dividing,  will  produce  the  same  result 
as  adding  to,  or  deducting  from,  the  quotieut. 

0.     Divide  2548230  by  24. 


OPERATION. 


i  of  24 
is        0 

op 


2548230 

84941 
212351 


10G17GJ  Ans. 


Explanation. — In  this  problem,  we  first  divide  by  30  and  then  add 
J  of  the  (iiiotient  to  itself.  The  reason  for  the  work  is  the  same  as 
that  given  in  the  above  problem. 


10.    Divide  831G  by  36. 


FIRST 
i   Of 

is 


OPERATION.      SECOND   OPERATION. 


30 
0 

30 


8310 
13S0  =  i 

0930 

231  Ans. 


30 


30 


831)5 


46 


231  Ans. 


Explanation. — In  the  first  operation,  we  de- 
duct from  the  dividend  and  divisor  \  of  them- 
selves, which  decreases  each  in  like  proportion, 
and  tlien  divide. 

In  tlie  second  operation  we  divide  by  30, 
which  gives  a  quotieut  ^  too  large,  because 
the  30  with  which  we  divide  is  \  smaller  than 
36,  the  true  divisor ;  hence  we  deduct  J  of  the 
quotient  by  30  from  itself,  and  obtain  in  the  re- 
mainder the  true  result. 


Note. — In  nearly  all  contractions  like  the  above,  it  is  easier  and  shorter  to  divide  by  the  fac- 
tors of  the  divisor.  Hence  we  give  but  few  problems,  merely  to  show  the  ajiplication  of  the  princi- 
ple. 

PROBLEMS. 


Divide  the  following  numbers  ; 


1.    489720  by  54. 


8725  by  35. 


3.    26748  by  48. 


4.     8316  by  36. 


SYNOPSIS  FOR  REVIEW  OF  DIVISION. 


Note.— The  numbers  before  the  words  and  phrases  refer  to  the  articles  treating  the  subjects 


named. 


236.     Define  the  following  words  and  phrases  ; 


207. 

Division. 

022 

209. 

The  Logic  of  Division. 

210. 

The  Dividend. 

223 

211. 

The  Divisor. 

224 

212. 

The  Quotient. 

213. 

The  Eemainder. 

225 

214. 

The  Sign  of  Division. 

215. 

Other  Signs  of  Division. 

21G. 

Trinciples  of  Division. 

220 

217. 

Proof  of  Division. 

218. 

How  many  ways  may  Division  be 
performed. 

227. 

219. 

Oral  Exercises. 

228 

220. 

Fractional  Numbers. 

229 

221 

The  Philosophy  of  Division. 

How  to  Divide  when  the  Divisor 
does  not  exceed  Twelve. 

Short  Division. 

To  Divide  when  the  Divisor  ex- 
ceeds Twelve. 

Give  the  Greneral  Directions  for 
Long  Division  and  all  the  De- 
tails for  the  Fidl  Operation. 

To  Divide  when  there  are  Naughts 
on  the  Eight  of  the  Divisor. 

To  Divide  by  the  Factors  of  the 
Divisor. 

Operation  in  English  Money. 

To  Find  the  True  Eemainder. 


(142) 


MISCnLI.A.NBOUS    PROBLErMS 

Involving  the   Principles   of  Addition,    Subtraction,  Multiplication, 

AHD  Division. 


237.     1.     The  subtrahend  is  21G,  and  the  remainder  is  184.    What  is  the  min- 
uend ?  •  Ans.  400. 

2.  A  grocer  paid  $350  for  some  tea  and  some  coffee,  and  for  the  tea  he  paid 
$50  more  than  for  the  coffee.     What  did  he  pay  for  each  ? 

Ans.  tea,  $200 ;  coffee,  $150. 

3.  The  sum  of  two  numbers  is  787,  and  one  of  them  is  298.  "WTiat  is  the 
other?  Ans.  489. 

4.  The  difference  between  two  numbers  is  97,  and  the  lesser  number  is  103. 
What  is  the  greater  ?  Ans.  200. 

5.  What  number  multiplied  by  4  will  give  the  same  product  as  16  multiplied 
by  12  ?  Ans.  48. 

6.  If  the  speed  of  the  steamer  J.  M.  White  is  15  miles  per  hour  in  still 
■water,  and  the  velocity  of  the  river  is  3  miles  per  hour,  how  far  will  she  run  \vp  the 
river  in  4  hours?  How  many  miles  down  the  river  in  4  hours  ?  How  far  if  she  runs 
in  still  water  4  hours  ?  Ans.  48  miles  up ;  72  miles  down  ; 

CO  miles  in  still  water. 

7.  If  two  men  start  from  the  same  point  and  travel  in  opposite  directions, 
one  at  the  rate  of  20  miles  per  day  and  the  other  at  the  rate  of  25  miles  per  day, 
how  many  days  will  they  travel  before  they  are  495  miles  apart?  Ans.  11. 

8.  The  greater  of  two  numbers  is  897  and  their  difference  598 ;  what  is  the 
lesser?  Ans.  299. 

9.  A  newsboy  sold  20  papers  at  of'  each,  and  with  the  money  bought  oranges 
at  4/  each.    How  many  oranges  did  he  get  ?  Ans.  25, 

10.  A  boy  sold  5  chickens  at  25/  a  iiiece,  and  8  ducks  at  50/  each.  He 
received  in  i^ayment  3  j)igeons  at  30/  each,  and  the  balance  in  money.  How  much 
money  did  he  receive  ?  Ans.  $4.35. 

11.  The  divisor  is  187  and  the  quotient  101,  what  is  the  dividend  ? 

Ans.  1S887. 

Note. — In  division  operations,  the  worker  must  remember,  1.  That  the  divisor,  while  it  is 
the  unit  of  measure,  is,  also,  one  of  the  factors  of  the  dividend.  2.  That  the  quotient  is  the  other 
factor  of  the  dividend.  3.  That  the  dividend  is  the  product  of  the  divisor  and  the  qnotient,  plus 
the  remainder  when  there  is  one,  and  hence,  dividing  the  dividend  by  either  factor  will  give  the 
other,  and  the  remainder  if  any. 

(143) 


144  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

12.  The  product  of  two  uumbers  is  42230  and  one  of  them  is  205 ;  what  is  the 
other?  Ans.  206. 

13.  A  grocer  bought  28  barrels  of  apples  at  $4  per  barrel;  at  what  price 
must  he  sell  them  per  barrel  to  realize  a  gain  of  $42?  Ans.  $5 J. 

14.  The  dividend  is  5484888  and  the  quotient  81864,  what  is  the  divisor  ? 

Ans.  67. 

15.  What  number  divided  by  33  will  give  the  quotient  30449,  and  remain- 
der 18  ?  Ans.  1004835. 

10.  J.  J.  Foley  bought  a  barrel  of  sirop  de  batterie  containing  43  gallons  at 
95/  per  gallon;  4  gallons  having  leaked  out,  he  sold  the  remainder  at  $1.05  a  gallon. 
How  much  did  he  gain  by  the  transaction  ?  Ans.  $.10. 

17.  E.  S.  Soul6  bought  354  barrels  of  flour  for  $2478,  and  sold  the  same  at 
$7.50  per  barrel.     How  much  did  he  gain  ?  Ans.  $177. 

18.  The  capital  stock  of  a  manufactory  is  $100000,  which  is  divided  into 
200  shares.     AMiat  are  5  shares  worth  ?  Ans.  $2500. 

19.  E.  D.  T.  Sherwood  sold  to  H.  Eies  25  barrels  of  ai>ples  at  $4  i>er  barrel, 
and  124  barrels  of  potatoes  at  $3.25  jier  barrel.  He  received  in  payment  1 
hogshead  of  sugar  containing  1143  jiounds  at  8/,  and  the  remainder  in  money. 
How  much  money  did  he  receive  ?  Ans.  $411.50. 

20.  A  specidator  bought  528  cords  of  wood  at  $6.50  per  cord.  He  re-corded 
the  wood  so  that  it  measured  579  cords,  wiiich  lie  sold  at  $6.75  a  cord.  How  much 
did  he  gain  ?  Ans.  $470.25. 

21.  The  dividend  is  7359,  the  quotient  198,  and  the  remainder  33 ;  what  is 
the  divisor  ?  Ans.  37. 

22.  A.  and  B.  were  partners  in  business,  equal  in  gains  and  losses;  A. 
managed  the  business,  for  wliich  lie  was  to  receive  from  the  gains  $1200.  The 
gains  of  the  business  were  $18524  and  the  losses  $7108.  What  amount  of  money 
was  due  each  on  settlement  ?  Ans.  A.  $6308,  B.  $5108. 

23.  A  fruit  dealer  bought  250  boxes  of  peaches  for  $375;  he  sold  94  boxes  at 
$2  per  box,  and  147  boxes  at  $3  per  box.  The  remainder  of  the  peaches  spoiled  in 
his  store;  his  expenses  in  buying  and  selling  were  $45.  How  many  boxes  were 
spoiled,  and  how  much  did  he  gain?  Ans.  9  boxes;  $209  gain. 

24.  A  cistern  holding  1500  gallons,  has  two  pipes,  one  of  which  will  fill  it  in 
10  hours,  and  the  other  empty  it  in  30  hoius.  If  both  pipes  are  left  open,  how  long 
will  the  cistern  be  in  iilling?  Ans.  15  hours. 

25.  A  cist-ern  holding  3000  gallons,  has  three  pipes ;  by  one  30  gallons  per 
minute,  .ind  by  another  20  gallons  jier  minute,  are  conducted  into  it,  and  by  the 
third  it  may  be  emptied  when  full  in  2J  hours.  If  aU  three  pipes  are  left  open,  iu 
what  time  will  the  cistern  be  filled  ?  Ans.  100  minutes 

NoTK. — There  are  60  minutes  in  an  hour. 


J 


*  MISCELLANEOUS    PROBLEMS.  145 

26.  Ecl^\anl  lias  $105.00.  One-fifth  of  the  amount  is  in  50-cent  i^ieces,  the 
remainder  is  in  25-ceut,  lO-cent,  5-eeut,  3-cent,  and  l-ceut  pieces ;  allowing  that  he 
has  an  equal  number  of  each  coin,  how  many  has  he  1 

Ans.  66  50-cent  pieces,  and  300  of  each  of  the  other  coins. 

27.  A  merchant  had  200  bbls.  of  potatoes  for  which  he  was  offered  $4.00  per 
barrel,  which  price  would  have  given  him  a  gain  of  $20.  Having  declined  this 
offer,  on  a  declining  market,  he  finally  sold  at  a  loss  of  $20.  How  much  jier 
barrel  did  his  potatoes  cost,  and  how  much  per  barrel  did  he  sell  them  for  ? 

Ans.  cost  $3.90,  and  sold  for  $3.80  per  bbl. 

28.  A  government  ration  consists  of  1  lb.,  4  oz.  of  beef,  and  1  lb.,  6  oz.  of 
flour;  if  beef  costs  10|('  a  lb.  and  tlour  3^/  a  lb.,  what  will  be  tlie  cost  of  10840 
government  rations,  and  how  many  lbs.  of  beef  and  flour  will  there  be  ? 

Ans.  $1983.38,  cost;  14905  lbs.  of  flour;  13550  lbs.  of  beef. 

Note. — There  are  16  ounces  in  a  pound. 

OPERATION   INDICATED. 

1  lb.,  4  oz.  =  1 J  lbs.  beef  ®  10|,<r  = 

I  "    6  "    =  1|    "    flour  ©  3T  c    = 
Cost  of  one  ration      .    -    -    -        =  $ 

II  X  10840  =  lbs.  beef. 
Ig  X  10840  =    "    flour. 

29.  A  specidator  bought  an  invoice  of  flour  aiid  jjork  for  $297  ;  the  flour  cost 
$7  a  barrel,  and  the  pork  $12 ;  there  were  3  times  as  many  barrels  of  flour  as  of  pork. 
How  many  were  there  of  each  1  Ans.  27  bbls.  flour,  9  bbls.  pork. 


piroperties  of  Numbers. 


^ 


DEFINITIONS.  * 

238.  The  Properties  of  numbers  are  those  qualities  which  belong  to  them. 

239.  An  Integer  is  a  whole  number ;  as  1,  5,  6, 18,  etc.  Whole  numbers 
are  divided  into  two  classes,  Prime  and  Composite. 

210.  A  Prime  Number  is  one  that  can  be  divided,  without  a  remainder, 
only  by  itself  and  1 ;  as  1,  2,  3,  5,  7,  11, 13, 17,  etc. 

241.  A  Composite  Number  is  one  that  can  be  divided,  without  a  remainder, 
by  some  other  whole  number  than  itself  and  1 ;  as  4,  9, 12,  15,  24,  etc. 

All  Composite  Numbers  are  the  product  of  two  or  more  other  numbers. 

Numbers  are  prime  to  each  other  when  they  have  no  common  factor  that 
will  divide  each  without  a  remainder;  as  6, 13,  29,  etc. 

212.  An  Even  number  is  one  that  can  be  divided  by  2,  without  a  remain- 
der ;  as  4,  8,  12,  56,  etc. 

213.  An  Odd  number  is  one  that  cannot  be  divided  by  2,  without  a  remain- 
der; as  1,  7, 19,  45, 133,  etc. 

211.  The  Factors  of  a  number  are  the  numbers  which  multiplied  together 
will  produce  it.  Thus,  4  and  4,  or  2  and  8,  are  factors  of  16 ;  2,  3,  and  4,  or  2,  2,  2, 
and  3  are  factors  of  24. 

Every  factor  of  a  member  is  a  divisor  of  it. 
*NoTE. — For  other  Definitions  of  Numbers,  see  pages  25  to  31. 

215.  A  Prime  Factor  of  a  number  is  a  prime  number  that  will  divide  it 
without  a  remainder.    Thus,  1,  2,  3,  and  5,  are  the  prime  factors  of  30. 

216.  A  Composite  Factor  of  a  number  is  a  composite  number  that  will 
divide  it  without  a  remainder.    Thus,  6  and  8,  are  composite  factors  of  48. 

217.  A  Perfect  Number  is  one  which  is  equal  to  the  sum  of  all  its  divisors, 
except  itself;  as  6  =  1  -t-  2  -f  3. 

218.  Ati  Imperfect  Number  is  one,  the  sum  of  whose  divisors  is  more  or  less 
than  itself.  Thus,  ]4  is  greater  than  1  -f  2  +  7,  its  divisors;  and  18  is  less  than 
1  +  2  -f-  3  +  6  -t-  9,  its  divisors.    This  last  is  called  an  abundant  number. 

(146) 


*  PROPERTIES    OK    NUMBERS.  1 47 

249.  An  Aliquot  part  of  a  number  is  sucli  a  part  as  will  divide  it  without  a 
remainder.  Thus,  1,  2,  3,  4,  6,  and  8,  are  aliquot  parts  of  24.  All  aliquots  are 
factors  of  the  number. 

250.  The  Reciprocal  of  a  number  is  the  quotient  of  1  divided  by  the 
number.     Thus,  the  reciprocal  of  S  is  1  -r  8  =  J- ;  and  the  reciiirocal  of  J  is  1  -^  ^  =  4. 

251.  Powers  of  a  number  are  as  follows:  The  first  power  is  the  number 
expressed  by  itself;  as  5  is  the  first  power  of  5.  The  second  power  is  the  product 
arising  by  using  the  number  as  a  factor  twice;  as  5  x  5  =  25,  which  is  the  second 
power  of  5.  The  third  power  is  the  product  obtained  by  using  the  number  as  a 
factor  three  times ;  as  5  x  5  x  5  =  125,  which  is  the  third  potcer  of  5.  And  in  like 
manner  higher  powers  of  numbers  are  obtained. 

252.  A  Multiple  of  a  number  is  one  or  more  times  the  number;  or  it  is  any 
product,  dividend,  or  number  of  which  a  given  number  is  a  factor,  or  which  is 
exactly  divisible  by  a  given  number;  as  25  is  a  multiple  of  5 ;  24  is  the  multiple  of  2, 3, 
4,  6,  8,  and  12.  A  number  may  have  an  indefinite  number  of  multiiDles ;  as  2  will 
divide  4,  6,  8, 10,  12,  14,  etc.,  indefinitely. 

253.  A  Coninion  Multiple  of  two  or  more  given  numbers  is  a  number 
divisible  by  each  of  them  without  a  remainder.  Thus,  24  is  a  common  multii)le  of 
1,  2,  3,  4,  6,  12,  and  24. 

25-lr.  The  Least  Common  Multiple  of  two  or  more  given  numbers  is  the 
least  number  that  is  divisible  by  each  of  them  without  a  remainder.  Thus,  12  is  the 
least  common  multiiile  of  1,  2,  3,  4,  6,  and  12. 

Note. — Since  every  number  is  divisible  by  1  and  itself,  the  factors  1  and  the  given  number 
are  not  usually  given  when  naming  the  multiples.     We  shall  not  hereafter  name  them  as  multiples. 


DIVISIBILITY  OP  imiVIBERS. 


255.  A  Divisor  or  measure  of  a  number,  is  any  number  that  will  divide  it 
without  a  remaindei-.  Thus,  4  is  a  divisor,  or  measure,  of  12,  and  5  is  a  divisor,  or 
measure,  of  20. 

One  number  is  said  to  be  Divisible  by  another  when  there  is  no  remainder 
after  di\iding. 

256.  A  Common  Divisor  of  two  or  more  numbers  is  a  number  that  will 
divide  each  of  them  without  a  remainder.  Thus,  2  is  a  common  divisor  of  12, 18, 
and  24. 

257.  The  Greatest  Common  Divisor  of  two  or  more  given  numbers  is  the 
greatest  luimber  that  will  divide  each  of  them  without  a  remainder.  Thus,  6  is  the 
greatest  common  divisor  of  12, 18,  and  24. 


148  soule's  philosophic  practical  mathematics.  * 

25S.  Every  number  is  divisible  by  2,  whose  unit  figure  is  divisible  by  2. 
Thus,  31,  176,  790  are  each  divisible  by  2. 

259.  Every  number  is  divisible  by  i  when  its  units'  and  tens'  figures  are 
divisible  by  4.     Thus,  150,  204,  34512,  501308,  are  each  divisible  by  4. 

260.  All  numbers  are  divisible  by  3,  the  sum  of  whose  figures  are  divisible 
by  3.     Thus,  114,  223,  4101,  are  each  divisible  by  3. 

261.  All  numbers  ending  in  0  or  5  are  divisible  by  5.  Tluxs,  10,  15,  and  33^, 
are  each  divisible  by  5. 

262.  All  numbers  whose  U7iit  figure  is  divisible  by  2,  aiul  the  sum  of  M'hose 
figures  is  divisible  by  3,  are  divisible  by  C.  Thus,  36,  102,  678,  15936,  are  each 
divisible  by  6. 

263.  Every  nund>er  is  divisible  by  S,  when  the  imits',  tens',  and  hundreds' 
figures  are  divisible  by  8.     Thus,  3824,  12512,  190720,  are  each  divisible  by  8, 

264.  All  numbers  are  divisible  by  9,  the  sum  of  whose  figures  are  divisible 
by  9.    Thus,  441,  3456,  123453,  are  each  divisible  by  9. 

265.  All  numbers  ending  in  naught  are  divisible  by  10.  Thus,  20,  380,. 
11750,  are  each  divisible  by  10. 

266.  A  number  is  divisible  by  12  when  the  sura  of  its  figures  is  divisible  by 
3,  and  the  uuits  and  tens  are  divisible  by  4. 

267.  The  combination  of  numbers  that  are  divisible  by  7,  and  11,  are  so 
varied  and  intricate  that  they  cannot  be  conveniently  classified  or  formulated  under 
any  brief  general  law,  and  being  also  of  very  little  i^ractical  value,  they  are  not 
presented. 


SVAOJ^SZS    JPOR    JRBVIBW. 


named. 


Note. — The  numbers  before  the  words  and  phrases  refer  to  the  articles  treating  the  subjects 


268.     Define  the  following  words  and  phrases : 


238. 

The  Properties  of  Numbers. 

250. 

239, 

An  Integer. 

251. 

240. 

A  Prime  Number. 

241. 

A  Composite  Number. 

252. 

242. 

An  Even  Number. 

253. 

243. 

An  Odd  Number. 

254. 

244. 

Factors  of  a  Number. 

255. 

245. 

A  Prime  Factor. 

256. 

246. 

A  Composite  Factor. 

257. 

247. 

A  Perfect  Number. 

358. 

248. 

An  Imperfect  Number. 

259. 

249. 

An  Aliquot. 

The  Reciprocal  of  a  Number. 
The   Powers   of  a   Number;    1st. 

2d,  3d,  etc. 
The  Multiple  of  a  Number. 
A  Multiple. 
A  Common  Multiple. 
Least  Common  Multiple. 
A  Divisor. 
A  Common  Divisor. 
Greatest  Common  Divisor. 
What  numbers  are  divisible  by  2, 

3,  4,  5,  6,  8,  9,  10,  12. 


(149) 


properties  of  Nines  and  Elevens. 


^ 


269.  The  luimber  nine  lias  been  called  the  "  Magical  Number."  It  pos- 
sesses a  great  many  peculiar  properties,  some  of  which  are  more  curious  than 
practical,  and  will  be  but  briefly  treated  here.  Among  the  Ancients,  the  special 
properties  of  9,  and  of  some  other  figures,  were  used  in  their  magic  squares  and 
circles,  and  in  their  numerical  combinations  and  jmzzles,  to  such  an  extent  as  to 
Induce  the  belief,  among  the  ignorant,  that  they  possessed  supernatural  power,  or 
that  numbers  possessed  material  virtues. 

On  page  55,  we  presented  a  very  peculiar  property  of  9,  applied  to  the  opera- 
tion of  addition. 

A  curious  principle  of  9  is  shown  in  the  following:  Take  any  number  of 
two  places  of  different  figures,  transpose  the  figures,  and  take  the  difference  between 
the  two  numbers,  and  name  one  of  the  figures  of  the  diflerence,  and  the  other  figure 
is  known.  Thus,  take  73,  transpose  the  figures,  and  we  have  37 ;  then  73  — 37  =  36 
the  difference  or  remainder,  in  which  we  see  that  the  sum  of  3  and  6  =  9.  In  all 
similar  cases,  the  sum  of  the  remainder  will  be  9,  and  hence  when  one  of  the  figures  is 
known  the  other  may  be  found  by  subtracting  the  one  known  from  9. 

Another  interesting  puzzle  based  upon  the  same  princii^le  is  to  write  any 
number  of  three  or  more  figures,  divide  by  9,  and  give  the  remainder;  then 
erase  any  one  of  the  figures  in  the  number  and  divide  by  9,  and  give  the  remainder. 
The  figure  erased  will  be  the  difference  beticeen  the  remainders,  when  the  second  re- 
mainder is  less  than  the  first;  but  when  the  second  remainder  is  greater  than  the 
first,  the  figure  erased  will  be  the  difference  of  the  remainders  subtracted  from  9. 

Thus,  4381  divided  by  9,  gives  7  remainder.  Then  cancelling  3  and  dividing 
481  by  9,  gives  4  remainder,  and  7  — 4  =  3  the  figure  erased. 

Again,  taking  4381  —  9  gives  7  remainder.  Then  ei'asing  8  and  dividing  431 
by  9,  gives  8  remainder.    Then  8  —  7  =  1  and  9  —  1  =8,  the  figui-e  erased. 

Again,  let  the  reader  take  any  number  and  divide  it  by  9,  and  give  the 
remainder.  Then  nmltiply  the  number  by  any  immber  given,  and  divide  the  prod- 
uct by  9,  and  the  remainder  may  be  told  before  the  work  is  performed. 

Thus,  428  -^  9  gives  5  remainder.  Now  nmltiply  the  428  by  4  and  divide 
the  product  by  9,  and  the  remainder  will  be  2.  To  deteriuine  this  remainder,  we 
multiply  the  first  remainder  by  the  number  used  tO  multiply  the  first  number 
and  divide  the  product  by  9 ;  the  remainder  thus  obtained  will  be  the  same  as  the 
remainder  in  the  second  division. 

But,  as  our  purpose  is  now  not  to  present  a  discussion  of  these  questions 
except  to  apply  one  of  the  properties  of  9  and  11  to  the  proof  of  addition,  sub- 
traction, midtiplication,  and  division,  we  will  not  occui)y  space  to  discuss  the 
philosophy  or  reason  of  these  puzzling  results. 

(150) 


I 


*  PROPERTIES    OF    NINES    AND    ELEVENS.  l5l 

The  basis  or  radix  of  our  numerical  system  being  10,  equal  to  9  +  1,  if  we 
multiply  tlie  radix  and  its  equivalent  by  the  same  digit,  thus  10  x  4:  =  40,  and 
9  +  1  X  4  =  36  +  4  =  40,  we  find  the  jiroduct  of  9  alone  to  lack  just  the  multiplier 
of  being  equal  to  the  product  of  10,  and  as  this  will  hold  good  of  any  multiple  of  10, 
we  see  that  if  we  divide  any  decujile  as  40,  60,  80,  etc.,  by  9,  the  remainder  is 
always  the  same  as  the  number  of  tens  divided,  viz :  4,  6,  8,  etc.,  and  if  we  multiply 
these  numbers  by  10,  or  any  power  of  10,  as  400,  6000,  etc.,  the  remainders  will 
still  be  the  same  digits  4,  0,  etc.  Therefore  we  may  conclude  that  any  number  as 
467853  can  be  resolved  into  as  many  multiples  of  10  as  it  contains  digits,  thus : 
400000  +  COOOO  +  7000  +  800  +  50  +  3,  and  since  the  excess  of  9's  in  each  of  these 
round  numbers  is  the  same  as  its  significant  figure,  it  is  plain  the  excess  of  9's  in 
the  entire  number  467853  is  equal  to  the  excess  of  9's  in  the  sum  of  its  digits,  thus : 
4  +  6+7  +  8  +  5  +  3=  33;  divide  this  by  9  gives  quotient  3,  and  a  I'emainder  or 
excess  of  6.  In  finding  the  excess  of  9's,  it  is  best  to  reject  the  9's  as  soon  as  its  sum 
occurs  iu  the  additions,  thus  :  4  +  6  =  10,  dropping  9  =  1  +  7  +  S  =  16,  dropping 
9  =  7  +  5  =  12,  dropping  9  =  3  +  3  =  6  excess. 

This  property  is  peculiar  to  9  by  reason  of  its  being  1  less  than  the  radix  of 
notation.  This  property  belongs  also  to  any  number  that  will  give  a  remainder  1 
when  dividing  the  radix  by  it,  as  3,  and  3  and  9  are  the  only  numbers  that  will 
divide  10,  with  1  for  a  remainder ;  if  the  basis  of  our  system  were  9,  then  8,  4,  and  2, 
would  possess  these  properties. 

The  number  11  has  several  special  properties,  one  of  which  is  similar  to  that 
of  9  and  results  from  the  fact  that  11  is  1  more  than  the  radix  10.  The  usual 
method  of  finding  the  excess  of  ll's  is  somewhat  different,  viz  :  by  adding  the  alter- 
nate figures ;  thus,  to  cast  the  ll's  out  of  93257458,  we  first  find  the  sum  of  the 
figures  in  the  odd  places,  beginning  at  the  right  hand,  thus :  8  +  4  +  5  +  2  =  19, 
casting  out  the  ll's  gives  us  an  excess  of  8;  we  then  add  the  figures  iu  the  even 
places,  thus :5  +  7+2  +  9  =  23,  giving  an  excess  of  1 ;  subtract  this  last  excess 
from  the  first,  8  — 1  =  7,  the  excess  required;  if  the  first  excess  was  the  smaller,  we 
Mould  increase  it  by  11,  then  subtract  the  second.  The  excess  of  ll's  may  be  cast 
out  at  the  partial  additions,  as  for  9's. 

These  two  properties  of  9  and  11  can  be  used  in  proving  addition,  subtrac- 
tion, nudtiplication  and  division,  depending  on  this  xirinciple,  that  any  number 
divided  by  9  or  1 1  leaves  the  same  remainder  as  the  excess  obtained  by  casting  out 
its  9's  or  ll's. 

These  methods  of  proof,  however,  are  not  absolutely  infallible,  as  will  be 
shovm  on  the  second  page  following.  By  reason,  therefore,  of  the  liability  to  error, 
the  great  majority  of  business  men,  in  verifying  their  calculations,  prefer  to  repeat 
or  go  over,  in  a  reversed  direction,  the  first  operation. 


l52 


SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


PEOOF  OF  ADDITION  BY  CASTING   OUT  THE   NINES   AND  ELEVENS. 


270.     1.    Add  the  following  numbers  and  prove  the  ■work  by  casting  out  the 
9's  and  ll's. 


OPERATION 


AND  PROOF 
OUT  THE  9's. 


BY    CASTING 


Excess  of  9's. 

365592  3 

286548  6 

64320  6 

94094  8 


23  =  5  excess  of  9's 
iu  the  numbers. 


OPERATION 


AND     PROOF     BY 
OUT    THE  ll's. 


CASTING 


Excess  of  ll's. 

365592     7 

286548    9 

64320    3 


94094    0  3 


=  19  =  8  excess  of  ll's 
iu  the  numbers. 


Sum  810554=8  excess  of  ll's  in  the  sum. 

Note. — Instead  of  casting  the  ll's  out  of  each 
number  separately,  tUey  may  be  cast  out  of  all 
tlie  numbers  collectively.  Thus  we  may  add  the 
figures  in  the  odd  places  of  all  the  numbers, 
making  56  =  1  excess  of  11 ;  then  add  the  figures 
in  the  even  places  of  all  the  numbers,  making 
48  =  4  excess  of  ll's  ;  then  subtract  the  last  ex- 
cess from  the  first  increased  by  11,  thus  1  +  11  = 
12  —  4^8  excess  of  ll's  iu  all  the  numbers. 


Sum  810554=5  excess  of  9's  in  the  sum. 

Note. — Instead  of  casting  the  9's  out  of  each 
number  separately,  they  may  be  cast  out  of  all 
the  numbers  continuously.  Thus  wo  may  say 
3  +  6  =  9,  which  is  dropped.  Then  5  +  5  =  10 
—  9  =  1  (omitting  the  9) +  2  =  3  +  2  =  5  +  8 
=  13  —  9  =  4  +  6  =  10  —  9  =  1  +  5  =  6+4  = 
10  —  9  =  1  +  8  =  9  —  9  =  0,  6  +  4  =  10  —  9  = 
1  +  3  =  4  +  2  =  6  (omitting  9)  +  4  =  10  —  9 
=  1,  (omitting  9)  +  4  =  5  excess  of  9's  in  all  the 
numbers. 

Explanation. — To  prove  problems  of  addition  by  casting  out  9's  or  ll's,  first  find  the  excess  of 
the  9's  or  ll's  in  all  the  numbers ;  then  find  the  excess  of  9's  or  ll's  in  the  sum  of  the  numbers  ;  if 
the  two  excesses  are  the  same  the  work  is  correct,  as  far  as  this  proof  can  determine. 

THE     CHECK    OR    KEY    FIGURE. 

271.  The  excess  of  9's  or  ll's  is  called  the  check  or  key  figure  by  account- 
ants, and  is  used  not  only  to  prove  the  operations  of  addition,  subtraction,  multipli- 
cation and  division,  but  to  prove  other  operations  of  accounting. 

CONTRACTED  METHODS  OF  FINDING  THE  CHECK  OR  KEY  FIGURE. 


272.      1st  CONTRACTED  METHOD, 

Add  the  following  numbers  and  prove 
by  the  check  figure  system : 

365592 

286548 

64320 

94094  =  104  =  5  check  figure  of  all  the 

luiiiibers. 

810554  =  23  =  5  check  figure  of  the  sum. 

Explanation. — By  the  first  contracted  method, 
add  the  figures  in  all  the  numbers  as  units  =  104  ; 
then  add  these  figures  as  units  ^  5  the  check  fig- 
nre  of  all  the  numbers.  Then  add  the  figures  in 
the  sum  as  units  =  23,  which  add  as  units  =  5  the 
check  figure  of  the  sum. 

The  addition  of  the  several  sums  is  to  be  con- 
tinued as  long  as  any  succeeding  sum  has  more 
than  one  figure. 

The  check  figure  of  all  the  numbers,  and  the 
check  figure  of  the  sum  l>cing  equal,  proves  the 
■work  as  far  as  the  method  can. 

The  first  of  these  contracted  methods 
ffs,  is  new,  and  is  now  published  for  the  firs 


2d   CONTRACTED   METHOD. 

Add  the  following  numbers  aud  prove  by 
the  check  figure  system : 

365592  =  30  =  3 
280548  =  33  =  6 

64320  =  15  =  6 

94094  =  26  =  8 

23  =  5  check  figure. 

810554  =  23  =  5  check  figure. 

Explanation — By  the  second  contracted  method, 
add  the  figures  of  each  number  as  units  and  then 
add  the  figures  of  the  result,  when  it  has  more 
than  one  figure.  The  results  .are  the  check  fig- 
ures of  eacii  number.  Then  add  the  check  figures 
and  the  sum  of  the  snnu',  and  thus  produce  5  as 
the  check  figure  of  all  the  numbers.  Then  add 
the  figures  of  the  sum  of  all  the  numbers  as 
units,  and  also  .add  this  sum  which  gives  5,  the 
check  figure. 

of  finding  the  check  figure,  or  the  excess  of 
t  time. 


PROOF    BY    CASTING    OUT   THE    9  S    OR    IIS. 


i53 


ERRORS  ilSrOT  DETECTED  BY  CASTING  OUT  THE   NHSTES,  OR   BY   THE 
NI2fE  CHECK  FIGURE  METHOD. 


273.  1.  Wlien  figures  have  been  transposed.  2.  "Wlien  the  value  of  one 
figure  is  as  much  too  great  as  that  of  another  is  too  small.  3.  When  a  fignre  or 
figures  have  been  omitted,  and  other  figures  having  the  same  sum  have  been  sub- 
stituted.   4.    When  9  or  9's  have  been  omitted. 


ERRORS  iSrOT  DETECTED   BY  CASTING  OUT   THE   ELEVENS,   OR   BY 
THE   ELEVEN   CHECK  FIGURE  METHOD. 


274.  When  alternate  figures,  i.  e.  figures  in  the  odd,  or  figures  in  the  even 
places,  have  been  transposed,  thus,  234567  =  3  excess ;  then  234705  =  3  excess ;  or 
230547  =  3  excess. 

2.  When  two  figures  of  the  same  unit  value  in  the  odd  and  even  places  have 
been  transposed  for  two  other  contiguous  figures,  whose  value  is  the  same  but  differ- 
ent from  the  value  of  the  other  two,  thus,  335044  =  1  excess ;  then  445033  =  1 
excess,  or  504433  =  1  excess. 

3.  When  figures  of  the  same  unit  value  in  the  odd  and  even  places  have 
been  omitted,  thus,  45507  =  5  excess ;  then  407  =  5  excess. 

By  the  foregoing,  we  see  that  the  method  of  jiroof  by  castiug  out  the  ll's, 
is  more  reliable  than  the  method  by  casting  out  the  9's. 


PROOF  OF  SUBTRACTION  BY  CASTING  OUT  THE  NINES  OR  ELEVENS. 


275. 

and  ll's. 


From  1748272  take  1451004.  and  prove  the  work  by  castiug  out  the  9's 


OPERATION  AND  PEOOP  BY  CASTING 
OUT  THE  9's. 

Excess  of  9*3. 

Minuend,       1748272    4)1  =  difference 
Subtrahend,  1451604    3  )       of  excess. 


Remainder,     296008  =  1  excess. 


OPERATION  AND  PROOF   BY   CASTING 
OUT  THE  ll's. 

Excess  of  ll's. 

1748272  =    9  )  „   ,.„      „ 
14''ie()4  —    0  ^  ^^''^nsrence  of  excess. 


296668  =  9  excess  in  remainder. 


Explanation. — To  prove  problems  in  suhtraction  by  casting  out  the  9's  or  ll's,  first  find  the 
excess  of  9's  or  ll's  in  the  minuend  and  subtrahend,  and  take  the  ditference.  Then  find  the  excess 
of  9's  or  ll's  in  the  remainder;  if  the  excess  in  the  remainder  is  the  same  as  the  difference  of  the 
excesses  of  the  minuend  and  subtrahend,  the  work  is  correct,  as  far  as  this  proof  can  determine. 


1 54 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


PROOF    OF    MULTIPLICATION    BY    CASTING    OUT    THE    NINES    OR 

ELEVENS. 

276.    Multiply  350412  by  29595,  and  prove  the  work  by  casting  out  tlie  9's 
and  ll's. 


OPERATION    AND    PROOF    BY    CASTING 
OUT  THE  9's. 

Excess  of  9's. 

INhiltipliciind,         350412     0)6x3=18= 


Multiplier, 


29595    3  )      0  excess. 


Product,       10370443140=0  excess. 


OPERATION 


AND    PROOF    BY 
OUT  THE  ll's. 
Excess  of  ll's. 


CASTING 


350412 
29595 


I  17x5=35=2 


excess. 


10370443140  =  2  excess. 


Ex}>lnnation. — To  prove  multijilication  by  casting  out  the  9's  or  ll's,  first  find  the  excess  of  9'8 
or  ll's  in  the  multiplicand  and  multiplier,  and  then  find  the  excess  of  the  ])rodnct  of  these  two 
excesses.  Thou  find  the  excess  JTi  the  product ;  if  the  excess  in  the  ])roduct  eqn.ils  the  excess  of  the 
product  of  the  excesses  of  the  factors,  the  work  is  correct,  as  far  as  this  proof  can  determine. 


PROOF  OF  DIVISION  BY  CASTING  OUT  THE  NINES  OR  ELEVENS. 


277. 
and  ll's. 


Divide  65358547823  by  2789,  and  prove  tlie  woik  by  tlie  excess  of  O's 


OPERATION  AND  PROOF  BY  CASTING 
OUT  THE  9's. 

Dividend,  65358547823  =  2  excess  of  9's. 


Divisor, 


2789  =  8  excess  of  9's. 


Quotient,        23434402  =  4  excess  of  9's. 
Remainder,  645  =  G  excess  of  9's. 

8  X  4  =  32  =  5  excess  of  9's. 

5  +  6  =  11  =  2  excess  of  9's. 


OPERATION   AND    PROOF    BY    CASTING 
OUT  THE  ll's. 

Dividend,  6.3358547823  =  8  excess  of  ll's. 
Divisor,  2789  =  6  excess  of  ll's. 


Quotient,         23434402  =  2  excess  of  ll's. 
Remainder,  645  =  7  excess  of  ll's. 

6  X  2  =  12  =  1  excess  of  ll's. 

1  +  7  =   8  =  8  excess  of  1 1's. 


lExplanation. — To  prove  division  liy  casting  out  9's  or  ll's,  first  find  the  excess  of  9'8  or  ll's  in 
the  dividend,  divisor,  quotient,  .and  remainder;  then  multiply  together  the  excesses  in  the  divisor 
and  quotient,  and  find  the  excess  of  9's  or  ll's  of  the  product ;  then,  to  this  excess,  add  the  excess 
of  the  remainder  and  find  the  excess  of  9's  or  ll's  in  this  sum;  if  the  excess  is  the  same  as  the 
excess  iu  the  dividend,  the  ■work  is  correct,  as  far  as  this  method  of  proof  can  determine. 

EXCESS  OF  ELEVENS,  OR  THE  CHECK  OR  KEY  FIGURE. 

278.  Tlie  excess  of  ll's  iu  a  number  may  be  found  iu  various  ■ways.  We 
here  indicate  tliree  of  the  methods : 

First  method.    By  dividing  the  number  by  11. 

Second  method.  By  casting  the  ll's  out  of  the  sum  of  the  figures  in  the  odd 
l>laces,  then  casting  the  ll's  out  of  Uta  sum  of  the  figui'es  iu  the  even  places,  and 


*  PROOF    BY    CASTING    OUT    THE    g's    OR    I  I 's.  1 55 

subtracting  tlie  excess  of  ll's  in  tlie  even  places  from  the  excess  of  ll'.s  in  tUe  odd 
places.  In  case  the  first  excess  is  smaller  than  the  second,  increase  it  by  11  and 
then  subtract. 

Third  method.  From  the  sum  of  the  figures  in  the  odd  places,  subtract  the 
siun  of  the  figures  in  the  even  jdaces.  In  case  the  first  sum  is  less  than  the  second, 
add  11  or  some  mixltiple  of  11  to  the  first  sum  and  then  subtract;  or  when  the  first 
sum  is  less  thau  the  second,  subtract  11  or  some  multiple  of  11  from  the  second  sum, 
aud  theu  take  the  remainder  from  the  first  sum. 

*      THE   CHECK  OR  KEY  FIGUEE   SYSTEM. 

279.  The  excess  of  ll's  is  used  lai'gely  by  accountants  in  locating  errors, 
proving  transfers,  etc.,  aud  in  books  of  accounts  as  above  stated,  is  kuo-mi  as  the  11 
Check  or  Key  Figure  System. 

EXERCISES   IN  FINDING  THE   CHECK  OR  KEY  FIGURE. 

280.  Find  the  check  or  key  figures  of  the  following  numbers : 

1.  64752.      Sum  of  figures  in  odd  i)laces,  15.      Sum  of  figures  in  even 

places,  9.    9  from  15  =  6,  check  figure. 
Note. — See  tbree  methods  above  for  finding  cheek  figure. 

2.  134091.      Sum  of  figures  in  odd  places,  4.     Sum  of  figures  in  even  places, 

14.     14  from  (4  +  11)  =  1,  check  figure.   Or,  (14  —  11)  = 

3  from  4  =  1  check  figure. 
Mentally  think  4  from  7,  3,  check  figure. 
Mentally  think  5  from  13,  8,  check  figure. 
Mentally  think  G  from  11,5,  check  figure. 
Mentally  think  9  from  15,  C,  check  figure. 
Mentally  think  0  from  12,  1  from  2,  1,  check  figure,  or  mentally 

think  11  from  12,  1,  check  figure. 

8.  20908.      Mentally  think  0  from  19,  1  from  9,  8,  check  fignire,  or  mentally 
think  11  from  19,  8,  check  figure. 

9.  50.      Mentally  think  5  from  11,  G,  check  figure. 

10.  7060.  Mentally  think  13  from  22,  9,  check  figure,  or  mentally  think  2 
from  11,  9,  check  figure. 

11.  34567.      Mentally  think  7,  12,  15,  and  6,  10,  5,  check  figure. 

12.  $  7428.95  Mentally  think  5, 13, 17,  and.  9, 11, 18,  7  from  17, 10,  check  figure. 

13.  $20000.00  Mentally  think  0,  0,  0,  2,  0,  0,  0,  2,  check  figure,  or  mentally 
think  0  from  2,   2,  check  figure. 

14.  $52978.10  Mentally  think  0,  8,  17,  22 ;  and  1,  8,  10,  from  22, 12 ;  1  from  2  , 
1,  check  figure,  or  mentally  think  0,  8,  17,  22  and  1,  8,  10, 
from  11,1  check  figure. 

15.  $  4872.27  16.  25607.14  17.  134560.05 
18.  289176.40  19.  10806.70  20.  100000.00 
21.    100708.05                          22.    50000.20                            23.    899887.69 


3. 

47. 

4. 

52. 

5. 

368. 

0. 

193. 

7. 

507. 

1 56 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


APPLICATION    OF    THE    ELEVEiST    CHECK    FIGURE    METHOD    OF 

LOCATING    EKROIiS    AND    OF    PROVING    TRANSFERS 

AND    POSTINGS. 

281.     Post  or  transfer  tlie  following  sales  book  items: 


LEDQEB   ACCOUNTS. 

JONES.  SMITH. 


SALES  BOOK. 


Bin 

?- 

Jrtnes  bot.  Mdse.    7 

487 

59 

Smith     "       "        6 

16541 

20 

Brown    "       "        7 

9078 

15 

Wood     "       "      10 

10862 

05 

1654100 


WOOD. 


8 

9087 

51 

0 

10862  50| 

Explanation. — Before  posting  the  item  of  Jones'  purchase,  find  and  write  the  check  figure  to 
the  left  of  the  folio  column  in  the  Sales  Book,  or  whatever  hook  is  used  as  a  jiosting  medium. 
Tlien  when  the  item  has  been  posted,  find  and  write  the  check  figure  of  the  posted  item  to  the  left 
of  the  folio  column  in  the  Ledger,  or  whatever  hook  the  item  was  transferred  to.  If  the  check  fig- 
ures are  the  same,  the  posting  is  correct.  If  they  are  not  the  same,  there  is  an  error,  and  it  has 
been  located  iu  this  posting. 

In  tha  first  posting  above,  the  check  figures  are  the  same  (7'8)  and  the  work  is  correct. 

In  the  second  item  (Smith's  purchase)  .the  check  figure  in  the  Sales  Book  is  6,  and  in  the  Led- 
ger account  it  is  8.     This  shows  an  error  and  locates  it  iu  this  posting. 

In  the  iliird  item  (Brown's  purchase)  the  check  figure  in  the  Sales  Book  is  7,  and  iu  the  Led- 
ger account  it  is  8,  hence  an  error  in  this  posting. 

The  check  figure  in  the  Sales  Book  of  the /our//i  item  is  10,  while  it  is  0  in  the  Ledger  posting. 
This  difference  in  the  check  figures  shows  an  error,  and  locates  it  in  the  item  posted. 

In  the  above  manner,  all  postings  and  transfers  may  be  proved. 

For  further  work  on  the  application  of  the  properties  of  9  and  11,  to  the 
detection  of  errors  in  trial  balances,  transfers,  postings,  etc.,  see  SoiiM's  New  Science 
and  Practice  of  Accounts. 


^i^:^ 


I 

I 


I  actoi-irig. 


-@Vn 


282.     Factoring  consists  in  separating  or  resolving  a  comi>osite  number  into 
its  factors.     The  operation  is  performed  by  division. 

Note. — Before  enterinfjupnn  this  workof  factorinji,  and  thn  two  subjects  follow  in ;;,  the  leariier 
should  tlioioufrhly  uiul'erstaiul  what  are  ]ii'iiiie  and  eimiiiosite  numbers,  factors  and  multiples  of  uuui- 
berSj  aud  the  divisibility  of  uunibers  as  shown  on  pages  146  to  150. 

WRITTEX  EXERCISES. 


283.     "Wliat  are  tlie  prime  factors,  or  divisors,  of  5160  ! 


Explanation. — In  all  problems  of  this  kind,  'we  first  divide  the 
given  number  by  any  prime  factor,  and  the  successive  quotients  by 
prime  factors,  or  divisors,  until  -n-e  obtain  a  quotient  that  is  a  prime 
number.  In  this  problem,  our  last  quotient  is  13,  which  not  being 
divisible,  is  a  prime  number.  Hence  the  divisors  2,  2,  5,  3,  7,  and  the 
quotient  13,  are  all  prime  factors,  or  divisors,  of  5160. 


OPERATION. 

o 

5460 

2 

2730 

5 

1365 

3 

273 

7 

91 

13 


PROBLEMS. 

284.     Find  tlie  iirime  factors  of  the  following  numbers : 

4.  6105 

5.  1083 

6.  3560 


1. 

3! 
285. 


84 
376 
864 


7.  25600 

8.  10376 

9.  71460 


10.  32460 

11.  13532 

12.  96033 


"What  are  the  common  prime  factors  of  28,  04,  and  72  ? 


OPERATION. 

Explanation. — In  all  problems  of  this  kind,  we  divide  the  given 
numbers  by  any  common  jirime  factor  of  all  the  numbers,  and  the 
quotients  thus  obtained  are  divided  in  the  same  manner,  till  they 
have  no  common  factor  or  divisor.  The  several  divisors  will  be  the 
common  prime  factors  of  the  numbers. 

Note.— A  number  that  is  a  factor,  or  divisor,  of  two  or  more  numbers  is  called  a  common 
factor  of  these  numbers. 

PROBLEMS. 

286.     Find  the  prime  factors  common  to  the  following  numbers : 


0 

28  04 

72 

2 

14  32 

30 

7  16 

18 

1.  18,  24,  and  36 

2.  54,  72,  and  84 

3.  506,  430,  and  308 


4.  44  and  280 

5.  148,  256,  and  320 

6.  325,  635,  and  550 

(157) 


7.  28  and  64 

8.  112  and  250 

9.  526,  400,  and  780 


i58 


SOULE  S    nilLOSOrHIC    TRACTICAL    MATHEMATICS. 

GREATEST   COMMON   DIVISOE. 


287.     For  the  definition  of  a  divisor,  a  common  divisor,  and  the  greatest  common 
divisor,  see  page  147. 

G.  C.  D.  is  the  abbreviation  for  the  greatest  common  divisor. 

1.    What  is  the  greatest  commou  divisor  of  42,  5G,  and  210 1 

OPERATION. 

Explanation. — In  all  problems  of  tliis  kind,  we  first  divide  by 
any  prime  factor  that  will  divide  all  the  numbers  ;  then  we  divide  in 
like  manner  the  successive  quotients  thus  obtained,  until  we  obtain 
quotients  that  have  no  common  factor,  or  are  prime  to  each  other; 
then  we  multiply  all  the  divisors  together,  and  in  the  product  we  have 
the  greatest  common  divisor. 

Note  1.  —When  there  is  no  number  greater  than  1,  that  will  divide  all  the  numbers  without 
a  remainder,  then  1  is  the  greatest  common  divisor. 

Note  2. — When  there  are  two  large  numbers,  the  operation  may  be  more  easily  performed 
by  first  dividing  the  larger  number  by  the  smaller,  and  if  there  is  a  remainder  divide  the  preceding 
divisor  by  it,  and  thus  continue  until  there  is  no  remainder. 

Note  3. — When  there  are  more  than  two  numbers,  proceed  as  with  two.  aiid  then,  with  the 
greatest  common  divisor  of  the  two  and  one  of  the  other  numbers,  and  thus  continue  until  through 
with  all  the  numbers.     The  last  divisor  will  be  the  greatest  comaiou  divisor. 

Problems  2  and  3  elucidate  this  operation : 


2 

42,     50,     210 

7 

21,    28,    105 

3       4        15 

2x7=14  Ans 

2.  What  is  the  greatest 
common  divisor  of  88  and 
24?  Ans.  8. 

OPERATION. 

24)88(3 
72 

10)24(1 
16 

8)10(2 
10 


3.  What  is  the  greatest 
common  divisor  of  195, 
285,  and  315?    Ans.  15. 

OPERATION. 

285)315(1 
285 

30)285(9 
270 

15)30(2 
30 


15)195(13 
15 

45 
45 


GEKERAL    DIRECTIONS    FOR    FINDING    THE    GREATEST    COilMON 

DIVISOR. 


288.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  finding  the  G.  C.  D. 

1.  Write  the  numhers  on  a  horizontal  line  and  divide  by  any  prime  number  that 
will  divide  all  tvithout  a  remainder,  and  write  the  quotients  in  a  line  below. 

2.  Continue   this  process  of  dividing  the  successive  quotients,  until  quotients 
are  optained  which  have  no  common  factor,  or  divisor. 


*  FACTORING.  I  Sq 

3.  2IuUiphj  lixjcllier  all  the  divinors,  and  ilieir  product  will  he  the  G.  C.  D. 

NoTK. — AVheii  there  are  two  iir  more  large  numbers,  it  is  often  more  convenient  to  work  by 
successive  divisions,  as  eluciilateil  in  problems  2  and  3. 

PROBLEMS. 

289.     Wliat  is  the  greatest  common  divisor  of  the  following  numbers? 

4.  Of  4-il  and  567  ?  Ans.  63. 

5.  Of  90,  315,  and  810?  Ans.  45. 

6.  Of  654,  216,  and  108?  Ans.     6. 

7.  Eobinson  has  25,  and  Blaise,  45  dimes.  How  shall  they  arrange  them  in 
packages,  so  that  each  will  have  the  same  number  in  each  package  ? 

Ans.  5  in  each  jiackage. 

8.  A  planter  has  697  bushels  of  corn  and  204  bushels  of  rough  rice,  which 
he  wishes  to  put  into  the  least  number  of  bins  containing  the  same  number  of 
bushels,  without  mixing  the  two  kinds.     How  many  bushels  must  each  bin  hold  ? 

Ans.  17  bushels. 

9.  A  commission  merchant  has  2490  bushels  of  wheat,  1886  bushels  of  corn, 
and  8438  bushels  of  oats,  which  he  M'ishes  to  ship  in  the  least  number  of  sacks  of 
equal  size,  that  will  exactly  hold  either  kind  of  grain.  How  many  sacks  will  he 
require?  Ans.  6407. 

OPERATION   INDICATED. 

Find  the  greatest  common  divisor  as  above,  (it  is  2). 
2)2490     1886    8438 
1245+  943+4219=6407  Ans. 


LEAST   COMMOX   MULTIPLE. 

290.     For  the  definition  of  a  multiple,  a  common  multiple,  and  the  least  common 
multiple,  see  page  147. 

L.  C.  M.  is  the  abbreviation  for  the  least  common  multiple. 

1.     What  is  the  least  common  multiple  of  5,  6,  8,  21,  28  ? 

OPERATION. 
2)   5,   6,   8,  21,   28  Explanation.— In   all    problems   of  this   kind,   we  first 

arrange  the  numbers  on  a  horizontal  line,  and  then  divide  by 
the  smallest  prime  number  that  will  divide  two  or  more 
■without  a  remainder,  and  write  the  quotients  and  undivided 
numbers  in  a  line  below ;  this  process  of  dividing  wo  con- 
tinue until  there  are  no  two  numbers  that  can  be  divided  by 

the  same  number  without  a  remainder;  then  we  multiply 

5    12      11 

the  divisors  and  the  numbers  in  the  last  line  together,  and 


2) 

5 

3 

4 

21 

14 

3) 

5 

3 

2 

21 

7 

7) 

5 

1 

2 

7 

7 

2x2v3x7x5x  2^840  Ans  ^^^  product  is  the  least  common  multixile. 

Note. — When  there  is  any  number  of  the  problem  that  will  divide  any  of  the  others  without 
a  remainder,  it  may  be  cancelled  before  commencing  to  divide. 


l6o  soule's  I'HiLosornic  practical  mathematics.  * 

GENERAL    DIRECTIONS    FOR    FINDING    THE    LEAST    COMMON 

MULTIPLE. 

291.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  linding  the  L.  C.  M. 

1.  Write  ilie  niiiiibcrs  in  a  line  and  then  diride  hij  the  smallest  prime  number 
that  irill  diride  two  or  more  without  a  remainder,  and  write  the  quotients  and  undi- 
vided numbers  in  a  line  beloic. 

2.  Continue  this  process  of  diriding,  until  there  are  no  tico  numbers,  with 
common  factors,  or  that  can  be  divided  by  any  number  (jreater  than  1. 

3.  Find  the  continued  product  of  the  divisors  and  the  numbers  in  the  last  line, 
and  it  will  be  the  L.  C.  M. 

PROBLEMS. 

'2.     What  is  the  least  common  multiple  of  4,  0,  12,  1.1,  and  24 1       Aiis.  3G0. 
What  is  the  least  common  multiple  of  the  following  numbers: 

3.  Of  8,  4,  9,  and  30  ?  ^  Ans.     300. 

4.  Of  50,  27,  3,  45,  and  G3  ?       *  Ans.  9450. 

5.  Of  21,  3G,  11,  and  22  1  Ans.  2772. 
G.  Of  800,  GOO,  10,  40,  and  12  1  Ans.  2400. 

7.  Of  8,  18,  20,  and  70  ?  Ans.  2520. 

8.  A  drayman  has  2  drays  and  2  tloats.  On  1  dray  he  can  haul  9  barrels  of 
flour,  and  on  the  other  12  barrels ;  on  1  float  he  cau  haul  18  barrels,  and  on  the 
other  21  barrels.  What  is  the  least  number  of  barrels  that  will  make  full  loads  for 
either  of  the  drays  or  the  floats?  Ans.  252. 

0.  A  fruit  dealer  desires  to  invest  an  equal  amount  of  money  in  oranges, 
peaches,  and  grapes,  and  to  expend  as  small  a  sum  as  possible.  The  price  of 
oranges  is  $2.40  per  box;  peaches,  $1.00;  and  grapes  for  a  medium  article,  90c'., 
and  for  first  quality,  $1.20;  of  these  two  qualities  the  fruit  dealer  took  the  cheaper. 
IIow  much  more  money  did  he  invest  than  he  Avoukl  have  done  had  he  taken  the 
grapes  at  $1.20  per  box?  ■  Ans.  $28.80. 

PARTIAL     OPERATION. 

L.  C.  M.  of  $2.40,  $1.00,  .90  =  $14.40. 

$14.40.  X  3  (kinds  of  fruit)  =  $43.20  spent  by  purchasing  grapes  ®  90/. 
L.  G.  M.  of  $2.40,  $1.00,  $1.20  =  $4.80. 

$4^0.  X  3  =  $14.40,  what  he  would  have  spent  by  taking  grapes  'S)  $1.20. 

$43.20  — $14.40  =  $28.80,  Aus. 

SYNOPSIS  FOB  RBVIEW  OF  FACTORING 


NoTK. — The  numbers  before  the  words  and  phrases,  refer  to  the  articles  treating  the  subjects 
named. 

292.     Define  the  following  words  and  phrases  : 
282.     What  is  Factoring  ? 
285.     A  Common  Factor. 

288.    General  Directions  for  Finding  Greatest  Common  Divisor. 
291.    General  Directions  for  Finding  Least  Common  Multiple. 


!/anoellation. 

5  H-H-^^X-H^^--',-'^;-HX^X,-HH-i-'/-';;i^xN 


293.  Cancellation  is  the  process  of  shortening  the  operations  of  division,  or 
of  the  indicated  result  of  multiplication  and  division  operations  combined,  by 
rejecting  equal  factors  from  both  dividend  and  divisor,  or  trom  both  increasing  and 
decreasing  numbers,  of  an  indicated  result. 

The  operation  is  performed  by  drawing  a  line  across  each  factor  cancelled^  or 
cut  out ;  thus,  p,  f,  2-;5. 

294.  The  Principles  of  Cancellation,  are  : 

1.  Rejecting,  or  cancelling  a  factor  from  any  niimber,  is  in  effect  dividing  the 
number  by  that  factor. 

2.  Eejecting,  or  cancelling  equal  factors  from  both  dividend  and  divisor,  or 

from  both  increasing  and  decreasing  numbers  in   an  indicated  result,  does   not 

change  the  quotient  or  result. 

Note. — A  p;ood  unilerstanfling  of  factors,   multiples,  and  the  divisibility  of  numbers,  will 
greatly  facilitate  the  student  in  the  operations  of  cancellation. 

STATEMENT    LINE. 

295.  A  statement  line,  as  used  in  this  work,  consists  of  a  vertical  line  upon 
whose  right  and  left  sides  are  placed  respectively  the  increasing  and  decreasing 
numbers  of  the  problem.  This  statement  is  made  to  facilitate  the  operation,  as  will 
be  shown  in  the  problems  following. 

In  the  solution  of  all  practical  problems,  when  writing  the  numbers  on  the 
statement  line,  whether  increasing  or  decreasing,  a  reason  must  be  mentally  given 
for  each  number  written,  as  is  elucidated  in  the  l-'th  I'roblemof  Article  297,  follow- 
ing. See  pages  77  to  81,  and  125  to  128,  for  a  full  elucidation  of  the  reasoning  given 
in  the  solution  of  xn-oblems,  and  of  the  philosophic  system  presented  in  this  work. 

PKOBLEMS. 

Divide  7  x  3  x  -1  by  7  x  J:. 

Operation  bv  Cancellation.  Explanation. — In  all  problems  where  we  liave  both  nnil" 

^   ij  tiplication  and  division  operations  to  perform,  we  use  a  vertical 

,'  or  perpendicular  line,  which  we  call  the  statement  line.     This 

'-'  line  is  used  'io  facilitate  the  work  by  separating  the  dividends 

i  and  the  divisors,  or  the  increasing  and  the  decreasing  numbers. 

The  dividends,  or  the  increasing  numbers,  are  always   placed 

~  '  upon   the   right   hand   side    of  the   line,     and   the    divisors   or 

o,  Aus.  decreasing  numbers,  are  always  placed  upon  the  left  hand  side. 

In  this  examide,  having  written  the  numbers  that  constitute  the  dividend  and  the  divisor, 

respectively  upon  the  right  and  left  hand  sides  of  the  statement  line,  we  cut  out,  ov cancel,  the  enual 

factors  7's  and  4's  in  the  numbers  constituting  the  dividend  and  the  divisor,  and  thus  obtain  3   as 

the  answer  to  the  iiroblem. 

To  perform  the  work  without  the  aid  of  cancellation,  we  wonld  be  obliged   to  make   the 
following  figures :  7  X  3  =  21,  which  X  1  =  84,  the  dividend ;  then  7  X  4  =  28,  the  divisor ;  then 

28)84(3,  Ans. 
84 

(161) 


/ 


l62 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2, 

Multiply  25,  48,  a 

of  10,  30,  and  8. 

Operation  by  Caucellatioii 

3     ^)5 

n  5 
n  11 

110 


,  48,  and  88  together,  and  divide  the  product  by  the  product 


Ex2>laiiation. — In  this  example,  we  write  the  numbers 
on  the  line  as  above  directed  and  then  cancel  10  and  25  by 
5 ;  then  36  and  48  by  12 ;  then  8  and  88  by  8 ;  then  4  and  2  by 
2.  This  is  all  that  can  be  cancelled,  and  we  then  multiply 
together  5,  2,  and  11,  divide  the  product  by  3,  and  thus 
obtain  the  true  result,  36|. 


36§,  Aus. 
3.     Divide  the  product  of  32  x  3  by  S  x  9  x  IG, 


Operation  by  Cancellation. 


12 


?M 


1  =  -i\,  Ans. 


Exphmation. — Having  written  the  numbers  on  the 
statement  line,  we  first  cancel  8  and  32  by  8;  then  9  and  3 
by  3 ;  then  IG  and  4  by  4.  Now  having  no  more  numbers  on 
the  increasing  side  of  the  line  to  cancel,  we  multiply 
together  the  remaining  numbers  on  the  decreasing  side  of 

,1, 


Note. — In  all  cases  where,  after  cancelling,  no  factor  appears  on  either  side  of  the  .statement 
line,  the  factor  1  is  always  understood  as  being  tliere.  Its  niin-aiipearance  is  in  conaeiiueuce  of 
not  having  written  it  when  we  cancelled  a  number  by  itself,  as  in  tlio  following  problem. 

4.     Multiijly  14,  5,  and  3  together,  and  divide  the  product  by  the  ijroduct  of 

2, 15,  and  7. 

OPERATION. 


f 

0  n 


HI 


1  t 

or         \  f 


1,  Ans. 


Uf  1 

.3  1 


1  =  1,  Ans. 


Explaiintion. — In  this  problem,  we  first 
cancellecl  2  and  11 ;  then  5  and  15 ;  then  the  3 
and  3  ;  and  lastly  the  7  and  7.  The  work  might 
have  been  abridged  by  multiplying  the  3  and  5 
and  then  cancelling  bntli  15's,  and  tlien  multi- 
plying the  7  and  2  and  cancelling  both  14's. 


GENERAL    DIRECTIONS    FOR    CANCELLATION. 


296.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  cancellation : 

1.  Cancel  all  ihe  factors  common  to  both  dividend  and  divisor,  or  to  the  increas- 
ing and  decreasing  numbers. 

2.  Divide  the  jtvoduct  of  the  remaining  factors  of  the  dividend  by  the  product 
of  the  remaining  factors  of  the  divisor. 

Note. — Wlien  a  factor  cancelled  is  equal  to  the  number  itself,  the  unit  1  remains,  since  a 
nnmber  divided  by  itself  gives  1  as  .a  quotient.  If  the  1  is  in  the  dividend  it  must  be  retained;  if  in 
the  divisor,  it  may  be  disregarded,  since  dividing  by  1  does  not  change  the  quotient. 


297.     Cancel  and  work  the  following  line  statements : 

(1)  (2)  (3)  (4) 


(5) 


2 

G 

12 

45 

31 

124 

0 

76 

10 

16 

3 

9 

9 

56 

5 

17 

20 

91 

135 

45 

18 

54 

7 

84 

51 

10 

70 

140 

24 

00 

27 

IG 

4 

4 

25 

1,  Ans. 

70,  Ans. 

§,  Ans. 

19211  Ans. 

14, 

*  CANCELLATION.  1 63 

6.  Divide  the  product  of  6,  7,  12,  and  22  by  the  product  of  11,  3, 14,  and  8. 

Ans.  3. 

7.  What  is  the  quotient  of  28  x  05  x  7  x  78  ^  56  x  130  x  42  x  13  ? 

Ans.  1 

8.  Multiply  21,  55,  and  128  together,  and  divide  the  product  by  14  x  25  x  04. 

Ans.  6f. 

9.  How  many  bushels  of  corn,  at   70/  a  bushel,  will  pay  for  140  gallons 
molasses  at  65  cents  a  gallon  1  Ans.  130  bushels. 

10.  Bought  420  pounds  of  sugar  at  6  cents  a  pound,  and  gave  in  payment 
360  jiounds  of  rice.     What  was  the  price  of  the  rice  per  pound  !        Ans.  7  cents. 

11.  Sold  to  a  drayman  64  bushels  of  oats  at  75  cents  a  bushel,  for  which  he  is 
to  pay  in  drayage  at  50  cents  a  load.     How  many  loads  must  he  haul  ? 

Ans.  96  loads. 

12.  Paid  65/  for  5  yards  of  calico.    What  will  27  yards  cost  at  the  same  rate? 

Analytic  Solution  by  Explanafton. — In  all  practical   prohlems   of  this  kind, 

(1...  „„|i    ij ,,,  we  give  a  reason  for  each  step  of  the  operation,  and  make 

^^.uioeii.itiuii.  ^^^   whole  statement  to  indicate  the  linal  result  without 

6^  13  performing  any  of  the  intermediate  work.     In  this  problem, 

27  we  place  the  65c.  on  the  increasing  side  of  the  statement  line 

as  our  premise,  and  reason  thus:  5  yards  cost  65c.     Since  5 
yards  cost  65c.,  1  yard  will  cost  ^  part  of  it,  and  27  yards  will 


f- 


!.51,  Ans.  cost  27  times  as  much  as  1  yard. 


13.  A  merchant  sold  25  boxes  of  candles  containing  36  pounds  each  at  16/ 
per  pouud,  and  received  in  i>ayment  starch  at  0  cents  per  pound.  How  many  boxes, 
each  containing  30  pounds,  did  he  receive  ?  Ans.  80  boxes. 

14.  How  many  pounds  of  butter  at  35/  per  pouTid,  will  pay  for  245  pounds 
of  rice  at  5  cents  per  pound  1  Ans.  35  pounds. 

15.  If  17  barrels  of  flour  cost  $110.50,  what  will  500  barrels  cost  at  the  same 
rate?  Ans.  $3250. 

16.  Sold  340  boxes  of  soap,  containing  45  i)ounds  each,  at  12/  per  pound, 
and  received  in  ])ayment  starch  at  9/  per  pound.  How  many  boxes,  each  contain- 
ing 24  pounds,  did  I  receive  1  Ans.  850  boxes. 


SYNOPSIS   FOR   BE  VIEW    OF    CANCELLATION. 


Note. — The  numbers  before  the  words  and  phrases,  refer  to  the  articles  treating  the  subjects 
named. 

298.     Define  the  following  words  and  phrases  ? 

293.  Cancellation. 

294.  Principles  of  Cancellation. 

295.  Statement  Line. 

290.    General  Directions  for  the  Operation. 


s^^^ 


raotions. 


^ 


299.  According  to  the  general  methods  prescribed  by  the  text  books  of  our 
country,  for  the  various  operations  of  fractions,  the  subject  is  correctly  considered 
by  both  the  teacher  and  the  pupil  to  be  the  most  diflicult  in  the  science  of  iiiimbers. 
But  by  our  fully  evolved,  logical,  and  philosophic  system  of  handling  fractional 
uumbcrs,  the  subject  is  simplitied,  rationalized,  and  rendered  pleasing  to  the  student. 
By  our  method  of  "work,  fully  onc-hul/oi  the  time  and  figures  required  by  the  ordi- 
nary methods  is  saved,  and  all  the  arbitrary  and  absuid  rules  which  overload  the 
organ  of  memory  and  prevent  the  expansion  of  the  higher  faculties  of  causality  and 
comparison,  are  abandoned  to  the  shades  of  the  dead  i)ast,  and  entombed  with  the 
ingenious  minds  which  gave  them  birth. 

By  our  system,  all  the  reasoning  faculties  of  the  mind  are  brought  into  action 
and  exercised  in  a  manner  to  give  logical  strength  and  acuteness  to  work,  not  only 
in  the  fields  of  mathematics,  but  upon  all  the  plains  of  life. 

In  behalf  of  truth,  education,  and  humanity,  we  lament  the  non-progressive 
methods  of  the  general  school  and  college  arithmetics. 

The  present  authors  of  school  arithmetics,  with  but  few  exceptions,  are  timidly 
following  in  the  obscure  paths  of  arithmetical  science  which  were  marlied  out  ages 
ago,  when  the  science  was  in  its  infancy — too  cowardly,  non-progressive,  and  con- 
tracted in  their  views  to  seek  and  explore  more  direct  and  comprehensible  routes  to 
the  fountains  of  mathematical  knowledge. 

300.  Tlie  Unit  is  the  universal  basis  of  numbers,  and  the  foundation  of  arith- 
metic. From  unity  arise  two  distin(!t  classes  of  numbers.  1.  Integers.  2.  Frac- 
tions. The  first  class,  integers,  has  its  origin  in  the  nraltiplication  of  the  unit; 
and  the  second  class,  fractions,  results  from  the  division  of  the  unit.  The  first  is 
synthetical,  the  second  is  analytical. 

DEFINITIONS. 

301.  A  Fraction  is  one  or  more  of  the  equal  parts  of  a  unit,  or  of  a  collection 
of  units.  Or  more  briefly,  it  is  a  part  of  anything,  or  a  numerical  expression  of 
a  i)art  of  a  unit;  thus,  1  half  and  3  fourths  are  fractions. 

302.  A  Fractional  Unit  is  one  of  the  equal  parts  into  which  any  integral 
unit  is  divided.  If  the  integral  unit  is  divided  into  two  equal  parts,  each  is  called 
a  half;  if  into  three,  ea('h  is  called  a  third  ;  if  into  four,  each  is  called  a  fourth  ; 
and  so  on,  according  to  the  number  of  parts  into  which  the  integral  unit  is  divided. 

303.  Fractions  are  divided  into  two  kinds.  Common,  or  Yulgar,  and  Decimal 
Fractions. 

(164) 


*  FRACTIONS.  .  1 65 

304.  Common  Fractions  are  expressed  by  two  numbers,  one  written  above 
the  other,  with  a  horizontal  line  between  them.  The  number  below  the  line  is  called 
the  Denominator,  and  the  number  above  the  line  is  called  the  Numerator.  Thus, 
^  {one-half),  f  (three-fourths),  f  (five-sixths)^  f  (seven-eighths),  and  }}  (thirteen-seven- 
feenths),  are  common  fractions,  the  denominators  of  which  are  respectively  2,  4, 
G,  8,  and  17.  The  numerator  and  denominator  together,  are  called  the  terms  of  the 
fraction. 

305.  The  Denominator  of  a  fraction  shows  the  number  of  equal  parts  into 
which  the  unit  is  divided.  Thus  in  the  fraction  |,  the  8  is  the  denominator  and 
shows  that  the  unit  is  divided  into  8  equal  parts  called  eighths. 

306.  The  Numerator  of  a  fractiou  shows  the  number  of  equal  parts  taken  to 
form  the  fraction. 

Thus  in  |,  the  numerator  is  5  and  shows  that  5  of  the  S  equal  parts  are  taken, 
or  exiiressed,  by  the  fraction. 

All  fractions  arise  from  division  and  are  expressions  of  unexecuted  division 
ill  which  the  numerator  is  the  dividend,  the  denominator  the  divisor,  and  the  fraction 
itself  the  quotient. 

307.  Decimal  Fractions  are  those  in  which  the  denominators  are  not  generally 
expressed,  but  are  always  10,  or  some  power  of  ten;  thus,  .5,  .75,  .821,  read  respec- 
tively five  tenths,  seventy-five  hundredths  and  eight  hundred  twenty-one  thousandths, 
are  decimal  fractions.  To  write  these  fractions  as  common  fractions,  tbey  would  be 
■written  thus,  -■%  -^,  and  -B-s^^. 

The  point  (  .  )  placed  before  the  5,  7,  and  8,  in  the  above  decimally  expressed 
fractions,  is  called  the  decimal  point,  aud  is  used  to  abbreviate  the  work. 


CLASSIFICATIOiSr   OF    FRACTIONS. 

308.  For  convenience,  fractious  are  classed  under  the  following  heads:  Proper 
Fractions,  Improper  Fractions,  ISimple  Fractions,  Mixed  Numbers,  Compound  Frac- 
tions,  and  Complex  Fractions. 

309.  A  Proper  Fraction  is  one  in  which  the  numerator  is  less  than  the 
denominator;  as  i,  |,  g. 

310.  An  Improper  Fraction  is  one  in  which  the  numerator  is  equal  to  or 
greater  than  the  denominator;  as,  *,  -j,  -g,  and  J-/. 

311.  A  Simple  Fraction  is  one  in  which  both  terms  are  whole  numbers,  and 
may  be  either  a  proper  or  an  improper  fraction  ;  as,  |,  ^,  W,  or  ^^. 

312.  A  Mixed  Number  is  a  number  composed  of  a  Avhole  number  and  a 
fraction;  as,  2^,  5a,  and  1l-h- 

313.  A  Compound  Fraction  is  a  fractional  part  of  a  fraction  or  mixed 
number;  as,  f  of  |,  J  of  -^f^^  of  12a. 


1 66  soule's  philosophic  practical  mathematics.  * 

311.     A  Complex  Fraction  is  one  that  has  one  or  more  of  its  terms  fractional ; 
as, 

f  CJ  3  I  GJ 

of        — ,  _        of        —        of        — 

f  5f  I  IJ  8 

315.     The  Reciprocal  of  a  Fraction  is  the  result  of  1  divided  liy  the  fraction. 
Thus,  tlie  recii)roeal  of  |  is  1  4-  -5-  =  f  =  IJ. 

31(5.     The  Yalue  of  a  Fraction  is  the  result  of  its  numeratoi-  divided  by  its 
denominator.    Thus,  f  =  4,  ^^  =  2f . 

GEXEEAL  PEIXCIPLES  OF  FKACTIOXS. 

317.     1.     Multiplying  the  numerator,   or  dividing  the  denominator,  multiplies 
the  fraction. 

2.  THviding    the   numerator,   or   multiplying  the    denominator,    divides  the 
fraction. 

3.  Multiplying  or   dividing  both   numerator  and   denominator  by  the  same 
number  does  not  change  the  value  of  the  fraction. 


SYNOPSIS  FOB  ME  VIEW  OF  FB  ACTIONS. 


Note. — The  numbers  before  tbe  words  anil  }ibrases,  refer  to  the  articles  treating  tlie  subjects 
named. 

318.     Define  the  following  words  and  phrases  : 

301.  A  Fraction.  310.  An  Improper  Fraction, 

302.  A  Fractional  Unit.  311.  A  Simple  Fraction. 

303.  How  Fractions  are  Divided.  312.  A  Mixed  Xumber. 

304.  Common  Fractions.  313.  A  Compound  Fraction. 

305.  Denominator.  314.  A  Complex  Fraction. 

306.  Numerator.  315.  The  Keciprocal  of  a  Fraction. 

307.  Decimal  Fractions.  316.  Value  of  a  Fraction. 

308.  Classification  of  Fractions.  317.  General  Principles  of  Fractions. 

309.  A  Proper  Fraction. 


Seduction  of  Fractions. 


319.  Reduction  of  Fractious  is  the  process  of  changing  their  form  mthout 
altering  their  value. 

320.  A  fraction  is  reduced  to  Higher  Terms  when  the  numerator  and  the 
denominator  are  expressed  in  hirger  numbers.      Thus,  A  =  f?  or  s>  or-/j;  §  =  1,  or 

321.  A  fraction  is  reduced  to  Lower  Terms  when  the  numerator  and  the 
denominator  are  expressed  in  smaller  numbers.     Thus,  1%  =  |,  or  §;  ||  =  a^  or  J. 

322.  A  fraction  is  reduced  to  its  Lowest  Terms  when  its  numerator  and  its 
denominator  are  jirime  to  each  other,  or  have  no  common  divisor.  Thus,  |,  J,  and 
IJ,  are  in  their  lowest  terms. 

323.  TMiole  Numbers  may  be  reduced  to  fractions  having  any  desired  denom- 
inator. 


Whole  line.  Half  lines. 


Third  lines.  Fourth  lines. 

ORAL  EXERCISES. 

324.     1.    If  a  line,  an  orange,  an  apple,  or  a  unit  of  any  kind  is  divided  into 
two  equal  parts,  what  is  each  iiart  called  ?  Ans.  J. 

2.  If  divided  into  three  equal  parts,  what  is  each  part  called  1         Ans.  J. 

3.  If  divided  into  four  equal  parts,  what  is  each  jiart  called  ?  Ans.  i. 

i.     When  divided  into  four  equal  parts,  what  are  three  of  those  parts  called  ? 

Ans.  5. 
5.     How  would  you  get  ^  of  an  apple  1 

Ans.  Divide  it  into  4  equal  parts  and  take  3  of  the  parts. 
C.    When  any  number  or  thing  is  divided  into  five  equal  parts,  what  is  one 
of  those  parts  called  ?  Ans.  -g. 

7.  What  are  two,  three,  and  four  of  the  parts  called  respectively  ? 

Ans.  i,  h  and  •*. 

8.  1  unit,  abstract,  or  denominate  of  any  kind,  equals  how  many  halves? 
thirds?  fourths?  fifths?  sixths?  sevenths?  eighths?  ninths? 

(167J 


i68 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


9.  2  =  Low  many  halves  ?  thirds  ?  fourths  ?  fifths  ?  sixths  ?  sevenths  ?  eighths? 
ninths  ? 

10.  3  =  how  many  halves?  thirds?  fourths?  fifths?  sixths  ?  sevenths  ?  eighths? 
ninths  ? 

11.  4  =  how  many  halves  ?  thirds?  fourths  ?  fifths?  sixths?  sevenths  ?  eighths  ? 
ninths  ? 

12.  5  =  how  many  halves  ?  thirds  ?  fourths  ?  fifths  ?  sixths  ?  sevenths  ?  eighths  ? 
ninths  ? 

13.  G  =  how  many  halves  ?  thirds  ?  fourths  ?  fifths?  sixths  ?  sevenths  ?  eighths  ? 
ninths  ? 

What  kind  of  numerical  Avork  is  the  above  called  ? 

325.    i  =  how  many  fourtlis  ?  sixths  ?  eighths  ?  tenths  ?  twelfths  ?  fourteenths  ? 

^  =  howmanysixths?iiintbs?twelfths?fifteenths?eighteenths?  twenty-firsts? 

J  =  how  many  eighths  ?  twelfths  ?  sixteenths  ?  twentieths  ?  twenty-fourths  ? 

1  =  liow  many  tenths  ?  fifteenths  ?  twentieths  ?  twenty-fifths  ?  thirtieths  ? 

1^  =  how  many  sixteenths  ?  twenty-fourths  ?  thirty-seconds  ?  fortieths  ?  sixty- 
fourths  ? 

-jL  =  iiow  many  thirty-seconds  ?  forty-eighths  ?  sixty-fourths  ?  eightieths  ? 

What  kind  of  numerical  work  is  the  above  called  ? 


326.     Answer  the  following  numerical  questions  : 


2     _ 

4      - 

t  = 

-1%  = 

=  ? 
=  ? 
=  ? 

f  =? 

-o\  =   ? 

if 
it 

=  ? 
=  ? 
=  ? 
=  ? 

X4 
1.'  8 

4  5 

=  ? 
=  ? 
=  ? 

=:? 

To3 

= 

? 

? 

? 
? 

-i¥o-  =  ? 

W  hat  kind  of  numerical  woi 

kis  tl 

le  above  ci 

ill 

ed? 

327. 

How  many  halves      =  a  unit  ? 

How  many  eighths        =  a 

unit  ? 

thirds      =       " 

u 

ninths          = 

fourths    =       " 

ii 

tenths          = 

fifths        =       " 

ii 

elevenths    = 

sixths     =       " 

ii 

twelfths       = 

sevenths  =       " 

ii 

thirteenths  = 

1  unit  equals  how  many  eighths? 

8     —    ? 
8     • 

n 

units 

equal  he 

w  many  thirds? 

A=? 

1 

u          a 

"             twelfths  ? 

ii=? 

n 

u 

"            sixths  ? 

J-/=? 

3  units  equal 

"             tliirds  ? 

t=? 

if 

u 

"            eighths  ? 

i/=? 

4 

a          u 

"             fourths  ? 

iS.  =  ? 

3i 

(( 

'•            fourths  ? 

J-/=? 

5 

a          u 

"             halves? 

12=1 

2S 

u 

"            sixteenths  ?f|=? 

n 

a          u 

"             halves? 

i   =? 

4i 

a 

"            eighths  ? 

¥=? 

n 

ii          a 

fourth 

s? 

i=? 

328.  What  is  the  reciprocal  of  1,  of  2,  of  3,  of  J,  of  a,  of  2| 

329.  Analyse  the  fraction  f. 

^4yiah/Ki,<!. — 5  is  a  proper  fraction,  sinro  the  numerator  is  Icsa  than  the  <lcnominator;  4  is  th© 
denominator,  and  shows  that  the  unit  is  divideil  into  4  equal  parts;  J- is  the  fractional  nnit,  since 
it  is  ONK  of  the  four  equal  parts  into  ivhich  the  unit  is  divided;  S  is  the  numerator  and  shows  that 
three  of  these  equal  parts  are  taken ;  3  and  4  are  the  terms  of  the  fraction,  and  its  value  is  less 
than  1,  or  unity. 


In  like  manner,  analyse  the  following  fractions :    |, 


12.     JLi     a     Al 

4    )     3  2)     6)      1  5) 


REDUCTIOX    OF    FRACTIONS.  1 69 

TO    llEDUCE    niACTIOXS    TO    HIGHER    TERMS. 
3:30.     1.     Cliange  15  to  a  fraction  whose  denominator  is  64. 

(IPERATION. 
64  4-  10  =  4  ExpJanailon. — In  all   problems  of  tins  kind,    we    first   (liviile 

the   required   denominator   by  the   denominator   of  the   given 
13  X  4  :=  52  fraction.    Then  -with  the  quotient  thus  obtained,  multiply  both 

—  Alls.  terms  of  the  given  fraction,  &nd  in  their  products  ■vre  have  the 

10  X  4  =  04  required  fraction. 


GEXERAL    DIRECTION    FOR    REDUCIXG    FRACTIONS    FROM    LOWER 

TO  HIGHER  TERMS. 

331.  From  tlie  foregoing  elucidations,  we  derive  tlie  following  general  direction 
for  reducing  fractions  from  lower  to  higher  terms  : 

Divide  the  required  denominator  by  the  denominator  of  the  given  fraction,  and 
multiply  both  terms  of  the  fraction  by  the  quotient. 

PROBLEMS. 

332.  Change  ji  to  a  fraction  whose  denominator  is      75 

"  AJto  "  "  "  176 

"  fito  "  "  "  384      , 

"        -iifVto  "  "  "  2180 


TO  REDUCE  FRACTIONS  TO  THEIR  LOWEST  TERMS. 

333.    Reduce  ff  to  its  lowest  terms. 

FIKST    OPERATION,  Explanation.— la  all  problems  of  this  kind,  vpe  divide 

o\5.e  2  8.  /i\28  7    A  i.Q  both  the  numerator  and  the  denominator  by  their  common 

factors.     Or,  aa  sho'n-n  in  the  second  operation,  we  may 

SECOND   OPERATION.  produce  the   same  result  with  less  figures,   by  dividing 

8)f  f^  =  H  Ans.  both  terms  of  the  fraction  by  their  greatest  common  divisor. 

By  this  reduction  we  change  the  form  of  the  fraction  f  f ,  but  we  do  not  alter, 
or  change,  its  Aalne,  for  the  fractional  unit  of  the  resulting  fraction  (J)  is  8  times  as 
great,  while  the  number  taken  is  J  as  great. 

When  the  terms  of  the  fraction  have  no  common  factor  greater  than  1,  the 
fraction  is  in  its  lowest  terms  and  is  called  an  irreducible  fraction. 

The  object  of  reducing  fractions  to  their  lowest  terms  is  to  enable  ns  to  understand  their 
value  more  easily  and  readily. 


I70  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

GENERAL    DIltECTION    FOK     EEDUCIXG     FliACTIOXS    TO    THELB 

LOWEST  TEKMS. 

334.  From  the  foregoing  elucidations,  we  derive  the  following  general  directiou 
for  reducing  fractions  to  their  lowest  terms : 

Cancel  all  ilie  factors  common  to  both  the  numerator  and  the  denominator  j 
Or,  divide  both  numerator  and  denominator  by  their  greatest  common  divisor. 

PROBLEMS. 

335.  Reduce  the  following  fractions  to  their  lowest  terms. 


lA     15.     S^i.     AS- 
■^S7     »5J     7aj     80? 

Ans. 

2)     7)     ij    40- 

7. 

fl72 

TTiTS) 

3  2  4.       2^3  8        2  3.1 
7  J  0?     3  00?     t)aa? 

Alls. 

-fo,  i,  i¥r- 

8. 

imh 

liliii 

2  620) 

Alls,      f. 

9. 

5 IS40> 

624  0 
1  5  U  JS} 

Ans.  tM- 

10. 

S575 
4  7  10? 

224 

4ao> 

Aus.    15. 

11. 

1  flS4  S 
S4992? 

*\  ^"4.       2^58        2:^1  ^1^a  ^4  77  C  13915  A  n  e      ^Z^A 

*^*  7li09     3G05      Uii;H  -rxiIO.        o  0  9      n1     S  1   1  ■  *-'•  Q  O  0.1  0  9  XXUO.     4  0  0  6* 

25. 
33' 

13. 
27' 


4.  im,  Alls.      f.  9.    leH,  Ans. 

5.  m^%  Ans.  M.  10.    mi,  Ans. 

6ejM  A  iisj        J.R  11  J.fi « IS  Ana 


TO  REDUCE  WHOLE  OR  ]V[IXED  NUMBERS  TO  IMPROPER 

FRACTIONS. 

336.     1.     Reduce  5§  to  an  imiiroiier  fractions,  or  to  thirds. 

OPERATION. 

58  l^xplanalion. — In  .ill  prolilpins  of  tins  Icind,  yre  rpason 

thus:  SiiKte  there  are  S  thirds  in   every   unit   or  whole 

,Y    ,  number,  in  5  units  there  are  5  times  as  many,  which  are 

-3-  All S.  i^3  _(.  the  J  make  \' . 

2.     Reduce  9  to  a  fraction  whose  denominator  is  6. 

OPERATION. 

9  X  C  =  \i  Ans. 


GENERAL    DIRECTION    TO    REDTXE    WHOLE    OR    MIXED    NUMBERS 

TO  IMPROPER  FRACTIONS. 


337.     From  the  foregoing  elucidations,  we  derive  the  following  general  direction 
for  reducilig  whole  or  mixed  numbers  to  improper  fractions  : 

Multiply  the  irhoJe  number  by  the  reqtiired  denominator  and  to  the  product  add 
the  numerator  of  ilie  fraction,  and  write  the  required  denominator  tinder  the  result. 


REDUCTION    OF    FRACTIONS. 

PROBLEMS. 
Eeduce  the  following  numerical  expressions  to  iiuxjroper  fractions : 


171 


3. 

H 

4. 

16J 

5. 

17f 

6. 

32i 

7. 

435| 

13. 

14. 

14. 

Ans. 

25 
3    • 

Ans. 

3Ji 

Ans. 

7  1 
4    • 

Ans. 

2111 

Aus. 

2  1  7  s 
5     • 

9. 


71* 


10.  21S3J 


11. 
12. 


03. 3_ 

losJij 

Rednce  14  to  a  fraction  whose  denominator  is  9. 
"  37  "  "  "  "  24. 

"         543  "  "  "  "  16. 


Ans. 

Ans. 
Ans. 
Ans,      -a/a 
Ans.  HffJ- 


±xa. 

7  • 

87:1  5 


TO  EEDUCE  IMPROPER  FRACTIONS  TO  WHOLE  OR  MIXED  ETTINrBEES. 


338.     Reduce  -f-  to  a  mixed  number. 


OPERATION, 

Jf  =  44  Ans. 
or 
17  -^  4  =  4J  Aus. 


ExjjJattation. — lu  all  problems  of  this  kind,  we  reason  thus : 
Since  there  are  4 /o«r(7is  in  limit,  or  whole  number,  in  17  fourths 
there  are  as  many  units  as  17  is  equal  to  4,  which  is  4  times  with  1 
remainder,  or  altogether  4^  as  the  proper  quotient,  or  answer. 


GENERAL    DIRECTIONS    TO    REDUCE     IMPROPER    FRACTIONS    TO 
WHOLE    OR    MIXED    NUMBERS, 


339.  From  the  foregoing  elucidations,  we  derive  the  following  general  direction 
for  reducing  improper  fractious  to  whole  or  mixed  numbers  : 

Divide  the  Numerator  hy  the  Denominator, 
PEOBLEMS. 

340.  Reduce  the  following  improper  fractions  to  whole  or  mixed  numbers : 


2, 

3, 

.4, 

5. 


ii..t 


222 
12 


Ans. 

4. 

6 

Ans, 

5|. 

7 

Ans. 

48. 

8 

Ans. 

18J. 

9 

1  :> 
4.sr! 

5  1' 
271H  0 

ao9 


Ans,    3§. 
Ans,    5/5-, 
Ans.    9i\, 
Ans.  34f2-|. 


172  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

TO  REDUCE   COMPOUND  FEACTIOXS  TO   SIMPLE  FRACTIONS. 
341.     1.    Reduce  i  of  f  of  ^  to  a  simi)le  fraction. 

OPERATION.  ICxplanatioii. — In  all  proMema  of  this  kind,  ivo  iiiiiltii>ly  togotUnr 

,T-   oj^       .      ^  all  the  numerators  fiir  a  new  numerator  and  all  the  deuomiuators 

2  X   -5-  X    g         vo     -'i-ll*»'        fur  a  new  denominator. 

When  a  comiiouud  fraction  contains  whole  or  mixed  numbers, 
they  mnst  first  be  reduced  to  improper  Cractions. 

When  there  are  common  factors  in  both  terms  of  a  compound  fraction,  they  should  be  can- 
celled before  mnltii)lyiMj;.  Jiy  i-aueelliuii  the  coiuuioii  factors,  the  woilv  is  shorteued  and  the  result 
nnclianjied  for  the  reason  that  di\'i<liug  l)oth  terms  of  a  fraetiou  l)y  the  same  number  does  not  alter 
its  value. 

2.     Reduce  f  of  7.J  of  j  of  4  of  -i\r  to  a  simple  fraction. 

OPERATION. 

3 

?       U       '^       :i        1         1 
-X—  X-X-X  —  =  -     Alls. 

^     ^     ^     1    y^     -d 

?  3 

GENERAL    DIRECTIONS    TO    REDUCE    COMPOUND    FRACTIONS     TO 

Si:\IPLE    FRACTIONS. 

3-42.     From  tlie  foregoing  elucidatiotis,  we  derive  tlie  following  general  direction 
for  reducing  compound  fractions  to  simjile  fractions  : 

Cancel  common  factors  if  they  occur  in  loth  terms  of  the  fractions  ;  then  multi- 
ply the  numerators  together  for  the  new  numerator  and  the  denominators  together  for 
the  new  denominator  of  the  fraction. 

PROBLEMS. 

343.     Reduce  the  following  compound  fractions  to  simple  ones : 

1.  Reduce  §  of  ^  of  |  to  a  simple  fraction.  Ans.  -^j. 

2.  Reduce  f  of  7J  of  |  of  4  of  -i\r  to  a  simple  fraction.  Ans.  J. 

3.  Reduce  §  of  2J  of  v-  of  f  of  ]  3  to  a  simple  fraction.  Ans.  |^. 


4.  f  of  ii  of -i\.  Ans,     fj. 

r..  ^  of   §  of  -fu.  Ans.     -^. 

0.  A  of  3 J  of  |.  Ans.     1;^, 

7.  i  of  8^.  Ans.  2  jif. 


8.  f,-  of  00.  Ans.  54. 

9.  f  of  11  of  17i.  Ans.     6. 

10.  §  X  f  X  t{.  Ans.  -^a,, 

11.  5  X  2  X  A-  Ans.  J-a 


TO  REDUCE  FRACTIONS  OF  DIFFERENT  DENOMINATORS  TO   EQUIV- 
ALENT FRACTIONS  OF  A  COMMON   DENOMINATOR,  OR  OF 
TUB  LEAST  COMMON  DENOMINATOR. 

344.  A   Common   Denominator  is  a  denominator  common  to  two   or  more 
fractions. 

345.  The  Least  Common  Denominator  of  two  or  more  fractions  is  the  least 
number  divisible  by  each  of  tlie  denonunators. 


REDUCTION    OF    FRACTIONS. 


173 


346.  A  Common  Denominator  of  two  or  more  fractions  is  a  common  imdtiple 
of  their  dcuoiiiiuators;  and  the  hast  common  denominator  of  two  or  more  fractions 
is  the  least  common  multiple  of  their  denominators,  for  the  reason  that  all  higher 
terms  of  a  fraction  are  multiples  of  its  corresponding  lower  or  lowest  terms. 

347.  Eeduce  ^,  f ,  and  J,  to  equivalent  fractious  having  a  common  denominator. 

OPERATION. 

13    1 

3)   41    8J 

3  X  J:  X  S  =  9G,  common  denominator. 

J  of  ^  =  31! ;  hence  ^|,  equivalent  of  j^. 
I  of  ^  90=  7li;  hence  1%  equivalent  of  ^. 
I  of   )       =  84 ;  hence  f  i,  equivalent  of  ^, 

ICxpJanalion . — In  all  proljlems  of  this  Itind,  we  obtain  the  common  denominator  by  mnltiplying 
togetliLT  the  denominatois  of  all  the  fractions.  Then  to  find  the  respective  numerators  we  take 
snch  a  part  of  the  common  denominator  as  the  resjiective  fractions  are  parts  of  a  nnit,  as  shown  in 
the  operation. 

Or,  divide  the  common  denominator  by  the  denominator  of  each  fraction,  and  multiply  the 
quotient  bj'  its  numerator. 


GENEEAL    DIEECTIOXS    TO     EEDTTCE    FEACTIOXS    TO     A    CO]VniON 

DEXOMINATOE. 


348.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  reducing  fractions  to  a  common  denominator: 

1.  Multiply  together  the  denominators  of  all  the  fractions  for  a  common 
denominator. 

2.  Then  to  find  the  respective  numerators,  tal;e  such  a  part  of  the  common 
denominator  as  the  respective  fractions  are  parts  of  a  nnit.  Or,  divide  the  common 
denominator  hy  the  denominator  of  each  fraction,  aiid  mitlti^dy  the  quotient  hy  its 
numerator. 

PROBLEMS. 

349.  Eeduce  the  following  fractions  to  equivalent  fractions  having  a  common 
denominator : 

1.    f ,  §,  and  f.  Ans.  -^^^,  ttJV,  and  -1%. 

o      .fl     1    r,,id  -S-  Ans    ^J-fi   ISO    n.,fi  J.60 

"•        T"5»    aUU   5f2'  -/VUS.  4807    "U"    4a0" 

A       1     s      .a     11    Qiirl  1  Aii<j    XXs.2.0    _72on_    203,3.0.    S.XI.ZO    and  -Sfilil- 

*•      2>   nf;   10?   12;  •tnii  ij.  .a-ua.    23040>  230405  23040)  23040;  ""^"^  23040" 

"•      17;  f;  4)  ana  o^.  .^us.    1224;  1224;  1224;  ""'-'^  i22i' 


174  SOULES    PIIII.OSOPIIIC    PRACTICAL    MATHEMATICS.  * 

350.     Reduce  ^,  |,   iiud  J  to  equivalent  fractions  having  tlie  least  common 
dcnominaior. 

OPERATION. 

li     3.  4.  8 


;?.  2.  4 

3.  1.  2 

2x2x3x2  =  24,  least  common  denominator. 
|-  of   ^         =    8,  hence  -^  is  the  equivalent  of  -J-. 
I  of   >  24  =  18,  hence  ^f  is  the  equivalent  of  ^. 
I  of  )        =  21,  hence  f  |  is  the  equivalent  of  %. 

Explanation. — In  all  problems  of  this  kind,  we  first  find  the  least  common  multiple  of  the 
denominators  of  all  the  fractious  as  explained  in  Article  290,  page  159,  which  is  the  least  common 
denominator.  Then,  having  the  least  common  denominator,  to  iind  the  respective  nnmerators,  we 
take  such  a  part  of  the  least  common  denominator  as  the  respective  fractions  are  parts  of  a  unit, 
as  shown  in  the  operation.  Or,  divide  the  L.  C.  B.  by  the  denominator  of  each  fraction  and  multiply 
the  quotient  l>y  its  numerator. 

Before  finding  the  L.  C.  D.,  reduce  mixed  numbers  to  improper  fractions,  and  the  fractions  to 
their  lowest  terms. 


GENERAL     DIRECTIONS     TO     REDUCE     FRACTIONS     TO    A     LEAST 

COMMON    DENOMINATOR. 

351.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  reducing  fractions  to  the  least  common  denominator. 

1.  Find  the  least  common  multiple  of  the  denominators  of  all  the  given  fractions. 

2.  Then  to  find  the  respective  numerators,  tal;e  sucli  a  part  of  the  least  common 

denominator  as  the  respective  fractions  are  parts  of  a  unit.     Or,  divide  the  L.  C.  D.  by 

the  denominator  of  each  fraction  and  multiply  the  quotient  hy  its  numerator. 

Note. — Mixed  numbers  must  be  reduced  to  improper  fractions,    and  the  fractions  to  their 
lowest  terms  before  finding  the  least  common  denominator. 

PROBLEMS. 

352.  Reduce  the  following  fractions  to  equivalent  fractions  having  a  least 
conmion  denominator. 

1.  %,  t,  and  I.  Ans.  i%  |f ,  and  fi. 

2.  T^,  i,  and  -^.  Ans.  U,  U,  and  ff. 

6.    5^,  a,  1-,%  and  J.  Ans.  ^Z/-,  f f,  «f,  and  if. 

6.  ^f,  8,  It,  and  1.  Ans.  U,  W,  |f,  and  |f. 

7.  3J,  a,  2,  and  f.  Ans.  If,  -^,  -^,  and  if. 

8Sn_     Q      1      „,,J    15.  Ans      liaJl      5328      _44.4_     qrifl     1  665 

^     ft)   (t)_g)Jt)_^^ 


* 


Edition  of  Fractions. 


^^ 


353.  Addition  of  Fractious  is  the  process  of  adding  two  or  more  fractional 
numbers  of  the  same  kind,  or  of  the  same  denomination. 

351.     Like  Fractions  are  those  -which  express  like  parts  of  Ulce  units  or  things. 
Thus,  J  yard  and  ^  yard,  also  f  and  f  are  like  fractions. 

355.  Unlike  Fractious  are  those  which  express  unlike  parts  of  Uke  units  or 
things,  or  parts  of  nnlike  units  or  things. 

Thus,  §  of  a  pound  and  f  of  a  i)ound  are  unlike  parts  of  like  units  or  things, 
and  J  and  ^  are  unlike  parts  of  unlike  units. 

In  addition  of  whole  luimbers,  we  learned  that  we  could  not  add  apples  to 
oranges,  po^mds  to  boxes,  or  units  to  tens  or  hundreds  ;  that  we  could  only  add  things 
that  were  of  the  same  unit  kind. 

This  same  principle  maintains  in  the  addition  of  fractional  numbers.  We 
cannot  add  halves  to  thirds,  fourths  to  fifths,  etc.  We  can  only  add  halves  to 
halves,  fourths  to  fourths,  etc. 

ORAL  EXERCISES. 
356.     1.    Add  1,  f ,  and  a.  Ans.  U. 

SoLUTiox. — Since  the  fractional  parts  are  alike,  we  have  only  to  add  the  numerators  to  oljtain 
the  Slim  of  the  fractions.     Thus,  J  +  J  +  f  =  J  =  1},  or  li,  Ans. 

2.  Paid  $  J  for  a  grammar  and  $  J  for  an  arithmetic.     What  did  both  cost  t 

Ans.  $li. 

SoLt'Tiox. — Since  the  halves  and  fourths  are  unlike  parts  of  the  unit  dollar,  we  cannot  add 
them  in  their  present  form.  We  must  first  reduce  the  i  to  the  fractional  unit  of  the  fourths,  and 
bj'  the  exercise  of  our  reason  and  knowledge  of  numbers,  we  see  that  i  is  equal  to  j,  and  J  added  to 
I  equals  5,  and  f  equals  $1J,  Aus. 

3.  What  is  the  sum  of  f  and  f  ? 

Solution. — Since  the  fractional  parts  are  unlike,  we  cannot  add  them  in  their  present  form. 
We  must  first  reduce  the  f  to  the  fractional  unit  of  eighths  ;  and  by  the  exercise  of  our  reason  and 
knowledge  of  numbers,  we  see  that  f  is  equal  to  j,  and  f  added  to  f  equals  -y-,  and  -V- equals  If,  Ans. 

4.  What  is  the  sum  of  f ,  J,  -f-g,  and  H  ? 

5.  Add  f , -3^,,  ii,  and  f  f , 
Mentally  add  the  following  fractions : 

1  +  1=  ?  i+^,'=^ 

(175) 


'r1 

Ans.  2i|. 
Ans.  2fa. 

T^+A=? 

^+i+^=^ 

t  +M  =  ? 

i  +  i+-h^'! 

i  +  i  +  ^  = 

=  ? 

i+^+^=i 

§  +  f  +A 

=  1 

l  +  5+ll  =  « 

176 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


357.    What  is  the  sum  of  §  aud  I  ? 


Ans.  1-fj. 


OPERATION. 

s    9 

-ii-=  lA  Ans. 


Explanation. — Since  the  fractional  units  are  unlike,  and  since 
neither  can  be  retluiud  to  the  unit  value  of  the  other,  we,  there- 
fore, tirst  reduce  both  to  a  comiuou  denominator,  by  muUi2)lyiu!5 
their  denominators  tiipiethcr,  wliich  gives  12,  as  the  least  coniuiou 
denominator.  Havini;  tlie  least  common  denominator,  '\ve  next 
find  the  numerical  fraction.al  equivalent  value  of  f  and  f  iu  the  unit  of  twelfths.  To  do  this,  we 
reason  tlius  :  +  =  A  and  f  =  yj  ;  (write  the  8  under  the  5-  as  a  memorandum).     J  =  -^^  and  f  =  I'A; : 

•  -       ■!       17-1-.* 


PKOBLEMS. 

2.  Wliat  is  the  sum  of  f  and  |  ? 

3.  AVIiat  is  the  sum  of  f  and  -ft-  ? 

4.  Add  I  and  -jV ;  |  and  f ;  ^  and  J, 

358.    Add  J,  §,  and  I, 


Ans.  1^^. 
Ans.  1-3=-. 

Ana     .a5_.    12^a.    i-L 

Ans.  2-.}^. 


OPERATION. 

X 

^ 

; 

f 

16 

MEMOEAJn'DUM 

NITS. 

FllACTIO.NS. 

1 
1 

^       ^^       1 

A  iifi 

Explanation. — In  all  problems  having  more  than  two  terms,  we 
select  and  add  only  two  terms  at  a  time.  Here  we  select  the  i 
and  ^  and  reason  tliiis  ;  i  =  i,  which  added  to  f  =  -y-  =  If.  AVe 
set  the  1  in  the  colnmu  of  units  and  the  |  in  the  colunm  of  mem- 
moranduin  fractions,  and  caucel  the  terms  added,  4  and  |.  AVe 
now  have  the  §■  and  f  to  add,  and  ))roducing  tlie  least  eonmion. 
denominator,  24,  we  reason  thus :  i=:  A  aud  f  =  ij.  (Write  the 
16  under  the  f  as  a  memorandum).  Then  ^  =  -;V  and  f  =  j^lf. 
(Write  the  9  under  the  |  as  a  memorandum).  Tlieu  i'}  and  i\  =  " 
1}  =1/4.  We  write  the  1  in  the  colunm  of  units  and  the  -./j-  in  the 
column  of  memorandum  fractions.  We  then  cancel  the  |J,  add  the 
column  of  units  and  to  the  sum  annex  the  remaining  fraction  and 
obtain  2j-f,  the  answer. 


Add  the  following : 

2.    i,  I,  I,  and  ^„ 

'^       J.     ji    ^-. 

'-'•       "1  a'7    7)    U) 

4.     f ,  f ,  and  il, 
359.    Add  i,  §.  a,  4,  f ,  and  |  together, 

OPERATION. 


Ans.  2-AV 
Ans.  lx|« 
Ans.    2?-f. 


Ans.    4^. 


Explanation. — .Selecting  the  ^  and  f  to  add  first,  we  reason  thus  : 
•J  :=  I  which  added  to  f  =  f  =  1^^.  We  write  the  1  iu  the  column 
of  units  and  the  J  iu  tlJe  column  of  memoranduui  fractious  and  can- 
cel the  J  aud  J.  Next  we  select  the  f  and  §  aud  reason  thus :  ^  =  f , 
and  i  =  i  wliich  added  to  £-  =  §  =  i;|  =  li.  We  write  the  1  in  the 
colnmu  of  units  aud  the  i  in  the  column  of  memorandum  fractions, 
and  cancel  tlie  f  aud  |.  Then  we  select  the  i  and  the  i  ( in  the  col- 
umn of  memorandum  fractions)  andreasou  thus:  -J^fwhich  added 
to  J  =  f ,  We  set  the  Jin  tlie  column  of  memorandum  fractions 
and  cancel  the  i  and  i.  Next  we  select  the  f  and  the  J  and  reason 
thus :  J  =  I  and  f  =  I ,  which  added  to  |  =  \^  =  If.  We  write  the 
1  iu  the  unit  column  and  the  f  in  the  column  of  memorandum  frac- 
tions, and  cancel  the  J  and  f .  |  and  f  remain  to  be  added ;  and  as 
they  are  not  of  the  same  imit  value  and  neither  can  be  reduced  to 
the  unit  value  of  the  other,  we  proceed  to  reduce  both  to  eriuivalent  fractions  of  the  same  unit,  orto 
the  least  common  denominator,  which  is  40.  Having  the  least  common  denominator,  we  then 
reduce  the  i  and  i  to  their  numerical  equivalent  fractional  values  in  fortieths.  To  do  this,  we  rea- 
son thus :  i  =  -4%  and  ^  =  Jj ;  i  =  -iV  and  f  =  |§,  which  added  to  f-S  =  ^  5  =  m.  AVe  write  the  1  iu 
the  column  of  units  aud  the  ^5  ia  the  column  of  memorandum  fractions  and  cancel  the  i  aud  f. 


fXITS. 

1 
1 
1 
1 

4i§  Ans. 


MEMORAXDUM 
FRACTIONS. 


ADDITION'    OF    FRACTIONS. 


177 


Lastly  Tve  add  the  columii  of  units  aud  to  the  sum  annex  the  remainmg  fraction,  and  thus  produce 
4ii  as  the  correct  result. 

Note. — The  foregoing  problems  illustrate  the  most  easy,  rapid,  and  rational  system  of  adding 
fractions  known,  aud  as  fractions  are  so  indispensable  and  .of  so  frequent  occurrence  in  practical 
life,  the  principles  iuvulved  in  the  system  should  bo  thoroughly  understood. 

In  pr;Kti<;il  work,  we  would  very  much  shorten  the  operation  by  adding  several  fractions  at 
once,  aud  nuntally  performing  the  most,  if  not  all  of  the  reduction  and  addition  work,  without 
stating  the  results.  Thus,  in  the  above  problem,  we  would  add  the  i,  i,  aud  i  at  once.  We  can 
instantly  see  that  their  sum  is  V",  or  -it  and  without  naming  or  setting  the  2^.  we  add  to  it  mentally 
the  result  of  J  aud  f ,  which  we  mentally  see  is  j,  or  li,  making  3j,  which  are  the  only  figures  we 
write.  Thus  all  the  fractions,  except  i  and  f,  are  added  at  one  mental  operation.  Then  we  men- 
tally add  the  sum  of  *  and  J  by  the  same  process  of  reasoning  as  given  in  the  illustration  of  the 
above  example,  and  obtain  the  correct  result,  1|J. 


Add   3, 


5)     91     -Ji' 


OPERATION. 


•^Tiio)  Alls, 


MEMOR.\XDU.M 
FK.\CTI()XS. 


,11    Xt 


2  5.'i 
360- 


Explannfion. — Selecting  the  }  and  i  to  first  add,  we  reason 
thus:  The  least  common  denominator  is  20;  i  of  20  ^  J,^  aud  J 
=  fj ;  then  j^  of  20  =  vV  and  J  =  iS,  which  added  to  ii  =  iA  = 
l^f,,  which  we  set  in  their  proper  colunuis  aud  cancel  J  and  |. 
We  next  select  J  and  H  and  reason  thus  :    The  L.  C.  D.  is  72 ;  A- 
of  72  =  ,8f  and  i  =  85  ;  then  ^(  of  72  =  ,V  and  AJ  =  f  i;  and  H 
+  ?}  ^  f  I  =  l\i,  w-hich  we  write  in  the  proper  columns,  and 
cancel  the  J  and  the  H.     We  now  have  i,'  and  -fi  to  add,  and 
having  obtained  3G0  as  the  least  common  denominator,  wo  rea- 
son thus  :^?)-j  =  ^V.f  and  -i;\  =  Ko;  ■>^/  =  3sa,  andH  =  i!Vj;  #J 
+  iSS  =  liJ  which  we  write  in  the  proper  column  and  cancel 
ijj  aud  +i.     Then  adding  the  units'  column  and  annexing  to  the 
sum  the  final  fraction,  we  produce  the  answer,  2^. 
Note. — To  contract  the  work  of  finding  the  least  common  denominator  of  two  fractions,  we 
see,  by  inspection,  what  is  the  greatest  common  divisor  of  the  denominators,  and  then  divide  either 
denominat(u-  by  it,  aud  with  the  quotient  multiply  the  other  denominator  the  product  will  be  the 
least  common  denominator. 

2.    Add  A  of  §  of  2^,  f ,  t,  aud  -^  together. 

OPERATION. 

Statement  showinij  the  reduction  of  the  fractions. 
3 

1  ^      ^     3     o      2 
—  X  —  X  — 

2  ?    4    s    0  :23 

Statetnent  showiny  the  result  of  the  reduction  and 
the  addition  of  the  fractions 


MEMOliAXDUM   1-RACT10.\S. 


^  3  ^    g       1    p^ 
20  216      3  529 


Explanation. — Here  we  have  compound  frac- 
tions and  mixed  munbers,  and  before  adding, 
we  reduce  the  mixed  numbers  to  improper  frac- 
tious, and  the  compound  fractions  to  simple 
ones.  Then  we  add  the  J  and  |,  which  are 
equal  to  1  and  | ;  then  the  §  and  i,  which  are 
equal  to  13  ;  then  the  5  J  and  j,'V,  which  are  e(iual 
to  1  and  ^'^3.  Then  adding  the  whole  numbers, 
aud  annexing  the  fraction,  we  have  2Jj'3  as  the 
correct  result. 


2^i,  Alls. 


GENERAL    DIEECTIOXS    FOR    THE    ADDITION    OP    FRACTIONS. 


360.    From  tlie  foregoing  elucidations,  \ye  derive  the  following  general  direc- 
tions for  the  addition  of  fractions : 

1.     Select  such  two  fractions  as  can  he  the  most  casih/  reduced  to  the  same  frac- 


178 


SOULk  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


tional  vnit ;  find  the  equivalent  value  of  both  the  fractions  in  this  fractional  unit, 
and  add  the  numerators  together ;  if  the  sum  equals  or  exceeds  a  unit  write  the  unit 
in  the  column  of  units,  and  the  fractional  remainder  in  the  column  of  memorandum 
fractions.  If  the  sum  is  less  than  a  unit,  write  the  fraction  in  the  column  of  memo- 
randum fractions.     Then  cancel  the  fractions  added. 

2.  In  like  manner  select,  reduce,  and  add  two  more  fractions,  and  thus  proceed 
until  all  are  added. 

3.  Vi^hen  there  are  compound  fractions,  redxice  them  to  simple  ones  before 
adding.  Vi^hen  there  are  mixed  numbers  tcrite  the  whole  numbers  in  the  column  of 
units,  cancel  them  from  among  the  fractions,  and  then  add  the  fractional  numbers. 
All  fractional  expressions  should  be  in  their  lowest  terms  before  adding. 


361. 


PEOBLEMS     IN    ADDITION     OF     FRACTIONS. 


1.  Add  i,  %,  f ,  f ,  I,  and  ^„-, 

2.  Add  5,  i  and  U, 

3.  Add  i,  f,  f ,  and  -,% 

4.  Addi,|i,f|,aud|f,  Ans. 
5    Add  ^  -^   lA   and  J-J^ 

6.  f  of  U  and  2  J  of  |  of  i^, 

7.  Add  2^,  6f ,  5|,  and  2J, 

8.  §,  I,  and  -f., 

9.  f ,  4,   and  J, 

10.  1-jV,  6f,  18M,  and  2,^, 


Ans.  4f^. 
Ans.  2-ift,-. 
Ans.     2-iL. 


01  140         5  7         1     1_ 

-"^'        3aor      bO)     -^10 


1  fl 
7       20> 


"64) 

Ans. 
Ans. 
Ans. 
Ans. 
Ans. 
Ans. 

and 


or  3.; 


"6  U* 
4  7 
92- 

17. 

2 1^2. 
1  2a_ 

-^10  5* 

28  Ai 

1  ti07 


11.  i,  .%,  and  M, 

12.  3^,  li,  2i,  and  4,%, 

13.  if,  -A,  U,  and  iji, 

14.  2^,  3i,  41,  and  5, 
l-'i-  i^hh  and  -i%, 
IG.  7J,  5|,  and  lOf, 

17      li.     2i     Qtlfl   -3. 

-•^ '  •  75>  3¥j  anu  YT, 

18.  14^,  3  a-,  1§,  and  M, 

19.  I,  If,-,  lOf,  and  5. 

20.  125A  327,^2,  and  25 J, 


Ans. 

im 

Ans. 

Hi 

Ans. 

2-AV 

Ans. 

15t^ 

Ans. 

1    37 
-■-IBS 

Ans. 

-"J  12 

Ans.l-A¥5 

Ans. 

21it 

Ans. 

ISA 

Ans.  478^ 

Ans. 

3|M 

22.  What  is  the  weight  of  10  sacks  of  wheat  which  weigh  respectively :  154J, 
149J,  IGOa,  157|,  152J,  141§,  ICSf,  15SJ,  139^,  and  IGIJ  pounds?       Ans.  1539J  lbs. 

23.  Add  J  of  a  of  f  and  2J  of  -,\-  of  1.  Ans.  IJ. 

24.  Add  f  of  f  of  4  and  §  of  J  of  |.  Ans.  2ff. 

25.  How  many  yards  in  8  bolts  of  domestic,  measuring  as  follows :  40^,  39J, 
43i,  42J,  43f ,  38J,  SOtS^,  and  41 J  yards  1  Ans.  328H- 

26.  14  bags  of  coffee  weigh  as  follows:  162;^,  1635, 161-^.,  1G4J,  lG5i,  IG45, 
165|,  1623,  iGoja ,  lG4i,  1653,  165^,  164H,  and  165J  pounds.  How  many  pounds 
in  all?  Ans.  2301. 

27.  A  merchant  bought  1153i  pounds  of  rice  for  $92^;  871^  pounds  of 
sugar  for  $87i;  580|  pounds  of  coflee  for  $lly|;  240;^  pounds  of  cheese  for  $43J; 
and  408 J  pounds  of  Graham  flour  for  $18§.  What  was  the  total  number  of  pounds, 
and  the  total  cost  of  all  he  purchased  ?  Ans.  3254f  pounds ;  $357i  cost. 


28.  Add  3,  f,  i,  -p,,  and  Sf  of  -f.  of  li. 

29.  Add  f,  §,  i,  H,  li,  and  i  of  3. 


Ans.  5ft. 
Ans.  5M%. 


FRACTIONS.     UNITS.    COMPLE.MENTS 
i         _  1        2. 


*  ADDITION    OF    FRACTIONS.  1/9 

ADDITION    OF    FEACTIOXS    BY    A    NEW    METHOD. 

363.  The  following  new  method  of  addition  of  fractions  may  be  used  to 
advantage  when  the  fractions  are  large,  and  the  denominators  are  prime  to  each, 
other. 

The  principles  involved  are  novel  and  should  serve  to  awaken  increased 
interest  in  the  operation  of  numbers. 

363.     1.     Add  the  following  fractions :  |,  f ,  and  |. 

OPERATION. 

f ,  I,  and  I,  added  by  the  Explanation.— in  the  operation  of  problems  by 

usual  method    gives  2^.^.  *^'^  method,  -n-e  are  governed  by  the  following  gen- 

eral principles:     1.   A  proper  fraction  is  less  than 

OPERATION   BY  THE   NEW  METHOD.         "°''y-     ^-  ^  complement  of  a  fraction  is  the  differ- 

ence  between  the  fraction  and  unity,  and  is  found 
by  subtracting  the  numerator  from  the  denominator. 
3.   The  sum  of  several  fractions  is  therefore  equal 

5        __        2      a  to  as  many  units  as  there  are  fractions,  less  the  sum 

of  the  complements  of  the  fractions. 
^  »  In  this  problem  the  sum  of  the  units  is  3,  and 

~"        •     ~_         ^  the  sum  of  the  complements  is  |gj,  and  the  differ- 

•^  foi  =  -^S6  0j -^US.        ence  is  2fjij,  the  answer. 

SECOND   OPERATION  BY   THE   NEW   METHOD. 
FRACTION.S.  UNITS.  COMPLEMENTS. 

I        =        1        —        I      [  360  the  C.  D.  x  the  3  units  =  1080  the  sum  of  the 
4        =        1        —        X     \  units  in  the  C.  D.  unit. 


Then  f  ^  ^  =    80  ^ 

I  >  of  300  =  [  =  13o  I  —  287  the  sum  of  the  complements. 
5-)  )=    72^ 

1080  —  287  =  793  =  the  sum  of  the  fractions.  793  -^  300  =  2/^%,  Ans. 

Note. — In  case  some  of  the  denominators  are  composite  and  not  all  prime  to  each  other,  the 
L.  C.  D.  may  be  found  and  the  operation  performed  as  above. 

PROBLEMS. 

Add  the  following  problems  by  the  above  method : 

2       -fi-    a    fiiid  IS.  A        2.3.     5M_    2    nnd   i"'? 

"■         3  1)     12»    27    aUU.    e^.  O.        l-QO",    tlUU    3-f^. 


ubtraction  of  Fractions. 


^ 


364.  Subtraction  of  Fractions  is  the  process  of   finding  the  difference 
between  two  fractional  numbers  of  like  units  and  of  like  i)arts. 

365.  The  Principle  is,  that  fractions  can  be  subtracted  only  when  they 
express  like  parts  of  like  unifn  or  when  tliey  have  a  common  denominator. 

ORAL    EXERCISES. 
Note. — See  Oral  Exercises  pages  167  and  168. 

366.  1.    One  unit  of  any  kind  equals  how  many  J's? 

2.  ^  taken  from  f  leaves  how  many  J's  ? 

3.  One  unit  of  any  kind  equals  how  many  ^'s  ? 

4.  ^  from  ^  leaves  bow  many  ^'s  ? 

5.  f  from  I  leaves  bow  many  ^'s  ? 

6.  f  from  I  leaves  bow  many  ^'s? 

7.  I  from  f  leaves  bow  many  ^'s  1 

367.  Answer  by  mental  work  the  following  numerical  questions : 

|-i  =  ?  A--^  =  ^  J-f  =?  *  +  §-§=? 

5_§  =  ?  %-i=^-  3i_l|=?  2^-1+^  =  1 

PROBLEMS. 

1.  What  is  the  difference  between  ^f  and  }}  1  Aus.  -^5. 

OPERATION. 

J-3.  Explanation. — In  this  example  tlie   denomina- 

xi-     or    — - -1  ^  -^    Ans. 

'^  '^        ^^        '^'  '  tions  being  the  same,  we  have  hut  to  take  {\ 

-f-g,   Ans.  from  |J  to  obtain  the  correct  difference. 

2.  What  is  the  difference  between  f  and  |  ?  Ans.  oV- 

OPERATION. 

Exphinalion  —Here  ■we  see  that  the  fractions  are  not  of  like 

or      3  i.  units,  and  that  neither  can  be  reduced  to  tlie  unit  of  the  other, 

I'rj        1  g  therefore,  we  must  reduce  tlie  tractions  to  a  common  dentmii- 

nator.     By  inspection,  .and  in  accordance  with  the  principles 

as  explained  in  the  first  four  problems  of  .addition  of  fractions, 

To;  Ans.  we  see  that  the  least  common  denominator  is  20;  then,  that  t 

are  equal  to  ia,  and  that  J  are  eciual  to  \l,  and  that  the  difler- 

ence  is  ^^. 

(180) 


1  =  15 
♦  =  16 


^0,  Ans, 


*  SUBTRACTION    OF    FRACTIONS.  l8l 

3.  From  2S|  take  7J.  Ans.  21J. 

OPERATION. 

285  Explanation. — In  this  problem  we  first  observe  that  the 

r  1  i   luay   be  reduced   to   8ths,  and  by   the  exercise   of  our 

*  reason  we  see  that  it  is  equal  to  f,  which  takeu  from  f 

leaves  i  ;  then  we  find  the  ditfereuce  between   the  whole 

21J  Alls.  numbers  as  in  simjilo  subtraction. 

4.  What  is  the  difiference  between  37fau(il2i?  Ans.  2.1Af. 

OPERATION. 

72  )  qq  Explanation. — ^^By  inspection,  we  here  see  that  the  frac- 

379  =  27  ^  tions  belonging  to  the  whole  numbers  are  not  of  like  units 

^•>T_  __  rg  or   of  the   same  denominator,    and  that  neither   can   bo 

""■*  ~"  reduced  to  an  equivalent  fraction  of  the  same  unit  as  the 

other,  and,  therefore,  we  must  reduce  both  to  fractions 

24:4-j,  Ans.  of  like  units  or  of  a  common   denominator,  before  we  can 

snhtract ;  and  by  multiplying  together  the  denominators,  we  produce  72  as  the  least  common 

denominator,  which,  for  convenience,  we  write  below  the  fractions,  and  by  the  same  reasoning  as 

given  in  addition  of  fractions,  we  see  that  J  are  equal  to  ^|  and  that  J  are  equal  to  f  J,  which,   for 

convenience,  we  carry  to  the  right  of  the  respective  fractions,  and  to  economize  time,  wo  write  only 

the  numerators.     We  now  observe  that  the  upper  fraction,  belonging  to  the  greater  number,  is  less 

than  the  lower  fraction,  belonging  to  the  lesser  number.     Therefore,  before  we  can  subtract  the 

fractions,  we  must  add  1,  reduced  to  72ds,  to  ^,  which  gives  us  ^J.     We  now  subtract  ?f  from  ?  J  and 

have  a  remainder  of  ^J  as  the  fractional  part  of  our  answer.     We  now  add  1  to  the  subtrahend,  to 

compensate  for  the  1,  previously  added  to  the  minuend,  making  it  13.  which  we  subtract  from  37 

and  have  a  remainder  of  24,  which  we  write  below  the  line  and  thus  complete  the  operation. 


GEl^TEEAL     DIEECTIONS     FOE     SUBTEACTING     FEACTIONS. 

368.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  subtracting  fractions : 

1.  Reduce  the  given  fractions  to  equivalent  fractions  of  like  fractional  units,  or 
of  the  least  common  denominator ;  then  take  the  difference  of  their  mimerators  and 
write  the  same  over  the  common  denominator. 

2.  When  there  are  mixed  numbers,  subtract  the  fractional  parts  first  and  then 
the  whole  numbers. 

PEOBLEMS    IX    SUBTEACTIOIS^    OF    FEACTIONS. 

369.     1.  What  is  the  difference  between  H  a^'l  I  ^ 

2.  What  is  the  difference  between  f  and  |  ? 

3.  What  is  the  difference  between  5i  and  3 J  ? 

4.  What  is  the  difference  between  7  and  S^,,  ? 

5.  What  is  the  difference  between  23f  and  14  ? 

What  is  the  difference  between  the  following  numbers : 

6.  f  and  a,  Ans.       H-  9-     *  of  f  and  i  of  f, 

7.  if  and -I,  Ans.      A.  10.    8i  and  3J, 

8.  fand^,  Ans.      ■^.  11.    12|  and  9^, 


Ans. 

A. 

Ans. 

A. 

Ans. 

2a 

Ans. 

3if. 

Ans. 

9f 

^ns. 

i*. 

A.ns. 

*18' 

^ns. 

26^ 

t82 

SOULE  S    PHILOSOPHIC 

PRACTICAL    MATHEMATICS. 

12, 

TT  aud  f , 

Aus. 

Ifl 

136' 

17. 

25-1  aud  9fo-, 

Ans.  16^^. 

13. 

•H  aud  f , 

Ans. 

1  7 

ar- 

18. 

9  aud  3 -/„ 

Ans.    Sfi 

14. 

TiS  and  -1^, 

Aus. 

il 

70- 

19. 

-M  aud  3-^, 

Ans.  Sg^. 

15. 

2f  aud  li, 

Aus. 

lA- 

20. 

7yf  and  4, 

Aus.    3if. 

16. 

9i  and  25, 

Aus. 

6iV. 

2L 

31J  aud  17§, 

Aus.     13J. 

22.     From  GJ 

ot-h+m 

take  J 

of 

3of  1 

5  — f  of  J. 

Aus.  13ff. 

OPEKATION 

INDICATED. 

2      X     13    -^  , 

2  +  13|  =  15|;  ix 

S 

Xl5  = 

=  i^=3f;  33- 

-(§ 

X 

1)  =  2^. 

Aus.  15|-2,V  =  13|t. 

23.  From  8J  +  C|  —  -^g  take  i  of  §  of  3  of  If  +  2J.  Aus.  lO^ij- 

24.  E.  J.  Jacquet  bad  $38f.  He  gave  $2J  for  a  pair  of  Indian  clubs,  $5|  for 
books,  $14  for  a  drawing  board,  aud  $§  for  ink  and  pencils.    How  uuicli  bad  lie  left? 

Aus.  $28f. 

25.  V.  G.  Orena  had  $7^  aud  bis  friend  gave  bim  $  J  more ;  R.  C.  Bush  bad 
$16J  and  be  spent  $5^.    How  mucb  more  has  E.  C.  Bush  tban  V.  G.  Creua  ? 

Aus.  $2|. 

26.  M.  Gundersbeimer  bought  2  bags  of  coffee,  each  weighing  163J  pounds. 
He  sold  27J  pouuds,  50|  pounds,  87J  pounds,  aud  45^  pounds.  How  many  ])ouuds 
has  he  left  ?  Aus.  116|  lbs. 

27.  S.  Delerno  bought  75J  gallons  of  molasses.  He  used  4^  gallons,  lost  by 
leakage  2|  gallons,  and  sold  22^  gallons.     How  mucb  has  he  left  ?  Ans.  46J  gallons. 

28.  What  is  the  difference  between  a  dozen  times  6,  jjIus  6J,  aud  6  times  a 
dozen  minus  one  dozen  and  a  half  dozen  ?  Aus.  24i. 

29.  J.  Astredo  bought  6  chests  of  tea  weighing  38i,  423,  44J,  41f ,  39i,  aud 
43f  pouuds.  He  sold  120§  pouuds  aud  used  53  pouuds.  How  mauy  pouuds  has 
ho  on  hand  ?  Aus.  123f . 

30.  J.  Birba  owned  the  steamer  Isabel,  aud  sold  |  of  it.  What  is  J  of  his 
present  interest  ?  Ans. 


5_ 
1  6* 


31.    From  the  sum  of  6J  and  8f ,  take  the  difference  between  14f  aud  9i  ? 

Aus.  10-ai 


30- 


32.  What  number  is  that  to  which  if  16J  be  added,  the  sum  will  be  44|  ? 

Aus.  27J. 

33.  P.  H.  Weiss  bought  J  of  §  of  a  vessel  and  sold  §  of  3  of  his  share.  How 
much  of  the  whole  vessel  has  be  left  1  Ans.  -J. 

34.  W.  Van  Beuthuysen  owned  f  of  the  steamer  Natchez.  He  sold  to 
J.  Maier  J  interest  in  the  steamer,  aud  to  Leo.  Winner,  ^  of  his  remaining  interest. 
What  is  the  present  interest  of  each  in  the  boat  ? 

Ans.    Van  Beuthuysen,  f ;  Maier,  J ;  and  Winner,  ^. 

35.  F.  Brinker  and  Henry  Pike,  were  each  J  owners  of  a  broom  and  brush 
factory.  F.  Brinker  sold  i  of  his  interest  to  G.  Wagner,  and  then  J  of  his  remaining 
interest  to  Henry  Pike,  who  subsequently  sold  J  of  f  of  his  whole  interest  to 
W.  Lacoume.    What  is  the  present  interest  of  each  owner  ? 

Ans.  F.  Brinker,  ^ ;  G.  W^agner,  J ;  H.  Pike,  |f  and  Wm.  Lacoume, 


1 5 

«4* 


ultiplication  of  Fractions. 


^ 


370.    Multiplication  of  Fractions  is  the  process  of  multiplying  when  one  or 
both  of  the  factors  contain  fractional  numbers. 

Note. — See  the  Philosopliy  of  Multiplication,  pages  77  to  80. 

In  the  multiplication  of  simple  numbers,  we  saw  that  the  result  of  multipli- 
cation operations  was  increasing,  but  iu  the  multiplication  of  fractions,  when  the 
multiplier  is  less  than  a  unit,  the  result  is  decreasing.  This  is  evident  from  the  fact 
that  multiplication  is  the  process  of  repeating  the  multiplicand  as  many  times  as 
there  are  units  in  the  multiplier,  aiid,  therefore,  Avhen  the  multiplier  is  less  than  a 
tinit,  the  multiplicand  will  be  repeated  only  a  part  of  a  time,  or  such  a  x^art  of  itself 
as  the  multiplier  is  a  part  of  a  unit. 


PEACTICAL    DEFISriTIOX    OF    MULTIPLICATION. 

371.  To  elucidate  the  principles  of  the  subject  and  render  clear  the  reasoning, 
we  present  our  first  questions  in  denominate  numbers ;  and  to  aid  still  further  in 
comprehending  the  work,  we  give  the  following  practical  definition  of  multiplication. 

372.  Multiplication  is  that  operation  in  the  practical  computation  of 
numbers,  of  finding  the  cost  of  either  a  j)a)-t  of  one,  or  of  viani/  pounds,  yards, 
barrels,  etc.,  when  the  cost  of  one  pound,  yard,  barrel,  etc.,  is  given.  On  the  prin- 
ciple or  fact  embraced  in  this  definition,  we  base  our  reasoning  for  the  solution  of 
every  practical  question  that  can  possibly  be  presented  in  multiplication,  either  of 
simple  numbers  or  of  fractions. 

METHOD     OF    REASOlSriNG. 

373.  Considering  the  foregoing,  we  see  that  in  all  multiplication  questions 
of  a  iiractical  nature,  we  must  necessarily  reason  from  one,  or  iinify,  to  a  part  of  one 
or  many.  Thus,  if  1  jiound  cost  50/,  \  of  a  pound  will  cost  ^  part  of  it;  and  if  1 
yard  cost  $2,  3  yards  will  cost  three  times  as  much,  or  3  times  $2. 

In  the  solution  of  questions  in  abstract  numbers,  we  apply  the  same  system 
of  reasoning  without  naming  the  factors,  and  thereby  avoid  all  of  the  arbitrary 
rules  given  in  other  arithmetics  of  the  day. 

(183) 


1 84 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 
STATEMENT    LINE. 


374.  Ill  all  problems  where  we  liave  both  multiplication  and  division  work 
to  perform,  we  use  a  vertical  line,  to  facilitate  the  operation. 

The  right  hand  side  is  the  increasing  or  multiplication  side,  and  the  left  hand 
side  is  the  decreasing  or  division  side.  At  the  top  of  the  right  hand  side,  we  first 
write  the  i^remise  of  problems,  or  the  number  to  be  multiplied  or  divided,  or  that 
which  the  conditions  of  the  question  require  the  answer  to  be  in.  Then,  continuing 
with  the  operation,  the  increasing  and  decreasing  numbers  are  written  respectively 
upon  the  increasing  and  decreasing  sides.  But  be  it  remembered,  we  never  place  a 
number  on  either  side  after  the  premise  or  nature  of  the  answer  is  written,  without 
giving  a  reason  therefor.  In  practice,  the  reason  is  instantly  seen  by  the  mind  and 
directs  the  placing  of  the  figures.  When  a  problem  is  fully  stated  on  this  line, 
cancellatioii  is  applied  and  the  operation  is  performed  more  readily  than  by  any 
other  manner  of  statement. 


NOTK. — See  page  161  for  a  full  elucidation  of  the  Statement  Line. 

DEAL  EXERCISES. 
375.     1.     What  will  6  pounds  cost  at  5/  per  pound  ? 


Ans.  30/. 


Solution. — According  to  Article  152,  page  77,  we  reason  as  follows:  1  pound  costs  5c.     Since 
1  ponnd  costs  5c.,  6  pouuds  will  cost  6  times  as  much,  which  is  30c. 


2.     What  will  6  pounds 
cost  ®  5J/  per  pound  ! 

Ans.  33/. 


SOLUTION   STATEMENT. 


11 
0  3 


33/,  Ans. 


Reason. — One  pound  costs  5Jc. 
Since  1  pound  costs  5ic.,  or  ^^c, 
6  poiinds  will  cost  6  times  as 
much,  which  is  33c. 


3.     What  will  Gi  pounds 
eost  at  5/  per  pound  ? 

Ans.  32J/. 


4.  One  pound  costs  5  J/. 
What  will  Gh  i>ounds  cost 
at  the  same  rate  ? 


SOLUTION   STATEMENT. 

/ 

13 


32.i/  Ans. 


SOLUTION   STATEMENT. 


/ 


11 

13 

143 


Reason. — One  pound  costs  5c. 
Since  1  pound  costs  5c.  |  jiound 
will  cost  i  as  much,  and  A?  lbs. 
will  cost  13  times  as  much, 
which  is  32Jc 


Explanation  andJieaaon. — Here 
we  see  by  considering  the  ques 
tion,  that  5ic.  is  the  number  to 
be  multiplied.  Accordingly,  we 
jilace  the  same  at  the  top  of  the 
statement  line ;  and  to  facilitate 
the  work,  to  free  the  operation  of 
fractions — we  reduce  the  5Ac.  to 


35|/,  Ans, 
halves,  making  V-  ^li©  denominator  of  which  we  place  on  the  decreasing  side,  and  the  numerator 
on  the  increasing  side,  of  the  statement  line.  We  then  reason  as  follows:  Since  1  pound  costs  V-. 
•J  of  a  pound  will  cost  }  part  of  it,  and  as  this  conclusion  is  a  decreasing  one,  we  write  the  2  on  the 
decreasing  side;  then,  since  |  of  a  ponnd  costs  the  result  of  the  statement  thus  far  made,  V>  ^i 
reduced,  will  cost  13  times  as  much,  which,  because  the  conclusion  is  an  increasing  one,  we  write  on 
the  increasing  side,  and  thus  complete  the  reason  and  the  statement,  which  worked  out,  gives  the 
answer,  35|c. 


MULTIPLICATION    OF    FRACTIONS. 


i85 


SOLUTION    STATEMENT. 


5.  One  pound  costs  oj  f'. 
What  will  6^  pounds  cost 
at  tbe  same  rate  ? 


n  i 


100 


33J/,  Ans. 


Season . — One  iiouiul  costs  5i 
or  -"-/c.  Since  1  pound  costs  ^fC. 
i  pound  Trill  cost  the  4tli  part 
and  "4-  pounds  will  cost  25  times 
as  much. 


THE  REASON,  WHY,  AND  WHEREFORE,  CONTINUED. 


of  a  yard  will 


Question.    How  do  you  know  that  if  1  yard  costs  \-  cents 
cost  the  4th  part  1 

Answer.  By  the  exercise  of  my  Judgment — by  the  use  of  the  reasouiug 
faculties  of  the  mind. 

Question.    What  do  you  mean  in  this  connection,  by  judgment  ? 

Answer.  The  conclusion  arrived  at  by  the  operations  of  the  miud  after  duly 
considering  the  premise,  the  facts,  and  the  conditions  of  the  problem. 

Question.    What  do  you  mean  by  premise  or  jiremises  ? 

Answer.  The  proposition,  declaration,  truth,  or  fact,  which  is  asserted  as 
the  basis  or  predicate  of  a  question.  In  this  jiroblem  the  premise  is,  one  pound  costs 
5Jor 


-3-  cents. 


Question. 
Answer. 
Question. 
Answer. 


Why  will  6J  pounds  cost  6J  times  as  much  as  1  pound  ? 


Because  C J  is  si.x:  and  one-fourth  times  as  much  as  J . 

What  kind  of  reasoning  is  the  foregoing? 
Analogical  and  axiomatical.    It  is  Analogical,   because  there  is 
analogy,  relationship,  or  likeness  existing  between  the  cost  of  1  pound  and  the  cost 
of  6J  pounds.    It  is  Axiomatical,  because  the  premise  and  question  considered,  the 
conclusion  is  self-e\ndent. 

Question.     What  Is  reason  ? 

Answer.  The  faculty  or  power  of  the  human  mind  by  which  truth  is  dis- 
tinguished from  falsehood,  right  from  wrong,  and  by  which  correct  conclusions  are 
reached  by  considering  the  logical  relationship  which  exists  between  the  premises, 
the  facts,  and  the  conditions  of  particular  statements  and  questions. 


SOLUTION   STATEMENT. 


6.    At  lC§c'   per    yard, 
what  will  22^  yards  cost  ? 


7.  What  will  i^  bush- 
els cost  at  66i/  per  bush- 
el? 


/ 


50 
4.5 


$3.75,  Ans. 


SOLUTION   STATEMENT. 


)^ 


133 

19 

$3.15J,  Ans. 


Reason. — One  yard  costs  165c. 
Since  1  yard  costs  A,'Jc.,  |  of  a 
yard  will  cost  the  i  p.irt,  and 
4f  yards  will  cost  45  times  as 
much. 


Season- — One  bushel  costs  66ic. 
Since  1  bushel  costs -'^J-'c.,  Jof  a 
bushel  will  cost  the  J  part,  and 
V  bushels  will  cost  19  times  as 
much. 


i86 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Question.    How  do  you  kuow  this  ? 

SOLUTION   STATEMENT. 


8.  Chickens  are  selling 
at  $3^  per  dozen.  "What 
will  4+  dozen  cost  ? 


15 


$15§,  Ans. 


Beason. — One  dozen  costs  $\^. 
Since  Idoz.costs  ^4^  dollars,  ^ofa 
dozen  will  cost  \  part,  and  ^ 
dozen  will  cost  25  times  as  much. 


Questions.    1.  How  do  you  know  this  %    2.  What  do  you  mean  by  judgment  T 

SOLUTION   STATEMENT. 


9.  What  will  8  gallons 
cost  at  42i  cents  per  gal- 
lon? 


10.    What  will  1-ih  lbs. 
cost  at  12  cents  per  pound  1 


11.    What  will  §  of   a 
dozen  cost  at  $f  per  doz  ? 


12.  A  mechanic  works 
15i  days  and  receives  $3J 
per  day.  Howmuchmoney 
is  due  him  ? 


85 
8 


^3.40,  Ans. 


SOLUTION  STATEMENT. 

2 


12 
29 


$1.74,  Ans. 

SOLUTION   STATEMENT. 


'21 


Ans. 


SOLUTION   STATEMENT. 


13 
31 

403 

$50|,  Ans. 


Reason. — One  gallon  costs  42Jc 
Since  1  gallon  costs  Sf  cents,  8 
gallons  will  cost  8  times  as 
much. 


Reason. — One  potind  costs  12c. 
Since  1  pound  costs  12  cents,  i 
of  a  pound  will  cost  the  +  part, 
and  ^^  pounds  will  cost  29  times 
as  much. 


Reason. — One  dozen  costs  $}. 
Since  1  dozen  costs  f  of  a  dol- 
lar, i  of  a  dozen  will  cost  the 
i  part,  and  f  of  a  dozen  will  cost 
2  times  as  much. 


Reason. — One  day's  work  is 
worth  $3i.  Since  1  day's  work 
is  worth  ^^  dollars,  ^  of  a  day's 
work  is  worth  i  part,  and  V" 
days'  work  is  worth  31  times  as 
much. 


Questions. — 1.  How  do  you  know  this  ?  2.  What  do  you  mean  by  judgment  ? 
3.  What  is  the  use  of  the  statement  line  ?  4.  Wliat  kind  of  numbers  do  you  place 
on  the  riglit  hand  side  ?  5.  Wliat  kind  on  the  leftl  6.  What  do  you  do  before  you 
place  a  number  on  either  side  ?  Answer. — Give  the  reason  for  so  doing.  7.  What 
do  you  mean  by  reason  ?  8.  What  is  Cancellation  ?  9.  Why  do  you  use  Cancel- 
lation ? 

GENERAL     DIRECTIONS     FOR     MULTIPLYING    FRACTIONS. 


376.  From  the  foregoing  elucidations,  we  derive  the  following  general 
directions  for  multiplying  fractions? 

1.  Write  0)1  ilie  upper  right  hand  side  of  the  statement  line  the  premise  of  the 
problem  ;  or  the  number  ichich  is  to  be  multiplied  or  divided  ;  or  the  number  represen- 


MULTIPLICATION    OF    FRACTIONS. 


187 


ting  the  answer.  Then,  reasoning  from  one,  or  itnity,  to  a  part  of  one  or  to 
many,  urite  the  other  numbers  upon  the  muUipUcation  or  the  division  side  of  the  line, 
according  as  the  conclusion  is  increasing  or  decreasing. 

2.     2Iijce'l  numbers  shouldbe  reduced  to  fractional  exjiressions,   and  the  reason 
given  for  writing  both  the  numerator  and  the  denominator. 

PROBLEMS. 


377.     What  will  g  of  a  yard  cost,  at  i 
of  a  dollar  per  yard  ?  Aus.  $^. 

OPERATION 


^ 


Write  the  reason. 


2.     What  will  oSJ  pounds  cost,  at  16§^ 
per  pound  ?  Ans.  $9.75. 

OPERATION 


XXt  39        Write  the  reason. 


I  J,  Ans. 

3.  What  will  3|  dozen  cost,  at  $3|  jier  dozen  ? 
Note. — Make  the  statement  ami  write  the  reason. 

4.  What  will  5|  bushels  cost,  at  15J/  per  pint  ? 


.75,  Ans. 


Aus.  $123. 


OPERATION. 

31 

4  =  124 
43 


Explanation. — Reducing  and  placing:  the  loic.  on  the  state- 
ment line,  we  reason  as  follows:  Since  1  pint  costs  ■'/c,  2 
pints  or  a  quart  will  cost  2  times  as  much;  and  if  1  quart 
costs  the  result  of  the  statement  now  nuide,  8  quarts,  or 
a  peck,  will  cost  8  times  as  much;  and  if  a  peck  costs  the 
result  of  this  statement,  4  pecks,  or  .1  bushel,  will  cost  4 
times  as  much;  and  if  a  bushel  costs  the  result  of  this 
statement,  \  of  a  bushel  will  cost  the  |  part  of  it,  and  V 
cost  43  times  as  much. 


$53.32,  Ans. 
5.    What  will  50J  pounds  of  tea  cost,  at  lOJ/  per  ounce  f 


OPERATION. 

21 

n  2 

^    201 


Explanation . — Here  we  reduce  and  place  the  lOic  on  the 
line,  and  reason  thus :  Since  1  ounce  costs  V-.  16  ounces  or 
1  pound  will  cost  16  times  as  much,  and  since  1  pound  costs 
the  result  of  this  statement,  i  ]iart  of  a  pound  will  cost  J- 
part  of  it,  aud  ^4-'^  will  cost  201  times  as  much.  This  com- 
pletes the  reasoning  and  statement,  which  worked,  gives 
$84.42,  answer. 


$84.42,  Ans. 

Make  the  solution  statement  and  write  the  reason  for  the  following  problems: 
G.    What  will  24  pounds  cost,  at  9 J/  per  pound  ?  Ans.  $2.28. 

7.  What  will  14§  dozen  cost,  at  $5  per  dozen  ?  Ans.  $73J. 

8.  What  will  6  dozen  aud  7  chickens  cost,  at  $4.87.J  per  dozen  ? 

Ans.  $32.00g. 

9.  What  wiU  42  jwunds  and  11  ounces  of  butter  cost,  at  22i/'  per  pound? 

"  Aus.  $9.60i|. 


SOLUTION    STATEMENT. 

2     45 
16     083 


$9.60ii,  Ans. 


Explanation  and  Heason. — As  usual,  we  here  place 
the  premise,  the  price  of  one  pound,  on  the  statement 
line,  and  reason  as  follows:  I  pound,  or  16  ounces, 
cost  ^c.  Since  1  pound  costs  V"..  1  ounce  will  cost 
the  16th  part  and  683  ounces  (which  is  42  pounds  and 
11  ounces)  will  cost  683  times  as  much. 


i8S 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


10.  What  -will  103  ponnds  cost,  at  ]Sf/  per  pound? 

11.  What  will  :.'5|  yards  cost,  at  IT^/  per  yard  ! 

12.  AVhat  Mill  11|  yards  cost,  at  12 J/  per  yard ? 

13.  What  will  2ia  yards  cost,  at  IG^J C  per  yard  ? 

14.  What  will  14J  pounds  cost,  at  12Je'  per  pound  ? 
13.  What  will  31 J  iiounds  cost,  at  11  J/  per  pound? 
16.  What  will  13  pounds  and  9  ounces  of  tea  cost,  at  87 J/  per  pound  1 

Ans.  Ill.BGff, 


Ans. 

$3.C0fj. 

Ans. 

$4.55^^5. 

Ans. 

$1.42  ,V 

Ans. 

$3.58J. 

Ans. 

$1,741^,. 

Ans. 

$3.40  ^|. 

TO     JIULTIPLY     ABSTRACT     FRACTIONAL     NUMBERS. 


378.     1.     Multiply  SJ  by  3.3. 


OPERATION. 


or 


Explanation  and  Heason. — In   this  problem,   both   factors 
are   abstract   numl)ers.     Hence    we   cannot   Rive   the   same 
analogical  reasoning  as  -we  gave  in  the  foregoing  i)roblems 
j^25  where  the  factors  were  denominate,  or  concrete    numbers; 

"  althongli  were  wo  to  do  so,  the  result,  so  far  as  tlie  figures 

~~~  are  concerned,  would  be  correct.     We  therefore  reduce  and 

314j  Ans.  place  the  H^,  the  number  to  be  multiplied,  on  the  statement 

V^,  and  reason  as  follows :  Axiomatically  1  time  8|  or  -3=^  is  '3-.     Since  1  time  -3^  is  -j^,  J  time  -^ 
is  i  part  of  4,^^  and  \^  are  15  times  as  many. 

Or,  by  iinalysis,  thus:  1  time  '^  is  *3^,  Since  1  time  ^^  is  ^^,  J  time  ^-  is  the  }  part  =  tf, 
and  J-/  are  15  times' ;!  =  ^i'^  =  31i,  answer. 

Note. — It  will  be  observed  that  we  found  our  reasoning  upon  one,  which  is  the  basis  of  all 
nnmbers.  as  explained  in  Article  155,  page  80.  Multiplication  is  the  process  of  repeating  one  number 
as  many  times  as  there  are  units — ones — in  another.  And  it  is  self-evid(^ntthat  one  time  any  number 
is  eciual  to  tlie  number.  This  self-evident  conclusion  is  the  premise  for  all  questions  in  multiplica- 
tion of  abstract  numbers. 


2.     Multiply  a  by  I, 

SOLUTION   STATEMENT. 
4 
3 


Ans.  J. 

Season. — One  time  f  is  |.  Since  1  time  i  is  f ,  J  time  J  is  the 
i  part,  and  f  times  f  is  2  times  as  many,  which  is  J. 

Or,  by  analysis,  thus :  Since  1  time  f  is  |,  J  time  |  is  the  J 
part  of  J  =  1^^,  and  f  times  |  is  2  times  i\  =  ni,  or  i,  answer. 


i,    Ans. 
Multiply  \i  by  |,  and  write  the  reason, 

SOLUTION   STATEMENT 


Ans. 


77 
108' 


4.  Multiply  G  by  |, 
write  the  reason. 


and 


8 


Ans.  3|. 


5.     Multiply  t  by  9,  and 
•write  the  reason. 


SOLUTION   STATEMENT. 

6    5 
9 


G.    5i  by  If,  and  write  the  reason. 
7.    ii  by  ai,  and  write  the  reason. 


Ans.  7J. 


Ans.  2i. 

X 

6* 


Ans. 


■*  MULTIPLICATION    OF    FRACTIONS.  1 89 

MISOELLAXEOTJS  PROBLEMS  IX  MULTIPLICATION  OF  FRACTIONS. 

379.     1.  Wliat  will  16  yards  cost,  at  l-4|p'  per  yard  1  Ans.  $2.36. 

2.  What  will  23f  pounds  cost,  at  35/  per  pound  ?  Ans.  $8.31J. 

3.  "What  will  f  of  a  yard  cost,  at  $|  per  yard  ?  Ans.  $if . 

4.  "VMiat  will  J  a  yard  cost,  at  $i  per  yard  ?  Ans.  $1. 

5.  "VVTiat  will  8i  pounds  cost,  at  7i/  per  pound  ?  Ans.  635/. 

6.  What  will  lOf  pounds  cost,  at  9|/  per  pound  ?  Ans.  99 1^/. 

7.  Multiply  ii  by  12,  Aus.  5|. 

8.  Multiply  11  by  t^-,  Ans.  1. 

9.  Multiply  Jj-  by  11,  Ans.  1. 

10.  Multiply  13  by  ^V,  Ans.  91. 

11.  Multiply  lOJ  by  3^,  Ans.  12. 

12.  Multiply  136  by  -3%%,  Ans.  U^. 

13.  Multiply  12  by  31f ,  Ans.  382. 

14.  Midtiply  II  by  a,  Ans.  if. 

15.  Midtiply  IH  by  11,  Ans.  18f, 

16.  Find  tlie  value  of  f,  of  J,  of  f,  of  ff,  of  4,  Ans.  ^. 

17.  ^^^Iat  is  the  product  of  1-0-,  4,  f,  and  J?  Ans.  -^g. 

18.  What  is  tlie  product  of  If,  f ,  2,  and  5J  ?  Ans.  ll^J. 

19.  What  is  tlie  product  of  -s^  of  2^  by  i  of  7^  ?  Ans.  l^. 

20.  What  is  tlie  product  of  12i  nuiltiplied  by  5i  times  6f  ?  Ans.  464-ij, 

21.  At  -j-Y  of  a  doUar  a  pound,  what  will  -p^  of  a  pound  of  tea  cost  ? 

Ans.  T^j.  of  a  dollar. 

22.  What  will  5^  dozen  buttons  cost,  at  ^J^  of  a  dollar  per  dozen  ? 

Ans.  ^  of  a  dollar, 

23.  Wbat  will  4f  yards  cost,  at  4^/  per  yard  ?  Ans.  20-i^/. 

24.  Wliat  will  14|  dozen  cost  at  $5  per  dozen  ?  Ans.  $73^. 

25.  What  will  15i  pounds  cost,  at  lOi/  per  pound  ?  Ans.  $1,023. 

26.  Wliat  will  40§  pounds  cost,  at  22f  /  per  pound  ?  Ans.  $9.1 9 Jl. 

27.  What  will  2812 J  gallons  cost,  at  $4.50  per  gal  ?  Ans.  $12656.25. 
26.  Wliat  cost  4713  gallons,  at  $31  per  gallon  ?  Ans.  $1592-3%. 

29.  Sold  937852  J  pounds  of  cotton  at  14^/  per  pound.    Wliat  did  it  amount  to  ? 

Ans.  $135095.53|f. 

30.  If  a  man  earns  $2J  in  a  day,  bow  much  will  he  earn  in  lOJ  days  ? 

Ans.  $41i. 

31.  A  contractor  pays  $1J  per  day  for  labor,  and  he  has  370  men  employed  for 
six  days.  How  much  money  will  it  take  to  pay  them  ?  Ans.  $2775. 

32.  P.  Machray  paid  t%  of  a  dollar  for  a  book,  and  for  paper  §  of  the  cost  of 
the  book.  How  much  did  he  iiay  for  the  paper  ?  Aus.     $|. 


igo 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


33.    Distillers  of  tlie  essence  of  rose  have  determined,  by  experience,  that  it 
requires  -ISOOO  pounds  of  rose  leaves  to  make  or  distill  one  pound  of  the  ottar  of 
How  many  pounds  of  rose  leaves  will  it  require  to  distill  SOJ  jioiinds  of  the 


roses. 


ottar  of  roses  ? 


/ 
"     3 
51 


Write  the  reason. 


Ans.  2442000  pounds. 

OPEEATION    INDICATED. 

34.  If  one  pound  costs  a 
cent  and  a  half,  what  will 
25 J  pounds  cost  ? 

Ans.  38i/.  38^;.. 

35.  G.  V.  Hooper  owned  xl  of  the  steamer  Katie,  and  sold  §  of  his  share  to 
Masehek.    What  part  of  the  whole  steamer  did  he  sell  ?  Ans.  f . 

36.  E.  Garner  can  work  the  problems  in  this  book  in  4f  months.    How  many 
months  would  it  take  him  to  work  §  of  them  ?  Ans.  3^  months. 

OPEEATION. 

MOS.  Season. — He  can  Tvork  all  the  problems  in 

^^  months.  Since  he  can  work  all  the  prob- 
lems in -"^i  months,  to  work  J  of  the  problems, 
he  would  require  J  part  as  many  months,  and 
to  work  f ,  he  would  require  twice  as  many 
months. 


19 


or,  J^  X  §  =  3i,  Ans. 


H, 


Ans. 

37.  W.  J.  Kearney  paid  $|  for  1  gallon  of  molasses.  What  is  5  of  a  gallon 
worth  at  the  same  rate  ?  Ans.  $|. 

38.  What  will  7 J  boxes  of  raisins  cost,  at  $2J  per  box  ?  Ans  $16J. 

39.  On  one  occasion  at  the  New  Orleans  Opera,  J  of  the  ladies  and  gentlemen 
present  were  French ;  J  of  the  remainder,  American ;  J  of  the  remainder,  German ; 
and  the  others  were  of  different  nationalities.  What  part  were  Americans,  what 
part  were  Germans,  and  what  part  were  of  different  nationalities  ? 

Ans.  i  Americans,  -^  Germans,  and  |  of  different  nationalities. 

40.  C.  Eeynolds  owned  §  of  a  plantation  and  sold  |  of  his  share  to  D.  C. 
Williams,  who  sold  ^  of  what  he  purchased  to  Frank  Soule,  who  sold  f  of  Avhat  he 
purchased  to  L.  B.  Keiflfer.    What  is  Keiflfer's  share  in  the  plantation  ?    Ans.  /a. 

41.  W.  Weiss  owned  |  of  2000  acres  of  land  and  sold  f  of  his  share  to  H. 
Marsden,  who  sold  |  of  what  he  purchased  to  J.  T.  Finney.  How  many  acres  have 
each  ?  Ans.  W.  Weiss,  400 ;  H.  Marsden,  450 ;  and  J.  T.  Finney,  750  acres. 

MENTAL  CONTEACTIONS  IN  MULTIPLICATION  OF  FEACTIONS. 

Note. — See  pages  88  to  120,  for  Contracted  Methods  of  Multiplication 
380.     In  problems  of  multiplication  of  fractions  in  which  the  factors  are  small, 
the  operation  may  be  performed  mentally,  or  with  but  few  memorandum  figures, 
without  reducing  or  making  the  statement  as  above,  and  the  result  produced  almost 
instantly. 

Ans.  103J. 


1.    Multiply  12i  by  8^, 

OPERATION. 

12^ 

84 


103J,  Ans. 


Explanation.  8  X  12  =  96  +  (}  of  12  =  3)  =^  99,  +  (I  of 
8  =  4)  =  103,  which  we  set  in  the  product  lino.  Then,  J  of 
^  =  ^,  which  is  annexed  to  the  103,"  and  completes  the  an- 
swer. 


f^t 


CONTRACTIONS    IX    MULTIPLICATION    OF    FRACTIONS. 


191 


2.    What  will  1-|  pounds  cost,  at  O^f  ijer  pound  ? 

OPEKATION. 

9| 


Ans.  $1,234. 


12^ 


$1.23^,  Ans. 

3. 

Multiply  13f  by  8J, 

OPEEATION. 

13a  i 

Si 

IICJ,  Ans. 

4. 

Multiply  llj  by  8J, 

OPERATION. 

96^,  Ans. 

5.  Multiply  12i  by    4^,  Ans. 

6.  Multiply    9i  by    G§,  Ans. 

7.  Multiply  lOf  by  15^,  Ans. 

8.  Multiply   54  by    oj,  Ans. 

9.  Multiply  ll|  by  12-iL,  Ans. 


Sxplanalion.     12  X  9  =  108  +  (|  of  9  =  6)  =  114,  +  (J 
of  12  =  9)  =  123,  +  (I  of  f  =  liV  =  i)  =  $1.23*,  answer. 

Ans.  IIC4. 

Explanation.  8  X  13  =  104,  +  (i  of  13  =  6i)  =  110,  with 
i  reserved ;  +  (f  of  8  ^  6)  =  116,  which  is  written  in  the  prod- 
uct line.  Then  A  of  f  =  -^  +  the  i  reserved  ^  |,  which  is  an- 
nexed to  the  116  and  completes  the  answer. 

Ans.  9GJ. 

Explanation.  8  X  11  =  88,  +  (|  of  11  =  oi)  =  93,  with  i 
reserved ,-  -f  ( J-  of  8  =  2j)  =  95,  with  §■  reserved ;  (then  A  of  J 
=  I  -{-  A  reserved  =  J  -]-  f  reserved  =  |  =  IJ,  which  is  added 
to  the  95  and  completes  the  answer. 


PROBLEMS. 


53J. 

61  §. 
IGOig. 

28i. 
142-A. 


iOJ, 


10.  Multiply  24|  by    9|, 

11.  Multiply  37i  by  •> 
12. 
13. 


Ans.      239H. 

Ans.  948f. 
Multiply  71 J  by  52^,  Ans.  37271 
Multiply  370J  by  43^,  Ans.  16024J. 


PEACTICAL     APPROXIMATIYE     CONTRACTIONS. 

TO  MULTIPLY  ANT  FRACTIONAL  NUMBERS  TO  THE  NEAREST  UNIT. 


381.     1.    Midtiply  9J  by  81 


OPERATION. 

Si 


Explanation. — In  this  practical  system  of  contraction,   we 
first  multiply  the  whole  numbers,  and  retain  in  our  mind  the 
product,  to  which  we  mentally  add,   first   the  product,  to  the 
^^  nearest  unit,  of  the  multiplicand  by  the  fraction  of  the  multi- 

II,  Ans.  plier,  and  then  the  product,  to  the  nearest  unit,  of  the  multi- 

plier by  the  fraction  of  the  multiplicand,  and  set  the  result  in 
the  product  line,  the  same  being  the  practical  answer. 

In  thi.s  example,  we  see  that  the  product  of  9  and  8  is  72;  that  the  product  of  9  by  the  J  to 
the  nearest  unit  is  2;  that  the  product  of  the  8  by  the  i  to  the  nearest  unit  is  3,  and  that  the  sum 
of  the  three  products  is  77. 

2.    Multiply  101  by  7J. 


OPERATION. 
lOi 

^ 

73,  Ans. 


Explanation. — Here  we  see  that  the  product  of  the  whole 
numbers  is  70 ;  that  i  of  10  to  the  nearest  unit  is  1,  which  added 
to  the  70  makes  71 ;  then  that  the  i  of  7  to  the  nearest  unit  is  2, 
which  added  to  the  71  makes  73,  the  practical  answer ;  13^  is 
the  exact  result. 


192 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Wliat  will  15f  i)ounds  cost,  at  13J/  per  pound  ? 

OrERATION. 

15?  Explanation.     15  X  13  =  195,  +  J  of  15  to  the  nearest  unit 

13|  is  4,  —  199^  +  f  of  13  to  the  nearest  unit  =  10,  =  209,  as  the 

practical  result.     The  exact  result  is  $208^^. 


$2.09,  Ans. 

Wluit  will  25g  yards  cost,  at  17^/  per  yard  ? 

OPERATION. 
258  Explanation.    25  x  17  ^  425,  +  f  of  25  to  the  nearest  unit 

1 '  4  is   19  ^  444 ;  +   §  of   17  to  the  nearest  unit  =  11,  =  $4.55,  the 

practical  answer.     The  exact  result  is 


$i.i 


Ans. 


:.00l^. 


Note.— In  multiplying  the  25  by  J,  we  gained  Jc,  and  in  multiplying  the  17  hy  f,  we  lost  ic; 
this  iV".  excess  of  loss,  aud  the  loss  of  |c.  by  reason  of  not  multiplying  the  fractions,  account  for 
I'/C.  deficit  iu  the  answer.  lu  practice,  it  is  easy  to  see  whether  the  loss  or  gain,  by  reason  of  the 
contractions,  exceeds  a  unit ;  and  if  so,  to  increase  or  decrease  the  result  accordingly. 

With  largo  numbers,  the  regular  system,  as  first  ])resented,  under  the  head  of  multiplication 
of  fractions,  ia  preferable  to  this  approximative  method. 

PROBLEMS. 


5. 

Multiply  9i  by  6i, 

Ans.  59|. 

6. 

Wliat  will  11|  yards  cost,  at  12 J/  per  yard  1 

Ans.   $1.42-1%. 

7. 

What  will  21 1  yards 

cost,  at  ICi/  per  yard  1 

Ans.  $3,585. 

8. 

What  will  Sf  pounds 

cost,  at  16/  i)er  pound  1 

Ans.  $1.40. 

382.      TABLE  OF    THIRTY-SECONDS  AND  THEIR  EQUIVALENT  VALUES  IN  DECIMALS. 

Fractions. 

I>ecim.il 

Fractions. 

Decimal 

Fractions.              Decimal 

Fractions.                     Decimal 

Equivalent. 

Equivalent. 

Equivalent. 

Equivalent. 

5V  = 

.03125 

-A  = 

.28125 

34  =                     .53125 

H  =                             .78125 

v?  =  A'  = 

.0625 

iS  =  A-  = 

.3125 

15  =  A  =            -5625 

if  =  ig=                    .8125 

A  = 

.09375 

ii  = 

.34375 

JJ  =                     .59375 

15  =                             .84375 

i,^  =  A  =  i 

=     .125 

^  =  A  =  f  = 

=     .375 

J?  =  iS  =  f=     .625 

lf  =  U  =  l=            .875 

^i  = 

.15625 

ii  = 

.40625 

U  =                     .65625 

M  =                             .90625 

A  =  A  = 

.1875 

ii  =  i\  = 

.4375 

¥i  =  U  =            .6875 

U  =  II  =                    .9375 

.V  = 

.21875 

it  = 

.46875 

H  =                     .71875 

U  =                             .96875 

5V  =  A=} 

=     .25 

i5=/,=i=i 

=   .5 

H=f|=e=f=   .75 

Ji=}e=s=J=|=    1. 

383,      TABLE  SHOWING  THE  PRICE    OF  A  SINGLE   ARTICLE    AVHEN  THE   PRICE  PEE 

DOZEN  IS  GIVEN. 

Per  Doz. 

Per  riece. 

Per  Doz. 

Per  Piece. 

Per  Doz.                Per  Piece 

Per  Doz.                Per  Piece. 

25c.  =. 

2tVc. 

.$3.75     = 

olic 

$7.25  =                eOftc. 

$10.75  =              89,'jc. 

50c.  = 

4ic. 

4.00    = 

33ic. 

7.50  =                62ic. 

11.00  =               91|c. 

75c.  = 

6}c. 

4.25    = 

35i\c. 

7.75  =              64iVc. 

11.25  =               93|c. 

$1.00     = 

Sic 

4.50    = 

37Jc. 

8.00  =                66Jc. 

11.50  =               95gc. 

1.25    = 

10,\c. 

4.75     = 

39i'jc. 

8.25  =                68|c. 

11.75  =              971ic. 

1.50    = 

12ic. 

5.00    = 

41fC. 

8.50  =               70Sc. 

12.00  =         $1.00 

1.75    = 

14t^c. 

5.25     = 

43ic. 

8.75  =              72|ic. 

12.25  =           1.02AC. 

2.00    = 

16|c. 

5.50    = 

45ec. 

9.00  =                 75c. 

12.50  =           1.04Jc. 

2.25    = 

18ic. 

5.75     = 

47|ic. 

9.25  =              77|A,c. 

12.75  =           1.06ic. 

2.50    = 

20Sc. 

6.00    = 

50c. 

9.50  =                79ic. 

13.00  =           l.OSif. 

2.75    = 

22[J,c. 

6.25    = 

52,\c. 

9.75  =               81ic. 

13.25  =           l.lO^c. 

3.00    = 

25c. 

6.50    = 

54^c. 

10.00  =               83Jc. 

13.50  =           1.12ic. 

3.25    = 

27  .\c. 

6.75     = 

56ic. 

10.25   ;=               85,4,c. 

13.75  =           1.14,Vci 

3.50    = 

29io. 

7.00    = 

58ic. 

10.50  =               87ic. 

14.00  =           l.lBfo. 

*  CONTRACTIONS    IN    MULTIPLICATION    OF    FRACTIONS.  193 

PECULIAR   COiTTEACTIONS   OP  MULTIPLICATION  OF  FEACTIOXS. 

381.  Tlie  preceding  problems,  and  the  explanations  given,  fiilly  illustrate 
tlie  work,  or  tlio  general  jirinciples  of  contraction  of  fractions,  without  regard  to 
any  peculiar  combination  of  numbers. 

The  principles  ui^on  which  the  work  is  based  are  so  few  and  simple,  the 
oi)erations  so  short  and  easily  performed,  and  the  i)ractical  advantages  of  the  work 
so  great,  that  we  specially  commend  it  to  the  earnest  attention  of  all  classes  of  com- 
mercial men  and  accountants. 

In  the  following  methods  of  contraction,  the  process  dejiends  somewhat  on 
the  peculiar  combination  of  numbers,  and  hence,  although  shorter  than  the  fii-st 
contractions,  is  less  valuable. 


TO  MULTIPLY  KHMBEES  OF  TWO  OE  MOEE  FIGUEES   EACH,  WHEU 

ONE  OR  MORE  OF  THE  EIGHT  HAXD  FIGURES  CONSIDERED 

AS    TENS,    HUNDREDS,    Etc.,    MAY   BE   REDUCED    TO 

FRACTIONS  OF  HALVES,  FOURTHS,  OR  EIGHTHS. 

385.     1.    Multiply  425  by  875. 

OPEEATION. 

4i  hundreds.  Explanation. — In  this  problem,  we  reduce  the  iinits'  and 

8=^  hundreds.  tens'  tigures  of  both  the  multiplicand  and  nmltiiilier  to 

"*  *  fractions  of  J  and  f  resi>ectively  and  thus  obtain  4i-  and 

~    ^__  8|  to  be  multiplied  together,  ^vhich  operation  we  perlonu 

371S75,  AnS.  according  to  the  work  ehicidated  in  Problem  1,  page  190, 

aud  obtain  a  product  of  37A'-  Then  we  reduce  the  frac- 
tional part,  -|\-,  of  this  product,  to  a  whole  number,  and  aunes  it  to  the  integral  part  of  the  i)rod- 
uct;  i\-  =  1875,  which  annexed  to  37  gives  371875,  the  correct  product. 

Thisi\-  is  reduced  to  a  whole  number  for  the  reason  that  it  represents  -pg  of  10000;  and  it 
represents  this  for  the  reason  that  we  first  reduce  four  figures,  two  in  each  factor,  to  fractions. 

Note. — See  Table  of  Thirty-Seconds,  on  page  192,  for  the  decimal  equivalents  of  fractions 
from  ^1  to  §|. 


2.  Midtiply  850  by  450. 

OPERATION. 

SJ-  Explanation. — Cy  reducing  the  units'  and  tens'  figures 

^i  of  both  numbers  as  explained  in  the  preceiUng  problem, 

^  we  have  8J-  and  4J  to  be  multiplied  together.     The  jjrod- 

~;;~3  uct  of  8i  X  4^-  is  38i.       This  J  represents,   for  reasons 

382500,   Ans.  given  in  the  prei'eding  problem,  J-  of  10000,  and  hence  we 

reduce  it  to  a  whole  number  aud  annex  the  same  to  the 
38.     J  of  10000  =  2500,  which  annexed  gives  382500,  the  correct  product. 

3.  What  will  2812  J  gallons  cost,  at  $4.50  per  gallon  % 

OPERATION. 

2SJ  Explanation. — The  reason  for  the  work  of  this  problem 

*  is  the  same  as  elucidated  in  the  two  preceding,  and  hence 


fl2G5G.25.  Ans.  is  omitted. 


194 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


4.    What  M-ill  1050  bbls.  pork  cost,  at  $13.75  per  bbl. 

OPEEATION, 


If 

ICi 


21!0S7.50,  Ans. 
Multiply  the  following  niunbers : 


Explanation. — In  tliis  problem,  -wn  nse  the  imits',  tens* 
and  hnntlreils'  ligures  of  the  price  as  |,  ami  the  units'  and 
tens'  figures  of  the  barrels  as  ^,  and  thus  produce  a» 
product  of  22ji;  the  -fj;  wo  roduee  and  auuex  as  above 
directed  and  obtain  $2L'687.50. 


5.  12250  by  4750. 

7. 

3456  by 

2125, 

6.  24250  by  8125. 

OPERATIONS 

8. 

379  by 

425. 

(5) 
12i 

4;: 

(6) 

8i 

(7) 
3456 

2J 

(8) 
379 

58187500,  Aus. 

197031250,  Ans. 

6912 
432 

1516 
94f 

7344000,  Ans.       161075,  Ans. 
9.     Multiply  775  by  425 ;  1850  by  475 ;  3250  by  875 ;  14625  by  16125. 

TO  SQUARE  NUMBERS  THE  SUM  OP  WHOSE  FRACTIONS  ADDS  ONE. 


386.     1.    Multiply  9^  by  9J. 

OPEKATION. 

9i 


Explanafion, — In  this  problem  yre  add  1  to  9,  making  it 
10 ;  then  we  multiply  10  times  9  are  90,  and  -J  of  ^  is  i, 
and  thus  produce  the  correct  result.     We  do  this  becausa 
one  half  of  9  taken  twice  and  added  to  its  square,  is  the 
Ans.  same  as  to  multiply  the  9  by  1  more  than  itself.     This 

principle  and  j)roces8  of  work  are  ai)i)licable  to  the  mul- 
tiplying of  any  two  like  numbers  whose  fractions  add  unity  or  1. 

PROBLEMS. 


901, 


2.  What  will  12  J  pounds  cost,  at  12^/  per  pound  ? 

3.  What  will  6^  jjounds  cost,  at  6^0'  per  pound  1 

4.  Wliat  will  8|  pounds  cost,  at  8J/  per  jiouiid  1 
4.  What  will  19|  pounds  cost,  at  19|/  jier  pouiid  ? 

6.  What  will  49^^  pounds  cost,  at  49 /r/  per  pound  1 


Ans.  $1.56J. 

Ans.  42-j%. 

Ans.  72|. 

Ans.  $3.80i|. 

Ans.  $24.50^2V 


TO  MULTIPLY  ANY   TWO  NUMBERS,   THE   DIFFERENCE   OF   WHICH 
IS  ONE,  AND  THE  SUM  OF  WHOSE  FRACTIONgl  IS  ONE. 


387.    1.    Multiply  5i  by  4f. 

OPERATION. 

H 

4| 


24|t,  Ans. 


Explanation. — In  all  problems  of  this  kind,  we  add  1  to 
the  larger  number,  and  then  nuiltiply  the  sum  by  the 
lesser  number,  and  sot  the  result  in  tlio  product  line  ;  then 
we  square  the  fraction  of  tlie  larger  number  and  subtract 
the  result  from  1,  and  annex  the  remainder  to  the  whole 
number  iu  tlio  product.    Ju  this  example,  we  added  1  to 


5,  which  made  6  :  this  we  multiplied  by  4,  and  jiroduccd  24 ;  wo  then  squared  J,  which  gave  us  - 
•we  then  subtracted  the  -^^  from  1,  and  obtained  |^,  which  we  annexed  to  the  24  and  completed  the 
correct  result. 


CONTRACTIONS    IN    MULTIPLICATION    OF    FRACTIONS. 


195 


rROBLEMS. 

2.  Wliat  will  8^  pounds  cost,  at  7^/  per  pound  ? 

3.  AVliat  will  lOa  ponnds  cost,  at  9^/  per  iiound  ? 

4.  Wliat  -will  19|  pounds  cost,  at  18f  /  per  pound  ? 


Ans.  633/. 

Ans.  OO/e/. 

Atis. 


$3.G0ff. 


TO  MULTIPLY  AJsTT  TWO  LIKE  NUJVIBEES,  WHOSE  FEACTIOXS  HAVE 

LIKE   DENOMIKATOKS,   OR  WHOSE   DEXOMIXATOES  MAY  BE 

EASILY  EEDUCED  TO  THE  SAME  DENOMIXATIOK. 


388.    1.    Multiply  8^  by  8§. 


OPEEATION. 


H 


8f 


Explanation. —  In  all  problems  of  this  kind,  ■we  adtl  to 
the  multiplicand  the  fraction  of  the  multiplier,  and  then 
multiply  by  the  ichole  number  of  the  multiplier  and  set 
the  result  iu  the  product  line;  then  we  multiply  the  frac- 
tions tojrether  and  annex  the  result  to  the  -nhole  numbers, 
and  obtain  the  answer.  In  this  problem,  we  added  f  to 
8},  which  gave  ns  9,  which  we  multiplied  by  8  and  obtained  72  ;  we  then  multiplied  the  fractions 
and  obtained  -^.  The  reasons  given  in  the  first  problem,  under  the  head  of  contractions,  cover 
all  these  special  cases,  and  hence  we  here  only  give  the  directions  for  performing  the  operation, 


72A,  Ans. 


PROBLEMS. 

2.  What  will  4^  yards  cost,  at  4J/  per  yard  ? 

3.  TMiat  -will  9 J  yards  cost,  at  OJ  c  per  yard  ? 

4.  What  -will  I25  yards  cost,  at  12J/  per  yard  ? 


Ans.  20(36/. 

Ans.  92f/. 

Ans.  $1.60if. 


TO  MULTIPLY  ANY  TWO  NUMBEES   WHOSE  FEACTIONS  AEE  OI^TE- 

HALF,  ONE-THIED,  OlSTE-FOUETH,  Etc.,  OE  WHOSE  FEACTIONS 

AEE  OF  THE  SAME  VALUE. 

389.    Multiply  7J  by  5J. 


OPEEATION. 


Erplanaiion. — Here  we  multiply  the  7  by  5,  and  produce 
35,  to  which  we  mentally  add  i  of  the  sum  of  7  and  5, 
and  obtain  41,  which  we  set  in  the  product  line  ;  then  we 
multiply  the  fractions  together  and  obtain  i,  which  W6 
annex  to  the  41  and  obtain  the  correct  result.  The  reason 
for  adding  J  the  sum  of  7  and  5  is,  because  i  of  either 

number  added  to  i  of  the  other  is  the  same  as  i  of  their  sum.    Were  the  flractions  i,  i,  i,  etc.,  we 

woald  add  J,  J,  J,  etc.,  of  the  sum. 


41J,  Ans. 


196 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEiMATICS. 


2.     Multiply  S|  by  4g. 


OPERATION. 


Fxplaimlion. — In  this  prohlom,  -vre  first  multiplied  the  8 
1>y  4,  and  obtained  32,  which  we  retained  iu  the  mind; 
then  we  added  the  8  and  4,  which  gave  ns  12  |  we  then 
nniltiplied  the  12  by  f,  and  obtained  8,  wliich  we  added 
to  the  32,  and  set  the  result,  40,  in  the  product  line;  then 
we  multiplied  the  tractions  aud  annexed  the  jiroduct  to 
the  product  of  the  whole  numbers,  and  thus  obtained  the  correct  result. 


8§ 


Alls. 


PROBLEMS. 


3. 
4. 
5. 
6. 

7. 
8. 


"WTiat  vrUl  9^  pounds  cost,  at  5^f  per  poiiiul  f 
Wliat  will  14 J  jiouTuls  cost,  at  1^^/  per  poiiud  ? 
"Wliat  Mill  31 J  pounds  cost,  at  11  J/  per  pound  1 
What  will  95  pounds  cost,  at  11 5 (^  per  pound? 
Wliat  will  15  J  pounds  cost,  at  10  J/  j)er  pound  ? 
What  will  40f  pounds  cost,  at  22f  /  per  pound  ? 

Note. — See  pages  88  to  120  for  other  Contractions  iu  Multiplication. 


Ans.  49^./ 

Ans. 

$l./4)*g. 

Ans. 

$3.4Gi|. 

Ans. 

$1.14A. 

Ans 

.  $l.C2f. 

Ans. 

$9.19^. 

I 


ivision  of  Fractions. 


390.  Division  of  Fractious  is  tlie  process  of  dividing  when  the  divisor  or 
dividend,  or  botli,  contain  fractional  numbers. 

In  tlie  division  of  simple  iinmbers,  we  saw  that  the  residt  of  division  opera- 
tions was  decreasing ;  but  in  the  division  of  fractious,  when  the  divisor  is  less  tliau 
a  unit,  the  result  is  increasing.  This  fact  is  plain,  for  the  reason  that  the  operation 
of  di\-ision  is  the  process  of  finding  how  many  times  the  dividend  is  equal  to  the 
divisor,  and,  hence,  when  the  divisor  is  less  than  1,  the  dividend  will  be  equal  to  the 
divisor  as  many  times  itself  as  1  is  equal  to  the  divisor. 

QUESTIOiJS    IN    DIVISION. 

391.  In  practical  operations,  we  usually  have  the  three  following  cases  or 
questions  in  division  of  fractional  numbers : 

1st.  To  find  the  cost  of  one  pound,  yard,  or  article  of  any  kind,  when  we 
have  the  cost  of  many  pounds,  yards,  or  articles  of  any  kind  given. 

2d.  To  find  the  cost  of  one  pound,  yard,  or  article  of  any  kind,  when  we 
have  the  cost  of  a  part  of  a  pound,  yard,  or  article  of  any  kind  given. 

3d.  To  find  the  nnmbcr  of  pounds,  yards,  or  articles  of  any  kind  that  can  be 
bought  with  a  siiecifled  sum,  when  we  have  the  i^rice  of  one,  or  apart  of  one  pound, 
yard,  or  article  of  any  kind  given. 

METHOD     OF    REASONING. 

392.  From  these  questions,  we  see  that  division  is  the  converse  of  multiplica- 
tion, and  that  from  the  nature  of  the  question,  we  must  reason  from  many  to  one  or 
from  apart  of  one  to  one.  Thus:  1st,  if  5  i>ounds  cost  50/,  1  pound  will  cost  the  -J- 
part  of  it ;  in  the  2d  case,  if  g  of  a  yard  cost  $2,  ^  of  a  yard  will  cost  the  ^  part 
of  it,  and  |,  or  a  whole  yard,  will  cost  4  times  as  much ;  and  in  the  third  case,  if  ^/ 
buy  1  yard,  or  any  other  thing,  ^/  <^ill  buy  the  -^g  part  of  it,  and  |,  or  a  whole  cent, 
will  buy  2  times  as  much. 

Note. — See  lutroductory  Remarks  and  Elucidations  of  Division  on  pages  122  and  123. 

fl97) 


198 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


393. 


1.    If 


books 


ORAL     AND     WEITTEN    PEOBLEMS. 

cost  X,  what  will  1  book  cost  ? 


Answer.     If  5  books  cost  x,  1  book  will  cost  tlie  5th  part  of  x. 

2.     ,—  cost  X.    Wliat  will  1  bat  cost  ? 
hats 

Answer.    Since  8  bats  cost  x,  1  bat  will  cost  the  8tb  part  of  x. 


3. 
4. 
5. 


pencils 

6 
oranges 

9 


cost  X.    What  will  1  pencil  cost  ? 
cost  X.     What  will  1  orange  cost  1 
-  cost  X.     Wbat  will  1  chicken  cost  ? 


chickens 

C.    J  of  a  yard  cost  x.     Wbat  will  1  of  a  yard  cost  ? 

7.  If  3  of  a  yard  cost  10/,  what  will  J  of  a  yard  cost  T 

Analytic  Solution.— It  f  of  a  yard  cost  10  cts.,  J  of  a  yard  will  cost  the  half  part  of  10  cts., 
■which  is  5  cts. 

Question.    How  do  you  know  this  ? 

8.  f  of  a  iiound  cost  15  cts.    What  will  ^  of  a  pound  cost  ? 

9.  I  of  a  pound  cost  20  cts.     AVliat  is  the  cost  of  J  ? 

10.  ^  of  a  pound  cost  12  cts.    Wbat  did  1  pound  cost  ? 

Anah/tic  Solution. — Since  }  of  a  pound  cost  12  eta.,  J  of  a  pound  will  co.st  i  of  12  cts.,  which 
is  4  cts.,  and  J,  or  one  pound,  will  cost  4  times  as  much,  which  is  16  cts. 

11.  f  of  a  yard  cost  40  cts.     Wliat  will  bo  the  cost  of  1  yard  1 

12.  §  of  dozen  cost  $10.     Wbat  is  the  value  of  |  of  a  dozen  1 

Analytic  Solution. — Since  f  of  a  dozen  cost  $116,  J  of  a  dozen  will  cost  |  part,  which  is  |8,  and 
J  or  a  whole  dozen  will  cost  3  times  as  much,  which  is  $24;  and  since  1  dozen  costs  $24,  i  of  a 
dozen  will  cost  i  part,  which  is  $6,  and  J  of  a  dozen  will  cost  3  times  as  much,  which  is  $18. 


SOLUTION   STATEMENT. 


13.  2^  yards  cost  $CJ. 
Wbat  will  3t  yards  cost 
at  the  same  rate? 


14.  8J  pounds  cost  62J 
cents.  Wbat  was  the  cost 
of  1  pound  ? 


25 
2 

10 


3,  Ans. 


SOLUTION    STATEMENT. 


/ 


125 


7J/,  Ans. 


Reason  or  the  Philosophic 
Solution.  2^  or  |  yards  cost 
$6i,  or  $-/.  Since  f  yards  cost 
^^  dollars,  ^  of  a  yard  will  cost 
the  5th  part,  and  f  or  a  whole 
yard  will  cost  2  times  as  much ; 
and  since  1  yard  costs  the  result 
of  this  statement,  7^  of  a  yard 
will  cost  the  5th  part,  and  ^^ 
will  cost  16  times  as  much. 


Reason. — ",5-  pounds  cost^f^c. 
Since  'f-  pounds  cost  -'■j^  cts.,  J 
of  a  pound  will  cost  the  25th 
part,  and  |,  or  a  whole  pound, 
will  cost  3  times  as  much. 


Questiona.  1.  How  do  you  know  this  ?  2.  What  do  you  mean  by  judgment? 
3.  Wbat  kind  of  reasoning  is  this?  4.  Wbat  is  the  premise  in  tliis  in-oblemf 
6.  What  do  you  mean  by  premise  ? 


DIVISION    OK    FRACTIONS. 


199 


15. 
cents, 
cost? 


35   yards   cost    37^ 
AVluit  did   1   yard 


SOLUTION   STATEMENT. 


o 
l.> 


/ 


■4 


10/,  Aus. 


Season. —  ^^  yds.  cost  '^/cts. 
Since  ^^  yards  cost  ^  ceuts,  J 
of  a  yard  'nill  cost  the  15th 
part,  and  J  or  a  whole  yard  will 
cost  4  times  as  much. 


SOLUTION   STATEIJIENT. 


43 
•4 


Reason. — "4^  days'  services  cost 
S^j^.  Since  he  receives  ^^  dol- 
lars for  ^  days'  services,  for  i 
of  a  day's  service  he  will  receive 
the  25th  part,  and  for  J,  or  a 
P3.44,  Aus.  whole  day,  4  times  as  much. 
Or,  since  ^4^  days'  services  are  worth  ^^  dollars,  i  of  a  day's  service  is  worth  the  25th  part, 
and  J  or  a  whole  day's  service  is  worth  4  times  as  much. 


16.  A  man  receives  $21  J 
for  C J  days'  services.  What 
was  the  rate  per  day  ? 


17.    S§  dozen  cost   $5 
What  -will  1  dozen  cost  ? 


SOLUTION   STATEMENT. 


26 


S,  Ans 


Season. —  -/  dozen  cost  $52. 
Since  ^  dozen  cost  $52,  J  of  a 
dozen  will  cost  the  26th  part, 
and  J  or  a  whole  dozen,  will 
cost  3  times  as  much. 


IS.    9  slieep   cost  f 33f. 
WTiat  will  1  sheep  cost  ? 


SOLUTION   STATEMENT. 


135 


$33,  Aus. 


Season. — 9  sheep  cost  $4'. 
Since  9  sheep  cost  ^J^  dollars,  1 
sheep  will  cost  the  9th  part. 


19.    If  f  of  a  pound  cost 
^J,  what  will  1  pound  cost? 


SOLUTION   STATEMENT. 
$ 


Li,  Ans. 


Season. —  f  of  a  16.  cost  |J. 
Since  f  of  a  pound  cost  J  of  a 
dollar,  i  of  a  pound  will  cost  the 
third  part,  and  }  or  a  whole 
pound,  will  cost  4  times  as  much. 


20.    Bought  12 J  dozen  for  $12 J.     Wliat  was  the  cost  per  dozen  ? 


Ans.  $1. 


SOLUTION   STATEMENT. 


.  21.  At  7J  cents  per 
pound,  how  many  pounds 
can  be  bought  for  83  J  cents  ? 


ft 


15 
3 


250 


11^  lbs.,  Aus. 


Season. — \^e.  buy  1  pound. 
Since  -^  cents  Will  buy  1  pound, 
I  of  a  cent  will  buy  the  loth 
])art,  and  |  or  a  whole  cent,  will 
buy  2  times  as  much  ;  and  since 
1  cent  will  buy  this  result,  J  of 
a  cent  will  buy  the  third  part 
and  ^'^  cents  will  buy  250 
times  as  much. 


22.    Chickens  cost  $f  a  piece. 


How  many  can  be  bought  for  $13J  ? 

Ans.  18  chickens. 


200 


SOULES   PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


eeasonhstg-  for  the   divisio:^'   of  abstract  nibibers. 

394.     1.     Divido  6  by  2. 


STATEMENT. 
C 


Bemarls. — The  real  question  in  this  problem  is  to  find 
how  many  times  6  is  equal  to  2.     Or,  in  other  words,  we 
are  required  to  measure  6  by  the  unit  of  measure,  2.     The 
jj    J.J.110.  basis  orxmit  of  all  numbers  is  1;  and  hence  in  our  rea- 

soning for  division  of  abstract  numbers,  we  use  1  as  our  unit  of  first  measure.  The  following  is  our 
premise,  reasoning,  and  conclusion :  6  is  equal  to  1,  6  times.  Since  6  is  equal  to  1,  6  times,  it  is 
equal  to  2,  instead  of  1,  i  as  many  times,  which  is  3. 


Ans. 


2.     Divide  §  by 


SOLUTION 

STATEMENT. 

8 

3 

7 
i 

2i 

2S 

1^,  Ans, 

Erplanntion  and  Iteason. — Here  wo  .ire  required  to  mea- 
sure I  Ijy  the  unit  of  measure,  f .  The  basis  or  unit  of  all 
numbers  is  1.  Hence,  as  explained  on  page  122,  in  our 
reasoning  for  division  of  whole  numbers,  wo  use  1  as  our 
first  unit  of  measure.  Tlie  following  is  our  premise,  rea- 
soning, and  conclusion:  ^  is  eqiuil  to  1,  -J  of  a  time.  Since 
1  is  equal  to  1,  I  of  a  time,  it  is  equal  to  J,  instead  of  1,  i 
times  as  many  times  ;  and  to  f,  instead  of  i,  the  i  part  of 
the  jinmber  of  times. 


SOLUTION   STATEMENT. 


3.     Divide  3f  by  2§. 


15 
3 

li|,  Ans. 


Reason. — ^^  is  equal  to  1  -"^ 
times.  Since  \^  is  equal  to  1, 
J^  times,  it  is  equal  to  |,  instead 
of  1,  3  times  as  many  times, 
and  to  f,  instead  of  J,  the  eighth 
part  of  the  number  of  times. 


Divide  5  by  j^. 


SOLUTION    STATEMENT. 


n 


91,  Alls. 


Reason. — Writing  5  on  the 
statement  line,  we  reason  thus : 
5  is  equal  to  1,  5  times.  Since 
5  is  equal  to  1,  5  times,  it  is 
equal  to  1^,-,  11  times  as  many 
times,  and  to  -j^,  the  6th  part 
of  the  number  of  times. 


5.     Divide  A"  %  8. 


SOLUTION    STATEMENT. 


1.5 
8 


Ans. 


Reason — In  this  example,  we 
write  the  dividend  on  the  state- 
ment line  and  reason  thus :  -ft- 
is  =  to  1,  1^  of  a  time.  Since 
•j'yis  =  to  1,  iV  of  a  time,  it  is 
=  to  8,  the  8th  part  of  the 
number  of  times. 


Note. — The  solution  of  the  five  preceding  problems  elucidates  the  only  correct  reasoning  for 
dividing  abstract  fractional  numbers.  But  for  practical  work  we  would  not  advise  a  ch.ange  from 
the  reasoning  given  where  the  numbers  are  denominate. 


6.  Divide  22a  i,y  Si,  Ans.  4-A. 

7.  Divide  3  by  |,  Ans.     4J. 


8.  Divide  14 1  by  9, 

9.  Divide  32  by  9, 


Ans.  If. 
Ans.  3|. 


*  DIVISION    OF    FRACTIONS.  20I 

GEKEEAL    DmECTIOKS    FOE    DIVISION    OP    FEACTIONS. 

395.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  division  of  fractions  : 

1.  Write  on  the  upper  right  hand  side  of  the  statement  line  the  number  u-hich  is 
to  he  divided  or  measured.  Then,  reasoning  from  the  basis  of  many  to  one,  or  from 
a  PAKT  OF  ONE  to  ONE,  icrite  the  other  numbers  on  the  division  or  the  multiplication  side 
of  the  statement  line,  according  as  the  conclusion  is  decreasing  or  increasing. 

2.  Mixed  numbers  should  be  reduced  to  fractional  expressions,  and  the  reason, 
given  for  writing  both  the  numerator  and  the  denominator. 

MISCELLANEOUS     ORAL    EXEECISES. 

396.    1.  If  f  of  a  yard  costs  $2^,  what  will  1  yard  cost?  Ans.  $6. 

2.  If  f  of  a  yard  costs  $2^,  what  will  J  of  a  yard  cost?  Ans.  $3. 

3.  f  of  a  number  is  15,    What  is  the  number  1  Ans.  20. 

4.  If  §  of  a  number  is  8,  what  is  If  times  the  number?  Ans.  21. 

5.  If  §  of  a  dozen  cost  $8,  what  will  f  of  a  dozen  cost  at  the  same  rate  ? 

Aus.  $9. 

6.  What  part  of  4  is  3  ?  Ans.    £. 

Analytic  Sohilion — Here,  by  the  terms  of  the  question,  we  have  3  to  divide  or  measure  by  4' 
and  hy  the  exercise  of  our  reason,  we  proceed  thus:  Since  3  is  equal  to  1,  3  times,  it  is  equal  to  4> 
J  of  3  times,  which  is  |.     Or  thus  :  Since  1  is  J  of  4,  3  is  3  times  |,  which  is  J. 

7.  What  part  of  5  is  §  ?  Ans.  -^. 

Analytic  Solution. — Since  '  is  equal  to  1,  J  of  a  time,  it  is  equal  to  5  the  i  part  of  j  of  a  time, 
■which  is  1%. 

8.  What  part  of  |  is  7  ?  Ans.  SJ. 

Analytic  Solution. — Since  7  is  equal  to  1,  7  times,  it  is  equal  to  J,  5  times  7  which  is  35,  and  to 
i  instead  oi'  |,  J  part  of  35,  which  is  8|. 

9.  What  part  of  f  is  |  ?  Ans.  1^. 

Anali/tic  Solution. — Since  §  is  equal  to  1,  f  of  a  time,  it  is  equal  to  i,  8  times  5  which  is  *-",  and 
to  I  instead  of  i,  i  part  of  V,  which  is  |?,  or  i^?. 


10.  What  part  of  J  is  4  ?        Ans.  |f . 

11.  What  part  of  3 J  is  2J  ?     Ans.  -^. 


12.  What  part  of  5  is  f  of  2  ?  Ans.  -1%. 

13.  What  part  of  f  is  f  of  f  f  Ans.  //,. 


14.  9  is  J  of  what  number  ?  Ans.  72. 
Analytic  Solution. — Since  9  is  J  of  a  number,  g  or  the  whole  number  is  8  times  9,  or  72. 

15.  13  is  i  of  what  number  ?  Ans.     91. 

16.  21-a,  is  i  of  what  number  ?  Ans.  106J. 

17.  3^  is  I  of  Avhat  number  ?  Ans.        f . 

18.  24  is  f  of  how  many  times  3  ?  Ans.    10. 

Analytic  Solution. — Since  24  is  #  of  the  number,  i  is  }  part  of  24  which  is  6,  and  f  or  the  whole 
inunber  is  5  times  6,  which  is  30 ;  and  as  30  is  equal  to  3,  10  times,  therefore,  24  is  }  of  10  times  3. 

19.  32  is  4  of  how  many  times  8  ?  Ans.      7. 

20.  28  is  "1^5  of  how  many  times  12  ?  Ans.      5. 


202 


SOULIi  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


21.     2  of  48  is  §  of  what  number  ?  Aus.    54. 

Analytic  i^olulioii. — Since  48  is  the  whole  of  a  number,  i  of  the  numUcr  is  J  part  of  48,  which 
is  12,  and  J  is  3  times  12,  which  is  36;  and  since  36  is  f  of  an  unknown  number,  i  of  it  is  i  of  36, 
which  is  18,  and  J  or  the  whole  number  is  3  times  18,  which  is  54. 


22.  f  of  G3  is  -1^1-  of  what  number  ? 

23.  I  of  §  of  G4  is  iV  of  wliat  number? 

24.  I  of  4^  of  42  is  t  of  what  number  1 

25.  f  of  32  is  §  of  4  times  what  number  ? 


Aus.  154. 
Ans.  104. 
Ans.    7i. 

Ans.    9. 


Anahitic  Solution. — Since  33  is  the  whole  of  a  number,  J  of  the  number  is  J  part  of  32,  which 
is  8,  and  i  is  3  times  8,  which  is  24 ;  and  since  24  is  J  of  4  times  an  unknown  number,  i  of  4  times 
the  number  is  i  of  24,  which  is  12,  and  ij  or  the  whole  of  4  times  the  number,  is  3  times  12,  which  is 
36 ;  and  since  36  is  4  times  the  number,  i  of  36,  which  is  9,  is  the  required  number. 


26.  J  of  40  is  I  of  7  times  what  number  ? 

27.  A  of  5G  is  A  of  C  times  wliat  number  ? 

28.  §  of  f  of  66  is  3§  of  3  times  what  number? 

29.  What  is  J  and  J  of  a  J,  of  §  of  15  ? 


Ans.  6. 
Ans.  12. 
Ans.  3. 
Ans.     5. 


MISCELLANEOUS    PROBLEiMS    IX    DIVISIOIf    OP    FEACTIOITS. 


397.     1.  Bought  4  yards  for  $14i.     What  was 

2.  Sold  8J  pounds  for  $1.87.     What  was 

3.  Paid  37 J  cents  for  6J  yards  of  calico. 

4.  At  $1|  per  gallon,  how  many  gallons 


5. 
6. 

7. 

8. 

9. 
10. 
11. 
12. 


DiAride 
Divide 
Divide 
Divide 
Divide 
Divide 
Divide 
Divide 


fby2. 
A-  by  3. 
Wbys. 

fbyS. 
Ti  by  9. 
2byf 
3  by  #,-. 
5  by  If. 


Ans. 
Ans. 
Ans. 
Ans. 
Ans. 
Ans. 
Ans. 


7' 

J-1 
76' 
6_ 

■35' 


2i. 

7. 


Ans.  5^^^-. 


13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 


the  cost  per  yard  ?  Ans.  $3|. 

the  i>rice  per  pound  ? 

Ans.  22  cents. 

WTiat  was  the  price  per  yard  ? 

Ans.  0  cents, 
can  be  bought  for  $148 J  ? 

Ans.  108  gallons. 

Divide  21  by  -f,-. 

Divide  105  by  J|. 

Divide  ■}?■  by  -A. 

Divide  f  L  by  ^. 

Divide  2J[  by  f. 

Divide  if  by  ^. 

Divide  3i  by  2^. 

Divide  f  of  ia  by  f , 


Ans. 

33, 

Ans. 

119. 

Ans. 

4. 

Ans. 

13 

Ans. 

3. 

Ans. 

'^^ 

Ans. 

IM 

Ans 

I 

21.  If  one  pound  of  tea  costs  f  of  a  dollar,  how  many  pounds  can  be  bought 
for  $25  ?  Ans.  30  lbs. 

22.  6  barrels  of  flour  were  divided  among  some  poor  families  in  such  a  manner 
that  each  received  §  of  a  barrel.    IIow  many  families  were  there  ?    Ans.  9  families. 


OPERATION  INDICATED. 


1  =  family, 
6 


or. 


9  families,  Ans. 


Write  the  reason. 


9  families. 


*  DIVISION    OF    FRACTIONS.  2O3 

23.  If  a  boy  can  earu  -i-j-  of  a  dollar  iu  one  day,  how  many  days  will  it  take 
him  to  earu  $21  ?  Aiis.  33  days. 

24.  Henry  walked  2.5  miles,  wliicli  was  |  of  the  distance  llobert  walked. 
How  many  miles  did  Robert  walk  ?  Aus.  30  miles. 

25.  At  the  battle  of  Germantown,  the  .British  lost  about  GOO  men  ;  this  was 
|-  of  the  immber  lost  by  the  Amei'icaus;  and  the  number  lost  by  the  Americans 
was  f  of  the  number  they  received  as  re-enforcements  just  before  the  battle.  How 
many  men  did  the  Americans  lose,  and  how  many  did  they  receive  as  re-cuforce- 
ments  ?  Ans.  1000  men  lost,  2500  i-e-enlbrcements. 

26.  A  man  had  his  store  insured  for  $9000,  which  was  f  of  ^s^.  of  its  value. 
What  was  the  store  worth  ?  Ans.  $12375. 

27.  Sulphur  will  fuse  at  232°  Fahrenheit,  which  is  7^  times  the  temperature 
•required  to  melt  ice.    At  what  temjierature  will  ice  melt  ?  Ans.  32°, 

OPEEATION   INDICATED.  232°  4-  7^  =  32°. 

28.  A  quantity  of  mercury  weighed  32062^  lbs.,  which  is  13J  times  the  weight 
of  an  equal  bulk  of  water.    What  would  an  equal  bulk  of  water  weigh  ? 

Ans.  2375  lbs. 

29.  A  pound  of  water  at  212°  F.  was  mixed  with  a  pound  of  powdered  ice  at 
32°.    WTiat  was  the  temperature  of  the  mixture  ?  Ans.  122°. 

OPERATION. 

212  +  32 
^- —  =  122°,  Ans. 

Note. — A  body  in  cooling  1°  gives  out  just  as  much  /leat  as  it  takes  to  heat  it  1^.  In  the  above 
mixture  one  pound  gains  exactly  the  temperature  that  the  other  loses. 

This  will  not  be  the  case,  however,  when  substances  of  diflerent  natures  are  mixed. 

30.  When  the  air  was  at  the  freezing  point,  a  cannon  27613J  feet  distant  from 
New  Orleans  was  discharged.  25^  seconds  elapsed  after  the  discharge  before  the 
sound  reached  Xew  Orleans.     How  many  feet  per  second  did  the  sound  travel  ? 

Ans.  1090  feet. 


31.    Divide  287f  by  5.  '  Ans.  57 


20- 


Operation  U'ithout  the  Explanation. — AVe  first  divide  the  287  by  the  process  of 

Statement  Line  short  division  and  obtain  a  quotient  of  57,  and  a  remain- 

der of  2;  this  remainder  we  reduce  to  a  fraction  whose 
KNOOya  denominator  is  the  same  as   that   of  the   fraction   to   be 

'        *  divided,  add  it  to  this  fraction,  and  then  divide  the  sum 

by  5  and  annex  the  result  to  the  quotient  57.     Thus  2  =  J 
57^,  Ans.  +i  =  Y;  and  -V-  —  5  =  ii. 

32.  Divide  1471^  by  9.  Ans.  163-^. 

33.  Divide  1044|  by  12.  Ans.  87iV. 

34.  E.  T.  Churchill  divided  14t^2-  dozen  apples  among  3  boys  and  2  girls ;  he 
gave  to  each  girl  twice  as  many  as  to  each  boy.  How  many  did  each  boy  and  each 
girl  receive?  Ans.  2-i\,  doz.  each  boy,  4^  doz.  each  girl. 

(Operation  on  other  side). 


204 


SOULES    I'HILOSOPHIC    TRACTICAL    MATHEMATICS. 


OPERATION    INDICATED. 


3  boys,  each  receives  1  apple,  wliich  makes  3  apples. 

2  girls,  eacli  receives  2  apples,  which  makss  4  apples. 

3  +  4  =  7,  the  sum  of  the  proportion  of  the  apples  due  the  3  boys  and  2  girls. 


Statement  to  obtain  the  amount 

Statement  to  obtain  the  amount 

due  each  hoy. 

due  each  girl. 

DOZ. 

DOZ. 

12 
7 
3 

175 
3 

12 

7 
2 

175 

4 

Divide  1  by  i 

Ans. 

5. 

3G. 

Divide  i  by  1 

Ans 

35. 

37.  If  4J  pounds  of  coffee  cost  90  cents,  what  will  22a  pounds  cost  ? 

Ans.  $4.55. 

38.  R.  E.  L.  Flcmming  owns  |  of  the  capital  stock  of  a  factory  valued  at 
$24000 ;  he  gives  J  of  ^  to  educational  societies,  and  the  remainder  he  divides 
equally  among  his  four  children.  How  much  does  he  give  to  educational  societies 
and  how  much  does  each  child  receive  1 

Ans.  $1500  to  educational  societies ;  $1875  each  child  receives. 


39.     S.  J.  "Weis  has  65J  yards  of  cloth,  2  yards  wide, 
lining  |  of  a  yard  wide  will  be  required  to  line  it  1 

OPERATION   INDICATED.  (65i  X  2)  -^-  ^ 


How  many  yards  of 
Ans.  19CJ  yards. 


40.    Divide  18  oranges  between  A.  and  B.  so  that  A.  will  have  J  more  than  B, 


What  number  will  each  have  ? 


B.  receives 
A.  receives 


1 


A.  and  B.  receive       2^ 


OPERATION. 

Statement  Bhowing 
what  13.  rec'd. 

18 
4 

8 


Ans.  A.  10:  B.  8. 


Statement  sb owing 
what  A.  rec'd. 


18 

4 

5 

10 


41.  Divide  18  oranges  between  A.  and  B.  so  that  A.  will  have  J  less  than  B. 
What  number  will  each  have?  Ans.  A.  1^;  B.  lOf. 

42.  A.,  B.,  and  C.  are  to  receive  $26  in  proportion  to  ^,  J,  and  J.  What  will 
each  receive  ?  Ans.  A.  $12 ;  B.  $8 ;  C.  $6. 

43.  Frank  can  work  100  problems  in  4  hours,  and  Lillie  can  work  the  same 
problems  in  5  hours.  How  many  hours  will  it  require  for  both  to  work  the  prob- 
lems ?  Ans.  2|-  hours. 

Solution. — Since  Frank  can  work  the  problems  in  4  hours,  in  1  hour  he  can  work  i  of  them ; 
and  since  Lillie  can  -work  the  problems  in  5  hours,  in  1  hour  she  can  work  ^  of  them.  Hence,  i-\-t 
=  ^  of  the  problems,  worked  by  both  in  1  hour ;  and  since  ^,j  of  the  work  required  1  hour,  jV  of 
the  work  will  require  -i  of  an  hour,  and  |8>  or  the  whole  work,  will  require  20  times  as  many  hours, 


*  DIVISION    OF    FRACTIONS.  205 

44.    A.  and  B.  can  do  a  jjiece  of  woi-k  in  14  days,    A.  can  do  5  as  niucli  as  B. 
How  many  days  will  it  take  each  to  do  it,  working  alone  ?       Ans.  24J  days  for  B. 


OPERATION    INDICATEB. 


32§     "      "    A. 


1  equals  the  work  done  by  B. 

If,  or  f ,  equals  the  work  done  by  B.  and  A. 
Hence,  B.  does  4-  of  the  work  iu  1  day,  and  A.  docs  f  of  the  work  in  1  day. 
14  ds.  -^-  A  =  24i  days,  for  B.  to  do  the  work  alone.     14  ds.  -^  f  =  32§ 
days,  for  A.  to  do  the  work  alone. 

43.    Three  persons.  A.,  B.  and  C,  do  a  piece  of  work ;  A.  and  B.  together  do 
•J  of  it,  and  B.  and  0.  do  ^i  of  it.    What  part  of  the  work  is  done  by  B.?      Ans.  |^. 

Solution. — As  A.  and  B.  do  J  of  it,  it  is  clear  that  C.  does  the  remaining  j ;  and  as  B.  and  C. 
do  iV  "f  it,  it  is  cli'ar  tliat  A.  does  tbe  remainiug  ■^.  Tlien,  as  A.  does  -,=V,  and  C.  t,  they,  together, 
do -,■*,-  +  f  =  5ii ;  and  if  A.  and  C.  do  fg  of  a  piece  of  work  done  by  A.,  B.,  and  C,  it  is  clear  that  B. 
does  the  ditiereuco  between  |]j  and  fg,  which  is  |^. 

46.    If  G  oranges  and  7  lemons  cost  33/,  and  12  oranges  and  10  lemons  cost 
64/,  what  was  the  cost  of  1  orange  and  1  lemon  each  ?  Ans.   1  orange,  2,t'. 

1  lemon,  3fi. 

OPERATION. 

6  oranges  antl    7  lemons  cost  33/. 
12  "  10  "  54/. 

12  "  14  "  06  f!'. 

12  "  10  "  54/. 


4  "  12/. 

12/  -^  4  =  Zf,  cost  of  1  lemon. 
3/  X  7  (lemons)  =  21/,  cost  of  7  lemons. 

33/  —  21<?'  =  \2(f,  cost  of  6  oranges. 
12/  -4-6  =  2/,  cost  of  1  orange. 

iJeasoM.— Since  6  oranges  and  7  lemons  cost  33  cents,  twice  as  many,  or  12  oranges  and  14 
lemons  will  cost  twice  as  much  which  is  66  cents;  and  since,  by  the  second  condition  of  the  jirob- 
lein,  12  oranges  and  10  lemons  cost  54  cents,  the  difl'erence  between  66  cents  and  54  cents,  w  hich  is 
12  cents,  will  be  the  cost  of  the  difference  between  (12  oranges  and  14  lemons)  and  (12  oranges  and 
10  lemons)  which  is  4  lemons.  And  since  4  lemons  cost  12  cents,  one  lemon  will  cost  the  fonrth  part 
which  is  3  cents.  And  since  6  oranges  and  7  lemons  cost  33  cents,  by  snlitractiug  the  cost  of  7 
lemons  which  is  21  cents,  wo  Lave  12  cents,  the  cost  of  6  oranges.  And  since  6  oranges  cost  12 
cents,  one  orange  will  cost  the  sixth  part  which  is  2  cents. 

47.  A  miller  invested  $54  in  grain  of  which  -,^  was  barley  at  62 J/  per  bushel ; 
f  was  wheat  at  $l.S7i  per  bu.;  and  the  balance  oats  ®  37^/  per  bu.  How  many 
bushels  of  grain  did  the  miller  buy  ?  Ans.  40  bus. 

OPERATION. 
Bushels.  BU. 

l^u  ®  i|ic.=  \-c.  =    .181  proportionate  cost  of  the  barley, 

f   ®  2-1^0.=  a-Pc.=  1.12i  "  "  wheat. 

■i\j  @   ^c.  =  if  c.=    .031  "  "  oats. 


1.35 


1 
54.00 


40  bus.  purchased,  Ans. 


$1.33  proportionate  cost  of  1  bu.  of  grain. 
48.    A.  and  B.  can  do  a  piece  of  work  in  10  days;  A.  alone,  can  do  it  iu  15 
days.    How  many  days  will  it  take  B.  to  do  it?  Ans.  30  days. 


iscellaneous  Problems, 


INVOLVING    THE     PRINCIPLES     OF     ADDITION,     SUBTRACTION,     MULTIPLICATION, 

AND    DIVISION    OF     FRACTIONS. 


398.     Find  tlie  difference  between  |  and  f ;  f  and  %;  §  aiid  i^;  3 J  and  2f ;  4§ 
and  J  of  3i  ?  Ans.  To  last,  3. 

2.  Find  the  sum  of  4  of  -„\  and  J  of  /f.  Ans.  if. 

3.  To  the  quotient  of  2f  divided  by  5^,  add  the  quotient  of  3§  divided  by  i\. 

Ans.  7J. 

4.  A  number  -was  divided  by  J,  and  gave  a  quotient  of  20.    AVhat  -w'as  the 
number?  Ans.  15. 

6.     What  number  is  that,  which  being  nuxltiplied  by  i^i,  gives  as  a  product  i|  ? 

Ans.  §. 

OPERATION    INDICATED.  ^^  ~  -j^i  =  §,    AuS. 

6.  Wliat  number  is  that,  from  which,  if  you  take  f  of  itself,  the  remainder 
■will  be  12  ?  Ans.  30. 

OPERATION  INDICATED.     1  =  |  —  f  =  |  J  if  |  =  12,  i  =  G,  and  I  equals  30. 

7.  What  number  is  that,  to  which,  if  you  add  |  of  itself,  the  sum  will  be  40? 

Ans.  25. 

OPERATION   INDICATED.  l  =  |+|  =  f;   iff  =  40,  1  =  5,  and  |  =  25. 

8.  A.  owns  f  of  a  store  which  is  worth  $25000,  and  sells  f  of  his  share. 
What  part  does  he  still  own,  and  what  is  it  worth  ?    Ans.  A,  owns  -i\-,  worth  $2500. 

9.  Smith  owns  -{\-  of  a  cotton  mill  and  sells  -^^g  of  his  share  to  Jones  for 
$33000.     What  is  the  mill  worth  at  that  rate  ?  Ans.  $242000. 

10.  John  has  5  cents,  and  James  5  of  8  cents.    What  i>art  of  James'  money 
is  John's  ?  Ans.  f . 

OPERATION  INDICATED.         J  of  8/  =  2/  and  3  =  0/;  then  5  -h  G  =  f ,      Ans. 

11.  The  sum  of  four  fractious  is  1|.     Three  of  the  fractions  are  j,  J,  and  §. 
What  is  the  fourth  ?  Ans.  /f. 

12.  What  number  is  that,  to  which  if  f  of  ||  of  li  be  added,  the  sum  will 
beli?  Ans.  1. 

13.  Two  boys  bought  a  bushel  of  oranges,  one  paying  2^  dollars  and  the 
other  4§  dollars.    What  ;i)art  of  it  should  each  have  ?    Ans.  First,  ir ;  second,  -jf . 

(20G) 


I 


*  MISCELLANEOUS    PROBLEMS    IN    FRACTIONS.  20/ 

14.  A  farmer  sold  *  of  bis  mules  ou  Monday;  on  Tuesday  lie  bought  f  as 
many  as  be  sold,  and  tben  bad  40.     How  many  mides  bad  be  at  lirst  ? 

Ans.  56  mules. 

OPERATION   INDICATED.  f  from  t  =  f •      f  of  f  =  ?•      f  +  t  =  f- 

s  =  40 ;  }  =  S,  and  +  =  oG,  Ans. 

15.  A  planter  gave  50  bales  of  cotton  at  $50-;^  per  bale  for  fluur  at  $7J  per 
barrel.     How  many  barrels  of  flour  did  be  receive  ?  Ans.  334  bbls. 

16.  A  merchant  bought  20  barrels  of  flour  at  $05  per  barrel.  Having  sold  5 
barrels  at  $7J  a  barrel,  and  6  barrels  at  $8|  per  barrel,  at  what  price  per  barrel  must 
be  sell  the  remainder  to  gain  $10  ou  the  whole  ?  Ans.  $6J. 

17.  If  it  requires  3 J  yards  of  cloth  to  make  a  coat,  2 J  to"  make  a  pair  of 
pantaloons,  and  1§  to  make  a  vest,  bow  many  suits  may  be  made  out  of  157  i  yards, 
allowing  ^  of  a  yard  waste  in  cutting  eacli  suit  ?  Ans.  21. 

18.  A.  W.  McLellan  gave  ^,  ^,  and  ^  of  bis  money  to  different  benevolent 
institutions,  and  bad  $1000  left.     How  much  had  be  at  first  1  Ans.  $20000. 

19.  C.  Manson  owning  /V  <jf  ^  rice  mill,  sold  |  of  his  share  for  $8800.  What 
was  the  value  of  the  mill  ?  Ans.  $24200. 

20.  A  book-keeper  worked  91 J  days,  and  after  paying  §  of  f  of  his  earnings 
for  board  and  washing,  had  $438  remaining.  How  much  money  did  he  receive  in 
all,  and  how  much  per  day  ?  Ans.  $730  in  all,  $8  per  day. 

21.  Prophet  can  do  a  piece  of  work  in  6  days,  and  Fisher  can  do  the  same 
work  in  8  days.    How  many  days  will  it  take  both  together  to  do  the  work  ? 

Ans,  3f  days, 

OPERATION  INDICATED,        i  +  i  =  Ti-    1  (^ay  4-  ^  =  3f  days. 

22.  Myers,  Le\'j',  and  Hoffman  can  do  a  piece  of  work  in  10  days ;  Myers  and 
Levy  can  do  it  in  15  days.    In  what  time  can  Hoffmaii  do  it,  working  alone  1 

Ans.  30  days, 

23.  A  man  died  and  left  bis  wife  $14400,  which  was  f  of  ff  of  bis  estate. 
At  her  death  she  left  f  of  her  share  to  her  daughter.  How  much  money  did  the 
daughter  receive,  and  what  part  was  it  of  her  father's  estate  ? 

Ans.  $12000,  If  of  her  fother's  estate, 

24.  A  man  engaging  in  trade  lost  f  of  the  money  he  invested;  be  then 
gained  $1000,  when  he  had  $3800.  What  did  be  have  at  first,  and  what  was  bis 
loss^  Ans.  $4900  at  first,  $2100  loss, 

OPERATION  INDICATED,         $3800  —  $1000  =  $2800.     $2800  -^  A  =  $4900, 

f  of  $4900  =  $2100, 


2o8  soule's  philosophic  practical  mathematics.  * 

25.    A  mule  aud  a  dray  cost  $210 ;  the  imde  cost  Ig  times  as  mucli  as  the 
dray.    What  did  each  cost  ?  Aiis.  $90  dray,  $150  mule. 

OPEEATION  INDICATED.  1  (cost  of  dray)+  1§  (cost  of  mule)=  2§  =  $240. 

$ 


8 


240  8 

3  3 


$90  cost  of  dray. 


240 

3 

5 


$150  cost  of  mule. 


26.  How  many  bushels  of  apples  at  $|-  a  bushel,  will  pay  for  -^  of  a  barrel 
of  oranges  at  $0|  a  barrel  1  Ans.  7 J  bushels. 

27.  Sweeney  paid  ^  of  Ms  year's  wages  for  board,  f  of  the  remainder  for 
clothes,  and  had  $80  left.     How  many  dollars  ilid  he  receive  for  labor  ?    Ans.  $500. 

28.  Forcheimer  lost  |  of  his  lish-line,  and  then  added  25J  feet,  when  it  was 
just  f  of  its  original  length.    What  was  its  original  length  1  Ans.  204  feet. 

OPERATION   INDICATED.  l  =  t  — 1  =  1-      S —  ^  =  i  =  25|  feet. 

25i  X  8  =  204  feet,  Ans. 

29.  PurceU,  having  a  certain  number  of  cents,  gave  one-half  of  them  and 
half  a  cent  over  to  one  beggar ;  one-half  of  what  he  had  remaining  and  half  a  cent 
over  to  a  second  beggar ;  and  to  a  third,  one-half  of  what  he  then  had  and  half  a 
cent  over,  aud  had  left  3  cents.    How  many  cents  had  he  at  first  ?    Ans.  31  cents. 

OPERATION    INDICATED. 

(  3  -+-  J /  over)  x  2  =    7  C,  had  before  making  3rd  gift. 
(7  +  1/  over)  x  2  =  15/,  "  "        2nd     " 

(15  + i/over)x  2  =  31/'',  "  "         1st     " 

30.  A  housekeeper's  weekly  purchases  from  her  butcher  were  as  follows :  on 
Monday,  7f  jiounds  of  meat ;  on  Tuesday,  6J  jiounds  of  meat  and  1  duck ;  on 
Wednesday,  8 J  poiinds  of  meat;  on  Thursday,  6|  pounds  of  meat;  on  Friday,  3 
pounds  of  meat  and  2  fish ;  and  on  Saturday,  9|  pounds  of  meat,  1  chicken  and 
2  dozen  crabs.  The  meat  was  12|c'  jier  pound ;  the  fish  were  75/  each ;  the  chicken 
$1.40 ;  the  duck  80/,  aud  the  crabs  40/  per  dozen.  What  sum  of  money  was  due 
the  butcher  at  the  close  of  the  week  1  Ans.  $9. 79 J. 

31.  John  lives  with  his  parents,  but  works  for  Mr.  Smith  who  pays  him  $210 
per  year.  His  parents  board  him,  but  he  has  his  clothes  to  buy.  He  spends  f  of 
his  wages  for  cigars,  -|  of  the  remainder  for  tlieater  tickets,  J  of  the  remainder  for 
wine,  and  J  of  what  he  then  has  for  novels.  How  much  has  he  remaining  at  the 
end  of  the  year  to  pay  for  his  clothes  ?  Ans.  $30. 

32.  Joseph  worked  on  the  same  conditions  as  John,  in  the  problem  abovti. 
He  gave  -^  of  his  wages  to  the  cause  of  charity,  -i\-  of  the  remainder  for  useful 
books,  1  of  the  remainder  for  evening  tuition,  paid  $100  for  clothes,  and  deposited 
the  balance  in  the  bank.    How  many  dollars  did  he  put  in  the  bank  ?      Ans.  $50. 


*  MISCELLANEOUS    PROBLEMS    IN    FRACTIONS.  209 

33.  A.  G.  Niehues  and  R.  G.  Joues  have  $1899,  and  Jones  has  3i  times  as 
much  as  Niehues.    How  much  has  each  ?       Ans.  Mehues  $422,  and  Joues  $1477. 

Note. — For  the  operation,  see  problem  25. 

34.  J.  G.  Beals  can  solve  25  pi-oblems  in  50  minutes  and  H.  H.  Barlow  can 
solve  them  in  30  miiuites.     In  what  time  can  both  solve  them?    Ans.  18^  minutes. 

35.  U.  Burlie  piirchased  200  barrels  of  flour  for  $1450,  and  sold  f  of  it  at  a 
profit  of  $.J  per  barrel,  and  the  remainder  at  $1-^  per  barrel.  How  much  did  he 
gain?  Ans.  $G7.50. 

3G.    What  is  the  numerical  value  of  4  J  —  ^ 


2i  +  n  Ans.  l^V- 

37,  M.  Ernst  bought  3841J  pounds  of  cotton,  at  7^  pence  per  pound.  What 
did  it  cost?  Ans.  £124,  Os,  ll^d. 

38.  E.  J.  Kennedy  has  3  dozen  oranges  which  he  wishes  to  divide  between 
Miss  Katie  and  Miss  Tillie,  so  that  Miss  Katie  will  receive  J  more  than  Miss  Tillie. 
How  many  will  each  receive  ?  Ans.  Miss  K.  20,  and  Miss  T.  IG. 

OPERATION    INDICATED 


1    =  Miss  Tillie's  proportional  share. 
IJ  =  Miss  Katie's  proportional  share. 


36 

4  9 

—  4 

16 


36 

4 

5 


2^  =  the  sum  of  the  proportional  shares. 

20 

39.  A  tree  110  feet  high,  had  f  of  it  broken  off  in  a  storm.     How  much  of  it 
was  left  standing  ?  Ans.  44  feet. 

40.  What  cost  22f  pouuds  of  coffee  at  2ia/  per  pound  t  Ans.  $4.94^. 

41.  If  183  yards  cost  $3.37J,  what  will  3J-  yards  cost  ?  Ans.  63  cents. 

42.  Miss  Cora  has  $000  of  which  she  wishes  to  give  to  A.  ^,  B.  J,  C.  ■^,  and 
D.  i.     How  much  will  each  receive  ?      Ans.  A.  $200,  B.  $150,  C.  $120,  and  D.  $100. 

43.  Miss  Mamie  has  $600  which  she  wishes  to  give  to  A.,  B.,  C,  and  D.  in  the 
l^roportiou  of  ^,  ^,  i,  and  ^.    How  much  will  each  receive  ? 

Ans.  A.  $210|^,  B.  $157i|,  C.  $126,!^,  and  D.  $105 1%, 

OPEEATION    INDICATED. 


3 


600 

20     =  $210fa  A's  share. 


44.  If  a  yard  and  a  half  cost  a  dollar  and  a  half,  what  will  twelve  and  a 
half  yards  cost  ?  Ans.  $12J. 

45.  What  part  of  6  cents  is  f  of  5  cents  ?  Ans.  f . 

OPEEATION  INDICATED.        (5x3)-^  C-       Write  the  reason. 

46.  A  planter  remits  his  factor  $500  to  invest  in  rice  and  coffee,  in  ecptal  sums. 
He  pays  9J/  per  pound  for  rice,  and  23|/'  per  pound  for  coifee.  How  many  pound* 
of  each  did  he  purchase  ?  Ans.  2702f-?  lbs.  rice.     1069-f/^  lbs.  coffee. 

47.  If  i  of  6  be  3,  what  will  i  of  20  be  ?  Ans.  7J. 

48.  If  3  is  the  third  of  6,  what  will  the  foui-th  of  20  be  ?  Ans.  SJ. 


2IO  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  ^ 

49.  E.  L.  nunt  owned  a  quantity  of  rice,  of  which  he  sold  |  for  $99.00.  What 
is  §  of  the  remainder  worth  at  the  same  rate  1  Ans.  $10. CO. 

oO.  M.  Landman  jiaid  $00  for  §  of  an  acre  of  hind.  What  is  the  value  of  | 
of  an  acre  ?  Ans.  $50. 

51.  J.  J.  Hauler  bought  937SoliJ  pounds  of  cotton  at  14ij/ per  pound.  What 
was  the  cost?  Ans.  $135095..53Ji. 

52.  E.  T.  Berwick  invested  J  of  his  money  in  sugar,  J  in  rice,  |  in  coft'ee,  and 
deposited  in  bank  $2045.     How  much  money  had  he  at  first  ?  Ans.  $03480. 

53.  L.  Meyer  spends  J  of  his  time  in  study,  J  in  labor,  J  in  rest  and  recrea- 
tion, and  the  remainder  in  sleep.  How  many  of  the  24  hours  of  a  day  does  he 
sleep  ?  Ans.  7  hours. 

54.  An  industrious  young  _ lady  spends  ^  of  her  time  in  the  performance  of 
household  affairs,  i  in  reading  good  books,  -i\r  in  physical  exercise  in  the  open  air 
and  sunlight,  J  in  the  practice  of  music,  singing  and  jiarlor  amusements,  or  social 
intercourse,  2  hours  per  day  in  eating,  and  the  remainder  of  the  day  in  sleeping. 
How  many  hours  per  day  does  she  devote  to  each  ? 

Ans.  G  hours  to  household  affairs ;  4  hours  to  reading ;  2  hours  to  exercise  f 
3  liours  to  music,  etc.;  2  hours  to  eating,  and  7  hours  to  sleeping. 

55.  A  loafer  spends  4  hours  per  day  sauntering  on  street  corners,  3  hours 
smoking  and  drinking,  ^  of  the  day  in  sleep,  ^  of  the  day  in  drunkenness,  --p,-  in 
eating,  -^r  in  quarreling,  and  the  remainder  of  the  day  in  gaming.  How  many  hours 
does  he  spend  in  gaming?  Ans.  3  hours. 

50.  A  fashionable  young  lady  spends  ^  of  her  time  in  dressing,  painting,  and 
making  her  toilet,  ^  in  reading  novels  and  papers  of  senseless  fiction,  ^  in  making 
calls  and  gossiping,  ^^  in  street  promenading,  4j-  in  criticising  industrious  young^ 
men  and  speculating  upon  the  qualities  and  fortune  of  an  anticipated  husband,  -^^ 
in  making  remarks  derogatory  to  the  character  of  those  who  labor,  while  her  own 
mother  is  perhaps  cooking  or  washing,  ^r,  in  entertaining  young  men,  and  the 
remainder  in  eating  and  sleeping.  How  many  hours  does  she  devote  to  useful  ser- 
vice, and  how  many  to  eating  and  sleeping? 

Ans.  0  hour's  to  useful  service;  8  hours  to  eating  and  sleeping. 
67.     A  man  willed  J  of  his  property  to  his  wife,  f  of  the  remainder  to  his 
daughter,  and  the  remainder  to  his  son;  the  difference  between  his  wife's  and 
daughter's  share  was  $8000.    How  much  did  he  give  his  sou  ?  Ans.  $4800. 

OPEKATION  INDICATED, 

^  =  wife's  interest;  -1^^  =  daughter's  interest;  ^f=son's  interest;  -/V  —  i  = 
^  =  $8000 ;  then  $8000  ■^■h=  $25000  the  whole  estate ;  ^^  of  which  is  $4800,  Ans. 

58.  E.  W.  Tyler  owned  a  ^  interest  in  a  factory,  and  sold  to  C.  Modinger  J  of 
his  interest  for  $15000.  What  interest  does  he  still  own,  and  how  much  is  it  worth 
at  the  rate  received  for  the  part  sold  ?  Ans.  He  still  owns  §,  worth  $15000. 

59.  A.  and  B.  own  a  certain  number  of  oranges ;'  A.  owns  f  and  B.  f .  C.  pays 
A.  50  cents  for  ^  of  his  oranges,  and  B.  40  cents  for  ^  of  his  oranges.  What  part  of 
the  whole  mrmber  do  A.,  B.,  and  C.  respectively  own,  and  what  is  the  remaiTider  of 
A.  and  B.  worth  at  the  same  rate  ?       Ans.  A.  o^vns   f.    A's  share  is  worth  $1.00. 

B.  owns  a.    B's  share  is  worth  $1.00. 
0.  owns  i>g. 


«ft 


MISCELLANEOUS    PROBLEMS    IN    FRACTIONS. 


211 


CO.  J.  Cassidy  owned  §  of  the  steamer  11.  E.  Lee.  He  sold  to  G.  Buesiug  ^ 
interest  ill  the  steamer  for  $20000 ;  and  to  J.  C.  Beals  ^  of  his  remaining  interest 
at  the  same  rate.  What  did  he  receive  for  the  last  sale,  and  what  is  his  remaining 
interest  in  the  boat?  Ans.  He  received  $30000 ;  -i%  remaining  interest. 

Gl.  N.  Puech  and  A.  Palacio  bought,  on  joint  account,  each  ^,  the  New  Orleans 
Cotton  Factory.  N.  Puech  sold  ^  of  his  interest  to  E.  Krone,  and  subsequently  J  of 
his  remaining  interest  to  A.  Palacio,  who  subsequently  sold  J  of  J  of  his  whole 
interest  to  E.  Lynd,  for  $7500.  What  is  the  factory  worth  at  the  same  rate,  and 
what  is  each  owner's  interest  ? 

Ans.  $32000  value  of  factory;  Puech  owns  |^; 
Krone,  J ;  Palacio,  -Jf  ;   and  Lynd,  if. 

62.  A.  has  $100 ;  he  gives  f  of  it  for  3  barrels  of  flour,  and  J  of  the  remainder 
for  4  barrels  of  potatoes,  and  with  the  remainder  he  buys  coffee  at  20^'  per  pound. 
How  much  coffee  did  he  buy?  Atis.  200  lbs.  coffee. 

03.  A  person  owns  -^  of  a  stock  of  goods ;  he  sells  |  of  his  share  for  $5000, 
and  J  of  the  i-emainder  for  $5000,  and  then  the  balance  of  his  interest  for  $15000. 
What  part  did  he  sell  the  last  time,  and  what  would  the  whole  stock  be  worth  at 
that  rate?  Ans.  |^  sold  last.    $27,777 J  value  of  stock. 


OPEEATION^S    IN    MIXED    NUMBEES,    COMPLEX,    COMPOUND,    AND 

IMPEOPEE  FEACTIONS,   IN^^OLVING  +  ,  — ,  x  and  ^  SIGNS, 

AND  ALSO  THE  PAEENTHESIS  OE  VINCULUM. 


Note. — For  directions  how  to  reduce  or  simplify  fractional   expressions  indicated  by  the 
+  ,  —  ,  X  and  -^  signs,  and  by  parentheses  or  vinculums,  see  Article  230,  page  134. 


H 


C4.     Multiply  li  by  - 


3 

oo 

11 

2 

4 

25 

10 

3 

or 


OPERATION    INDICATED. 


3 
11 

25 
3 


22 
o 

4 
10 


or 


Ans.  ft. 


3 
11 


Ans.  2J. 
65.    Divide  ^  by  3- 


22 
2 

14 


4 
10 


3 


OPEEATION  INDICATED, 


3 

11 


22 
2 


4 
10 


25 
3 

n 


3 
15 


4 
15 


2J,  Ans. 


4 
8 

ih  Ans. 


212  SOULP:  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

66.    Multiply  ||  by  ^^      Ans.  §.  67.    Divide  ||  by  jl| 


68. 


21         31 
Eediice  ^  of  -~  of  8^  -^  7i  to  a  simple  fraction. 

O  J — in 


6Q.  Wh.it  is  the  result  ofJ  +  §x  J  +  i-^-f+J? 

70.  Find  the  value  of  ^i  X  il  +  7J  —  G|. 

71.  ij  +  tV  -^  M  —  6  X  a  r  ecpials  what  number  ? 

72.  5  —  ^  X  if  -^  4-  =  what  number'? 

73.  What  is  the  value  of  A-  +  §  —  I  X  (f|  +  1^3)  -^  A  ? 

74.  Add  i  of  §,  ^,  f,  I,  aud  7i  —  5|. 

75.  What  is  the  product  of  f  of  12  J  by  §  of  C|  —  J  ? 


Ans 

0251 

Aiis 

.    ^. 

Ans. 

0   1    3 

Ans. 

Iff. 

Ans. 

lA. 

Ans. 

ig. 

Ans. 

32 
'1  9  5" 

Ans. 

A    11. 
*1  20* 

Ans.  34f  J. 

Ans. 

*1  6- 

76.  What  is  tlie  product  of  -|  of  13^  +  1§  by  f  of  (If  —  J)  ■? 

77.  Divide  the  sum  of  J,  §,  f ,  aud  3,  by  the  difference  between  3^  and  ^  of  3. 

Ans.  3-1%. 


78.  Multiply  [CO  divided  by  (;^of  7 J  of  3)]  by  f  of  7.  Ans.  12. 

'  7 

Note. — ^Vllen  there  are  double  parentbeses,  perform  the  operation  indicated  by  the  interior 
parenthesis  first.  See  page  134,  Article  230,  for  full  instructions  regarding  Parentheses  and 
Vinculums. 

79.  Simplify  the  fraction  j^"*"  "  ~  ^  ^^''^  't  "'~°  ti^^  ^^s.  8if. 

OPERATION   INDICATED. 

ai2. ii.sv.a s..  AX s. s..    a   j:_   «    s.a.    2.j.2.fl, iss  .    1  y  24 

5~7    35>S'^10    5>35  5    7J       15^^20     a;       7~9      U3J       B^a5 

.^     i_  ,     _4        •_  ^   2.  .     Xi.  2.   52.  .5.2  J.7    1  J.  15  3.     •      XX  S14        Alio 

35;     35      •       7    57       i)  5    Jo)     43  ■.'  U    36'  li  3      "^    3  (i    "777     -^"O. 

80.  Divide  (||-  of  g-)  by(^  of  i)  Ans.  f. 

81.  L.  Kaiser  bought  §  of  f  of  28 J  barrels  of  apples,  and  sold  to  S.  L. 
Crawford  ^  of  9  barrels  for  $20J,  which  was  $1.50  more  than  the  same  cost.  AVhat 
was  the  cost  of  the  whole,  and  how  m.any  barrels  has  he  unsold  ? 

Ans.  $39-i^o  cost;  7 J  barrels  unsold. 

82.  What  is  the  difference  between  f  of  f  of  eleven  thousand  eleven  hundred 
and  eleven,  and  a  third,  plus  one  half  a  third  of  two  thirds  of  150J  ? 

Ans.  7249-iW 


ireatest  Cominon  Divisor  of  Fractions. 


^ 


399.  Tlie  Greatest  Common  Divisor  of  two  or  more  fractions  is  the  greatest 
fraction  tliat  vrill  divide  eacli  of  them,  without  a  remainder. 

lOO.  One  fraction  is  divisible  by  another  when  the  numerator  of  the  divisor 
is  a  factor  of  the  numerator  of  the  dividend,  and  the  denominator  of  the  divisor 
is  a  mnltiple  of  the  denominator  of  the  dividend. 

Thus  i§-  is  divisible  by  ^% ;  for  if  =  M ;  and  ft  4-  h  =  4. 

401.  The  Greatest  Common  Divisor  of  several  fractions,  is  a  fi-action 
whose  numerator  is  the  greatest  common  divisor  of  all  the  numerators,  and  whose 
denominator  is  the  least  common  multiple  of  all  the  denominators. 

Thus  the  G.  C.  D.  of  A  and  W  is  s^j- 

1.    What  is  the  greatest  common  divisor  of  J,  -IJ,  and  -^f  ?  Ans.  yg^. 


Operation  to  find  the  G.  C.  D, 

of  the  numerators : 

3  13        9        18 


3        0 
3  =  G.  C. 


D. 


Operation  to  find  the  L.  C.  M. 

of  the  denominators : 

2  I  4        2        25 


2         1        25 
2  X  2  X  25  =  100  L.  C.  M 


GENERAL     DIRECTION?"     FOR     FIXDIjSTG    THE     GREATEST    COMMON" 

DIVISOR     OF    FRACTIONS. 


402.    From  the  foregoing  elucidations,  -we  derive  the  following  general  direc- 
tion for  finding  the  greatest  common  divisor  of  fractions : 

Find  the  G.  C.  T>.  of  the  numerators  of  all  the  fractions  and  write  it  over  the 
L.C.  M.  of  their  denominators. 

Note.— All  the  fractions  must  be  lu  their  simplest  form  before  commencing  the  operation. 

2.    What  is  the  greatest  common  divisor  of  §,  |,  f ,  and  |f  ?  Ans, 

What  is  the  greatest  common  divisor  of  SJ  and  \^  ? 
What  is  the  greatest  common  divisor  of  5,  3J,  Cf,  and  -^  1 


10  6' 


Ans. 


What  is  the  greatest  common  divisor  of  ^,  §,  f,  4,  and  f  ? 


Ans.  -,-V. 
Ans.  'Q-Q. 


3. 

4. 

5. 

6.  A  grocer  has  three  kinds  of  molasses,  which  he  wishes  to  ship  in  the  least 
number  of  full  kegs.  Of  the  1st  quality,  he  has  115^  gallons;  of  the  2d  quality, 
12SJ  gallons ;  and  of  the  3d  quality,  1343  gallons.  How  many  gallons  will  there  be 
in  each  keg,  and  how  many  kegs  will  be  required  ? 

Ans.  6-j^  gals,  in  each  keg;  59  kegs  required. 

(213) 


least  Coininon  Multiple  of  Fractions. 


403.  The  Least  Common  Multiple  of  two  or  more  fractions  is  the  least 
number  that  each  fractiou  will  divide  without  a  remainder. 

Note  1. — The  G.  C  D.  of  several  fractions  is  always  a  fraction ;  but  the  L.  C.  M.  of  several 
fractious  may  he  a  fractiou  or  a  whole  uiimber. 

404.  A  fraction  is  a  multiple  of  a  given  fraction  when  its  lUTmerator  is  a 
mitltiplo  of  the  given  numerator  and  its  denominator  is  a  divisor  of  the  given 
denominator. 

Thus  i  is  a  multiple  of  -fg ;  for  4  is  a  midtiple  of  2,  and  5  is  a  divisor  of  15 
Hence  ^^-^  =  6;  or  thus  |  =  Hj  and  if  -j-  -ft-  =  G. 

405.  A  fraction  is  a  common  mnltiple  of  two  or  more  given  fractions  when 
its  numerator  is  a  common  multiple  of  the  numerators  of  the  given  fractions,  and 
its  denominator  is  a  common  divisor  of  the  denominators  of  the  given  fractions. 

406.  A  fraction  is  the  least  common  mnltiple  of  two  or  more  given  fractions, 
when  its  numerator  is  the  least  common  niTiltiple  of  the  given  numerators,  and  its 
denominator  is  the  greatest  common  divisor  of  the  given  denominators. 

1.    What  is  the  L.  C.  M.  of  §,  -A,  and  -ts  1  Ans.  C§. 

Operation  to  find  the  L.  C.  M.  Operation  to  find  the  6.  C.  B. 

of  the  numerators  :  of  the  denominators  : 

2  I  2  5  4  3  I  3  12  15 


15  2  14 

L.  C.  M.  is,  2  X  5  X  2  =  20.  G.  C.  D.  is  3. 

Hence  the  L.  C.  M.  of  the  fractions  is  \^  =  6§. 


GENERAL     DIREGTIOX     FOR    FINDING     THE     LEAST     COMMON 
MULTIPLE    OP    FRACTIONS. 

407.     From   the  foregoing  elucidations,   we  derive  the  following    general   'i 
direction  for  finding  the  least  common  multijile  of  fractions : 

Find  the  least  common  multiple  of  the  numerators  and  the  greatest  common 
divisor  of  the  denominators,  and  then  divide  the  L.  C.  M.  of  the  numerators  by  the 
O,  G.  D.  of  the  denominators. 

IfOXE. — The  fractions  must  he  in  their  sijnplest  form  hefore  commencing  the  operation. 

(214) 


*  LEAST    COMMON    MULTIPLE   OF    FRACTIONS.  2l5 

2.  What  is  the  L.  C.  M.  of  4,  if,  and  i^  1  '  Ans.  4f . 

3.  What  is  tlie  L.  C.  M.  of  §,  f ,  *,  f ,  and  f  f  Ans.  60. 

4.  What  is  the  L.  C.  M.  of  5.J,  7J,  If,  and  |f  ?  Ans.  44. 

5.  There  is  an  island  15  miles  in  circuit,  around  which  A.  can  travel  in  ^ 
of  a  day,  B.  in  §  of  a  day,  and  a  horse  car  in  -,aj  of  a  day.  Suppose  all  start 
together  from  the  same  point  to  travel  around  it  in  the  same  direction,  how  long 
must  they  travel  before  coming  together  again  at  the  place  of  departure,  and  how 
many  miles  will  each  have  traveled  ? 

Ans.  lOJ  days ;  A.,  210  miles ;  B.,  180  miles ;  Horse  Car,  525  miles, 

PARTIAL    OPERATION. 

2)  4    8    10  Denominators. 

3)  3     7     3  Numerators. 

2  the  Greatest  Common  Divisor. 


17     1 


3x7=21-h2  =  lOi  days  before  they  all  meet;  then  the  following  propor- 
tional statements  give  the  miles  traveled  by  each  : 


A. 

B. 

H.  Car.     or  thus,     15  -^  a  X  lOJ  =  210  miles 

15 

15 

15                          traveled  by  A. 

3 

4 

7 

8 

3 

10                     15  ^  §  X  10.^  =  180  miles 

o 

21 

o 

21 

*> 

21                          traveled  by  B. 

— 

— 

— 

15  -^  -^%-  X  lO.i  =  525  miles 

210  m. 

Ans. 

180  m. 

Ans. 

525  m.,  Ans.       traveled  by  Horse  Car. 

G.  What  is  the  smallest  sum  of  money  for  which  I  could  purchase  a  number 
of  bushels  of  oats,  at  $f^  ^  bushel;  a  number  of  bushels  of  corn,  at  $f  a  bushel;  a 
number  of  bushels  of  rye,  at  $1^  a  bushel;  or  a  number  of  bushels  of  wheat,  at 
$2^  a  bushel ;  and  how  many  bushels  of  each  could  I  purchase  for  that  sum  ? 

Ans.  $22^;  72  bushels  of  oats;  36  bushels  of  corn;  15  bushels 

of  rye ;  10  bushels  of  wheat. 

PARTIAL    OPERATION. 

5)  5    5    3    9  Numerators. 

2)  16    8    2    4  Denominators. 


3)  1     1    3    9 


1113 


2  the  Greatest  Common  Divisor. 


5  X  3  X  3  =  45  the  least  common  multiple  of  the  numerators,  which  divided 
by  2,  the  greatest  covimon  divisor  of  the  denominators,  gives  $22J,  the  smallest  sum  ; 
then  22i  divided  by  $1^,  $f ,  $f,  and  $f  gives  respectively  the  number  of  bushels 
of  each  article  represented  by  the  different  prices. 

7.  In  December  1875,  the  Earth,  Mars,  and  Saturn,  were  in  conjunction. 
The  period  of  the  revolution  of  Mars  is  1|  years,  and  of  Saturn  29  J  years.  When 
will  they  be  again  in  conjunction  at  the  same  point  of  their  orbits  ? 

Ans.  In  1003  years. 


ecimal  Fractions. 


^ 


408.  A  Decimal  Fraction  is  one  or  more  of  the  equal  i)arts  of  a  unit,  wbicli 
is  divided  into  tenths,  hundredths,  thousandths,  etc.,  according  to  tlie  decimal  scale  j 
hence  the  denominator  of  decimal  fractions  is  always  10  or  some  jxiwer  of  10.  The 
word  decimal  is  derived  from  the  Latin  word  decern,  which  means  ten. 

409.  The  Decimal  Point  (  .  )  is  used  to  distinguish  decimals  from  whole 
numbers.  When  there  are  mixed  numbers,  it  also  separates  the  whole  numbers 
from  the  decimals. 

The  following  are  decimal  fractions:  tt,-,  i\\,  i-Wo?  and  toooo-  They  are  here 
wintten  as  common  fractions,  but  generally  the  denominator  of  decimal  fractions  is 
omitted  and  the  value  is  indicated  by  writing  the  decimal  point  before  the  numerator. 

To  winte  the  above  fractions  according  to  the  decimal  notation,  they  would 
be  written  thus : 

1^0-  decimally  expressed  is  .7  --^i^g  decimally  expressed  is       .137 

-i¥o  decimally  expressed  is  .15  toIo ir  decimally  expressed  is     .0423 

410.  Notation  of  Decimals.  Whenever  decimal  fractions  are  expressed 
decimally,  the  numerator  must  have  as  many  decimal  places  as  there  are  naughts 
in  the  denominator.  Thus,  -^^  —  -4  ;  -nro  =  -1^  j  -iWuT)  =  .1-456.  When  the  number 
of  naughts  in  the  denominator  is  greater  than  the  number  of  figures  in  the  numer- 
ator, naughts  must  be  prefixed  to  the  numerator  until  the  number  of  places  is 
equal  to  the  naughts  in  the  denominator.  Thus,  ^o o  =  -^^ >  t^os  =  -007  ;  toWoo 
=  .00125 ;  etc. 

411.  A  Pure  or  Simple  Decimal  consists  of  a  decimal  fraction,  decimally 
expressed  or  written.  Thus  .5,  .42,  .875,  and  .1256  are  pure  decimals,  and  are  read 
respectively  5  tenths ;  42  hundredths ;  875  thousandths ;  and  1256  ten  thousandths. 

412.  A  Mixed  Decimal  consists  of  a  whole  number  and  a  decimal.  Thus, 
24.5  and  41.25  are  mixed  decimals.  They  are  read  respectively,  24  and  5  tenths;  41 
and  25  hundredths. 

Note. — A  mixed  decimal  may  be  read  as  an  improper  fraction.     Thus,  24.5  =24i%  =  -,*,f. 

413.  A  Complex  Decimal  consists  of  a  decimal  with  a  common  fraction 
annexed.  Thus,  .15^  and  .005^  are  complex  decimals.  They  are  read  respectively, 
15f  hundredths ;  5J  thousandths. 

414.  A  Terminate  Decimal  is  one  which  ends,  as  ^  =  .125.  And  an 
interminate  decimal  is  one  that  does  not  end,  as  ^  =  .3333+. 

Note. ^Decimal  fractions  were  invented  by  Regiomontanus,  in  1464 ;  but  tliey  were  not  in 
general  use  until  the  latter  i)art  of  the  sixteenth  century,  when  the  first  treatise  on  decimals  was 
written  by  Stevinus,  in  1585. 

(216) 


I 


*  DECIMAL    FRACTIONS.  21/ 

CIRCULATmG  DECIMALS. 

415.  A*  Circulating  Decimal  is  one  in  which  a  figure  or  set  of  figures  con- 
stantly repeats  itself.     Thus,  J  =  .3333+  ,f^  =  .181818+,  I  =  .142857  +  . 

416.  A  Repetend  is  the  figure  or  set  of  figures  which  repeats  itself,  and  it 
is  expressed  Lj- placing  a  dot  over  the  figure  repeated,  thus:  ^  =  .3333+  =  .3. 
When  the  repetend  consists  of  more  than  one  figure  the  dot  is  placed  over  the  first 
and  last  figure,  thus:  -,^  =    .181818+  =  .iS;  and  ^  =  .112857+  =  il2857. 

417.  A  Pure  Circulating  Decimal  is  one  which  contains  only  the  repetend, 
as§  =  G;   i  =  .il2S57;   ^  =  i. 

418.  A  Mixed  Circulating  Decimal  is  one  which  contains  other  figm-es  than 
the  repetend,  as  ^  =  .16  ;  ffj  =  .G17. 

419.  A  Simple  Repetend  contains  one  figure,  as  .6. 

420.  A  Compound  Repetend  contains  two  or  more  figures,  as  .18  and  .342. 

421.  Similar  Repetends  are  those  which  begin  and  end  at  the  same  decimal 
place,  as  .536  and  .427. 

422.  Dissimilar  Repetends  are  those  which  begin  and  end  at  different  decimal 
places,  as  .205  and  .3765. 

423.  A  Perfect  Repetend  is  one  which  contains  as  many  decimal  places, 
less  1,  as  there  are  units  in  the  denominator  of  the  equivalent  common  fraction ; 
thus,  \  =  .142857. 

424.  Conterminous  Repetends  are  those  which  end  at  the  same  decimal 
place,  as  .50397  and  .42618. 

425.  Co-originous  Repetends  are  those  which  begin  at  the  same  decimal 

place,  as  .5  and  .124. 

Circulating  decimals  had  their  origin  in  reducing  common  fractions  to  deci- 
mals. 

The  general  law  for  the  formation  of  repetends  will  be  seen  from  the  follow- 
ing: 


1.  i    =.1111+         =.1. 

2.  -sV   =  .01010+       =  .61. 

3.  ^^  =  .001001+     =  .OOl. 

4.  WW  =  .00010001  +  =  .oooi. 


5.  I     =  .4444+         =  .4. 

6.  II   =  .2727+         =  .27. 
7-  iU  =  .135135+     =  .135. 
8.  Wfl  =  .17281728+=  .1728. 


2x8  soule's  philosophic  practical  mathematics.  * 

426.  Decimal  fractions,  like  whole  numbers,  decrease  toward  the  right  and 
increase  toward  the  left  in  a  ten-fold  ratio;  and  hence  the  i)reflxing  of  naughts 
between  the  decimal  figures  and  the  decimal  point,  or  the  removal"  of  the  decimal 
point  towards  the  left  diminishes  their  value  ten-fold,  or  divides  the  decimal  by  ton 
for  each  order  or  place  removed.     Thus :  .5  =  -f^ ;  .05  =  j|}j ;  .005  =  ^o^o  ;  etc. 

4:27.  The  removal  of  the  decimal  point  to  the  right,  increases  the  value 
tenfold  or  multiplies  the  decimal  by  ten  for  each  place  removed.    Thus  :   .005  — 

T(H)  0  5    •        ^^  "10  0  '    '"^  ^^  "i^^o  5  ^ 

428.  Annexing  naughts  to  decimals  does  not  change  their  value,  because 
the  significant  figures  are  not  thereby  removed  nearer  to  or  farther  from  the  decimal 
l^oint.    Thus :  .5  =  1% ;  also  .50  =  ^^^ ;  .500  =  -jinnj,  all  of  which  are  equal. 

429.  Decimal  orders  are  also  called  decimal  ^jtoces,  each  order  being  counted 
as  one  place.  Thus,  in  .0043  there  are  four  decimal  places,  although  the  3  is  of  the 
fifth  decimal  place  ft-oni  unity,  the  base  of  the  system. 

430.  The  following  table  will  illustrate  more  fully  the  relation  of  whole 
luimbers  and  decimals,  with  their  increasing  and  decreasing  orders  to  the  left  and 
right  of  the  decimal  point : 

TABLE. 
WHOLE    NUMBERS.  DECIMALS. 

EC 

»  1  55 


CO 


o  2  »  "^ 

5-^  .  ./a 


o  _ 


^5     H0  -g  .;g2      2g 

c  S      c  ®  ■©      » .=  5  .a     '^  [q 

WhSShhWh   P  QeHtrlHHtrlSHW 
98765433     1     .234507  8  9 

Orders  of  ascending  scale.         Orders  of  descending  scale. 

This  number  is  read  987  million  654  thousand  321,  and  23  million  456  thousand 
789  hundred  miUionths. 

431.  In  order  to  clearly  understand  decimals,  we  must  bear  in  mind  that 
one  is  the  basis  of  all  numbers,  integral  and  fractional,  abstract  and  denominate, 
and  that  all  mathematical  operations  have  this  fundamental  principle  for  their 
origin,  and  every  number  is  but  a  multiple,  either  ascending  or  descending  of  unity 
or  one. 

432.  The  names  of  the  decimal  orders  are  derived  from  the  names  of  the 
orders  of  whole  numbers.  Thus  the  names  of  the  orders  in  the  ascending  scale,  are, 
after  units,  tfns,  hundreds,  etc. ,  and  the  orders  in  the  de.scending  scale,  are,  after  units, 
tenths,  hundredths,  etc.,  the  decimal  orders  being  the  reciprocal  of  the  orders  of  whole 
numbers  equally  distant  with  themselves  fiom  the  units. 


*  DECIMAL    FRACTIONS.  219 

433.  Jfumeration  of  Decimals.  In  reading  decimal  fractions,  the  entire 
decimal  is  regarded  as  reduced  to  units  of  the  lowest  order  expressed,  and  the  name 
of  this  order  is  given  to  tlie  entire  number  of  decimal  units.  Thus,  .25  is  read 
twenty-Jive  hundredths. 

434.  Before  reading  a  decimal,  wo  nnxst  determine  1st.  How  many  units 
are  expressed.  To  do  this,  we  numerate  and  read  the  significant  figures  of  the 
decimal  as  in  whole  numbers.  2nd.  We  nmst  determine  tlie  name  of  the  lowest 
order  iu  the  decimal.  To  do  this,  we  numerate  the  number  decimally.  Thus  to 
read  .001073,  we  commence  at  the  3  and  numerate  to  the  1  thousand,  and  thus  find 
that  1073  units  are  expressed  ;  then  we  commence  at  the  decimal  point  and  numerate 
decimally  to  the  3,  and  thus  find  that  milliouths  is  the  lowest  order;  we  then  read 
1073  milliouths. 

EXERCISES. 

435.  Eead  the  following  numbers : 

1.  10.008  reads  thus:  Sixteen  units  and  eight  thousandths. 

2.  .94|  reads  thus:  Ninety-four  and  three-eights  hundredths. 

3.  50G7.4005  reads  thus:   50G7  units  and  400.5  ten-thousandths. 

4.  Write  and  read  107.8 ;  4.0S9()7  ;  .()()073;  48.70!)14(). 

5.  Write  and  read  2.401;  IO.OIOIOSDKm  ;  582.40041000.5. 

G.     Write  and  read  5841.291 1 ;  8000.0000000217  ;  987G541.1000001. 

436.  "Writing  Decimals.  In  writing  decimals,  we  write  down  the  given 
number  as  if  it  were  a  wliole  number  ;  then,  to  facilitate  the  ojieration,  we  numerate 
from  right  to  left,  beginning  the  numeration  with  tenths,  and  continue  until  we  come 
to  the  required  place  or  order,  always  writing  O's  to  fill  the  places  not  occupied  by 
significant  figures.  Thus,  to  write  25  ten  thousandths,  we  first  write  the  25;  then 
we  begin  at  the  right  and  numerate  thus,  tenths,  hundredths,  thousandths,  ten- 
thousandths  ;  by  this  we  find  that  four  places  are  required  and  as  there  are  but 
two  figures  in  the  number,  we  prefix  two  O's  and  obtain  the  correct  result  .0025. 

1.    Write  104  hundred  thousandths.  Aus.  .00104. 

Explanalion .—AccoTA'iDg  to  the   above  directions,   we  write 

npTTTjATTnAi  the  104  and  then  commence  on  the  right   and  numerate   thns: 

(lOKU  Tenths,   hnndrudtlis,    thonsandths,    ten-thonsandths,    Imndred- 

.00104.  thonsandths.      Tliis   numeration    shows    that   five    i)laces    are 

required,  and  .as  we  have  but  three,  we  therefore  prefix  two  O's. 

EXERCISES. 

437.     1.     Write  10101  hundred  billionths.  Ans.  .00000010101. 

AVrite  decimally,  numerate,  and  read  the  following: 


2.  314  milliouths. 

3.  12  tliousandths. 

4.  107  billionths. 

5.  1  trillionth. 


14. 
15. 


5iiA9n 

100000000 

1000000 


G.  1205  ten  milliouths. 

7.  897  hundred  billionths. 

8.  1  sextillionth. 

9.  21001  teu  vigintillionths. 


IG. 
17. 


1  0* 

2  6_ 

i  0  0 ' 


10.  ?t 

11.  -?-o\ 
-to  jr  4  2__ 

-L-.  1  0  U  U" 

13.  ''''* 


1000000000 


87 

10000 


100000 


1^-                               100 
1 Q 99.00  as 

-'■"'•     lOOOUOOOOOOOO 


220 


SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS. 


rillNCIPLES. 

438.    From  the  foregoing  M^ork  we  recapitulate  the  following  principles  : 

1.  Decimals  are  governed  by  the  same  laws  of  notation  as  whole  numbers ; 
hence  the  value  of  any  decimal  figure  depends  upon  the  place  it  occupies. 

2.  Each  removal  of  the  decimal  point  one  place  to  the  right,  is  equivalent  to 
multiplying  the  decimal  by  10. 

3.  Each  removal  of  the  decimal  point  one  place  to  the  left,  is  ecpiivalent  to 
dividing  the  decimal  by  10. 

4.  Annexing  or  rejecting  naughts  at  tlie  right  of  any  decimal  does  not 
change  its  value. 


REDUCTION  OF  DECIMALS. 


TO  EEDUCE  DECIMAL  FRACTIONS  TO  A  COMMON  DENOMINATOR. 

439.     1.     Reduce  .7,  .18,  .2581,  and  .045,  to  a  common  denominator. 


OPERATION. 

.7000 
.1800 
.2581 
.0450 


Explanation. — To  reduce  decimals  to  a  common  denom- 
inator, -n-o  have  Ijiit  to  annex  a  sufficient  number  of  O's  to 
give  each  decimal  the  same  number  of  places. 


TO    REDUCE    A    DECIMAL    TO    A    COMMON    FRACTION. 


440.     1.     Reduce  .25  to  a  common  fraction. 


OPERATION. 

-nro  =  h  ^"S- 


Explanation. — In  all  problems  of  this  kind,  we  simply  write 
the  <lecimal  as  a  common  fraction  and  then  reduce  it  to  its 
lowest  terms. 


2.  Reduce  .125  to  a  common  fraction. 

OPERATION.  --^^-g  =  J,   AUS. 

3.  Reduce  .59|  to  a  common  fraction. 


FIRST  OPERATION. 
59a       AJLa.  Explanation. — To  rednce  complex  decimals 

^      nnd  ^-^^  -^  ---  =  --—  =  J-^    Ans        ^'^'  '''"'Pl*'  fractions,  we  first  write  the  decimal 

aaa'  '  B-»  800        3  a;  •        as  a  common  fraction;  then  we  reduce  both 

100        -s-f  j^jjg  numerator  and   the  denominator   to   the 

fractional  unit  of  the  denominator  contained 
in  the  numerator  term  of  the  fraction,  and  thus  obtain  a  complex  fraction,  which  we  reduce  to  a 
simple  fraction. 


*  REDUCTION    OF    DECIMALS.  221 

SECOND  OPERATION.  Explanation  — Here,   when  reducing  the   fraction  to    the 

r:tjn  fractional  unit  of  the  denominator  contained  in  the  nuuiera- 

475         13      A     ^  tor  term  of  the  fraction,  we  shorten  the  worlv  by   omitting 

=  sTJo  =  Tt')   Ans.  the  denominator  (8)  in  both  terms  of  the  complex  fracticm, 

100  and  writing  the  result  as  a  simple  fraction.     By  this  process, 

we  save  the  operation  of  division,  the  result  of  which  is  the 
cancelling  of  the  denominator  in  both  terms  of  the  complex  fraction. 


GEXEEAL    DIEECTIOX    FOR    EEDUCIXG    DECEVIAL    FEACTIOXS    TO 

COMMON    FEACTIOXS. 


441.     From  the  foregoing  elucidatioiLS,  we  derive  the  following  general  direc- 
tion for  reducing  decimal  fractions  to  common  fractions : 

Write  tlie  decimal  as  a  fraction,  then  reduce  the  fraction  to  its  loicest  terms. 

Eeduce  the  following  decimals  to  common  fractions : 

10.  .07, 

(.  11.  .005, 

12.  .10-15, 

13.  .88, 
•      14.  .909, 

15.     .0002; 


TO  EEDUCE  COMMON"    FEAGTIOXS    TO    EQUIVALENT    DECIMALS. 


442.     1.    Eeduce  |  to  a  decimal. 

OPERATION.  Explanaiion. — To  reduce  common  fractions  to  decimals,  we  an- 

nex naughts  to  the  numerator  and  divide  by  the  denominator,  and 
S\   "?  nnn  then  point  otf  as  many  spaces  for  decimals  as  there  were  O's  annexed. 

O)   o.UUU  When  a  remainder  continues  beyond  four  or  six  places,  we  discou- 

tinue  dividing  and  write  the  sign  -|-  to  the  right  of  the  last  figure 

•375,   AnS.  obtained,  which  indicates  that  the  quotient  is  not  complete.     The 

annexing  of  O's  to  the  numerator  is  equivalent  to  multiplying  it  by 

10  for  each  naught  annexed,  consequently  the  quotient  obtained  is  as  many  times  10  too  great  as 

there  were  O's  annexed ;  and  hence  the  reason  for  pointing  ofl"  aa  many  places  in  the  quotient  as 

there  were  O's  annexed  to  the  numerator. 

2.    Eeduce  f  to  an  equivalent  decimal. 

OPERATION.  7)  5.000000 


4. 

.8, 

Ans. 

5. 

.05, 

Aus. 

6. 

.25, 

Ans. 

7. 

.125, 

Ans. 

8. 

.075, 

Ans. 

9. 

.105, 

Ans. 

Ans. 

1  00- 

le. 

.48i, 

Ans. 

■No- 

Ans. 

1 

200- 

17. 

.0553, 

Ans. 

_2ai_ 

4000' 

Ans. 

2¥o^d- 

18. 

.008 1, 

Ans. 

Ans. 

fi- 

19. 

.00054tI^. 

Ans. 

Ans. 

20. 

.999, 

Ans. 

Ans. 

21. 

.4007-1-, 

Aus. 

2  2  n  3  9 
65000 

.714285-f,  Ans. 


222 


SOULE  S    PHILOSOPHIC    PP  ACTICAL    MATHEMATICS. 


3.    Eediice  yf^-  to  an  equivalent  ileemial. 


FIRST   OPERATION. 


SECOND    OPERATION. 


725)  3.000000(4137+  725)  3.    (.004137+,  Ans. 

2900        .004137+,  Ans.  30  tenths. 


1000 
725 

2750 
2175 


300  Immlredths. 

3000  tliousandths. 
2900 


5750 


1000  ten  thousandths. 
725 


Explaitation, — Here,  in  the 
first  operation,  we  annex  sis 
O's  and  obtain  but  4  ligures  in 
tho  quotient.  Therefore,  iu 
order  to  point  off  as  many  dec- 
imal phices  as  wo  annexed  O's, 
we  prefix  two  O's  and  thus  ob- 
tain the  correct  result.  The 
reason  for  til  is  will  ii])]iear  clear 
if  we  consider  each  step  of  the 
work  as  performed  in  the  sec- 
ond operation.  We  are  to  di- 
vide or  measure  3  liy  725,  and 
we  first  see  tliat  3  is  not  e<iual 
to  725  any  wliole  or  unit  num- 
ber of  times;  we,  therefore, 
write  the  decimal  point  in  the 
quotient,  annex  a  0  to  the  3 
units  and  thus  reduce  it  to  30 
tenths,  which  we  also  see  is  not 
equal  to  725  any  tenth  times, 
and  hence  we  write  0  in  the  tenths'  place  of  the  quotient ;  we  then  annex  another  0  .and  thereby  reduce 
the  30  tenths  to  300  hundredths,  which  we  see  is  not  equal  to  725  any  hundredths  times,  aud  hence  we 
write  0  in  tlio  hundredths'  place  of  the  quotient;  we  then  annex  another  0  and  thereby  reduce  the 
300  hundredths  to  3000  thousandths,  which  we  see  is  equal  to  725,  4  times,  with  a  remainder.  We 
have  now  obtained  the  first  significant  figure  of  the  decimal,  .and  we  continue  the  division  in  the 
usual  manner  to  tlie  sixth  decimal  ])lace  and  .annex  the  +  sign  to  indicate  that  there  is  still  a  remain- 
der. Tlie  sign  —  is  sometimes  used  to  indicate  that  the  last  figure  is  too  great.  Thus,  ^  ^ .  1666+ ; 
.167  — 


2750  hundred  thousandths. 
2175 


5750  millionths. 
5075 

675  remainder. 


4.    Eediice  C^  to  a  decimal. 

FIRST    OPERATION. 

6%  =  -/  aud  ^  =  4)  27.00 


SECOND   OPERATION. 

Gf  =  6  and  f ;  and 
=  4^3.00 


G.75,  Ans. 


.75  +  G  =  G.75,  Ans. 


ge:n"eral  direction-  for  redugixg  common   fractions  to 

equivalent  decimals. 


443.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tion for  reducing  common  fractions  to  equivalent  decimals: 

Annex  naughts  to  the  numerator  and  divide  by  the  denominator.  Then  point 
off,  from  the  right  of  the  quotient,  as  many  decimal  figures  as  there  are  naughts  annexed. 
Reduce  the  following  fractions  to  equivalent  decimals  not  exceeding  G  places : 
5. 
G. 
7. 
8. 


3  H' 

Ans. 

.71875 

9. 

1, 

Ans. 

.025 

1  1  0 

Ans. 

.336 

10. 

.1 

1  3> 

Ans. 

.07C923+ 

4 
1  257 

Ans. 

.032 

11. 

.37A, 

Ans. 

.370G25 

3of+, 

Ans. 

.107142+ 

12. 

47.18f, 

Ans. 

47.1875 

13. 


14. 
15. 


Reduce  §  to  a  comjilex  decimal  of  3  places. 

OPERATION. 

3)  2.000 


.G00§,  Ans. 
Reduce  f  to  a  complex  decimal  of  4  places. 
Reduce  f  to  a  complex  decimal  of  6  jilaces. 


Ans.  .4285f. 
Ans.  .2222221 


ADDITION    OF    DECIMALS,  223 

ADDITION  OF  DECIMALS. 


.785 

.93 

1GG.8 

72.5487 

4.17 

245.2337, 

Aus, 

444.    Addition  of  Decimals  is  flnding  the  sum  of  two  or  more  decimals. 

Since  decimals  increase  from  right  to  left,  and  decrease  from  left  to  right  in 
a  ten-fold  ratio  as  do  simple  whole  numbers,  they  may  be  added,  subtracted,  mul- 
tiplied, and  divided  in  the  same  manner. 

1.    Add  .785,  .93, 166.8,  72.5487,  and  4.17. 

OPERATION, 

Explanation. — In  all  problems  of  this  kind,  Tre  write  the 
numbers  so  th;vt  units  of  the  same  order  stand  in  the  same 
columu,  aud  the  decimal  points  are  iu  a  vertical  line ;  then  we 
add  as  iu  simple  whole  numbers.  When  the  addition  is 
completed,  we  point  ott'  iu  the  snui,  from  the  rifjlit  hand,  as 
many  places  for  decimals  as  equal  the  greatest  number  of 
decimal  places  iu  .any  of  the  numbers  added.  If  there  are 
complex  decimals  they  must  be  reduced  to  pure  decimals,  as 
far,  at  least,  as  the  decimal  places  extend,  in  the  other  numbers. 


GENERAL    DIRECTIONS     FOE    ADDITION    OF    DECIMALS. 


445.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  addition  of  decimals  : 

1.  Write  the  numbers  so  that  units  or  figures  of  the  same  order  stand  in  the 
same  column. 

2.  Then  add  as  in  simple  numbers,  and  point  off,  in  the  sum,  from  the  right,  as 
many  decimal  places  as  equal  the  greatest  number  of  decimal  places  in  any  of  the 
numbers  added. 

Add  the  following  numbers  : 

2.  3.25,  42.348,  748.4,  and  29.32.  Ans.  823.318. 

3.  .0049,  47.0426,  37.041,  and  360.0039.  Ans.  444.0924. 

4.  1121.6116,  61.87,  46.67, 165.13,  and  676.167895.  Ans.  2071.449495. 

5.  .8,  .09,  34.275,  562.0785,  and  1.01.  Ans.  598.2535. 

6.  81.61356,  6716.31,  413.1678956,  35.14671,  3.1671,  and  314.6. 

Ans.  7564.0052656. 

7.  l.Olf,  240.06i,  999.9,  80.6051,  and  .17.  Ans.  1321.7576. 

8.  What  is  the  sum  of  the  following  numbers:  twenty-five,  and  seven 
miUionths;  one  hundred  forty-five,  and  six  hundred  forty-three  thousandths;  one 
hundred  seventy-five,  and  eighty-nine  hundredths ;  seventeen,  and  three  hundred 
forty-eight  hundred  thousandths.  Ans.  363.536487. 

9.  A  farmer  sold  at  one  time  3  tons  and  75  hundredths  of  a  ton  of  hay ;  at 
another  time,  11  tons  and  7  tenths  of  a  ton;  and  at  a  third  time,  16  tons  and  125 
thousandths  of  a  ton.     How  much  did  he  sell  in  all  ?  Aus.  31.575  tons. 


224  SOULES    PHILOSOI'HIC    PRACTICAL    MATHEMATICS. 

SUBTRACTION  OF  DECIMALS. 


446.     Subtractiou  of  Decimals  is  finding'  the  difference  between  two  decimals. 

1.  From  345.304G  subtract  92.1435847. 

OPERATION. 

345.3046  Explanation. — In  all  problems  of  this  kind,  vre  ■write  the 

1)2.1435847  nnmliers  so  tlint  nnits  of  tUe  same  order  stand  in  the  same 

' colnmn,  and  the  decimal  j)oiuts  are  in  a  vertical  line;  then  we 

o-TTT^iivT-Q      A  snlitract  as  in  simple  whole  nnmbers,  and  point  off  in  the 

•jO.j.lolOlyo,   Alls.  difference,  from  the  right  hand,  as  many  ]ilace.s  for  decimals 

as  equal  the  sreiitest  unmlier  of  decimal  jilaces  in  cither  the  minuend  or  subtrahend. 

When  tlie  decimal  places  in  the  subtrahcnil  exceed  those  in  the  minuend,  naughts  are  uuder- 
stood  to  occujiy  the  vacant  places,  aud  may  be  tilled  iu  if  it  is  desired. 

2.  From  142.6§  subtract  51.11111 

OPERATION. 
rC\\  1111  Explanation. — In  all  problems  of  this  kind,  we  reduce  the 

Ol.llllg- 

complex  decimals  to  pure  decimals  of  equal  places  aud  then 


91.5o55g,  AnS.  subtract  as  in  subtractiou  of  fractions. 


GENEEAL    DIRECTIONS    FOE    SUBTEACTION    OF    DECIMALS. 


447.     From  the  foregoing  elucidations,   we  derive  the    following    general 
directions  for  subtractiou  of  decimals. 

1.  Write  the  numbers  so  that  the  units  of  the  same  order  stand  in  the  same 
column. 

2.  Then  subtract  as  in  simple  numbers,  and  icrite  the  decimal  point  as  in  addi- 
tion of  decimals. 

Note  1. — If  there  are  complex  decimals  of  unequal  places  in  either  or  both  of  the  given  dec- 
imals, reduce  them  to  pure  decimals  of  equal  places,  and  then  subtract  as  in  subtraction  of  fractions. 

Note  2. — If  there  are  not  as  many  decimal  places  iu  the  minuend  as  there  are  in  the  subtra- 
hend, annex  decimal  naughts  to  it,  until  the  decimal  places  are  equal. 

PROBLEMS. 
(3)  (4)  (5) 

SI. 04089  121.25  532.8 

14.587  109.05438  9.00451681 


66.45389,  Ans.                               12.19562,  Ans.  623.79548319,  Ans. 

6.  From  461.072  take  427.125,  Ans.  33.947, 

7.  From  17.5  take  4.19,  Ans.    13.31. 

8.  From  4000.0004  take  4.3,  Ans.  3995.7004. 

9.  From  three  million  take  three  millionths,  Ans.  2999999.999997. 

10.  From  11  take  1  and  9  trillionths,  Ans.  9.999999999991. 

11.  From  24000  subtract  2.078,  Ans.  23997.922. 

12.  From  886.333  subtract  98.5427,  Ans.  787.7903. 


MULTIPLICATION    OF    DECIMALS. 


225 


MULTIPLICATION  OF  DECmALS. 


448.    Multiplication  of  Decimals 

both  of  the  factors  coutaiu  decimals. 


is   fliuliug  the  product,  when  either  or 


1.    Multiply  26.58  by  4.3. 

PIKST    OPERATION. 

20.58 
4.3 


7974 
10632 


114.294,  Alls. 


SECOND   OPERATION. 


100 
10 


2068 
43 

114294 


Aus. 


1000 

or  decimallv  written 

114.294,  Aus. 


Explanation. — In  .ill  problems  of  this  kind,  we  multiply  as  in 
whole  numbers,  .and  point  off  on  the  right  of  the  product  as  many 
places  for  decimals  as  there  are  decimal  places  in  both  the  multipli- 
cand and  multiplier.  The  reason  for  thus  pointing  off  the  'A  deci- 
mal places  in  this  problem  is  obvious  from  the  fact,  that  in  the 
multiplicand,  we  have  2  decimal  places  or  hundredths,  which  we 
used  as  whole  numbers,  and  thereby  ])roduced  a  product  100  times 
too  great;  and  in  the  multiplier  we  have  I  decimal  jdace  or  tenths 
which  we  also  )ised  as  a  whole  number,  and  thereby  produced  a 
product  10  times  too  great ;  and  both  together,  give  a  product  1000 
times  too  great;  hence,  to  obtain  the  correct  product,  we  divide  by 
1000,  or  point  off  3  decimal  places. 

Explanation. — In  this  operation,  we  reduce  the  factors  to  com- 
mon fractions,  and  then  multiplying  them  together,  we  obtain  a 
product  of  Huoil'',  which,  written  decimally,  is  114.294.  This  pro. 
cess  shows,  in  another  waj-,  why  we  point  oft"  on  the  right  of  the 
product  as  many  places  for  decimals  as  there  are  decimal  places  in 
both  factors. 


2.    Multiply  4.024  by  .0056. 

OPERATION. 

4.024  Explanation.— In  all  problems  of  this  tind,  where  the  num- 

.0050  ber  of  figures  in  the  product  is  not  equal  to  the  number  of 

decimal  places  in  the  two  factors,  we  must  prefix  a  sulfieient 

24144  number  of  O's  to  supply  the  deficiency.     In  this  example,  we 

20120  prefix  one  0.     The  reason  of  this  will  ap^iear  evident  by 

1 —  working  the  example  as  a  common  fraction  as  shown  in  the 

.0225344,  Aus.  second  operation  of  the  first  problem. 


GENERAL    DIRECTIONS    FOK    MULTIPLICATION"    OF    DECIMALS. 


449.     From  the  foregoing  elucidations,  vre  derive  the  following  general  direc- 
tions for  the  iimltiplication  of  decimals : 

1.  Multiply  as  in  whole  numbers  and  from  the  right  of  the  product  i)oint  off  as 
many  figures  for  decimals  as  there  are  decimal  places  in  the  multiplicand  and  midtiplier, 

2.  If  the  product  does  not  contain  as  many  decimal  places  as  both  factors, 
supply  the  deficiency  by  prefixing  naughts. 

PROBLEMS. 

Ans.  24.3. 

Ans.  3.04. 

Ans.  .3150. 

Ans.  .0000018. 


3. 
4. 
5. 
0. 


[Multiply  27  by  .9, 
Multiply  .38  by  8, 
Multiiily  .75  by  .42, 
Multiply  .006  by  .0103, 


2  26  soule's  philosophic  practical  mathematics.  * 

7.  Multiply  OnnO.  by  .9990,  •  Ans.  9998.0001. 

8.  Multiply  7.0007  by  .()o(t3,  Aus.  .212l::il'l. 

9.  Multiply  340.012  by  G1.23,  Aus.  2081S.9347(!. 

10.  Multiply  .12;U  by  12;54.  Ans.  152.27o(;. 

11.  Multiply  1500  by  .00014,  Aus.     21. 

12.  What  is  tbe  product  of  one   tlionsaud   tvrenty-five   uiultiplied  by  tlirce 
hundred  tweuty-sevpn  ten-thousandths  ?  Ans.  33.5175. 

13.  What  is  the  product  of  seventy-eight  milliou  two  hundred  live  thousand 
two,  multiplied  by  fifty-three  hundredths?  Ans.  41448G51.0G. 

14.  Multiply  one  hundred  flfty-three  thousandths  by  one  hundred  twenty- 
nine  niillionths.  Ans.  .000019737. 

15.  Multiply  1  thousand  by  1  thousandth.  Ans.  1. 

16.  Multiply  2  milliou  by  ^  billionths.  Ans.  .004. 

17.  What  will  37.23  tons  of  hay  cost  at  $20.75  per  ton?         Ans.  $772.52-f. 

18.  What  will  428.431  bushels  cost  at  $1,125  per  bushel  ?      Ans.  $481.98+ . 

TO    MULTIPLY    A    DECIMAL    OE    MIXED    NUMBER    BY    TEN,    ONE 
HUNDEED,  ONE  THOUSAND,  Etc. 

450.     1.    Multiply  428.375  by  100. 

OPERATION.  ExplnnntioiK—Jn  all  jiroblems  wliere  the  iiuiltiplier  is  10, 

100,  I'tc.  wc  Kun])ly  remove  the  decimal  j)()int  as  iiiiiiiy  places 
4L8o<.i>,   Ans.  to  the  right  :i,s  tlJere  are  naughts  iu  the  multiplier,  annexing 

naughts  if  reciuired,  as  shown  in  Articles  426  aud  438. 

2.  Multiply  271.32  by  1000,  Ans.  271320. 

3.  Multiply  .756  by  lOc,  Ans.        75.0. 

4.  Multiply  .0-'5  by  10,  Ans.  .25. 

5.  Multiply  01.052  by  10000,  Ans.  610520. 


DIVISIOIT  OF  DECIMALS. 


451.     Division  of  Decimals  is  the  process  of  finding  the  t[uotieut  when  the 
divisor  or  dividend,  or  both,  contain  decimals. 
1.    Divide  17.094  by  8.14. 

FIKST   OPERATION.  Explanation. — In  all  problems  of  tliis  kind,  we  divide  as  in  TvLola 

numbers,  and  then  ]>oint  off  as  many  places  for  decimals  from  the 

8.14)  1 1 .09-i  {.j.l,  Ans.        right  of  the  quotient  us  the  decimal  places  in  the  dividend  exceed  those  in 

16  28  the  divisor,  oliserring  to  supply  uni/  deficiency  hy  prefixing  naughts.     In 

this  problem,  the  excess  is  one,  aud  we  therefore  piiiut  olf  ouo  dec- 

ot  I  imal  place  in  the  ([Uotient.     The   reason   for   tlius   pointing   olf  is 

^7*  obvious  from  the  fact  that  in  the  dividend  we  had  3  decimals  or 

°1*  THOi'SANDTiis,  and  in  the  divisor  we  had  2  decimals  or  hundredths, 

iiud  thousandths  divided  by  hundredths  give  tenths  as  a  quotient. 

The  reason  will  also  .appear  plain  if  we  observe  that  the  dividend  is  the  product  of  the 
divisor  and  quotient  multiplied  together,  and  hence  we  point  off  enough  decimal  places  in  the 
quotient  to  make  the  number  in  the  two  factors  equal  to  the  number  in  the  i)roduct  or  dividend, 
according  to  the  priuciples  shown  in  the  iirst  problem  of  multiplication  of  decimals. 


DIVISION    OF    DECIMALS. 


227 


SECOND   OPERATION. 

1000      17094:  Explanation. — In  tliis  operation,  wo  reduce  tlie  decimals  to  eonnnoa 

814:      100  fractions  and  tlipn  procfcd  as  in  tlie  division  of  mixed   numbers. 

The  Tt'ductiou  of  tlie  dividend  anil   divisor   to  conimou  fractions, 
'~       ,  and  then  tlie  mixed  nunil)ers  to  improper  fractions,    is   performed 

-.1,   Alls.  thus:  Thedividenil  17.()Ut  =  17TnTW  =  -r'iM,''„*;  tliedivisor  8.14  =  8,^^ 

Decimally  Wlitten2.1  AllS.  =  lA^.    'f  liis  niethod  also  .shows  the  reason  for  pointing  oft',  and  may- 
be used  for  all  jiroblems  in  decimal  fractions. 


PRINCIPLES    OF    DIVISION    OF    DECIMALS. 

452.     From  tlie  foregoing  elucidations,  we  derive  the  following  principles : 

1°.  The  dividend  ninst  contain  at  least  as  many  decimal  places  as  tlie 
divisor;  anil  wlieu  both  contain  the  same,  the  quotient  is  a  whole  number. 

2°.  The  dividend  is  the  product  of  the  divisor  and  qnotieut,  and  hence  con- 
tains as  many  decimal  places  as  both  the  divisor  and  quotient, 

30.  The  quotient  must  contain  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceeds  the  number  iu  the  divisor. 

2.    Divide  7898.5G  by  2.4G83. 


OPERATION. 

2.4683)  7808.5600  (3200,  Ans. 
74049 


49366 
49366 


00 
3.     Divide  7.0761  by  687. 

OPERATION. 


Explanation.— Here  -we  have  an  excess  of  decimals  in 
the  divisor,  and  iu  all  cases  of  this  kind,  we  first  make 
them  equal  hy  annexing  naughts  to  the  dividend,  and 
the  quotient  will  be  a  whole  number.  The  reason  for 
annexing  the  naughts  will  appear  more  obvious  hy 
solving  the  problem  in  the  form  of  a  common  fraction. 


687)  7.0761  (103 

687         .0103,  Ans. 


2061 
2061 


10000 

687 


or  thus : 
0761 


.0103,  Ans. 


Explanation. — In  this  problem,  there 
are4decimalplacesinthe  dividend  and 
none  in  the  divisor;  hence,  according 
to  the  foregoing  instructions,  we  must 
point  otf  4  decimal  places  in  the  quo- 
tient, and  as  there  are  but  3  figures  in 
the  quotient,  we  prefix  1  nauglit.  In 
all  problems  of  this  kind,  O's  are 
prefixed  to  supply  any  deficiency  of 
figures  that  may  occur. 


4.    Divide  47.789  by  39.27. 

OPERATION. 

39.27)  47.789  (1.2168+,  Ans. 
39.27 


8519 
7854 

6650 
3927 

27230 
23562 

36680 
31416 


Explanation. — In  this  problem  we  have  a  remainder, 
after  dividing  the  dividend,  of  6C5 ;  to  this  and  the  two 
successive  remainders,  we  annex  O's  and  continue  the 
division  until  we  have  produced  4  decimal  places.  The 
annexing  of  O's  reduces  the  successive  remainders  to  the 
next  lower  order  of  tenths,  and  hence  all  quotient  figures 
produced  by  annexing  O's,  are  decimals.  We  therefore 
point  oft'  from  the  right  of  the  quotient  as  many  places  for 
decimals  as  the  number  of  decimals  iu  the  dividend  exceeds 
those  of  the  divisor  plus  the  number  of  O's  annexed. 
This  is  done  in  all  division  problems  where  O's  are  an- 
nexed, and  .a  sufficient  number  of  O's  should  he  annexed 
to  produce  4  or  6  decimal  places.  'V^'hen  there  is  a  remain- 
der after  the  last  division,  the  plus  (+)  sign  should  be 
annexed  to  the  answer  to  indicate  that  the  quotient  is 
incomplete. 


228 


soule's  philosophic  practical  mathematics. 


GENEKAL    DIEECTIOi!^S    FOE    DIVISION    OF    DECIMALS. 

453.    From   the  foregoing  elucidations,   we  derive  the  following    general 
directions  for  dividing  decimals : 

1.  Divide  as  in  ichole  numbers  and  point  off  as  many  decimal  places  in  the 
quotient  as  those  in  the  dividend  exceed  those  in  the  divisor. 

2.  Whin  there  is  a  remainder,  anncjc  naughts  to  the  dividend  and  carry  the 
trorJv  as  far  as  may  he  desired. 


5.    Divide  .112233  by  12. 

OI'EllATION. 

12)  .112233 

9352+=.009352+,  Ans. 

7.    Divide  11.2233  by  12. 

OPERATION. 

12)  11.2233 


,0. 


Divide  1.122.33  by  12. 

OI'KKATION. 

12)  1.12233 


9. 


.9352+,  Ans. 
Divide  .00048G9  by  396. 

OPERATION. 

396)  .0004869(12+ 

396  =.0000012+,  Aus. 

909 

792 


9352+ =.09352+,  Ans. 
.     Divide  112.233  by  12. 

OPERATION. 

12)  112.233 


117 


9.3527+,  Ans. 
10.  Divide  .0004809  by  3.96. 

OPERATION. 

3.96). 0004869(12+ 

396     =  .00012+,  Aus. 

909 
792 


12. 
13. 
14. 
15. 
16. 


117 

Divide  67.8632  by  32.8.  Ans.  2.009. 
Divide  983  by  6.6.  Ans.  148.939+ 
Divide  13192.2  by  10.47.  Ans.  1260. 
Divide  07.56785  by  .035.  Ans.  1930.51. 
Divide  .00125  by  .5.        Ans.  .0025. 


11.  Divide  .0004869  by  .0396 

I'IKST    OPHUATION. 

.0396).0004S69(  122+ 

396  =.0122+,  Ans. 

909 

792 

1170 
792 

378 

SECOND   OPERATION. 

396)4869(122+ =.0122+,  Ans. 
396 

909 
792 

1170 
792 

378 


17.  Divide  7.482  by  .0006.       Ans.  12470. 

18.  Divide  1  by  999.  Ans.  .001001  +  . 

19.  Divide  84375  by  3.75.         Ans.  22500. 

20.  Divide  1081  by  39.50.  Ans.  27.3255+. 

21.  Divide  35.7  by  485.         Ans.  .0736+. 


22.  If  rice  costs  $.0775  i^er  pound,  how  many  pounds  can  be  bought  for 
$40.64875?  Ans.  524.5  pounds. 

23.  Sold  14.75  acres  of  laud  for  $191.75.     MTiat  was  the  price  per  acre? 

Ans.  $13. 

24.  Divide  four  thousand  three  hundred  twenty-two   and  four  thousand  five 
hundred  seventy-three  teu-thousandths,  by  eight  thousand  and  nine  thousandths. 

Ans.  .5403+. 

TO    DIVIDE    DECIMAL    FRACTI0:N^S    BY    TEN,    ONE    HUNDEED,  OXE 

THOUSAND,    Etc.,    Etc. 

454.     1.     Divide  48.76  by  10. 

OPERATION.  Explanation. — In  all  problems  of  this  Icinil,  we  simply  remove 

the  ileeiiu.il  point  as  many  ])laces  to  the  left    as  there  are  O's  in 

4.876,  Ans.  the  divisor.     The  reason  for  this  was  fully  shown   in  Articles 

426  aufl  438.     When  there  are  not  a  sufficient  number  of  figures 

in  the  dividend  to  allow  this  to  be  done,  naughts  must  lie  prefixed  to  supply  the  deficiency. 

2.  Divide  875.25  by  100.   Ans. 

3.  Divide  .5231  bv  1000.    Ans. 

4.  Di^dde  72  by  10000.      Ans. 


8.7525. 

5. 

Divide  9.85  by  100. 

Ans.  .0985. 

.0005231. 

6. 

Divide  .025  by  200. 

Ans.  .000125 

.0072. 

7. 

Divide  412.99  by  10. 

Ans.  41.299. 

Compound  Denominate  Numbers. 


DEFINITIONS. 


455.  A  Denominate  Number  i«  a  concrete  number  which  expresses  a 
particular  kind  of  unit  or  quantity,  either  simple  or  compound. 

156.  A  Simple  Denominate  XumLer  is  a  number  which  expresses  a  unit 
or  units  or  quantities  of  but  one  kind  or  denomination :  as  5  dollars,  15  boxes,  8 
pounds,  i  days,  etc. 

457.  A  Compound  Denominate  Number  is  a  number  which  expresses  units 
or  quantities  of  tico  or  wore  denominations,  under  one  land  of  measure:  as  8  dollars, 
20  cents;  1:2  pounds,  10  ounces;  7  days,  18  hours,  21  minutes,  etc. 

In  u-hole  numbers  and  in  decimals,  the  law  of  increase  and  decrease,  between 
units  of  lower  and  of  higher  orders,  is  by  the  uniform  scale  of  10 ;  but  in  compound 
numbers  the  scale  varies  according  to  the  kind  of  measure  employed. 

MEASURES. 

458.  A  Measure  is  a  standard  unit  established  by  law  or  custom,  by  which 
quantity,  such  as  extent,  dimension,  capacity,  amount,  or  value,  is  measured  or 
estimated. 

There  are  eight  kinds  of  measure : 

1st.  Length.  2d.  Surface,  or  area.  3d.  Solidity,  or  cajjacity.  4th.  Volume. 
6th.  AYeight,  or  force  of  gravity.    Gth.  Time.     7th.  Angles.    8th.  Money,  or  Value. 

WEIGHT. 

459.  Weiglit  is  that  i)roperty  of  matter  by  virtue  of  which  a  body  teuds 
toward  the  centre  of  the  earth;  and  the  resistance  required  to  overcome  this 
centralizing  pressure,  or  gravitating  tendency  of  matter  is  named  weight.  Weight 
varies  according  to  the  quantity  of  matter  and  its  distance  from  the  centre  of  the 
earth. 

VALUE. 

460.  Talue  is  the  ratio  or  unit  of  measure  of  wealth  existing  between 
different  commodities  with  reference  to  an  exchange.  It  is  the  sole  condition  of 
■wealth  and  the  universal  name  given  to  the  inherent  quality  or  poicer  of  one  thing  to 
command  another  in  exchange. 

Briefly  expressed,  value  is  the  worth  of  one  thing  as  compared  with  some 
other  thing. 

(229) 


230  SOULE  S    I'lIILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

461.  Money  if*  stiimped  metal  called  coin,  or  printed  bills  or  notes  called 
pai^er  money,  issned  by  the  general  government  of  States  or  Nations,  and  snpi)licd 
to  the  people  to  facilitate  trade  and  commerce.  It  is  the  standard  measure  of  the 
valne  of  things  and  services,  and  hence,  it  is  the  universal  medinm  of  exchange 
and  settlement  among  all  civilized  peoples.     See  Domestic  Exchange. 

462.  Currency  (from  cnrro,  I  run),  is  a  term  aijplied  to  the  money  of  a 
nation,  whether  it  be  coin  or  i>aper  money. 

463.  Bullion  is  the  name  given  to  uncoined  gold  and  silver,  and  includes 
bars  or  ingots. 

464.  Legal  Tender.  By  tliis  term  is  meant  money  which  is  required  l)y 
lavs-  to  be  accejited  in  payment  of  debts. 

465.  Gold  Coins  are  a  legal  tender  for  any  amount,  when  not  below  the 
standard  weight  prescribed  by  law.  "When  gold  coins  are  reduced  by  natural 
abrasion,  below  the  standard  weight,  they  are  a  legal  tender  at  a  valuation  propor- 
tional to  their  actual  M'eight. 

466.  Standard  Silver  Dollars  are  by  Act  of  Congress  of  1S78,  a  legal 
tender  for  all  debts,  except  in  contracts  where  other  money  is  specified. 

467.  United  States  £ank  Notes  are  a  legal  tender  for  all  debts  except 
duties  on  imports  and  the  interest  on  the  public  debt. 

468.  Silver  Certificates  are  a  legal  tender  for  all  debts  and  are  also 
receivable  for  customs,  taxes  and  all  x^ublic  diies. 

469.  National  Bank  Notes  are  not  a  legal  tender ;  but  as  they  are  secured 

by  a  deposit  with  the  Comptroller  of  Currency  at  Washington,  and  are  redeemable 

in  lawful  money  by  the  National  Banks,  and  the  Treasurer  of  the  United  States, 

they  are  perfectly  good,  and  are  accepted  without  question  by  individuals,  firms 

corjiorations,  States  and  tlie  general  government,  in  payment  of  all  kinds  of  debts, 

except  interest  on  the  public  debt,  throughout  the  United  States  of  America. 

Note. — Standard  Silver  Dollars  m.'iy  ho  deposited  in  amounts  not  less  than  $10  witli  the 
Treasurer  or  any  Assistant  Treasurer  of  the  United  States,  and  Silver  Certificates  received 
therefor. 


470.  IVIONETARY  UNITS  OP  DIFPEEENT  AGES,  NATIONS  AND  PEOPLES. 

A  great  variety  of  articles  and  objects  have  been  made  to  serve  as  money  in  different  ages  and 
different  countries.  The  money,  or  measure  of  value  of  the  ancient  Greeks  was  an  ox.  We  leara 
from  Homer  that  the  armor  of  Diomede  cost  nine  oxen,  and  that  of  Glaucus,  one  hvmdred. 

History  tells  us,  also,  that  when  Pythagoras  made  his  first  great  discovery  iu  Mathematics — 
that  the  square  described  upon  the  hypotenuse  of  a  right  angled  triangle,  -was  equal  to  the  sum  of 
the  squares  described  upon  the  other  two  sides — he  sacrificed  100  oxen,  in  honor  of  the  discovery 

Evidence  is  .abundant  to  show  that,  in  the  most  distant  regions  and  at  different  times,  cattle 
formed  a  currency  for  the  early  pastoral  .and  agricultural  nations.  The  ancient  Germans  used 
cattle;  and  cattle  and  slaves,  or  "living  money,"  were  iu  commou  use  among  the  Anglo-Saxons. 


*  MONETARY    UNITS    OF    DIFFERENT    AGES,    NATIONS    AND    PEOPLES.         23 1 

With  the  early  Italians  and  Romans,  sheep  and  oxen  formed  the  oldest  medinm  of  exchange,  ten 
sheep  equalling  one  ox.  lu  the  early  English  and  Irish  civilization,  the  sheep  and  ox  were  the 
units  of  exchange.  The  Zulns  and  Kaffres  estimate  their  wealth,  to-day,  in  the  unit  of  cattle,  and 
use  them  as  money  in  all  transactions  which  involve  payment. 

The  savage  tribes  of  Australia  use  greenstone,  serviceable  for  making  hatchets,  and  red  ochre, 
serviceable  for  painting  their  bodies,  as  a  general  medium  of  exchange  or  money. 

The  Indians  of  British  Columbia,  iise  for  money  haiqua  shells ;  while  the  Indians  of  Alaska 
nse  the  .skins  of  animals.  The  early  C'arthagenians  and  Spartans  used  for  money,  pieces  of  leather 
and  skins,  marked  with  a  government  stamp.  In  Chinese  Tartary,  cubes  of  tea  pressed  together; 
in  Ilindostan  and  the  Western  coast  of  Africa,  shells  called  cowries,  and  in  Abyssinia,  blocks  ol 
rock  salt  are  to  this  day  used  as  money.  In  Iceland,  dried  fish,  and  among  the  North  American 
Indians,  the  helt  of  u-anqxim,  or  Indian  beads  made  of  shells,  are  used  as  money.  The  monetary 
history  of  some  of  the  American  colonies,  from  1620  to  1700,  shows  that  almost  every  commodity  pro- 
duced or  manufactured,  was  used  as  money. 

The  early  settlers  of  Virginia  made  Tobacco  a  legal  tender  for  the  payment  of  debts ;  and 
hundreds  of  Virginians  bought  their  wives  from  English  wife  sellers  with  tobacco,  paying  from  120 
to  150  pounds  per  wife.  This  was  undoubtedly  the  best  use  ever  made  of  body  and  brain  injuring 
tobacco. 

Following  these  various  rude  forms  of  money,  iron,  lead,  copper,  tin,  and  later,  silver  and 
gold,  were  introduced  by  the  ancients,  and  copper,  bronze,  silver  and  gold,  and  paper  bills  are  used 
by  most  modern  nations.     Russia  at  one  time  used  platinum  for  money,  but  found  it  unsuitable. 

Nearly  all  modern  civilized  nations  have  adopted  gold  and  silver  as  the  general  standards  for 
money.  They  are  among  the  most  imperishable  of  all  substances,  contain  great  value  in  small 
volume,  and  are  easily  transported  or  secreted,  which  are  important  considerations  in  business,  and 
in  case  of  unjust  or  insecure  government. 

Our  own  country,  and  many  others,  find  it  convenient  to  use  copper,  nickel  and  bronze  for  the 
subdivisions  of  the  monetary  unit  in  connection  with  gold,  silver,  and  paper  money. 

MEASURE     TABLES. 

471.  A  Measure  Table  is  a  reguliirly  arranged  statement  showing  Low 
many  times  a  higher  denomination  of  a  system  of  measurement  equals  the  next 
lower  denomination  of  tlie  same  system.  Or,  in  other  words,  it  shows  how  many 
units  of  each  lower  denomination  are  required  to  equal  one  unit  of  the  next  higher 
denomination. 

472.  TABLE    OF    UNITED     STATES    MONET. 

10  Mills  (m)  =  1  Cent,  C. 

10  Cents  =  1  Dime,  d. 

10  Dimes  =  1  Dollar,  8. 

10  Dollars  =  1  Eagle,  E. 

20  Dollars  =  Double  Eagle. 

Note  1. — The  mill  is  not  coined.     It  is  used  only  in  computations. 

Note  2.— The  word  dollar  is  derived  from  Thai,  a  dale  or  valley ;  or  from  Joachim's  Thaler, 
first  coined  in  the  mines  of  the  Bohemian  Valley  of  Saint  Joachim,  from  whence  it  was  borrowed  from 
the  Austrians  bv  Spain,  during  her  union  with  Austria,  under  the  Empire  of  Charles  the  V.,  and 
•which  was  known  as  the  old  Spanish  Carolns  pillar  dollar,  which,  for  ages,  had  been  used  by- 
bankers  as  a  standard  of  value  for  the  English  monetary  pound,  during  its  many  changes. 

The  diyne  is  from  the  French  word  disme  meaning  one-tenth ;  the  cmt  is  from  the  Latin  centum, 
a  hundred  ;  the  mill  is  from  the  Latin  wille.  a  thousand  ;  the  ear/le  the  §10  coin,  is  the  name  of  the 
national  bird.     See  page  38,  for  the  Origin  of  the  Symbol  of  the  Dollar  (  $  ). 


E. 

$    d.     ct.     m. 

1  = 

10  =  100  =  1000  =  10000 

1  =  10  =  100  =  1000 

1  =   10  =   100 

1  =    10 

232 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


473.  United  States  Money  is  tlie  legal  monetary  measure  of  value  in  tlie 
TJnited  States  of  America.    The  unit  of  the  measure  is  the  gold  and  the  silver  dollar. 

47-1.  The  U.  S.  monetary  unit,  the  Dollar,  was  established  by  the  Continental 
Congress,  August  8,  1780,  Avith  the  proviso  that  it  should  be  decimally  divided, 
■w'hich  gives  great  facility  in  commercial  computations. 

475.  The  coin  of  the  TJnited  States  consists  of  gold,  silver,  nickel,  and 
bronze.  The  following  table  shows  the  name,  value,  composition  and  weight  of  each 
coin,  as  now  issued  (1894)  by  the  Mints : 

Note. — ParaRraph  5  of  Section  8,  of  Article  I,  of  tlio  Constitution  of  tlie  United  States,  gives 
Conpcress  power  "to  coin  money,  regulate  tbo  value  thereof,  anil  of  foreign  coins,  and  to  tix  the 
'Btaiulard  of -neights  and  measures ;"  and  Section  10  of  Article  I,  says  no  State  shall  coin  money, 
emit  Bills  of  Credit,  or  make  any  thing  but  gold  or  silver  coin  a  tender  in  payment  of  debts. 

By  Act  of  Congress,  of  February  25th,  1862,  the  issue  of  United  States  Treasury  Notes  was 
authorized  and  made  "lawful  money  and  a  legal  tender  in  payment  of  all  debts,  public  and  jiri vate, 
within  the  United  States,  except  duties  on  imports  and  interest  ou  bonds  and  notes  of  the 
United  States." 


476. 


TABLE  OF  UNITED   STATES   COIN. 


COIN. 

VALUE. 

COMPOSITION. 

WEIGHT. 

GOLD. 

Dollar 

100  cents 

90  "narts  ""old    10  Darts  allov 

25.8  grains  Troy. 

64  5       "          " 

Quarter  Eagle 

Three   Dollars.... 

90     "         "        10     "         "      

3  dollars 

90     "         "        10     "         "      

77  4      "          " 

Half  Eagle 

Eai'le 

5  dolLirs 

90     "        "    ,10     "        "      

129        "          " 

10  dollars 

90     "        "    ,10     "        "      

258        "          " 

20  dollars 

90     "        "     ,10     "        "      

516        "          " 

SILVER. 

Dime 

10  cents 

90  parts  silver,  10  parts  alloy 

38.58  grains  Troy. 
96  45       "        " 

Quarter  dollar 

Half  dollar 

25  cents 

90     "         "     ,10     "        "      

90    "         "     ,10     "        "      

50  cents 

192.9        "        " 

Dollar 

100  cents 

90     "          "     ,10     "        "      

412.5         "         " 

NICKEL. 

3-cent  piece 

5-cent  piece 

3  cents 

75  parts  copper,  25  parts  nickel 

30       grains  Troy. 
77.16      "          " 

75      "          "       ,  25     "           "       

ISKOXZE. 

One  cent 

1  cent 

95  parts  copper,  5  parts  tin  and  zinc 

48  grains  Troy. 

Note  1.— From  1786  to  1834  the  $10  gold  piece  weighed  270  grs.,  [i  pure,  and  all  other 
denominations  were  in  like  proportion. 

From  1834  to  the  present  time  1894,  it  has  weighed  258  grains,  -j^.j  pure. 

Note  2. — The  silver  dollar  weighed,  from  1786  to  1837,  416  grains,  |i  pure,  and  all  other 
denominations  in  silver  were  in  like  proportion.  From  1837  to  18.53,  it  weighed  412.5  grains.  -iV 
pure,  and  the  50c,  2.5c,  10c,  and  5c.  pieces,  in  like  proportion.  In  1851,  the  3c.  piece  was  coined,  f 
pure.  From  1853  to  the  ])resent  time,  the  Dollar  has  weighed  412.5  grains,  but  the  subdivisions 
thereof  were  reduced  in  1853,  to  the  weights  shown  in  the  above  table. 

477.  Silver  Coin  of  the  denominations  of  50/,  25/,  10/,  and  5/,  are  all 
made  under  the  standard  weight  to  prevent  their  being  shipped  abroad  and  are 
legal  tender  only  for  amounts  not  exceeding  $10. 

478.  The  Nickel  5  and  3-cent  pieces,  and  the  bronze  1-cent,  are  legal  tender 
only  for  amounts  not  exceeding  25  cents. 

Note. — The  5c.  nickel  weighs  5  grams,  or  77.16  grains,  and  is  nearly  ^  of  a  meter  in  diameter"; 
48.6,  laid  side  by  side,  measure  one  meter. 

See  Report  of  Director  of  the  U.  S.  Mint  of  1893,  for  all  the  laws  relating  to  the  coinage  of  U. 
R.  monev,  and  the  weights  and  purity  of  the  various  coins  from  the  founding  of  the  government  to 
July  1893. 


COMPARATIVE    VALUES    OF    GOLD    AND    SILVER.  233 

COMPARATIVE     VALUES     OF     GOLD     AND     SILVER. 

As  sliown  in  tlie  above  table  of  coins,  25.8  grains  of  uine-tentUs  pure  gold 
coin  and  412.5  grains  of  nine- tenths  pure  silver  coin  are  eacli  $1.00  United  States 
money.  Hence  tlie  monetary  ratio  between  gold  and  silver  in  the  United  States  is 
as  1  to  15.9S8.  i.  e.  silver  being  1,  gold  is  15.988. 

lu  England  this  ratio  is  1  to  14.287;  in  France  it  is  1  to  15.5;  and  in 
Germany,  1  to  13.95. 

ALLOWANCE    MADE    FOR    DEVIATION    IX    THE    WEIGHT    OF    COIN. 


479.  In  the  coinage  of  the  United  States  money  the  following  allowance  is 
made  by  law  for  a  deviation  in  M-eight : 

^  grain  for  the  Double  Eagles  and  Eagles; 
i  grain  for  all  other  gold  pieces ; 
l.i  grains  in  all  silver  pieces ; 

3  graiTis  in  the  nickel  Scent  piece,  and  2  grains  in  the  nickel  3-cent 
piece,  and  bronze  1-cent  piece. 

480.  The  old  silver  lialf  dime  and  3-cent  i)ieces,  the  bronze  2-cent  pieces, 
and  the  nickel  1-cent  pieces  are  not  now  coined. 

481.  The  Trade  Dollar  was  coined  for  Asiatic  commerce,  and  not  for 
currency.    The  weight  is  420  grains,  -i%  pure. 

482.  The  Alloy  of  a  coin  is  some  harder  metal  mixed  with  the  gold  or 
silver  to  harden  it  moderately  and  thus  lessen  the  wear  or  abrasion.  Gold  and 
silver,  in  a  pure  state,  being  very  soft,  would  rapidly  wear  away,  were  they  not 
alloyed. 

The  alloy  for  American  gold  coin  is  composed  of  about  i\  silver  and  tV  cop- 
per. The  difference  in  color  of  our  gold  coins  is  because  of  the  diflerent  quantity 
of  silver  in  the  alloy. 

The  alloy  for  silver  coin  is  pure  copper.     Coin  thus  alloyed  is  called  standard. 

WEIGHT    OF    COIN. 

483.     $10000  gold  =  258000  gr.  =  44  lbs.  9  oz.  10  pwt.  0  gr.  Troy. 

$1000  standard  silver  dollars  =  412500  gr.  =  71  lbs.  7  oz.  7  pwt.  12  gr.  Troy. 
$1000000  gold  weighs  53750  ounces  Troy  or  3085.71  avoirduj^ois  pounds. 
$1000000  standard  silver  dollars  =  412500000  gr.  Troy  or  =  58928.55+ 

pounds  avoirdupois  =  3. 08+  car  loads  of  10000  pounds  each. 
$1000000  silver  trade  dollars  weigh  875000  ounces  Troy  or  GOOOO  pounds 

avoirdupois. 
$1000000  silver,  half  and   quarter  dollars,   20-cent  pieces   and  dimes, 

weigh  803750  ounces  Troy  or  55114.28  avoiidupois  xjounds. 


234  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.    ■      '  * 

VALUE     OF     GOLD     AXD     SILVEIl. 

48-1:.     The  following  values  are  based  unon  the  i>reseiit  United  States  mone- 
taiy  v:da;itiou  (1894)  of  these  metals: 

1  ounce  Troy  of  pure  gohl  is  wovth  $20.07+ 

1  pennyweight  Troy  of  jiure  gold  is  worth  1.03  + 

1  ounce  Troy  of  pure  silver  is  worth  1.29  + 

1  itennjnveight  Troy  of  pure  silver  is  worth  .0(545+ 

CANADA     MOXEY. 

485.  Canada  Money  is  the  legal  currency  of  the  Dominion  of  Canada.    It 
consists  of  gold,  silver,  and  bronze  coin  and  of  paper  money. 

486.  The  silver  coins  are  the  SOc',  25f',  10c',  and  5C  pieces.    The  bronze 
coin  is  the  1/'  piece.     The  gold  coins  in  use  are  the  sovereign  and  half  sovereign, 

ENGLISH    MONEY. 

487.  English,  or  Sterling  Money,  is  the  legal  currency  of  Great  Britain. 

Note. — Sterling  is  derived  frmu  Ktisterlinj;,  the  uame  given  to  tbe  early  Geriiiau  traders 
■whose  money  was  called  Easterlinjj  money. 

488.  The  Monetary  Unit  of  Great  Britain  is  the  pound  sterling  which  is  a 
gold  coin  weighing  123.274  grains,  \h  pure.     It  is  equivalent  to  $4.8065  U.  S.  money. 

JJOTE. — For  exchange  purposes  hetween  the  United  States  and  England,  the  pound  sterling 
is  valued  by  li:iiikers  at  •'Si.SOi  and  tlie  rate  of  exchange  is  (juoted  in  dollars  aud  cents,  $4.86J,  more 
or  less,  according  as  premium  is  charged  or  discount  is  allowed. 

TABLE     OF    ENGLISH    MONEY. 

4  Farthings  (far.  or  qr.)  = 
12  Pence  = 

20  Shillings  = 

21  Shillings  = 
1  Crown                            = 


1  Penny 

d. 

ff.      £ 

s.        d.       far. 

1  Shilling 

s. 

1   =  12^0- 

=  21  =  252  =  1008. 

(  1  Sovereign 
;          or 

sov. 

1 

=  20  =  240=  900. 

1=    12=     48. 

(  1  Pound 

£. 

1=       4. 

1  (iuinea 

S- 

5  Shillings 

c. 

489.    The  money  of  Great  Britain  consists  of  gold,  silver,  copper,  and  Bank 
of  England  notes,  or  bills. 


VALUE    OF    FOREIGN    MONIES.  235 

VALUE  OP   EXGLISn   MOXEY  IX   THE  IT.  S.   MO:^fETAEY  UXIT. 


Pt 

esont  legal  value. 

Value 

previous  to  1834. 

1 

Pound  or 

Sov 

ereign 

n= 

$4:.SG6j 

UAH 

1 

Shilling 

= 

.1»433 

00  2. 

1 

Penny 

= 

.0202 -f. 

•oiff 

1 

Oi-owii 

= 

1.21CG+ 

1 

Guinea 

= 

5.1098+ 

Note. — By  Act  56,  GeorjioIII.,  ■n-liioh  is  still  in  forfp.  one  pound  of  standard  Troy  gold  (\i  pnrej 
Is  coined  into  £46.  14s.  6d.  The  lull  weight  of  one  Rold  pound  or  sovereign  is  therefore  123.274 
grains  of  standard  gold,  or  113.001  grains  of  ]>ure  gold.  Bnt,  allowing  for  the  .abrasion  or  wear, 
a  sovereign  weighing  but  122.75  grains  of  standard  gold,  is  in  England  a  legal  tender  for  the 
payment  of  debts. 

A  pound  of  silver  ^  pure,  is  coined  Into  66  shillings.  The  full  weight  of  the  shilling  is 
therefore  87 1-*!-  grains  standard  silver,  or  iiO-,\  grains  of  jiure  silver. 

A  pound  of  copper  is  coined  into  24  pennies. 

A  pound  of  bronze,  955!o  copper,  4%  tiu,  and  1^^  zinc,  is  coined  into  40  jiennies  or  80  half 
pennies,  or  100  farthings. 

Bank  of  England  Notes  are  .i  legal  tender  for  any  sum  exceeding  £5.  Gold  is  .-i  leg.al  tender 
for  any  amount;  silver,  not  exceeding  408.;  and  copper  not  exceeding  12d.,  when  in  pennies  or  half 
pennies,  and  not  exceeding  6d.,  when  in  farthings. 

The  gold  coins  are  the  sovereign  and  the  half  sovereign.  The  silver  coins  are  the  crown, 
half  crown,  4s.,  2s.,  Is.,  and  6d. 

See  English  Exchange  fur  the  history  of  English  money,  and  a  full  elucidation  of  the  subject. 

FRENCH    MONEY. 

490.  French  Money  is  the  legal  currency  of  France.  It  is  based  on  the 
decimal  system,  and  the  Unit  is  the  silver  Franc,  which  equals  19.3  cents  U.  S. 
money. 

NoTK. — In  exchange  transactions  between  the  United  .States  and  France,  the  rate  of  exchange 
is  the  variable  number  of  francs  and  centimes  allowed  for  SI.  The  basis  for  the  rate  is  5.20  francs 
for  $1.  This  rate  is  the  par  of  exchauyv,  and  is  quoted  more  or  less  as  premium  is  declared  or  discount 
is  allowed. 


TABLE     OF 

FRENCH 

MONEY. 

fr.     dc.      ct.         m. 

10  Millimes  (m. 

)=  1  Centime 

ct. 

1  =  10  =  100  =  1000 

10  Centimes 

=  1  I^ecime 

dc. 

1=    10=    100 

10  Beciines 

= 1  Franc 

■fr. 

1  =      10 

Note. — The  Millime  is  not  coined;  the  term  simply  means  the  tenth  part  of  a  centime. 

491.  The  money  of  France  consists  of  gold,  silver,  bronze,  and  XationaJ 
Bank  notes.     All  French  coin  is  ba.sed  upon  the  unit  of  weight — the  grnnime, 

A  kilogramme  of  standard  gold,  -^  pure,  is  coined  into  155  Napoleons  (20 
franc  pieces),  or  3100  francs.    The  gold  coins  are  the  100,  50,  20, 10,  and  5  franc  pieces. 

A  kilogramme  of  silver  is  coined  into  200  francs. 

The  silver  coins  are  the  5,  2, 1,  J,  and  ^  franc  pieces. 

The  copper  or  bronze  coins  of  the  French  are  the  10,  5,  2,  and  1  centimes, 
which  weigh  respectively  10,  5,  2,  and  1  grammes. 

The  Franc  is  used  in  Switzerland  and  Belgium,  and  under  different  names, 

in  Spain,  Italy,  Greece,  and  Venezuela. 

Note. — See  the  subject.  Foreign  Exchange,  in  this  work,  for  a  full   elucidation  of  French 
Exchange.     For  the  history  of  French  money,  see  the  Metric  System,  in  this  work. 


236  SOULES    rilll^OSOPHIC    PRACTICAL    MATHEMATICS.  ^ 

GEEMAX    MONEY. 

492.  German  Money  is  the  legal  currency  of  the  German  Empire. 

In  1S71  the  German  Empire  established  a  new  and  uniform  system  of  money 
of  which  the  gold  ".l/rt./-A"  (Reichsmark),  is  the  Unit.  The  Marie  is  equal  to  23.S 
cents  United  States  money. 

493.  The  coin  of  the  Empire  consists  of  gold,  silver,  and  nickel.  Pajier 
money  is  also  used. 

TABLE     OP     GERMAN"    MONEY. 

100  xifennige,  marked  Pf.,  make  1  mark,  marked  EM. 

Note. — In  oxfhnngo  transactions  with  tlie  GiTinan  Kiii]iir('.,  for  convenience,  bankers  Ijase 
the  rate  of  exchange  upon  the  equivalent  value  of  4  marks  expressed  in  dollars  and  cents. 

The  exchange  7iar  of  4  marks  is  9.")^  cents.  The  rate  of  exchange  is  9.ji,  more  or  less,  accor- 
ding as  i)reniiuniis  charged  or  discount  is  allowed. 

See  the  subject,  Foreign  Exchange,  in  this  work  for  fnrthcr  information  of  Oerman  money 
and  a  full  ehicidation  of  (Jerman,  Englisli,  French,  Italian,  Austrian,  Belgian,  Russian,  Spanish, 
Grecian,  Portugese,  Japanese,  Chinese,  and  Brazilian,  and  other  South  American  Exchange. 

MEASURE    OP    TIME. 

494.  1. — Time  is  a  measured  portion  of  duration. 

495.  2. — The  Unit  of  measure  is  the  mean  solar  day. 

496.  3. — A  Tear  is  the  time  of  the  revolution  of  the  earth  around  the  sun. 

497.  4. — A  Day  is  the  time  of  the  revolution  of  the  earth  ou  its  axis. 

498.  5. — The  Solar  Day  is  the  interval  of  time  between  two  successive 
passages  of  the  sun  across  the  same  meridian  of  any  place,  and  they  are  of  unequal 
length  on  account  of  the  iiuequal  orbital  motion  of  the  earth  and  the  obliquity  of 
the  ecliptic. 

499.  6. — The  Mean  Solar  Day  is  the  mean,  or  average  length  of  all  the 
liolar  days  in  the  year.     Its  duration  is  twenty-four  hours. 

500.  7. — The  Civil,  or  Legal  Day  used  for  ordinary  purposes,  and  which 
corresponds  with  the  Mean  Solar  Day,  commences  at  midnight  and  closes  at  the 
next  midnight. 

501.  8. — The  Astronomical  Day  commences  at  noon,  and  closes  at  the 
next  noon. 

502.  9. — The  Solar  Year  is  305  days,  5  houi's,  48  minutes,  49.62+  seconds. 

503.  lO.-r-The  Common,  or  Civil  Year  consists  of  365  days  for  3  successive 
years,  every  foiu'th  year  containing  360  days,  one  day  being  added  for  the  excess  of 
the  Solar  Year  over  305  days.  This  intercalary  day  is  added  to  the  month  of 
February,  Avhich  then  has  29  days,  and  the  year  is  called  bissextile  or  leap  year. 

504.  11. — Leap  Y'ear.  To  determine  what  years  are  leap  years,  the  following 
regulation  has  been  adopted : 

Every  year  that  Is  divisible  by  4  is  a  leap  year,  uidess  it  ends  with  two 
naughts,  in  which  case  it  must  be  divisible  by  400  to  be  a  leap  year.  Thus,  1884, 
1776,  and  1600  were,  and  2000,  2400,  and  2800  will  be  leap  years;  but  1885, 1794,  and 
1800  were  not,  and  1900,  2100  and  2200  will  not  be  leap  years. 


*  BRIEF    HISTORY    OF    THE    YEAR    AND    ADJUSTMENT    OF    CALENDAR.        237 

505.  A  BRIEF  HISTORY  OP  THE  YEAR  AND  OF  THE  ADJUSTMENT 

OF  THE  CALENDAR. 

The  necessity  of  some  division  and  measnrement  of  time  was  early  felt,  and  in  all  ages  the 
subject  has  received  the  attention  of  the  most  learned  men. 

The  phases  of  changes  of  the  moon  snpplied  a  natnral  and  very  obvious  mode  of  dividing  and 
reckoning  time,  and  hcuce  the  division  into  months  of  29  or  30  days  was,  perhaps,  the  earliest  and 
most  universal.  But  it  would  soon  be  observed  that,  for  many  purposes,  the  changes  of  the  seasons 
were  more  serviceable  as  marks  of  division ;  and  thus  arose  the  division  into  years,  determined  by 
the  motions  of  the  sun.  It  was,  however,  soon  discovered  that  the  year,  or  larger  division,  did  not 
contain  an  exact  number  of  the  smaller  divisions  or  months,  and  that  an  accommodation  was  neces- 
sary ;  and  various  not  very  dissimilar  expedients  were  employed  for  correcting  errors  that  arose. 

The  ancient  Egyptians  had  a  year  determined  by  the  changes  of  the  seasons,  without  refer- 
ence to  the  changes  of  the  moou,  and  containing  365  d.ays,  divided  into  12  mouths  of  30  days 
each,  with  5  supplementary  days  at  the  end  of  the  year.  The  Jewish  year  consisted,  in  the  earliest 
periods,  as  it  still  does,  of  twelve  lunar  months,  a  thirteenth  being  from  time  to  time  introduced, 
to  accommodate  it  to  the  sun  and  seasous;  this  was  also  the  case  with  the  ancient  Syriaus,  Mace- 
donians, etc. 

The  Jewish  months  have  alternately  29  and  30  days;  and  in  a  cycle  of  19  years  there  are  7 
years  having  the  intercalary  month  ;  some  of  these  years  having  also  one,  and  some  two  days  more 
than  others  have,  so  that  the  length  of  the  year  varies  from  353  to  385  days. 

The  Greeks,  in  the  most  ancient  periods,  reckoned  according  to  real  lunar  months,  twelve 
making  a  year;  and  about  594  B.  C,  Solon  introduced  in  Atheus  the  mode  of  reckoning  alternately 
30  and  29  days  to  the  month,  accommodating  this  civil  year  of  3o-t  days  to  the  solar  year  by  the  occa- 
sional introduction  of  an  intercalary  mouth.  A  change  was  afterwards  made,  by  which  three  times 
in  eight  years  a  month  of  30  days  was  intercalated,  making  the  average  length  of  the  year  365^ 
days. 

The  divisions  of  calendars  of  time  now  in  use  in  America,  and  throughout  Europe,  are  the 
same  as  those  of  the  Romans.  It  is  supjiosed  that  Romulus,  the  founder  of  Rome,  lirst  uudertook 
to  divide  the  year  in  such  a  manner  that  certain  epochs  should  return  periodically  after  a  revolu- 
tion of  the  sun  ;  but  the  knowledge  of  astrouomy  was  not  then  sufficiently  advanced  to  allow  this 
to  be  done  with  much  precision.  He  placed  the  commencement  of  the  year  in  spring,  and  divided 
it  into  10  months:  March,  April,  May,  June,  Quiutilis,  Sextilis,  September,  October,  November,  and 
December.  March,  May,  Quiutilis,  aud  October,  coutained  thirty-one  days  each;  the  other  six 
contained  only  thirty.  The  names  Quiutilis  and  Sextilis  remained  in  the  calendar  till  the  end  of 
the  republic,  when  they  were  changed  into  July  and  August;  the  former  in  houor  of  Julius  Citsar, 
and  the  latter,  of  Augustus. 

The  year  of  Romulus  contained  only  304  days.  Numa  Pompilius,  672  B.  C,  added  two 
months  :  January  to  the  beginning  of  the  year,  aud  February  to  the  end.  About  the  year  452  B.  C, 
tliis  arrangemeut  was  changed  by  the  Decemvirs,  who  placed  February  after  January  ;  since  that 
time  the  order  of  the  mouths  has  remained  undisturbed.  In  Num.Vs  year,  the  months  consisted  of 
29  and  30  days  alternately,  to  correspond  with  the  synodic  revolution  of  the  moon.  The  year 
would  therefore  consist  of  354  days;  but  one  day  was  added  to  make  the  number  odd,  as  being 
more  lucky.  In  order  to  produce  a  correspondence  with  the  solar  year,  Numa  ordered  an  intercalary 
month  to  be  inserted  every  second  year,  between  the  23d  and  24th  of  February,  consisting  alternately 
of  22  and  23  days.  Had  this  regulation  been  strictly  adhered  to,  the  mean  length  of  the  year  would 
have  been  365i  days,  and  the  months  would  have  continued  for  a  long  time  to  correspond  with  the 
same  seasons. 

But  a  discretionary  power  over  the  intercalary  mouth  was  exercised  by  the  pontifis,  who  fre- 
quently abused  it  for  the  purpose  of  hastening  or  retarding  the  days  of  the  election  of  magistrates  ; 
and  the  Roman  calendar  continued  in  a  state  of  uncertainty  and  confusion  till  the  time  of  Julius 
Caesar,  when  the  civil  equinox  differed  from  the  astronomical  by  three  months. 

Under  the  advice  of  the  astrouomer  Sosigeues,  Ciesar,  46  B.  C,  abolished  the  luuar  j'ear,  aud 


23S  SOULE's    nilLOSOPHIC    PRACTICAL    MATHEMATICS.  * 

roguliitcd  the  civil  year  entirely  by  the  sun.  lie  decreed  that  the  comiiiou  year  should  consist  of 
3(i.')  days,  but  tliat  every  I'imrtU  year  should  contain  3(36.  In  distributing  the  days  among  the  diH'erent 
inontlis,  he  ordered  that  tlie  odd  months,  that  is  the  first,  third,  til'th,  seventh,  ninth,  and  eleventh 
should  contain  each  iil  days,  and  the  other  months  30;  excepting  February,  which  in  common  years 
■\vas  to  contain  only  2y  days,  but  every  fourth  year  30  days.  This  natural  and  convenient  arrange- 
ment was  intcrrnided  to  gratify  the  frivolous  vanity  of  Augustus,  by  giving  August,  th&  mouth 
named  alter  liiui,  an  eijnal  number  of  daj's  with  July,  which  was  named  after  the  first  Ciesgr. 

The  intercalary  day,  which  occurred  every  fourth  year,  was  inserted  between  the  /4th  and 
25th  of  February.  According  to  the  peculiar  and  awkward  manner  of  reckoning  adopted  by  the 
Ronuins,  the  24th  of  February  was  called  the  sixth  before  the  calends  of  March,  sexto  calendas,  i.  e. 
six  days  before  March  1st. 

In  the  intercalary  year,  this  day  was  repeated  ami  called  bis-sexlo  caleudas  ;  whence  the  term 
bissextile,  which  means  two  days,  both  of  which  are  six  days  before  March  1st.  The  corresponding 
English  term  leap  year  appears  less  correct,  as  it  seems  to  imply  that  a.  day  was  leapt-over  instead  of 
being  thrust  in.  It  may  Ije  remarked,  that  in  the  ecclesiastical  calendar,  the  intercalary  day  is  still 
inserted  between  the  24th  and  25th  of  February. 

The  Julian  year  consisted  of  365i-  days,  and  consequently  differed  in  excess  by  11  minutes 
10.38  seconds,  from  the  true  solar  year,  which  consists  of  365  days,  5  hours,  48  minutes,  49.62-|-  seconds. 
In  consequence  of  tliis  difference,  the  astronomical  equinox  in  the  course  of  a  few  centuries, 
sensibly'  fell  back  towards  the  beginning  of  the  year.  In  the  time  of  Julius  Ciesar,  it  corres. 
ponded  to  the  25th  of  March;  in  the  sixteenth  centurj^,  it  had  retrograded  to  the  11th. 
The  correction  of  this  error  was  one  of  the  purposes  sought  to  be  obtained  by  the  reforma- 
tion of  the  calendar  effected  by  Pope  Gregory  XIII.,  in  1582.  By  sujipressing  10  days  in  the 
calendar,  calling  October  5th  the  15th,  Gregory  restored  the  equinox  to  the  21st  of  March,  the  day 
on  which  it  fell  at  the  time  of  the  Council  of  Nice,  in  325;  the  place  of  Easter  and  the  other  mov- 
able church  feasts  in  the  ecclesiastical  calendar  having  been  prescribed  at  that  Council. 

And  in  order  that  the  same  inconvenience  might  be  prevented  in  future,  he  ordered  the  inter- 
calation which  took  place  every  fourth  year  to  be  omitted  in  centennial  years,  excepting  such  as 
could  be  divided  by  400  without  a  remainder,  as  shown  above  in  Article  504. 

The  Gregorian  method  of  intercalation  thus  gives  97  intercalations  in  400  years ;  consequently, 
400  years  contain :  400  X  365  +  97  =  146097  days,  and  therefore  the  length  of  one  year  is  365.2425 
days,  or  365  d.  5  h.  49  m.  12  sec,  which  exceeds  the  true  solar  year  by  22.38  sec,  an  error  which 
amounts  only  to  one  day  in  3866  years. 

The  Gregorian  calendar  was  received  immediately  or  shortly  after  its  promulgation  by  all  the 
Eomau  Catholic  countries  of  Europe. 

The  Protestant  States  of  Germany  and  the  kingdom  of  Denmark  a<lhered  to  the  Julian  Calen- 
dar till  1700;  and  in  England  and  the  English  American  colonies,  the  alteration  was  successfully 
opposed  by  i>opular  prejudice  till  1752.  In  that  year  the  Julian  calendar,  or  old  style,  as  it  was 
called,  was  formally  al)olislu!d  by  the  Act  of  Parliament,  and  the  date  used  in  all  public  transactions 
rendered  coincident  with  that  followed  in  other  European  countries,  by  enacting  that  the  day  follow- 
ing the  2d  of  Scptemlicr  of  the  year  1752,  should  be  called  the  14th  of  that  month.  When  the 
alteration  was  made  by  I'ope  Gregory,  it  was  only  necessary  to  drop  10  days  ;  the  year  1700  having 
intervened,  which  was  a  common  year  in  the  Gregorian,  but  a  leap  year  in  the  Julian  calendar,  it 
■was  now  necessary  to  drop  11  days. 

The  year  thus  adjusted  is  called  the  new  stijle ;  and  the  Julian  year,  before  the  adjustment,  is 
called  the  old  style.  The  old  style  is  still  adhered  to  in  Kussia  and  the  coutitries  following  the  com- 
munion of  the  Greek  church;  the  diflerenco  of  date  in  the  present  century  amounts  to  twelve  days. 
Thus,  when  it  is  January  1st  in  Russia  and  Greece,  it  is  January  13th  in  other  countries. 

The  civil  or  legal  year,  in  England  and  the  English  American  colonies,  formerlj'-  commenced 
on  the  25th  of  March,  the  day  of  the  Annunciation,  though  the  historical  year  began  on  the  1st  of 
January,  the  day  of  the  Circumcision.  By  the  Act  of  Parliament  for  the  alteration  of  the  style,  in 
.1752,  the  beginning  of  the  year  was  transferred  to  the  1st  of  January. 


*  BRIEF    HISTORY    OF    THE    YEAR    AND    ADJUSTMENT    OF    CALE.VDAR.        239 

A     CALENDAR. 

506.  A  Calendar  is  a  division  of  time  into  certain  periods  adojited  to  tlie 
pm-poses  of  civil  life,  as  hours,  days,  weeks,  months,  years,  etc. 

The  names  and  orders  of  the  months,  and  the  number  of  days  contained  in 
each,  are  now  as  follows : 

Xaraes.  Xo.    Xo.  days. 

Septend)er,    ittlj,  30 

October,       loth,  31 

November,  11th,  30 

December,  12th,  31 

The  number  of  days  in  each,  may  be  readily  remembered  by  committinjj:  to 
memory  the  following  lines  : 

"Thirty  days  Lath  September, 
April,  June,  anil  November; 
And  all  the  rest  have  thirty-one, 
Excejiting  February  alone  ; 
To  whieh  we  twenty-eight  assign, 
Till  Leap  Y'ear  gives  it  twenty-nine." 

507.  The  Seasons,  ]\rarch,  April,  and  IMay,  are  called  Spring.  June,  July, 
and  August,  are  called  tSx miner.  September,  October,  and  November,  are  called 
Autumn,  and  December,  January,  and  February  are  called  Winter, 


Xames. 

Xo. 

"Ko.  (lays. 

Xames. 

N... 

Xo.  <l^^■s. 

January, 

1st, 

31 

:^ray. 

5th, 

31 

February, 

i-'d, 

28 

June, 

6th, 

30 

March, 

3d, 

31 

July, 

7th, 

31 

April, 

4th, 

30 

August, 

8th, 

31 

508.  TABLE     OF    TIME 

60  Seconds  (see.)         =  1  Jlinute min. 

60  Minutes  =  1  Hour hr. 

24  Hours  =  1  Day d.      1  =  12 

7  Days  =1  "Week wk. 

365  Days  =  1  Common  year. .  .yr. 

366  Days  =  1  Leap  Year yr. 

12  Calendar  months  =  1  Civil  Year yr. 

10  Years  =  1  Decade d. 

10  Decades  or  100  Years  =  1  Ceutiuy c. 


MEASURE. 

■n-k.        ds.              his. 

min.                   sec. 

(  365    =    8700 

—  ;  366  =  8784 

1  =        7   =     168 

1   =       24 

1 

=  525600  =    31536000 
=   527040   =    31622400 
=     10080  =        604800 
=       1440  =          86400 
=           60   =           3600 
1   =               60 

509.     DEEIYATION   OF  THE  YEAR,  MONTH,  WEEK,  DAY,  Etc. 


The  term  Year  is  derived  from  the  Saxon  Gear,  and  is  the  time  of  the  earth's  revolution 
around  the  Sun. 

The  term  Cen^iiri/ is  derived  from  the  Latin  Centuria,  and  means  a  collection  of  a  hundred 
things. 

The  term  Month  is  derived  from  the  Saxon  Monath,  from  Mnna,  the  Moon,  and  means  the  time 
of  one  revolution  of  the  moon  around  the  earth. 

Week.  The  origin  of  the  term  week,  is  not  known.  Jt  i:^  more  ancient  than  the  writings  of 
Moses,  and  without  proper  evidence  it  has  been  thought  by  some  to  have  its  origin  in  the  phases 
of  the  moon,  or  the  number  of  planets  known  to  the  Chaldeans  and  Egyptians,  or  to  the  motion  of 
celestial  bodies.  Seven  days  have  been  used  as  a  division  or  period  of  time  by  most  eastern  nations 
from  time  immemorial.  The  Greeks,  however,  did  not  use  it,  and  it  was  not  introduced  into  Rome 
until  after  the  reign  of  Theodosius. 

The  Egyptian  week  commenced  with  Saturday.  AVlien  the  Jews  took  their  flight  from  Egypt, 
from  their  hatred  to  their  ancient  oppressors,  they  made  Saturday  the  last  day  of  the  week,  and  the 
Christians  have  so  continued  it. 


240  SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS.  * 

The  term  Day  is  derived  from  the  Saxon  Dacg,  and  means  the  time  of  tUo  revolution  of  the 
earth  ujion  its  axis. 

The  term  Hour  is  derived  from  the  Latin  hora,  and  means  a  definite  space  of  time  fixed  liy 
natural  laws. 

The  terra  J/(n»(e  is  from  the  Latin  minulitm,  (very  small).  The  sixtieth  jiart  of  an  hour  of 
time;  also  the  sixtieth  jiart  of  a  defjroe  of  angular  space. 

Second,  from  the  Latin  svciinilas,  (the  next  after  tlio  first).  The  sixtieth  ])art  of  a  minute  of 
time,  or  of  a  degree  of  angular  space;  that  is,  the  second  regular  subdivision  of  the  hour  or  of  the 
degree. 

510.  DEEIVATIOX  OF  THE  NAMES  OF  THE  MONTHS. 

January,  was  so  named  in  honor  of  Janus,  the  Roman  deity  who  presided  over  gates  and  doors, 
and  who  was  naturally  presumed  to  have  something  to  do  with  the  opening  of  the  year.  This  deity 
was  represented  with  two  faces,  one  looking  backward,  the  other  forward,  implying  that  ho  stood 
between  the  old  j^ear  and  the  new,  with  a  regard  for  each. 

February  derives  its  name  from  the  circumstance  that  during  the  month  occurred  the  Roman 
religious  festival  called  the  LiipcrcaVia,  and  also  Fehrualia,  i'nim  februare,  to  purify. 

March,  the  third  month  of  the  year,  but  originally  the  first,  was  named  Martius,  in  honor  of 
Mars,  the  god  of  war  and  the  reputed  father  of  Romulus,  the  founder  of  Rome. 

April.  The  Romans  gave  this  mouth  the  name  of  Jprilis,  from  aperire,  to  open,  because  it 
was  the  season  when  the  buds  began  to  open  and  the  earth  to  produce  vegetation. 

May.  The  common  belief  is  that  May  was  named  in  honor  of  Maia,  the  mother,  by  Jupiter, 
of  the  god  Hermes,  or  Mercury;  but  the  most  reliable  authority  is  that  it  was  named  in  honor  of 
the  majores,  or  maiorcs,  the  senate  in  the  original  constitution  of  Rome. 

June.  The  name  of  this  imuith  has  been  claimed  in  honor  of  Juno;  but  better  evidence 
proves  that  it  wiis  dedicated  a  junioribus,  that  is,  to  the  junior  or  inferior  branch  of  the  original 
legislature  of  Rome. 

Jh7// was  originally  designated  Quintilis,  in  Teferenco  to  its  fifth  place  in  the  calendar;  but 
having  been  the  mouth  in  which  Julius  CiESar  was  born,  after  the  death  of  this  emperor,  the 
Romans  changed  the  name  to  July,  in  honor  of  th.it  great  warrior. 

Aufiust,  in  the  old  Roman  calendar  was  called  Sextilis,  it  being  the  sixth  month  in  the  ten 
which  constituted  the  year.  But  it  was  subsequently  named  August,  in  honor  of  the  emperor 
Augustus,  and  one  day  was  taken  from  February  and  added  to  August  to  make  31,  the  same  as  July, 
the  month  named  in  honor  of  Ca;sar. 

Seplemhcr,  October,  November,  and  December,  are  named,  respectively,  from  the  Latin  numerals; 
Sepfem,  Octo,  Xovem,  and  Decern,  as  when  the  year  consisted  of  but  ten  months  and  began  in  March, 
they  were  the  seventh,  eighth,  ninth  and  tenth  months,  as  their  Latin  names  indicate. 

511.  DERIVATION    OF    THE    DAYS    OF    THE    WEEK. 

Sunday  is  so  called  because  it  was  anciently  dedicated  to  the  sun  by  the  sun  worshipers. 
Eom,ans,  Saxons,  etc.  In  the  4th  century,  A.  T>.,  it  was  made  the  Christian  Sabbath  by  Constantine, 
who  left  the  sun  worshipers  .and  became-a  Christian. 

Monday,  from  the  A.  Saxon,  monan'i-daey,  moon  day ;  a  day  sacred  to  the  moon  as  Sunday  was 
to  the  sun. 

Tuesday  derived  from  Tuisco,  the  Saxon  god  of  war. 

Wednesday  derived  from  the  Scandinavian  deity,  Woden,  whose  functions  corresponded  to  those 
of  Mercury,  in  the  Greek  and  Roman  Mythology. 

Thursday,  named  for  the  Scandinavian  god  Thor,  the  god  of  Thunder.  Thor  was  equivalent 
to  Jupiter  in  the  Greek  and  Roman  Mythology. 

Friday,  so  named  in  honor  of  Freya,  a  Saxon  goddess,  the  wife  of  Woden  or  Odeu. 


MEASURES    OF    EXTENSION.  24 1 

Saturday  is  so  called  from  being  dedicated  by  the  Romans  to  Saturn,  the  god  of  Agriculture, 
and  ■who  was  the  father  of  Jupiter,  Neptune,  and  Pluto,  and  in  whose  honor  -were  held  the  Saturn- 
alian  festivals. 

Thus  we  see  that  the  English  names  of  the  week  are  derived  from  the  ancient  Saxons,  who 
borrowed  the  week  from  some  eastern  nation,  and  substituted  the  names  of  their  owu  divinities 
for  those  of  the  gods  of  Greece  and  Rome. 

51*2.     The    following    are   the  na^ees  of  the  week   in  Latin,    Saxon 

AND  English: 


Dies  Solos,         Dies  Mercurii,  Sun's  day,  Woden's  day,  Sunday,         Wednesday, 

Dies  Lunae,       Dies  Jovis,  Moon's  day,      Thor's  day,  Monday,        Thursdaj^, 

Dies  Martis,      Dies  Veneris,  Tiw's  day,         Friga's  day,  Tuesday,       Friday, 
Dies  Saturui.                                             Seterne's  day.  Saturday. 

MEASUEES  OF  EXTENSION. 

513.  Extension  is  that  property  of  matter  by  which  it  occiai)ies  space.  It 
may  have  one  or  more  of  the  three  dimeusions — length,  breadth,  and  thickness. 

514.  The  Yard  is  the  United  States  and  English  nuit  of  measure  of  exten- 
sion, whether  for  a  line,  surface  or  solid.  A  yard  is  a  rod  or  tape  whose  length  is 
equivalent  to  ffrfa  3  of  the  length  of  a  pendulum  vibrating  seconds  of  mean  time 
in  the  latitude  of  London  in  a  vacuum  at  the  level  of  the  sea,  when  the  temperature 
is  62°  Fahrenheit;  or  differently  stated,  a  pendulum  that  vibrates  seconds  under 
the  conditions  given,  is  divided  into  391393  equal  parts,  and  360000  of  these  parts 
are  taken  for  a  yard,  i.  e.  the  second's  jiendulum  is  39.1393  inches  in  length  of  which 
the  yard  consists  of  36  of  the  39  inches. 

Note  1. — The  yard  was  thus  determined  in  England,  during  the  reign  of  William  IV.,  and 
has  been  adopted  by,  and  is  now  in  general  use  throughout,  the  United  States. 

2. — By  joint  resolutiou  of  Congress,  June  14th,  1836,  the  Secretary  of  State  was  directed  to 
furnish  to  the  Governors  of  each  State,  for  the  use  of  the  States,  a  complete  set  of  all  the  weights 
aud  measures  adopted  by  the  United  States;  this  was  done  and  the  statutory  standards  of  all  the 
States  have  been  made  conformable  thereto,  except  the  bushel  of  New  Vork  State,  and  the  barrel 
of  Massachusetts. 

3. — The  metric  system  was  legalized  by  the  United  States  iu  186G,  but  its  use  was  left 
optional  with  the  i)eople,  and  is  used  only  by  certain  professional  and  scientific  men 

Note. — See  Metric  System,  in  this  book. 

515.  A  Line  has  only  one  dimension — ^length. 

516.  A  Surfiice  has  two  dimensions — leng-th  and  breadth. 

517.  A  Solid,  or  volume  has  three  dimensions — length,  breadth,  and, 
thickness, 

LINE,  OR  LINEAE  MEASUEE. 

518.  Line,  or  Linear  Measure  is  used  to  measure  distances,  or  length, 
in  any  direction. 


One  iuch.  Two  iuclies. 

Three  iiichea. 

Note. — For  the  operations  of  linear,  surface,  and  solid  measure,  see  "  Mensuration  of  Surfaces 
and  Solids,"  in  this  book. 


2^2  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

TABLE. 

L.  m.    fur.    rd.        yd.  ft.  in. 

12     Inolies(iu.)              =  1  Foot I"t.  I=3=24=9(j0=51i80=15840=iyu08» 

3     ¥vL- 1                           =  1  Yard yd.  1=  8=320=1700=  5280=  633(30 

5^  Yards,  or  16.}  feet  =  1  l{od,  or  Poie rd.,  or  P.  1=  40=  220=     660=     7920 

40     Kods                          =  1  Kurlong fur.  (ob.solete)  1=     5i=     161=       198 

8     Furlongs  (320  rds.)=  1  Mile  (Statute  Mile),  mi.  1=         3=         36 

3    Miles                        =  1  League L.  1=        12 

Note.— The  units  of  length  are  nearly  all  dorired  from  different  parts  of  the  huiuau  body, 
and  from  other  objects :  Thus,  the/oot  was  originally  the  length  of  the  king's  foot,  and  consequently 
varied  as  a  king  with  a  long  or  a  short  foot  chanced  to  reigu.  The  span  was  the  si)ace  between  th'& 
end  of  the  thumb  and  little  finger,  when  both  were  extended. 

A  haiiil  was  the  width  of  the  hand  across  the  palm.  The  ctihit  was  the  length  of  the  arm 
from  the  elbow  to  the  end  of  the  middle  fiuger.  The/«//i()m  was  originally  the  distance  between 
the  ends  of  the  two  middle  fingers,  when  the  arms  and  fingers  witi^  extended  to  their  farthest  limit. 
Henry  I.,  king  of  England,  declared  that  the  ell,  or  ynrd,  should  lie  the  length  of  iiis  own  arm,  from 
the  end  of  the  longest  finger  to  the  middle  of  the  breast,  anil  that  the  other  measures  should  bo 
derived  from  this.  An  acre  was  originally  as  much  laud  as  could  be  ploughed  iu  a  day  hy  a  yoke 
of  oxen. 

Furlong,  now  obsolete,  is  from  fur.  meaning  furrow;  and  long,  i.  e.  the  length  of  a  furrow. 
Sod  (16+  ft.)  came  from  a  rod  or  stick  used  for  measuring.  An  inch  was,  with  Edward  IL,  iu  1324, 
the  length  of  3  barley  corus,  sound  and  di-y ;  and  12  of  such  inches  was  a  foot. 

See  Mariners',  Cloth,  and  Shoemakers'  Measures ;  also  Table  of  Comparative  Weights, 
Measures  and  Values,  and  the  old  French  and  Spanish  Measures,  iu  this  book. 


MARINEES'  MEASURE. 

519.  Mariners'  Measure  is  used  to  measure  distances  at  sea  and  also  to 
measure  the  depths  of  seas. 

TABLE. 

6  Feet  =  1  Fathom. 

120  Fathoms  =  1  Oable-leugth. 

880  Fathoms,  or  7^  Cable-lengths  =  5280  ft.  =  1  mile. 

520.  A  Miiiute  of  the  earth's  circumference  is  called  a  geographical,    or 

nautical  mile,  or  laiof,  which  is  ,fd  of  ^Ig  =  ^ihoo  of  the  circumference  of  the  earth. 
The  circumference  of  the  earth  at  the  equator  is  24899  miles,  which,  divided  by 
21600,  gives  1.15273+  statute  miles. 

The  length  of  a  degree  at  the  equator  is  1.15273x00  equals  G9.1C3S  statute 
miles, 

SHOEMAKEES'  MEASURE. 

521.  Slioeiiiakers'  Measure  is  used  by  shoemakers  to  measure  the  human 
feet  and  in  the  manufacture  of  boots  and  shoes; 

522.  The  Unit  of  Measure  is  J  of  an  inch  which  is  the  same  as  the  former 
unit  of  1  barley  corn,  when  3  barley  corns  made  1  inch. 

No.  1,  small  size,  is  4J  inches;  and  every  succeeding  number  increases  J 
of  an  inch  to  13. 

No.  1,  large  size,  is  S^\  inches  ;  and  every  succeeding  number  increases  J 
of  an  inch  to  15. 


523. 


TABLES    OF    MEASURES. 

miscella:n^eous  units  of  lineae  measuee. 


243 


NoTK. — The  harlcii  earn  v^ns  formerly  a  unit  of  measure  iu  England;  3  well  matured  barley 
corns  placed  together,  lengthwise  =  1  inch. 

TABLE. 


•i\r  of  an  Tiich 
-i^o  of  an  luch 
4    Iiic-lies 
3    luclies 


=  A  Line  (American). 
=  A  Line  (French). 
=  A  Hand. 
=  A  Palm. 


9    Inches 
3    Feet 
2J  Feet  (28  in.) 
IS  Inches 


=  A  Span. 

=  A  Pace. 

=  A  Military  Pace. 

=  A  Cubit. 


XoTE.— The  cubit  of  the  Egyptians,  about  3000  B.  C.  was  20.63  inches.  The  cubit  of  Moses 
j  varied  at  different  periods  from  1«  to  2.5.2  inches.  Egypt,  Chaldea  and  other  nations  had  regular 
'     though  varjing  systems  of  weights  and  measures. 


CLOTH  MEASUEE. 


524.     Cloth  Measure  is  used  to  measure  all  kinds  of  goods  sold  by  the  yard. 


TABLE. 

yd.     qr. 

na.        in. 

2^  Inches  (in.)          =     1  Xail  -    -    na. 

1  =  4 

=  16  =  36 

4    Nails  (9  inches)   =     1  Quarter  -    qr. 

1 

=    4=9 

4    Quarters              =     1  Yard      -    yd. 

1=    2i 

This  table  formerly  contained : 

NoTK  1. — The  Flemish  ell,  or  yard,  which  equaled  3  quarters  or  27  inches;  the  English  ell,  or 
yard,  which  equaled  5  quarters  or  45  inches;  the  French  ell,  or  yard,  which  equaled  6  quarters  or 
54  inches. 

Note  2. — All  of  the  above  units  of  lueasure  are  now  out  of  use  except  the  yard,  which  is 
divided  into  hnhrs,  quarters,  eiflltths,  sixieenlhs,  etc.,  in  place  of  feet  and  inches.  At  the  Custom- 
house, the  yard  is  decimallj'  divided. 

Note  3. — The  nail  of  2i  inches  is  derived  from  the  six-penny  nail  of  commerce,  which  is,  or  was 
formerly,  2^  inches  long. 

The  word  penny  when  applied  to  nails  refers  to  the  size  and  is  a  corruption  "of  pound.  Thus 
by  a  six-]ienny  nail  is  meant  a  nail,  1000  of  which  will  weigh  6  pounds.  And  by  the  4,  8,  10  and 
double  10-pen'ny  is  meant  that  1000  will  weigh  respectively  4,  8,  10  and  12  pounds. 


525. 


SQUAEE,  OE  SUEFACB  MEASUEE. 

3  Feet. 


One 

s 

(luar 

e 

Yard 

3  ft.  X  3  ft.  =  9  sq.  ft.  =  1  sq.  yd. 
Square,  or  Surface  Measure  is  used  in  computing  surfaces  or  areas. 
A  Surface  has  length  and  breadth,  but  no  measurable  thickness. 
The  Area  of  a  surface  is  the  quantity  of  surface  it  contains,  aud  is 
expressed  by  the  product  of  the  length  by  the  breadth. 

529.     A  Square  is  a  plane  figure  bounded  by  four  equal  sides,  and  having 
four  ritrht  auffles. 


526. 
527. 
528. 


H4 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


530. 


TABLE. 


lU    (12  X  12)  Square  Inches  (sq.  in.) 
9    (3  X  3)  iSquaie  Feet 
30J  (5J  X  5^)  Square  Yards 
or  272J  sq.  ft. 
160    Square  Eods 
G40    Acres 


mi. 

a. 

sq.  rd. 

1      = 

iiiO 

= 

102400 

1 

= 

IGO 
1 

sq.  in.)  =     1  Square  Foot  - 

-     sq.  ft. 

=     1  S(iuare  Yard 

-     sq.,  yd. 

_  (  1  Scjuare  Hod    - 

-     sq.  rd. 

~  \          or  Perch 

-     P. 

=     1  Acre      -     -     - 

-     A. 

=     1  Sq.  Mile,  or  Section,  .sq.  mi.,  or  sec. 

sq.  v<l.                        sq.  ft. 

sq.  ill. 

3097000     =     2787840U     = 

4014489600 

4810     =           43560     = 

6272640 

301  =               2721  = 

39204 

1     =                  9     = 

1296 

1     = 

144 

sq. 


Note. — The  term  Perch  is  from  the  French  perc/ic,  a  pole  or  rod,  used  for  measuring 
surface  equal  to  a  square  rod. 


and  is  a 


531.  Architects,  carpenters,  slaters,  and  some  other  mechanics  frequently 
measure  their  work  by  the  square,  which  is  a  space  10  ft.  by  10  ft.,  equaling  100 
square  feet. 

533.  A  square  foot,  yard,  or  mile,  is  a  square  each  side  of  which  is  one 
foot,  yard,  or  mile. 

3  Feet,  The  number  of  small  squares  contained  in  any  larfje  square  is  equ.il  to 

the  product  of  tlio  number  of  units  in  one  side,  multiplied  by  the  number 
of  units  in  the  other  side.  Thus,  in  figure  2,  each  side  of  which  is  three 
feet,  there  are  9  sqn.are  feet. 

The  difference  between  square  feet  and  feet  square,  square  viiles  and 
miles  square,  etc.,  for  any  unit  of  measure,  is  not  generally  understood, 
and  because  of  its  practical  importance  we  solicit  special  attention  to  it. 
l!y  3  yards  square  is  meant  a  square  figure,  each  side  of  which  is  3  yards; 
but  l)y  3  square  yards  is  meant  3  small  squares,  each  a  yard  long  and  a 
yard  wide. 

Figure  2  is  3  feet  square,  and  contains  9  square  feet. 

The  difference  between  500  rods  square  and  500  square  rods  is  249500  square  rods. 
There  is  no  difference  between  1  yard  square  and  1  scpiare  yard,  1  foot  scjuare  and  1  square 
foot,  etc.,  of  any  unit  of  measure,  but  increase  the  measure  above  1  unit  .ind  tliere  is  a  difference. 

Note. — Formerly,  but  now  obsolete,  40  square  rods  or  perches  =  1  rood,  i  roods  =  1  acre. 
CHAIN  MEASUEE,  OE  SUEVEYOES'  AND  ENGINEEES'  MEASUEE. 


633.     Chain  Measure  is  used  by  surveyors  and  topographical  engineers,  in 
measuring  land,  laying  out  roads,  etc. 

Note. — The  chain  here  used  is  called  Gunter's  Chain  and  is  n.amed  after  Edmund  Gunter,  an 
eminent  English  mathematician,  born  1581. 


534. 


TABLE. 


7.92  Inches 
25       Links 

4       Eods,  or 
66       Feet 
80      Chains 


I     = 


1  Link       -      - 
1  Eod,  or  Pole 

-  1. 

-  rd.,  or  P. 

1  Chain     -     - 

ch. 

1  Mile       -    -      - 

mi.  (5280  ft) 

iii. 

cb. 

rd. 

1 

=     80 

= 

320 

1 

= 

i 

1 

*  UNITED    STATES    GOVERNMENT    LANDS.  246 

Engineers  commonly  use  another  chain  or  tape  line,  which  consists  of  100 
links,  each  1  foot  long. 

1.  in. 

=     8000  =  03300 

=z       100  =  7013 

=         25  =  198 

1  =  7.92 

635.  Surveyors'  Square  Measure  is  ^ised  by  surveyors  in  measuring  the 
area  or  surface  of  laud.     The  acre  is  tlie  measui-ing  unit  for  land. 

TABLE. 

625  Square  Links  (sq.  1.)  =  1  Square  Eod,  or  Pole.  sq.  rd.,  or  P. 

16  Square  Kods  =  1  Square  Cham     -     -     sq.  ch. 

10  Square  Chains  =  1  Acre A. 

610  Acres  =  1  Square  Mile,  or  Sec.  sq.  mi.,  or  sec. 

36  Square  Miles  (6        )^  Township      -     -  -  Tp. 
nules  square)          )                            '■ 

also  10000  Square  Links,  sq.  L.      =1  Square  Chain  -  sq.  ch. 

10  Square  Chains  =  1  Acre      -    -      -  -  A. 

Tp.  sq.  mi.  A.  sq.  ch.  sq.  rd.  sq.  1. 

1  =  36  =  23010  =  230400  =  3686400  =  2304000000 

UOTTED  STATES  GOVERNMENT  LANDS. 

636.  In  1802,  Col.  Jared  Mansfield,  the  Surveyor  General  of  the  North- 
western territories,  instituted  a  very  convenient  system  of  laying  out  government 
lands.    The  system  adopted  is  as  follows : 

PiBST.— A  line  is  rim  North  and  South  and  is  caUed  the  Principal  Meridian, 
P.M. 

Second. — Other  lines  are  then  run  6  miles  apart  and  parallel  to  the  P.  M. 

Third. — A  line  is  run  on  a  parallel  of  latitude  East  and  West  which  is  called 
the  hase  line,  B.  L. 

Fourth. — Other  lines  are  then  run  6  miles  apart  and  parallel  to  the  B.  L., 
thus  dividing  the  land  into  squares,  6  miles  each  way,  called  townships,  each  of 
which  contains  36  square  miles. 

A  line  of  townsliij)s  running  North  and  South  is  called  a  Range  which  is 
designated  by  its  number  East  or  West  of  the  P.  M. 

Townships  are  also  designated  by  their  nimiber  North  or  South  of  the  B.  L. 

Toicnships  are  also  di\'ided  into  square  miles,  called  sections,  which  are 
numbered  from  1  to  36,  beginning  on  the  North-east  corner  of  each  township. 

Each  section  is  subdivided  into  half  sections,  quarter  sections,  half  quarter 
sections,  and  quarter  quarter  sections. 

537.  TABLE. 

1  Township  =  6  mi.  Square  =  36  Sq.  mi.  =  23040  A. 

1  Section  =  1    "        "       =    1      "        =      640    " 

1  Half  Section  =  1    "  by  J  mi.  =    ^      "        =      320   " 

1  Quarter  Section  =  i    "   "  ^   "    =    ^      "        =      160   " 

1  Half  Quarter  Section  =  |    "   "  |  "    =    J      "        =        80   " 

1  Quarter  Quarter  Section  =  ^    "   "  J  "    =  -^^      "       =        40   " 


246 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


The  followinff  diafrraiiis  will  iiiiike  dear  tlie  foregoing  statements  regarding 
towusliijjs  aud  sections,  and  tlie  manner  of  designating  tlie  subdivisions  of 
sections : 


538. 


DiAGRA:\r  snowi:NG  the  division  of  the  government 

LANDS  OF  the  united  STATES  INTO  TOWNSHIPS. 


Figure  1. 
N 


W 
ci 

M 

T.  3  N. 

3 

3 

1 
A 

2 

2 

3 

1 

1 

2 

3 

3 

2 

1 

1 

2 

3 

T.  1  S 

T.  2  S. 

2 

2 

3 

3 

B. 

Explanalion. — In  fig- 
ure 1,  \V.  E.  is  tlie  base 
line.  N.  S.  is  tlie  prin- 
cipal meridian.    R.  1 

E.  is  range  1,  east;  R. 
2  E.  is  range  2,  east ; 
R.  3  W.  is  range  3, 
•E  west;  T.  3  N.  is  town- 
sliip  3,  north ;  T.  1  S. 
is  townsliij)  1,  south  ; 
A.  is  township  2,  north, 
range  2,  ^^■cst ;  B.  is 
town.sliip  3,  south, 
range  2,  east. 


530. 


A  TOWNSHIP, 

Six    Miles    Square. 

N 


6 

5 

4 

3 

2 

1 

7 

8 

9 

10 

11 

12 

18 

17 

16 

15 

14 

13 

19 

20 

21 

22 

23 

24 

30 

29 

28 

27 

26 

25 

31 

32 

33 

34 

3,5 

36 

3 

tq 


A  SECTION  ENLARGED, 
One    Mile    Squ.are. 
N 

N.  i  Sec. 
320  Acres. 

S.  W.  i  Sec. 
160  Acres. 

W.i. 
of 
S.  E.  i 
80  A. 

N.  E.  i 

of 
S.  E.  i 
40  A. 

PI 


The  above  section  represents  one  of  the  36  sectiona 
of  a  txjwuship. 

jjOXE. — 1.  In  some  parts  of  Louisiana  and  Missouri,  that  were  settled  by  the  French,  the  old 
French  arpeni  is  still  used,  instead  of  the  acre,  as  the  unit  in  measuring  land.  2.  An  arpeiit  (Paris) 
contains  100  sq.  rods,  (Paris)  or  36804.12+  S([.  ft.  (English),  and  is  191.1844  ft.  (English)  on  each  side. 
3.  An  acre  contains  160  sq.  rods,  or  4.3.560  sq.  ft.,  and  is  208.7103+  ft.  on  each  side.  These  figures 
Bhow  that  an  arpent  (Paris)  is  .8449+  per  cent.,  a  little  more  than  f  of  an  acre. 

See  Table  of  French  and  Spanish  Measures,  and  Table  of  Comparative  Weights,  Measures,  and 
Values,  in  this  book. 


*  .  SOLID,    OR    CUBIC    MEASURE.  247 

SOLID,  OE  CUBIC  MEASURE. 

540.  Solid  or  Cubic  Measure  is  u.sed  iu  measuring  the  contents,  or  volume, 
of  solids. 

541.  A  Solid,  or  Body,  has  length,  breadth,  and  thickness. 

512.  A  Cube  is  a  solid,  bounded  by  six  equal  square  sides,  or  faces ;  hence 
its  three  dimensioii.s  are  equal  to  each  other. 

543.  The  Contents,  or  Volume,  of  a  regular  body  is  expressed  by  the 
product  of  its  length,  breadth,  and  thickness. 

544.  TABLE. 

1728    Cubic  Inches  (cu.  in.)         =     1  Cubic  Foot     -     -     -     -     cu.  ft. 
27    Cubic  Feet  (3x3x3)  =     1  Cubic  Yard   -     -     -     -     cu.  yd. 

10    Cubic  Feet  =     1  Cord  Foot      -     -     -     -     cd.  ft. 

8    Cord  Feet,  or  128  cubic  ft.    =     1  Cord  of  Wood  -     -     -     cd. 
243  Cubic  Feet,  or  KJi  feet   ) 

long,  1.}  ft.  high,  and     =     1  Ferch  of  Stone  -     -     -.    Pch. 
1  foot  wide  ) 

40    Cu.  Ft.  in  Uiiiteil  States,  and  42  Cu.  Ft.  in  England,     1  Shipping  Ton. 
100    Cu.  Ft.,   1  Register  Ton,  used  in  estimating  tlie  tonnage  of  vessels. 

cii.  yd.     CM.  ft.  cu.  in.  ed.     c<l.  ft.     cu.  ft.         eu.  in. 

1    =    27    =    46656  1  =  8  =  128  =  221184 

Note. — A  cord  of  icood  is  so  named  from  the  fact  that  originally  it  was  measured  by  a  cord  or 
string. 

545.  A  Square  of  earth  is  a  cube  0x6x6  =  216  cubic  feet. 

54G.  In  civil  engineering,  the  cubic  yard  is  the  unit  for  measuring  excava- 
tions, embankments,  and  levees.     A  cubic  yard  of  earth  is  called  a  load. 

547.  In  commerce,  the  cubic  foot  is  often  the  unit  for  computing  freight 
charges. 

LIQUID,  OR  WINE  MEASURE. 

548.  Liquid  or  Wine  Measure  is  used  in  measm-ing  molasses,  wine,  oil, 
etc.,  and  in  estimating  tlie  capacities  of  cisterns,  reservoirs,  etc. 

549.  The  Unit  of  measure  for  liquids  is  the  gallon,  which  contains  231 
cubic  inches. 

550.  The  Crallon  of  the  United  States  is  the  standard  or  Winchester  wine 
gallon  of  231  cubic  inches,  and  contains  8.3388822  avoirdupois  lbs.,  or  58,372.1754 
Troy  grains  of  distilled  water  at  39.83°  F.,  the  barometer  being  at  30  inches.  It  is 
equal  to  3. 785207  litres.  The  gallon  of  the  State  of  Xew  York  is  of  the  capacity  of 
8  lbs.  of  pure  water  at  its  maximum  density  39.2°  F.,  or  221.184  cubic  inches.  It  is 
equal  to  3.02436  litres. 

551.  TABLE. 

4    Gills  (gi.) 
2    Pints 
4    Quarts 
31J  Gallons 
2    Barrels,  or  63  gallons  =  1  Hogshead  hhd. 

Note. — 1.  In  the  old  tables,  42  gallons  =  1  tierce;  84  gallons  =  1  puncheon  ;  2  hogsheads  = 


1  Pint     - 

-     pt. 

gal.       qt.         lit.         gi. 

1  Quart 

-   qt- 

1  =  4  =  8  =  32 

1  Gallon 

-  gal. 

=  231  cubic  in 

1  =  2=8 

1  Barrel 

-  bbl. 

1=4 

248  soule's  philosophic  practical  mathematics.  * 

1  pipe  or  butt ;  and  2  pipes  ^  1  ton.  But  these  measures  are  no  longer  used  as  units.  They  are 
casks  of  variable  capacity,  and  in  commerce,  are  usually  gauged  separately  and  have  their  capaci- 
ties in  gallons  marked  on  tlicm. 

2.  The  barrel  is  used  as  a  unit  of  measure  only  in  estimating  the  capacity  of  wells,  cisterns, 
vats,  etc. 

3.  In  Massachusetts,  the  barrel  contains  32.28  gallons. 

4.  This  standard  unit  for  liquid  measure  is  called  the  Wincliesier  ijallon,  from  the  standard 
having  been  kept  at  Winchester,  Kuglaud.  It  was  called  the  wine  measure,  because  wine  was 
measured  by  it,  wliile  beer  was  measured  liy  a  larger  measure. 

5.  The  gallon  is  from  the  Latin  gelo,  a  vessel  for  holding  wine.  The  quart  is  derived  from'" 
the  Latin  qiiartus,  a  fourth  ;  the  pint  is  derived  from  the  Greek  pinto,  to  drink;  and  the  gill  is  from 
the  Latin  gilla,  a  ghiss  to  drinlc  from. 

Note. — The  units  of  tlio  old  beer  and  milk  measure,  now  nearly  obsolete,  are  : 
2  pints  =  1  quart;  4  quarts  =  1  gallon  =  282  cubic  inches;  36  gallons  =  1  barrel;  54  gallons 
=  1  hogshead. 

DKY    MEASUEE. 

652.     Dry  Measure  is  used  to  measure  grain,  fruit,  vegetables,  etc. 
553.     The  Ullit  of  dry  measure  is  tbe  bushel,  which  contains  2150.42  (prac- 
tically 2150.4)  cubic  inches. 

Note. — This  United  States  staiulard  unit  for  dry  measure  is  the  old  English  Winchester 
bushel,  so  called  from  the  standard  being  kept  at  Wincliester.  It  is  a  cylinder  18^  inches  in  diame- 
ter and  8  inches  deep.  It  contains  543391.89  standard  Troy  grains,  2150.42  cubic  inches,  or  77.6274 
avoirdupois  pounds  of  distilled  water  at  the  temperature  of  39.83'^  Fahrenheit,  and  barometer 
30  inches. 


TABLE. 

2  Pints  (pt.)                        =  1  Quart    -     -     - 

-      qt. 

bu. 

pk.           qt. 

pt. 

8  Quarts                             =  1  Peck      -     -     - 

-     pk. 

1 

=  4  =  32 

=  64 

4  Pecks                              =  1  Bushel       -    - 

-     bu. 

1=8 

=  16 

8  Bushels  (480  pounds)    =  1  Quarter      -     - 

-    qr- 

1 

=     2 

30  Bushels                            =  1  (Jhaldron 

-      ch 

Note. — 1.  The  heaped  bushel,  the  cone  of  which  is  6  inches  above  the  brim  of  the  measure, 
contains  2747.7  cubic  inches. 

2. — When  corn  is  bought  or  sold  in  the  shuck,  2i  to  4  cubic  feet  as  per  agreement,  are  allowed 
for  a  bushel,  according  to  the  quantity  of  shuck  on  the  corn  and  the  density  of  the  corn  in  the 
crib.     If  the  corn  is  shucked  or  husked,  2  cubic  feet  are  allowed  for  a  bushel. 

3. — In  many  sections  of  the  country,  2  heaped  bushels  of  corn  "in  the  ear"  are  called  1  bn.  of 
shelled  corn.     In  other  places,  2  even  bu.  of  "  ears  "  are  allowed  for  1  bu.  of  shelled  corn. 

4. — In  Maryland,  5  bushels  of  corn  make  a  barrel  of  corn. 

654.  The  dry  gallon,  or  half  peck,  contains  2G8.8  cubic  inches ;  and  10  dry 
gallons,  or  2688  cubic  inches  make  a  bushel  of  coal  in  Pennsylvania,  Ohio,  Kentucky, 
and  several  other  states,  when  gauging  coal  barges,  or  when  measuring  heaps 
or  piles  of  coal. 

The  dry  gallon  of  268.8  cu.  in.  is  also  one-eighth  of  the  American  standard 
bushel. 

In  Louisiana,  the  custom,  when  selling  coal,  is  to  allow  2.6  bushels  of  2688  cu. 
in.  =  6988.8  (6988.865)  cubic  inches  for  a  barrel.     See  page  409. 

Article  3912,  of  Weights  and  Measures,  Revised  Statutes  of  Louisiana,  says:  "That  Weights 
and  Measures  .are  to  be  jirocnred  by  the  Governor,  like  those  of  the  United  States,  and  deposited  in 
the  oiSce  of  the  Secretary  of  State." 

Article  3925,  says :  "  There  shall  bo  in  this  State  a  dry  measure  to  be  known  under  the  name 
of  barrel,  which  shall  contain  3^  bushels,  according  to  American  standard." 


■*•  MEASURES     OF    WEIGHT.  249 

Article  3926,  says :  "  Coal  shall  be  sold  by  the  barrel  or  bushel  measure."  Accordingly,  in 
Louisiana,  a  barrel  of  coal  is  3J  times  2150.42  cubic  inches,  =  6988. 80.5  cubic  inches. 

In  England,  the  standard  Imperial  bushel  contains  2218.192  cubic  inches;  the  Imperial  quar- 
ter is  480  pounds;  and  tlie  Imperial  gallon,  both  dry  and  liquid  measure,  contains  277.274  cubic 
inches,  which  is  oue-eighth  of  the  English  bushel. 

The  Imperial  standard  gallon,  both  for  dry  and  liquid  measure,  contains  10  pounds  avoirdu- 
pois of  distilled  water,  temperature  62*^  Farenheit,  barometer  30  inches.  In  volume  its  capacity  is 
277.274  cubic  inches. 

See  Mensuration  of  Solids  for  practical  problems  in  measuring  or  gaugiug  barges,  barrels,  etc. 


MEASURES  OF  WEIGHT. 


TEOT,     OR    MINT    WEIGHT. 

655.  Troy  Weight  is  used  iu  weighing  gold  and  silver,  aud  in  physical  and 
chemical  experiments. 

656.  The  Standard  Unit  of  Aveight  in  the  United  States  is  the  Troy  pound, 
which  contains  57G0  grains. 

TABLE. 

EQUIVALENT.S. 

lb.  oz.        ]iwt.  gr. 

24  Grains  (gr.)          =  1  Pennyweight      -    pwt.  or  dwt.          1  =  lii   =  240  =  5760 

20  Pennyweights     =  1  Ounce oz.  1   =     L'O  =  480 

12  Ounces                 =  1  Pound lb.  1  =  24 

Note  1. — The  Troy  pound  is  identical  with  the  Imperial  Troy  pound  of  England,  and  derives 
its  name  from  Troy  Novant,  the  ancient  name  of  London.  Some  derive  its  name  from  Troyes,  a 
town  in  France,  where  this  weight  was  tirst  u.sed  in  Europe.  It  was  brought  from  Cairo,  Egypt,  in 
the  12th  Century.  It  is  equivalent  to  the  weight  of  22.79442  cubic  inches  of  distille<l  water  at  its 
maximum  density,  or  of  22.8157  cubic  inches  62^  Fahrenheit,  barometer  at  30  inches  in  both  cases. 

2. — The  term  ;)OMnrf  is  derived  from  tlie  Latin  word  ^pnrfo,  meaning  to  bend  or  weigh.  The 
term  ounce  is  from  the  Latin  word  niicia,  meaning  one-twelfth  part  of  a  pound.  The  pivniiweight 
■was  the  weight  of  the  old  English  silver  penny,  which  was  formerly  used  as  a  weiglit.  The  term 
jraift  is  derived  from  grains  of  wheat  which  were  formerly  used  for  weighing.  At  tirst  32  grains 
taken  from  the  middle  of  the  ear  or  head,  aud  well  dried,  was  equal  to  a  pennyweight.  But  subse- 
quently the  pennyweight  was  divided  into  24  parts,  each  part  being  called  a  grain,  though  it  is 
heavier  than  a  grain  of  wheat. 

3. — The  symbol  Ih.  is  from  libra,  the  Latin  word  for  pound  ;  oz.  is  from  the  Spanish  word  onza, 
meaning  ounce';  pwt.  is  a  compound  of  p.  meaning  penny  and  wt.  meaning  weight ;  dwt.  is  from  d. 
in  the  Latin  word  denarius,  which  means  penny  or  the  240th  of  a  pound,  and  wt.  in  weight ;  gr.  is 
the  abbreviation  of  grains. 

AVOIRDUPOIS,    OR   COMMERCIAL   WEIGHT. 

557.    Avoirdupois  or  Commercial  Weight,  is  used  in  weighing  all  coaxse 
articles :  as  gi'oceries,  cotton,  iron,  etc. 


25o  soule's  philosophic  practical  mathematics. 


658. 

TABLE 

-'Ti- 

r Grains 

— 

1  Dram      .... 

dr. 

le 

Drams 

= 

1  Ounce     -      .      -    . 

oz. 

16 

Ounces 

= 

1  Pound    .... 

lb. 

25 

Pounds 

= 

1  Quarter     -     -     -     - 

qr. 

4 

Quarters,  or  100  pounds 

= 

1  Hundredweight     - 

cwt. 

20 

Hundredweight,  or  2000 

pounds 

= 

1  T(m 

T. 

480 

ir'ounds 

= 

1  Imperial  Quarter  - 

Im.  ( 

100 

Pounds  is  also  called 

1  Cental     .     -      .     - 

c. 

T.           cwt.              11.. 

oz. 

dr. 

1     =     20     r=     2000     = 

32000 

= 

512000 

1     =       100     = 

1000 

— 

25000 

1     = 

16 

= 

250     =     7000  gr 

1 

=: 

10 

qr. 


Note. — 1.  The  cwt.  in  England,  .and  in  some  cases  in  the  United  States,  is  112  ponnds,  or  4 
quarters  of  28  ponnds.  The  ton  English  is  2240  pounds.  This  is  called  the  long  ton,  and  2000 
pounds,  the  short  tun. 

2. — The  long  ton  is  used  in  estimating  duties  at  the  U.  S.  Customhouse,  and  also  at  the  mines 
in  weighing  coal,  ores,  etc. 

3. — The  ai-oirdnpois  pound  is  equivalent  to  the  weight  of  27.7015  cubic  inches  of  distilled 
•water  at  its  maxiiinim  density,  or  of  27.7274  cubic  inches  at  62°  F.ahronheit,  barometer  in  both  cases 
at  30  inches.     The  avoirdupois  pound  of  England  .and  the  Ihiited  State.s  is  the  same. 

4.— 1  ft.  avoirdupois  =  J?gS  of  1  16.  Troy,  or  Hf  <>*"  1  I''.  Troy.  Or  144  16.  avoirdupois  =  175 
}fe.  Troy.  But  the  Troy  ounce  is  heavier  than  the  avoirdupois  ounce.  175  Troy  ounces  =  192  avoir- 
dupois ounces. 

5. — The  term  aroirdniwin,  i.s  derived  from  the  French  word  avoirdupois,  to  hare  (a  fixed  or 
standard)  iceiV//i?.  The  term  (ore  is  derived  from  the  Saxon  word  tnnne,  a  cask.  The  origin  of  the 
pound,  ounce  and  grains  was  given  in  the  Troy  table.  The  symbol  cwt.  is  compounded  from  the 
words,  centum  and  weif/ht.  The  term  dram  is  from  drachma,  originally  a  small  Greek  weight;  it  is 
now  ol)solete  as  the  ounce,  in  commerce,  is  divided  into  halves  and  quarters. 

See  Miscellaneous  Table  for  old  English  units  of  stone,  i)ig,  fother,  etc. 

APOTHECARIES'     WEIGHT. 

559.  Apothecaries'  Weight  is  used  by  physicians  and  apothecaries  in 
weighing  and  compounding  dry  medicines.  Medicines  are  bought  and  sold  by 
avoirdupois  weight. 

TABLE. 

20  Grains  (gr.)  =  1  Scruple  -  scr.  or  9 

3  Scruples  =  1  Dram  -  dr.    <■'  ^ 

8  Drams  =  1  Ounce  -  oz.    "  5 

12  Ounces  =  1  Pound  -  lb.    "   lb 

Note. — 1.  The  grain,  the  ounce,  and  the  pound  of  this  weight  are  the  same  as  those  of 
Troy  weight,  the  ounce  being  ditierently  divided. 

2. — The  terms  })ound,  ounce,  dram  and  grains  were  defined  in  the  Troy  and  avoirdupois 
tables.  The  term  scruple  is  from  the  Latin  word  scrupulus,  a  small  stone.  The  symbols  used  in 
this  weight  had  their  origin,  according  to  Chanipollion,  in  the  Egyptian  hieroglyphics. 

NoTK. — Physicians,  when  writing  prescriptions  use  the  Koman  numerals  in  sm.all  letters, 
prefixing  the  symbols  and  writing.)  for  i  when  it  terminates  a  number.  Thus,  22  gr.  is  written  gr. 
xsij  ;  11  scruples,  Jxj  ;  6  ounces,  ^vj. 

B.  is  an  abbreviation  for  recipe,  or  take;  a.,  aa.,  for  equal  quantities;  ss.  for  semi,  or  half  as 
5  viss.,  means  6i  drams;  gr.  for  grain ;  P.  for  j^articula,  or  little  part;  P.  Jeq.  for  equal  parts;  q. 
p.  quantum  placet,  as  much  as  you  please. 

Apothecaries  and  Physicians,  of  earlier  times  designed  to  conceal  from  others  all  knowledge 
of  the  ingiedieuta  given  as  medicines,  and  hence  the  articles  used  as  medicine  were  named  in 
Latin,  indicated  in  Latin  numerals,  and  arbitrary  signs  employed  to  express  the  quantity. 


lb. 

oz.          dr.          scr.             fjr. 

1 

=  12  =  90  =  288  =  5760 

1  =     8  =     24  =     480 

1  =       3  =       60 

1  =       20 

*  TABLES     OF     WEIGHTS.  25 1 

560.  Apothecaries'  fluid  measure  is  used  in  compouuding  liquid  medicines. 

TABLE. 

60  Minims  (m.)  =  1  FluidracLm     -     -     -     -     fj 

8  Fluidrachms  =  1  Fluidomice     -     ■     -     -     f  5 

IG  Fluidounces  =  1  Pint O. 

S  Pints  =  1  Gallon        ....     cong. 

Cong.         O.  fg  fj  m. 

1   =  8  r=  12s  =  1024  =  61440  One  M.  =  about  1  drop  of  water. 

1   =     16  =     128  =     7680 
1  =         8  =       480 
1  =         60 

Note. — Minim  (M.)  is  from  the  Latin  minimus,  tlie  least;  the  minim  is  the  smallest  fluid 
measure  in  use ;  one  minim  is  equal  to  about  a  drop  of  water. 

Fluidrachms  and  Fluidounces  are  formed  by  prefixing  fluid  to  the  terms  drachms  and  ounces 
of  the  apothecaries'  weight. 

0.  is  an  abbreviation  of  vctans,  the  Latin  for  one-eight ;  meaning  one  pint,  or  the  eighth  part 
ot  a  giillon. 

Cong,  is  from  congiarium,  the  Latin  for  gallon. 

Drops,  are  indicated  in  physicians'  prescriptions  by  Gtt.  which  is  from  the  Latin  gutlae. 

561.  A  single  common  tcaspoonful,  or  4o  drops,  makes  about  one  fluidraclim; 
4  teaspoonfuls  make  1  tablespoonful,  wliicli  is  about  half  a  fluidounce;  a  wine 
glass  equals  about  one  and  a  half  fluidounces;  a  common  teacup  holds  about  4  fluid- 
ounces;  a  pint  of  water  weighs  1.0423+  pounds;  practically  1  pound. 

DIAMOXD     WEIGHT. 

562.  Diamond  Weight  is  u.sed  in  weighing  diamonds  and  other  precious 
stones. 

TABLE. 

16  Parts  =     1  Carat  Grain  =       .792  Troy  grains. 

4  Grains  =     1  Carat  =     3.168      "         " 

Note. — The  term  carat  is  derived  from  Knara,  the  name  of  a  tree  in  Abyssinia,  which  bears 
a  bean  which  varies  but  little  in  weight  from  the  time  it  is  gathered.  In  early  ages,  it  was  used 
as  a  weight  for  gold  iu  Africa.     lu  India,  it  was  used  as  a  weight  for  diamonds  and  gold. 

ASSAYEES'     WEIGHT. 

563.  Assayers'  Weight  is  used  by  assayers  in  determining  the  quantity 
of  any  particular  metal  in  ores,  or  metallic  compounds. 

TABLE. 

1  Carat  graiu  =  2  Pwts.  and  12  grains,  or  60  grains  Troy. 

1  Carat  =10  Pwts.  Troy. 

24  Carats  =  1  Pound  Troy. 

Note. — 1.     This  assay  carat  is  entirely  different  from  the  carat  in  di.amond  weight. 

2. — The  term  carat  is  also  used  by  assayers  to  express  the  proportional  purity  or  the  fineness 
of  gold,  each  carat  meaning  the  twenty-fourth  pari .  Thus,  gold  16  carats  fine  has  16  parts  gold  and 
8  parts  alloy. 


252  soule's  philosophic  practical  mathematics.  * 

CIRCULAE,     OK    ANGULAR    MEASURE. 

564.  Circular,  or  Angular  Measure  is  used  in  measuiing  angles,  latitude 
and  longitude,  the  location  of  vessels  at  sea,  of  planets,  stars,  etc. 

565.  The  Staudard  Unit  for  nieasui-ing  augles  is  the  degree,  which  varies 
■with  the  size  of  the  circle. 

566.  A  Degree  is  the  angle  measiired  by  the  arc  of  ^io  part  of  the  circum- 
ference of  a  circle. 

567.  A  Circle  is  a  plane  figure  bounded  by  a  curved 
line,  every  part  of  which  is  equally  distant  from  a  certain 
point  within  called  the  center. 

568.  The  Circumference  of  a  circle  is  the  curved 
line  by  which  it  is  bounded,  as  A.  E.  D.  B.  G.  H.  F. 

569.  The  Radius  of  a  circle  is  a  line  extending 
from  its  center  to  any  point  in  its  circumference,  as  C.  E. 
and  C.  D. 

570.  The  Diameter  of  a  circle  is  a  straight  line  passing  through  its  center 
and  terminating  at  each  end  in  the  circumference,  as  A.  B. 

571.  An  Arc  of  a  circle  is  any  portion  of  the  circumference,  as  B.  D., 
D.  E.,  etc. 

572.  A  Chord  of  a  circle  is  a  straight  line  drawn  within  a  circle  and  ter- 
minating in  the  circumference,  but  not  jiassing  through  the  center,  as  F.  G. 

573.  A  Segment  of  a  circle  is  any  part  cut  off  by  a  chord,  as  F.  H.  G. 

574.  A  Sector  of  a  circle  is  any  part  of  a  circle  bounded  by  two  radii  and 
the  arc  included  between  them ;  as  the  space  between  C.  D.,  C.  B.,  and  B.  D. 

575.  An  Angle  is  the  opening  or  space  between  two  lines  or  surfaces  which 
meet  in  a  common  point,  called  the  vertex.  Thus,  A.  C.  E.,  E.  C.  D.,  and  D.  C.  B. 
are  augles,  and  C  is  their  vertex. 

576.  A  Semi-Circumference  is  one-half  of  a  circumference,  or  180°. 
677.     A  (Quadrant  is  owe-/ot<r</i.  of  a  circumference,  or  90°, 

578.  A  Sextant  is  one-sixth  of  a  circumference,  or  60°. 

579.  A  Sign  is  one-twelfth  of  a  circumference,  or  30°. 

TABLE. 

60  Seconds  (marked  ")  =  1  Minute  -  '. 

60  Minutes                       =  1  Degree  -  °. 

30  Degrees                      =  1  Sign    -  -  s. 

12  Signs,  or  360°            =  1  Circle  -  •  c. 


c. 

1 

8. 

=  12 



SCO 



21000 



129C00O 

1 

^z 

30 

= 

1800 

= 

108000 

1 

= 

GO 

1 

__ 

3600 
60 

MEASURES    OF    LENGTH,    SURFACE    AND    SOLID. 


253 


THE  OLD  FEENCH  A]SrD  SPANISH  MEASUEES  OF  LENGTH,  SUEFACE, 

AND  SOLID. 


580.  Louisiana  having  been  both  a  French  and  a  Spanish  province,  the  old 
French  and  Spanish  units  of  measure  are  often  met  with  in  private  and  public 
records ;  and  to  aid  iu  understanding  such  units,  the  followiug  table  is  iiresented  : 

581.  TABLE. 

English  and  American  Measure. 


Old  French  System. 


.0074    English  inches. 
.08884  English  inches. 


1  Point  = 

1  Line  =  12  points  = 

1  Inch  =  12  lines  = 

1  Foot  =  12  inches  =        12.7802 

1  Ell     =  43  inches  10  lines  =        40.716 

1  Toise=    Gleet  =        70.7352 

1  Perch  or  Rod,  (Paris)  =  18  feet  =         19.1838 

1  Perch  or  Rod,  (Royal)  =  22  feet  =        23.447 

1  League  (common)  =  25  to  a  degree  =  2280  toises  =  14579.688  English  feet  =  2.761 

miles. 
1  League  (post)  =  2000  toises  =  12789.2  English  feet  =  2.422  miles. 
1  Fathom  (brass)  =  5  feet  French  =  03.946  English  inches. 
1  Cable  length  =  100  toises  =  639.46  English  feet  =  106.58  English  fathoms. 


1.0(i577  English  inches. 
English  inches. 
English  inches. 
English  inches. 
English  feet. 
English  feet. 


582. 


Old  Spanish  System. 


1  Foot  =  11.1284  English  inches. 

1  Vara  =  3  feet  =  0.9274  English  yard 

33.3864  English  inches. 
1  Common  League  ^  19800  Spanish  feet. 
1  Judicial  League  =  15000  Spanish  feet. 


583.   Old  French  Square  &  Cubic  Measure. 

1  8<iaaro  inch  =  1.13587  English  Square  inches. 
*1  Arpent   (Paris)    =    100    square    perches,   = 

36804.120330  sq.  feet,  English. 
1  Arpent  (Woodland)  =  100 sq.  perches  (Royal) 

=  54978.994576  sq.  feet,  English. 
1  Cubic  inch  =  1.2106  cubic  inches,  English. 
1  Cubic  foot  =  2091.85  cubic  inches,  English. 
♦Arpent  is  the  old  French  name  for  acre. 


584.     COMPARISON  OF  THE  FOOT  OF  DIFFERENT  NATIONS,  AS  NOW 

IN  USE. 


Russian  foot  (16  ver8hok)=  28  Inches  English  or 

U.  S.  measure. 
Prussian  foot  (fuss)=  12.3567  in.  English  or  U. 

S.  measure. 
Bavarian  foot  (fuss)=  11.4204  in.  English  or  U. 

S.  measure. 
Hanoverian  foot  (fuss)=:  11.4504  in.  English  or 

U.  S.  measure. 
Saxon  foot  (fuss)=  11.1494  iu.  English  or  U.  S. 

measure. 
Austrian  foot  (fuss)=  12.4455  in.  English  or  U. 

S.  measure. 
Spanish  foot  (pi^)=  11.1288  in.  English  or  U.  S. 

measure. 
French  foot,  old,  (pied  de  roi)=  12.7892  in.  Eng- 
lish or  U.  S.  measure. 
XoTE. — France  and  Spain  now  use  the  metric  system. 
Egyptian  foot  (cubit)=  21.3,  22.667,  25.13,  and 

26.5. 
Persian  foot  (gaz)=  25  common,  37.5  Royal,  in. 

English  or  U.  S.  measure. 


Turkish  foot  (pic)=26.8  or  27.06  in.  English  or 

U.  S.  measure. 
Bohemian  foot  (fuss)=  11.88  in.  English  or  U.  S. 

measure. 
Bremcrian  foot  (fuss)=  11.386  in.  English  or  U. 

S.  measure. 
Danish  foot  (fod)=  12.3576  iu.  Engli.sh  or  U.  S. 

measure. 
Swedish  and  Norwegian  foot  (fot)^  11.6904  in. 

English  or  U.   S.  measure. 
Note. — Sweden  and  Norway  now  use  the  metric  system. 

Hambxirgian  foot  (fuss)=  11.2896  in.  English  or 
U.  S.  measure. 

Japanese  foot  (.sbaku)  =  11.948  in.  English  or 
U.  S.  measure. 

Mexican  vara  or  yard  =  32.97  in.  English  or 
U.  S.  measure. 

Portugese  foot  (palmo  or  span)=  8.6568  in.  Eng- 
lish or  U.  S.  measure. 

Arabian  foot  (guz)=25  in.;  (covid)  19  in.  English 
or  U.  S.  measure. 


!54 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


585.     TAELE  OF  COMPAEATIVE  WEIGHTS,  MEASUEES,  AND  VALUES 


or 


or 


Avoirilu 
7000  gr. 

111). 
144  lbs. 

1  oz. 
192  oz. 


pois. 

=     1  lb. 


A  Wine  Gallon 

The  Old  Beer  Gallon 

A  Dry  Gallon 

An  Inq)erial  Gallon 

U.  S.  Bushel 

U.  S.  Bushel  heaped 

An  English  Bushel 

Diameter  of  Circle  = 

Area  of  a  Square      = 

diameter  of  which 

square 


Troy. 
5760  gr.     = 

1-M  lbs. 
175  lbs. 
Wioz. 
175  oz. 
cubic  inches 


cubic  inches. 


=      231 

=       282 
=       2ti8.« 
=       277.274 
=     2150.42 

=     2GSS.  "  " 

=     2218.192       "  " 

1,  Circumference  =  3.1416 
1.  Area  of  Circle 
=  one  side  of  the 

=    .7854 


Apothecaries. 
1  lb.  57GO  gr.     =     1  lb. 

=        1-M-  lbs. 
=         175  lbs. 

=      Hi  oz. 

=  175  oz. 

Solidity  of  Cube  =  1,  Solidity  of  Sphere 
diameter  of  which  =  one  side  of  Cube 

1  oz.  jiure  gold = 

1  oz.  juire  silver = 

1  p wt.  pure  gold = 

1  pwt.  pure  b  Iver = 

1  i>int  of  waier  weighs  1.0431  lbs. 

1  Gallon      "  "         8.338882  lbs. 

1  Cubic  footof  water  weighs  62.425  lbs.  at 

The  Common  year  has  365  days. 

The  Leap  year  has  366  days. 

The  Solar  year  has  365  days,  5  hrs.,  48  min., 


=    .5236 

$20.67+ 
$  1.29+ 
1.03 
.0645 


39.2°  F. 
49.7  sec. 


A  Yard. 


=  36 


A  Vara  of  Spain. . .  =  33.3864  " 
A  Vara  of  Mexico. .  =  .32.97     " 

A  Meter =  39.37     " 

$1 =100       cts. 

1  Franc  of  France.  .=19.3 
1  Mark  of  Germany.  =23. 8         " 
1  £  of  England.  .'...=1814.8665  " 
1  Crown  of  Austria. =20.3        •' 


1  Milreis  of  Brazil..  =  54.6 
1  Kouble  of  Russia 

gold =  77.2 

1  Yen  of  Japan  gold  =  99.7 

1  Lira  of  Italy =  19.3 

1  Peseta  of  Si)ain. .  =  19.3 
1  Crown  of  Sweden.  =  26.8 
1  Rupee  of  India 

silver =  25.2 


cts. 


1  Drachma  of 

Greece =19.3 

1  Peso  of  Cuba.  ..  .  =  92.2 
1  Peso,  or  dollar  of 

Mexico  silver. ..  ^  57.7 
1  Piaster  of  Egypt.  =:  4.9 
1  Piaster  of  Turkey  =:  4.4 
1  Sol  of  Peru  silver.  =  53.1 


cts. 


Note. — These  monetary  values  are  in  terms  of  United  States  gold,  October  1st,  1893. 
Note — See  Table  of  Values  of  Foreign  Coins. 

A  Statute  mile  =  .5280  ft.  A  Geographical,  or  Nautical  mile,  or  knot,  =  6086.41  ft. 

A  Statute  mile  being  1,  a  Geographical  mile  is  1.15273+ 

586.     All  acre  contains  160  sq.  rds.,  or  43560  sq.  ft.,  and  is  208.7103+  ft.  on 
eacli  side. 

An  arpent  (Paris)  contains  100  sq.  rds.  (Paris),  or  36S04.12033G  sq.  ft.  (English), 
and  is  191.1844  ft.  (Eugiisli)  on  each  side. 

An  arpent  (Woodland)  contains  100  sq.  rds.  (Eoyal),  or  54978.994576  sq.  ft. 
English,  and  is  234.476  English  ft.  on  each  side. 
NoTK. — See  Metric  System  for  other  comparisons. 


587. 


BOOKS    AND    PAPEE. 

SIZE   OF   PAPER. 


Inches. 

Flat  Letter 10x16 

Law  Blank,  or  Small  Cap  ...  13x16 

Flat  Cap 14x17 

Crown 15x19 

Demy 16x21 

Folio  Post 17x22 

Check  Folio 17x24 

Medium— Writing 18X23 

Medium 18x24 


Inches. 

Extra  Size  Folio 19x23 

Medium — Writing  and  Printing.  19x24 

Royal— Printing 20x24 

Medium — Printing 20x25 

Double  Cap 17x28 

Super  Royal- Writing 20x28 

Card  Board  &  Sup.  Royal,  Print'g  22x28 

Imperial — Writing 22x30 

Imperial— Printing 22  x  32 


Note. — Printing  paper  is  made  of  many  sizes,  according  to  the  requirements  of  the  printer. 
In  book  printing  24  X  38  inches,  called  double  medium,  is  used  very  largely. 


WEIGHT    OF    GRAIN    AND    PRODUCE    PER    BUSHEL. 


255 


5SS.  In  copying  legal  papers,  recording  deeds,  etc.,  clerks  are  usually  paid 
by  tlie  folio.  Tims:  100  words  nialce  1  folio  in  ^cw  York,  and  New  Orleans;  713 
words  make  1  folio  in  Com.  Law;  90  words  make  1  folio  in  Cliaucery.  In  printing 
books  240  impressions,  or  120  sheets  printed  on  both  sides,  make  1  token. 


A  shtet  (iiieiliuni)  folded  in  2  leaves  is  called  folio. 


12 
16 
18 
24 
32 


quarto  or  4to. 

octavo  or  8vo. 

duodecimo  or  12mo. 

16mo. 

18mo. 

24uio. 

32mo. 


24  Sheets  = 

480  Sheets  = 

2  Keanis  = 

12  Units  = 

144  Units  = 

12  Gross  = 

20  Units  = 

56  lbs. 

100  lbs.  = 

196  ll)s.  = 

200  lbs.  = 

200  lbs.  = 

280  lbs.  = 

100  lbs.  = 

14  lbs.  Iron  or 
A  Stone  of  Fisb 


1  Quire. 
20  guires  =  1  Ream. 

1  Bundle ;  5  Bundles  =  1  Bale. 

1  Dozen. 
12  Dozen  =  1  Gross. 

1  Great  Gross. 

1  Score. 

1  Firken  of  Butter. 

1  Quintal  of  Dried  Fish. 

1  Barrel  of  Flour. 

1  Barrel  of  Flour  in  California. 

1  Barrel  of  Beef,  Pork,  or  Fish. 

1  Barrel  of  Salt. 

1  Cask  of  Raisins. 
Lead  =  1  Stone, 
or  Meat  =  8  lbs. 


A  Stone  of  Glass 
A  Seam  of  Glass 
21i  Stones 

8    Pigs 

7    lbs.  Wool 

2    Cloves 

2    Stones 

64  Tods 

2    Weys 

12    Sacks 
A  Truss  of  Hay 
A  Truss  of  Straiv 
256  Pounds  of  Soap 

25  Pounds  of  Powder 

12  Barrels  of  Wheat 


=     5  lbs. 

=  24  Stones. 

=     1  Pig. 

=     1  Pother. 

=     1  Clove. 

=     1  Stone. 

=     1  Tod. 

=     IW^ey. 

=     1  Sack. 

=     1  Last. 

=  56  lbs. 

=  36  lbs. 

=     1  Barrel. 

=     1  Keg. 

=     7  English  Quarters. 


589.     *WEIGnT    OF    GKAIN   AIS^D    PEODUCE    PEE   BUSHEL, 

AS  USED  IN  NEW   ORLEANS  WHEN  THERE   IS  NO   AGREEMENT   TO   THE   CONTRARY. 


Wheat bush.  60  lbs.    Clover  Seed bush.  60  lbs.     Dried  Peaches b.ihs.  33  lbs. 


Corn 

Eve 

Oats 

Barley 

Irish  Potatoes 

Sweet  Potatoes. . . 

Beans 

Bran 


56 
56 
32 
48 
60 
60 
62 
24 


Timothy  Seed. 
Barley  Malt.... 

Peas,  "Split 

Small  Hominy. 

Flaxseed 

Hempseed 

Buckwheat. . . . 
Castor  Beans. . . 


45 
34 
60 
50 
56 
44 
52 
46 


Dried  Apples. 

Onions , . . . . 

Coarse  Salt 

F^ine  Salt 

Stone  Coal 

Corn  Meal 

Plastering  Hair. . . 
Blue  Grass  Seed. . . 


24 
56 
50 
50 
80 
44 
7 
10 


*  In  several  States  the  weight  of  some  of  these  articles  is  tUflereut  from  the  fi-nires  here  given. 


eduction  of  Denominate  Numters. 


Note. — For  the  Metric  System  of  Weights  and  Measures,  see  Appendix. 

690.  Reduction  is  the  operation  of  chauging  au  expression  in  one  or  more 
denominations  to  an  equivalent  expression  in  some  other  denomination  or  denomin- 
ations of  tlie  same  kind  of  measurement:  Thus,  $10  may  bo  changed  to  its  equiva- 
lent 1000  cents ;  36  inches  may  be  changed  to  its  equivalent  3  feet,  and  5  lbs.  8  oz. 
avoirdupoi.s  may  be  changed  to  its  equivalent  5J  lbs.,  or  88  ounces. 

Reduction  is  of  two  kinds,  descending  and  ascending. 

691.  Reduction  Descending  is  changing  the  forms  of  denominate  quan- 
tities from  a  higher  to  a  lower  order  of  units,  or  denomination ;  as  in  changing  or 
reducing  dollars  to  cents,  pounds  to  grains,  feet  to  inches,  etc. 

592.  Reduction  Ascending  is  the  converse  of  reduction  descending,  and 
hence  it  is  the  changing  of  the  form  of  denominate  quantities  from  a  lower  to  a 
higher  denomination ;  as  cents  to  dollars,  grains  to  pounds,  inches  to  feet,  etc. 

EEDUCTIOX    DESCENDING. 


TO  REDUCE  A  SIMPLE  DENOMINATE  NUMBER  TO  A  LOWER  DENOMINATION,  (IN  THE 

SAME  SYSTEM  OF  MEASURE). 


593.     1.    Eeduce  C  feet  to  inches. 

OPEEATION. 

0  feet. 
1'2  inches. 


72  inches,  Ans. 


ft. 
6. 


or, 
in. 
0 

12 
6 


Explanation  and  Jteason. — By  considering  the  conditions 
of  the  in-oblem,  we  see  that  we  are  required  to  reduce  6 
feet  to  inches,  1.  e.  to  find  the  equivalent  of  6  feet  iu  the 
unit  inches.  Before  we  can  perform  the  ojicration,  we 
must  know  the  units  of  the  lower  denominations  from 
feet  to  inches.  And  by  referring  to  the  Table  of  Linear 
Measure,  we  fiud  that  1  foot  =  12  inches.  This  gives  the 
premise  for  the  solution  and  from  it  we  reason  as  follows : 
Since  1  foot  is  =  to  12  inches,  6  feet  are  =  to  6  times  as 
many,  which  is  72  inches,  the  answer. 


bn. 

3 


72  inches,  Ans. 
Eeduce  3  bushels  to  pints. 

FIRST  OPERATION. 

Explanation  and  Ueason. — In  this  problem,  we  are  re- 
quired to  reduce  buslicls  to  pints;  but  before  we  can 
jn'rfdrni  the  o]M'ration  we  must  know  either  tlie  different 
units  of  tlio  lower  denominations  from  bushels  to  pints, 
or  the  equivalent  of  one  bushel  in  pints.  By  reference  tc. 
the  Table  of  Dry  Measure,  we  see  tliat  1  bushel  =  4  pecks ; 
1  peck  =  8  quarts;  and  1  quart  =  2  pints.  These equiva^ 
lO''  nts  Ans  lents  fiirnish  our  premises,  and  from  them  we  develoji  the 
"'■'''  '        solution. 

We  first  write  the  problem  as  shown  in  the  operation,  filling  all  vacant  denominations,  from 

(25G) 


pks. 
0 

•1 
3 


qts. 

0 


8 
12 


pts. 

0 


2 
90 


12       90 


REDUCTION    DESCENDING. 


257 


bushels  to  the  denomination  required,  pints,  with  naughts ;  then  helow  each  denomination  vpe  draw 
a  line  and  write  thereunder  the  number  of  units  of  each  order  which  malie  one  of  the  next  higher 
order.  Having  thus  stated  the  problem,  we  reason  as  follows :  Since  1  bushel  is  =:  to  4  pecks,  3 
bushels  are  =  to  3  times  as  many,  which  is  12  pecks;  then  since  1  peck  is  =  to  8  quarts,  12  pecks 
are  =  to  12  times  as  many,  which  is  'Jti  quarts;  theu  since  1  quart  is  =  to  2  pints,  96  quarts  are  ^ 
to  96  times  as  many,  which  is  192  ))ints,  the  answer. 

If  we  work  from  the  basis  of  the  number  of  pints  in  a  bushel,  we  wonld  reason  thus  ;  1  bushel 
is  =  to  64  2)iuts :  Since  1  bushel  is  =  to  64  pints,  3  bushels  are  =  to  3  times  as  many,  which  is 
192  pmts. 


SECOND   OPEKATION. 


3bu. 

4 

12  pks.    or 
S 

96  qts. 


192  pts.,  Aus. 


3bu. 
4  pks. 

Sqts. 
2  pts. 


192  pts.,  Ans. 


Explanation. — In  this  operation,  we  reason  as  follows : 
In  one  bushel  there  are  4  pecks,  and  in  3  bushels  there 
are  three  times  as  many  pecks,  which  is  12 ;  then  in  1 
peck  there  are  8  quarts,  and  iu  12  pecks,  there  are  12 
times  as  many,  which  is  96;  then  iu  1  quart  there  are  2 
pints  and  in  96  quarts  there  are  96  times  as  many,  which 
is  192  pints. 


3. 
4. 
5. 
6. 
7. 
8. 
9. 
10. 


Eeduce  $7  to  mills. 

Reduce  3  shillings  to  farthings. 

Reduce  4  i^ounds  Troy  to  pennyweights. 

Reduce  5  bushels  to  qitarts. 

Reduce  3  days  to  minutes. 

Reduce  2°  to  ". 

Reduce  2  square  feet  to  square  inches. 

Reduce  16  pounds  avoirdupois  to  grains. 


Ans.  7000  m. 

Ans,  144  for. 

Ans.  960  pwt. 

Ans.  IGO  qts, 

Ans.  4320  m. 

Ans.  7200". 

Ans.  288  sq.  in, 

Ans.  112000  grains. 


TO  REDUCE  A  COMPOUND    DENOMINATE  NUMBER    TO  A  LOWER  DENOMINATION,  (IN 

THE  SAME  SYSTEM  OF  MEASURE). 


bu. 
3 


594.     1.    Reduce  3  bu.,  2  pks.,  and  1  pint,  to  pints. 


pk: 


OPERATION. 


qts. 

0 


8 
14 


pts. 

1 


112 


Explanation  and  Heason. — Here  we  are  required  to  deter- 
mine the  number  of  pints  in  the  whole  expression.  We 
first  state  the  problem  as  shown  in  the  operation,  filling 
the  vacant  units  or  denominations  iu  the  scale  with  a 
naught  and  writing  under  each  term  the  number  of  units 
of  that  order  which  make  one  of  the  next  higher  order. 
Having  thus  stated  the  problem,  we  reason  as  follows: 
Since  1  bushel  =  4  pecks,  3  bushels  are  =  to  3  times  as 
many  which  is  12  pecks  +  tin'  2  pecks  =  14  pecks;  then 
since  1  peck  =  8  quarts,  14  pecks  are  =  to  14  times  as 
many,  which  is  112  quarts;  then  since  1  quart  =  2  pints,  112  cxuarts  are  =  to  112  times  as  many, 
which  is  224  pints  +  1  pint  =  225  pints,  answer. 

In  all  problems  of  like  character  to  the  three  preceding,  the  form  of  operation  and  the  pro- 
cess of  reasoning  here  given  should  be  used. 


14      112      225  pts.,  Ans. 


2.  Reduce  3  £.  2s.  8d.  3  far.  to  farthings. 

3.  Reduce  0  lbs.  12  oz.  13  drs.  to  drams. 

4.  Reduce  5  lbs.  2  oz.  12  grs,  to  grains. 


Ans.  3011  far. 

Ans.  1741  drs. 

Ads.  29772  grs. 


258 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


5.  Reduce  3  rods  20  links  6  inches  to  inclies. 

(5.  Eeduce  3  f.oz.  4  f.dis.  20  m.  to  minims. 

7.  Rediice  3  cable-lengths  5  fathoms  to  feet. 

8.  Eeduce  3  rds.  12  ft.  6  in.  to  inches. 

9.  Eeduce  C  hhds.  42  gals.  3  qts.  3  gills  to  gills. 


Ans.  7r)8.4  or  ToSf  in. 

Aus.  1700  m. 

Ans.  2190  ft. 

Aus.  744  in. 

Aus.  13407  gills. 


TO  EEDUCE  A  FEACTIONAL  DENOMINATE  NUMBER    TO    A  LOWER    DENOMINATION, 
(IN  THE  SAME  SYSTEM  OP  MEASURE). 


695.    Reduce  f  of  a  gallon  to  gills. 


OPERATION. 

3  gallons. 

4  quarts. 
2  i)ints. 
4  gills. 


ExpTanation  and  Benson. — In  this  problem,  -n-e  are  re- 
qijired  to  find  tho  eciuivalent  of  f  of  ;i  gallon  iu  the  unit 
of  gillH.  By  reftTriiig  to  the  Table  of  Wine  Measure,  wo 
see  that  4  gills  =  1  jiiiit;  2  pints  =  1  quart;  4  quarts  = 
1  gallon.  Then  with  this  data  as  our  premises,  we  writ© 
the  I  of  a  gallon  on  the  statenient  line  and  reason  as 
follows:  Since  1  gallon  ^=  4  quarts,  there  are  4  times  a» 
many  quarts  as  gallons;  then  since  1  quart  =  2  pints, 
there  are  two  times  as  many  pints  as  quarts;  then  since  1 
pint  =  4  gills,  there  are  4  times  as  many  gills  as  pints.  This  completes  the  reasoning,  and  by 
working  out  the  statement,  we  obtain  12  gills  as  the  equivalent  of  f  of  a  gallon. 


12  gills,  Ans, 


2.  Eeduce  f  of  a  dollar  to  cents. 

3.  Eeduce  f  of  a  iiouiul  (£)  to  farthings. 

4.  Eeduce  -^  of  a  Troy  pound  to  grains. 

5.  Eeduce  -fx  of  a  yard  to  inches, 
fl.  Eeduce  f  of  a  bushel  to  jtints. 

7.  Eeduce  §  of  a  ■ncek  to  seconds. 

8.  Eeduce  f  of  a  franc  to  millimes. 

9.  Reduce  ^  of  a  mark  to  pfenniges. 
10,  Eeduce  ^  of  an  acre  to  sq.  in. 


Ans.  CO  cts, 

Ans.  800  far. 

Ans.  1728  grs. 

Ans.  13iii-  in. 

Ans.  27f  pts. 

Ans.  403200  sec. 

Ans.  28of  m. 

Ans.  75  pf. 

Ans.  54S85C0  sq.  iu. 


p» 


I 


medflction  of  Denominate  Fractions. 


^ 


596.    A  Denominate  Fraction  is  a  fraction  whose  integral  unit  is  a  denom- 
inate number. 

Thus,  I  of  a  day,  .9  of  a  mile,  are  denominate  fractions. 

TO  REDUCE  DENOMINATE   FRACTIONS,  OR  FRACTIONAL  DENOMINATE   NUMBERS  TO 
LOWER  DENOMINATIONS,  OR  TO  COMPOUND  DENOMINATE  NUMBERS. 


597.     1.     Eeduce  f  of  a  bushel  to  lower  denominations. 

OPERATION. 


4              8  2 

5    2  5     3  5     4 

8              24  S 

If            H  If- 


bu.     pks.  qts.  pts.  Explanation  and  Reason. — According  to  the  conilitions  of 

f        0  0  0  tbis  problem,  we  are  to  liud  not  tbe  eiiuivaleut  of  f  of  .1 

bushbl  ill  peeks,  quarts,  or  piuts,  but  wo  are  to  determine 
■what  compound  denominate  number  composed  of  tbe 
denominations  of  pfcfcs,  9«ai(s,  and  pints,  will  be  equiva- 
lent to  I  of  a  bushel,  la  tbe  operation,  we  first  write 
the  J  of  a  bushel  and  fill  each  lower  denomination  with 
a  naught,  below  which  we  draw  a  line  and  underneath 
write  tbe  number  of  units  of  each  order  which  make  one 
of  tbe  next  biglier  order.  We  then  draw  a  vertical  lino 
to  the  left  of  the  4  pecks  and  reason  as  follows :  Since  I 
bushel  =  4  pecks,  ^  of  a  bushel  =  the  i  part,  and  f,  bushels 
,      ,  ,      .,       1  .,      ,  ^  2  times  as  many,  which  is,  as  shown  by  the  operation, 

1  Y}K.  i  qih.      O-g-  pis.       jj  pecks.     Tbe  1  peck  is  now  written  below  a  long  hori- 

zontal line  which  we  call  the  answer  line,  and  is  the  first  denomination  iu  the  number  sought. 

Then  wo  draw  a  Tertieal  line  to  the  left  of  the  8  quarts  and  reason  thus :  Since  1  peck  =  8 
quarts,  -t  of  a  peck  =  tbe  ^  part,  and  i  =  3  times  as  many,  which  work  gives  4}  quarts. 

Tbe  4  quarts  are  w  ritten  below  the  answer  line  and  is  the  second  denomination  of  the  com- 
pound number  required  by  tlie  terms  of  the  problem.  Lastly,  drawing  a  vertical  line  to  the  left 
of  the  2  jiiuts,  we  reason  thus:  Since  1  quart  =  2  pints,  i  of  a  quart  =  -}  part  and  |  =  4  times  as 
many,  which  is  If  pints.  This  being  the  last  denomination  iu  the  Dry  Measure  Table  of  Measure- 
ment, we  write  it  in  full  below  the  answer  line  and  thus  complete  the  solution — having  obtained 
1  pk.,  4  qts.,  aud  IJ  pts.,  as  the  equivalent  compound  denominate  value  of  |  of  a  bushel. 

2.  Eeduce  J  £  to  a  compound  denominate  number.  Ans.  6s.  8d. 

3.  Eeduce  |  yard  to  a  compound  denominate  number.         Ans.  1  ft.  2f  iu. 

4.  Eeduce  y  lbs.  Troy  to  a  compound  denominate  number. 

Ans.  5  oz.  2  pwt.  20f  grs. 
o.    Eeditce  ^  mile  to  a  compound  denonunate  number. 

Ans.  G  fur.  26  rds.  3  yds.  2  ft. 
G.     Eeduce  §  degree  to  a  compound  denominate  number.  Ans.  52',  30". 

7.  Eeduce  f  Cong,  to  a  compound  denominate  number. 

Ans.  4  O.  12  11.  oz.  G  H.  dr.   24  m. 

8.  Eeduce  1^3-  hhd.  wine  to  enuivaleut  integers  in  lower  denominations. 

Ans.  33  gals.  3  qts.  1  pt.  lj%  gills. 

9.  Eeduce  £.,%  to  its  value  in  integers  of  lower  denominations. 

Ans.  6s.  4d.  3^  far. 
(259) 


26o  soule's  philosophic  practical  mathematics.  * 

TO     REDUCE     DENOMINATE     DECIMAL     FRACTIONS,      OR     DECIMAL     DENOMINATE 
NUMBERS,  TO  LOWER  DENOMINATIONS  (IN  THE  SAME  SYSTEM  OF  MEASURE). 

598.     1.     Keduce  .875  of  a  busbel  to  pints. 

OPERATION. 

1)U.         jiks.         qts.  pts. 

.875           0             0           0  Explanation,  and  Reason. — Tliis  iirobleiu  is  very  similar 

~7          T~  ~  to  tlie  first  and  second  problems  in  denominate  numbers, 

875       3  5  "8  P^g^  256.     AVe  first  write  the  problem  as  shown   in  the 

operation,  filling  all  vacant  denominations  from  bushels 

3.500     28.0  5G  Xlt.S.,  AnS.      to  pints  with  naughts;  then,  below  each  lower  denomi- 
C 

75 

4 


"^;  nation  we  draw  a  line  and  write  thereunder  the  number 

875 

of  units  of  each  order  which  make  one  of  the  next  higher 


order.     Then,  from  these  premises  we  reason  as  follows: 

3.500  Since  1  bushel  =  4  pecks,  .875  of  a  bushel  =  .875  times  as 

"  many,  which  is  3.5  pecks;  then,  since  1  peck  =  8  quarts, 

cyo  f\  3.5  pecks  =  3.5  times  as  many,  which  is  28  quarts;  then, 

<>  since  1  quart  =  2  jiints,  28   quarts  =  28  times   as   many, 

which  is  56  pints,  answer. 


56  pints.,  Ans. 

2.  Eeduce  0.755  of  a  gallon  to  gills.  Aus.  2A.16,  or  24„a-  gills. 

3.  Eediice  0.375  of  a  rod  to  inches.  Ans.  74.25,  or  74J  in. 

4.  Eeduce  0.25  of  an  acre  to  sq.  ft.  Ans.  10890  sq.  ft. 

5.  Eeduce  0.42  of  a  ton  to  drauis.  Ans.  215040  drs. 
C.  Eeduce  0.3  of  a  yd.  to  nails.  Ans.  4  J  nails. 
7.  Eeduce  0.16  of  a  chain  to  inches.  Ans.  126.72  in. 

TO    REDUCE    A    DENOMINATE    DECIMAL    FRACTION,    OR    A    DECIMAL    DENOMINATE 
NUMBER,   TO   LOWER  COMPOUND  DENOMINATE  NUMBERS. 

599.     Eeduce  .374  lbs.  apothecaries'  weight,  to  a  compound  denominate  number. 

OPERATION. 

Explannlion  and  Heason. — This  prob- 
lem is  very  similar  to  the  elucidated 
problem  on  page  259.  In  the  oi)era- 
tion,  we  first  write  the  .374  of  a  pound 
and  fill  tlie  place  of  each  lower  de- 
nomination with  a  naught ;  then  draw- 
ing a  line  below  each  naught,  we 
write  tlie  number  of  units  of  each 
order  of  apothecaries'  weight,  which 
4  OZ.        3  dl'.       2  SO.       14.24  grs.     Ans.        it  takes  to  make  one  of  the  next  higher 

'  order.     Having  thus  stated  the  prob- 

lem, we  reason  as  follows:  Since  1 
pound  =  12  ounces,  .374  of  a  pound  =  .374  times  as  many,  which  is  4.488  ounces.  We  now  draw 
the  answer  line  and  write  the  4  ounces  below  it.  Then,  since  1  ounce  =  8  drams,  .488  of  an  ounce 
=  .488  times  as  many,  which  is  3.904  drams.  The  3  drams  we  write  below  the  answer  line,  and 
continue  thus:  Since  one  dram  =  3  scruples,  .904  of  a  dram  =  .904  times  as  many,  which  is  2.712 
scruples.  The  2  scruples  we  write  below  the  answer  line,  and  continue  thus:  Since  1  scruple 
=  20  grains,  .712  of  a  scruple  =  .712  times  as  many,  which  is  14.24  grains. 

This  being  the  lowest  denomination,  it  is  written  below  the  answer  line  and  completes  the 
.solatiou. 


lbs. 
.374 

OZ. 

0 

dr. 
0 

sc. 
0 

gr. 
0 

12 
.374 

8 
.488 

3 
.004 

20 

.712 

4.488 

3.004 

2.712 

14.240 

REDUCTION    OF    DENOMINATE    FRACTIONS.  26 1 

2.  Eediice  0.27  of  a  bu.  to  a  compouud  deuominate  number. 

Alls.  1  pk.     0  qts.     1.28  pts. 

3.  Eeduce  0.7  of  a  lb.  Troy  to  a  compound  deuominate  number. 

Ans.  8  oz.     8  i)wt. 

4.  Eeduce  0.35  of  a  mi.  to  a  compouud  denominate  number. 

Ans.  2  fur.     32  rds. 

5.  Eeduce  0.32  of  a  day  to  a  comi^ound  deuominate  number. 

Ans.  7  lirs.     40  miu.     48  sec. 

6.  Eeduce  0.3  of  a  cu.  yd.  to  a  comiiouud  denominate  number. 

Aiis.  8  cu.  ft.     172.8  cu.  in. 

7.  Eeduce  0.G5  of  a  gal.  to  a  compound  denominate  number. 

Ans.  2  qts.     1  pt.     0.8  gi. 

8.  Wliat  is  the  value  of  .6875  of  a  circle.  Ans.  8  s.     7°  30'. 

9.  What  is  the  value  of  .7525  of  a  day.  Ans.  18  h.    3  min.    36  sec. 


^^eduction  Ascending. 


^^. 


TO    EEDUCE    A    SIMPLE     DENOMINATE    NUMBER    TO    A    COMPOtTND    DENOMINATE 
NUMBER    OF    HIGHER    DENOMINATIONS. 


000.     1.    lleduce  942  gills  wine  measure  to  a  compouud  denominate  number. 


OPERATION. 


942  gills 


Explanation  and  liennon. — In  tUis  prol)lem,  vre  are 
to  determine  the  equivalent  of  942  gills  in  higher 
(lenoniiuations.  According  to  the  Table  of  Wine 
Measure,  4  gills  =  1  pint;  2  pints  =  1  quait,  and  4 
quarts  =  1  gallon.  Being  iu  possession  of  these 
117  quarts  +  1  X)t.  remainder.         equivalent  denominations,  we  first  reduce  the  gills 

to  the  next  higher  denomination,  pints;  then  we 
reduce  the  pints  to  the  next  higher  denomination, 
quarts;  and  thus  continue  to  reduce  each  lower 
denomination  to  the  next  higher,  iintil  the  highest 
29  gal.  1  qt.  1  Jlt.  2  gi.   Ans.  denomination  in  the  Tahle    of   Wine    Measure    is 

reached,  or  until  the  last  quotient  is  less  than  the 
next  higher  denomination.  Our  reasoning  is  as  follows:  Since  4  gills  =  1  pint,  there  are  J  as 
many  pints  as  gills,  which  is  235  pints  and  2  gills  remainder,  as  shown  iu  the  oi)eration. 


235  pints  +  2  gi.  remainder. 


29  gallons  +  1  (|t.  remainder. 


pint 

29 

2 


Then,  since  2  pints  =  1  quart,  there  are  -}  as  many  quarts  as  pints,  which  is  117  quarts  +  1 
lint  remainder.  Then,  since  4  quarts  =  1  gallon,  there  are  i  as  many  galhuis  as  quarts,  which  is 
9  gallons  and  1  quart  remainder.  This  comi>letes  the  reasoning  and  gives  29  gal.,  1  qt.,  1  lit.,  and 
gills  as  the  equivalent  of  942  gills. 

2.  Reduce  3721  pints  to  a  compound  denoniirate  number,  (dry  measure). 

Ans.  58  bu.  0  pk.  4  qt.  1  pt. 

3.  Reduce  1391  inches  to  a  compound  denominate  number,  (linear  measure). 

Ans.    G  rds.  5  yds.l  ft.  11  in. 

4.  Reduce  2756  grains,  Troy,  to  a  comjiound  denominate  number. 

Ans.  5  oz.  14  iiwt.  20  gr. 

5.  Reduce  3457  drams,  avoirdupois,  to  a  compound  denominate  number. 

Ans.  13  lbs.  8  oz.  1  dr. 

C.     Reduce  17353  fiirtliings  to  a  conqiound  denominate  number. 

Ans.  £18  Is.  Cd.  1  far. 

7.  Reduce  345G7  inches,  chain  measure,  to  a  compotind  denominate  number. 

Ans.  43  ch.  G4  links  4.12  inches. 

NoTK. — In  chain  measure,  the  chain  consists  of  100  links. 

8.  Reduce  4315100  seconds  to  weeks. 

Ans.  7  w.  0  d.  22  h.  38  m.  20  sec. 

9.  Reduce  a  million  mills  to  dollars.  Ans.  $1000. 
10.     What  is  the  value  of  300502  inches  in  higher  denominations  ? 

Ans.  5  m.  5  fur.  20  rods  3  y.  2  ft.  10  in. 
(262) 


REDUCTION    ASCENDING. 


263 


TO  REDUCE  A  SIMPLE  OR   A  COMPOUND   DENOMINATE   NUMBER  TO  A   DENOMINATE 

FRACTION  OF  A  HIGHER  UNIT. 


601.     1.    Eeduce  3  pints  to  the  fraction  of  a  bushel. 

OPERATION. 


3  pints. 


Explanation  and  Heason. — Since  we  are  to  find  the  cqnivalent 
of  3  pints  In  the  fraction  of  a  Ijusbel,  wo  mnst  first  know 
either  the  scale  of  units  from  pints  to  bushels,  or  the  number 
of  pints  in  a  bushel.  Hy  referring  to  the  Table  of  Dry  Measure, 
we  find  that  2  ])ints  =  1  quart ;  H  quarts  ^=  1  peck  ;  and  4  peeks 
=  1  bushel.  With  this  knowledge,  wo  place  the  3  pints  on  the 
right  of  the  statement  line  and  reason  as  follows:  Since  2 
l)ints  =  1  quart,  there  are  |  as  many  quarts  as  pints,  which  we 
indicate  by  writing  the  2  on  the  division  side  of  the  statement 
line.  This  gives  ns  the  value  of  3  pints  in  the  fraction  of  quarts.  Then,  since  8  quarts  :=  1  peck, 
there  will  be  -J-  as  many  ])ecks  as  quarts,  which  we  indicate  by  writing  the  8  on  the  division  side  of 
the  statement  line.  This  gives  us  the  value  of  3  pints  in  the  fraction  of  pecks.  Then,  since  4 
pecks  =  1  bushel,  there  will  be  i  as  many  bushels  as  pecks,  which  we  indicate  by  writing  the  4  on 
the  division  side  of  the  statement  line.  This  gives  us  what  the  condition  of  the  itroblem  required, 
the  value  of  3  pints  in  the  fraction  of  a  bushel. 


-g-f  bushel,  Ans. 


2.    Reduce  8s.  4d.  to  the  fraction  of  a  pound  sterling. 

OPERATION    INDICATED. 

Ss.     4d.  =  lOOd.  £i£a  =  £-frr,  Ans.         or, 


12 

20 


lOOd. 


3. 

4. 
5. 
6, 

7. 

8. 

9. 
10. 
11. 
12. 


£-1%  Ans. 
Ans.  ^  of  a  gal. 
Ans.  -sTo  0  0  of  a  ton- 


Eeduce  5  gills  to  the  fraction  of  a  gallon. 

Eeduce  72  drams  to  the  fraction  of  a  ton. 

Eeduce  10"  to  the  fraction  of  a  degree.  Ans.  ^45  of  a  degree. 

Eeduce  2  hours  and  20  minutes  to  the  fraction  of  a  day. 

Ans.  -^  of  a  day. 
Eeduce  3  minims  to  the  fraction  of  a  pint.  Ans.  -.^^0  of  a  pint. 

Eeduce  25  inches  to  the  fraction  of  a  yard.  Ans.  f|  of  a  yard. 

Eeduce  §  of  f  of  f  of  an  inch  to  the  fraction  of  a  yard.      Ans.  -i,-  yard. 
Eeduce  J  of  an  hour  to  the  fraction  of  a  week.  Aus.  yj^  w. 

Eeduce  3  E.  25  P.  to  the  fraction  of  an  acre.  Aus.  |f  ac. 

Eeduce  3  oz.  4  pwt.  10  grs.  to  the  fraction  of  a  pound  Troj'. 

OPERATION   INDICATED. 


3  oz. 


24 
20 

12 


1552 


4  i)wt. 
or  thus,  if  If 


10  grs. 


=  1552  grs. 


grs. 


=  s^oTj  of  a  pound,  Ans. 


•iiftr  ct  SI  pound,  Ans. 

13.  Eeduce  1  v.u.  ft.  200  cu.  in.  to  the  fraction  of  a  cu.  yd. 

Aus.  v/aV  of  a  cu.  yd. 

14.  Eeduce  3'  5"  to  the  fraction  of  a  degree.  Ans.  -,-/„  of  a  deg. 

15.  Eeduce  5  cable-lengths  3  fathoms  to  the  fraction  of  a  mile. 

An.s.  |§§  of  a  mi. 

16.  Eeduce  0  rds.  4  yds.  2  ft.  9  in.  to  the  fraction  of  a  furlong. 

Au.s.  /^4  of  a  fur. 

17.  Eeduce  50  min.  30  sec.  to  the  fraction  of  an  hour.       Aus.  -^f^  of  an  hr. 


264 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TO  REDUCE  A  SIMPLE  OR  A  COMPOUND  DENOMINATE  NUMBER  TO  A  DENOMINATE 

DECIMAL  OP  A  HIGHER  UNIT. 

602o     Ic    lleduce  10  inches  to  the  decimal  of  a  yard. 

OPERATION 


12 

3 


12 
3 


10  inches. 


.83i^  =  decimal  of  a  foot. 


.27^  decimal  of  a  yard. 

or  thus : 
10  inches. 


Explanation  and  Reason. — This  problem  is  closely 
related  to  the  one  under  Art.  601,  ami  the  operation 
and  reasoning  are  somewhat  similar.  We  first 
■write  the  10  inches  on  the  statement  line,  and 
considering  that  12  inches  =  1  foot,  and  3  feet  =  1 
yard,  we  reason  thus :  Since  12  inches  =  1  foot, 
there  are  -jV  as  many  feet  as  inches,  which  is,  as 
shown  by  the  operation,  .S3i  of  a  foot.  Then,  since 
3  feet  =  1  yard,  there  are  i  as  many  yards  as  feet, 
which  is  .27J  of  a  yard,  answer. 


27|  of  a  yd.,  Ans. 
or  thus : 
a  of  a  yd.  =  .27J  of  a  yd. 
.    Reduce  3  quarts,  0  pints,  and  1  gill  to  the  decimal  of  a  gallon. 
OPERATION. 

1.  gill. 


0.  25  pints, 

3.125  quarts. 

.78125  of  a  gallon, 
or  thus : 
3  qts.  0  pts.  1  gill  =  25  gills. 
25  gills. 
4 
2 
4 


.78125  of  a  gal.,  Ans. 
or  thus : 
H  of  a  gal.  r=  .78125  of  a  gal., 


Explanation  and  Beason. — For  convenience,  we 
here  write  the  different  denominations  in  column 
on  the  statement  line,  and  remembering  that  4  gills 
=  1  pint,  2  pints  =  1  quart,  and  4  quarts  =  1  gallon, 
■we  reason  as  in  the  above  problem,  and  divide  the 
gills  by  4,  which  gives  a  decimal  of  .25  of  a  pint. 
This  decimal  is  written  to  the  right  of  the  0  pints. 
Then  ■we  divide  the  .25  of  a  pint  by  2  and  produce 
.125  of  a  quart,  which  is  written  to  the  right  of  the 
3  quarts.  The  3.125  quarts  are  then  divided  by  4, 
and  the  required  decimal  of  .78125  of  a  gallon  is 
obtained. 


Ans. 


3.  Eeduce  3  gills  to  the  decimal  of  a  gallon.  Ans.  .09375  of  a  gal. 

4.  Iteduce  2  ounces  and  5  grains,  Troy,  to  the  decimal  of  a  pound. 

Ans.  .1675f|  of  a  lb. 

5.  Reduce  3"  to  the  decimal  of  a  degree.  Ans.  .OOOSi^  of  a  deg. 

6.  Reduce  7  seconds  to  the  decimal  of  an  hour.  Ans.  .0019|^  of  an  hr. 

7.  Reduce  108  square  inches  to  the  decimal  of  a  sq.  yd. 

Ans.  .08^  of  a  sq.  yd. 

8.  Reduce  12  drams  to  the  decimal  of  a  ton.        Ans.  .0000234375  of  a  ton. 

9.  Reduce  3  dimes,  4  cents,  2  mills,  to  the  decimal  of  a  dollar.  Ans.  $.342. 

10.  Reduce  3  oz.  2  drams  to  the  decimal  of  a  pound.  Ans.  .1953125  of  a  lb. 

11.  Reduce  3  qts.  1  pt.  2  gills  to  the  decimal  of  a  gallon. 

Ans.  .9375  of  a  gal. 

12.  Eeduce  3  sq.  ft.  72  sq.  in.  to  the  decimal  of  a  sq.  yd. 

Ans.  .38f  of  a  sq.  yd. 


*  REDUCTION   ASCENDING.  265 

IS.    Eeduce  3  quarters,  2  nails  to  the  decimal  of  a  yd.        Aus.  .875  of  a  yd. 

14.  Eeduce  6  minutes,  40  seconds  to  the  decimal  of  an  hour. 

Ans.  .11^  of  an  hr. 

15.  Eeduce  3  days  15  hours  45  minutes  48  seconds  to  the  decimal  of  a  week. 

Ans.  .5224  +  week. 

16.  Eeduce  240  A.  2  E.  25  P.  to  the  decimal  of  a  square  mile. 

Ans.  .3760  +  sq.  m. 

TO   EEDUCE   A  DENOMINATE  FRACTION  TO    A    FRACTION    OP   A    HIGHER     DENOM- 
INATION. 

603.     1.    Eeduce  f  of  an  ounce  avoirdupois  to  the  fraction  of  a  ton. 

OPERATION. 


16 

25 

4 

20 


5 

16 

2000 

or  thus :      


4  Explanation  and   Rea- 

son.— This  problem  is 
also  very  similar  to  the 
one   under   Article  601, 

and  requires  about  the 

40  00  0  of  a  ton,   Ans.       same  process  of  reason- 


ing- 

J  f       .  A      .  ^^^  ^^^^  write  the  f- 

JooTJ  Ot  a  ton,   Ans.  of  an  ounce  on  the  state- 

ment line,  and  then  re- 
duce it  to  the  next  higher  denomination,  pounds,  by  dividing  it  by  16;  and  thus  by  successively 
dividing  by  that  number  of  the  lower  order  which  makes  one  of  the  next  higher,  we  reduce  the  ^  of 
an  ounce  to  the  fraction  of  a  ton,  the  denomination  required. 

The  reasoning  for  the  work  is  as  follows:  Since  16  ounces  =:  1  pound,  there  are  -^^^  as  many 
pounds  as  ounces;  then,  since  25  pounds  =  1  quarter,  there  are  -^g  as  many  quarters  as  pounds; 
then,  since  4  quarters  =  a  hundredweight,  there  are  J  as  many  hundredweight  as  quarters ;  then, 
since  20  hundredweight  =  1  ton,  there  are  ijV  ■■>*  many  tons  as  hundredweight. 

If  it  is  desired,  after  dividing  by  16,  the  reasoning  may  be  given  as  follows :  Since  2000 
pounds  =  1  ton,  there  are  ioVu  ^s  many  tons  as  pounds. 

2.  Eeduce  f  of  a  gill  to  the  fraction  of  a  quart.  Ans.  /o  of  a  quart. 

3.  Eeduce  J  of  a  millime  to  the  fraction  of  a  franc.     Ans.  ^oo  of  a  franc. 

4.  Eeduce  |  of  a  fluid  dram  to  the  fraction  of  a  gallon. 

Ans.  -eTTj  of  a  gal. 

5.  Eeduce  §  of  a  pennyweight  to  the  fraction  of  a  pound  Troy. 

Ans.  3^u  of  a  jjound. 

6.  Eeduce  ^  of  a  square  rod  to  the  fraction  of  a  square  mile. 

Ans.  7-(ft 00  of  a  sq.  mi. 

7.  Eeduce  f  of  an  inch  to  the  fraction  of  a  quarter,  (cloth  measure). 

Aus.  -f-g  of  a  qr. 

TO  EEDUCE  A  DENOMINATE  FRACTION  TO  A  DENOMINATE  DECIMAL  OF  A  HIGHER 

DENOMINATION. 


4 
12 
20 


60-1:.     Eeduce  ^  of  a  penny  to  the  decimal  of  a  pound  sterling, 

OPERATION  Explanation  and    Reason. — This    problem    is    very 

3  Bimilar  to  the  one  under  Article  602.     Remembering 

the  Table  for  English  Money,  we  write  the  Jd.  on  the 
statement  line  and  reason  thus:  Since  12  pence  =  1 
shilling,  there  are  -[V  as  many   shillings   as  pence, 

which  we  indicate  by  writing  the  12  on  the  division 

_i_  __   ()03l'i5  of  a  £  ""^^  "^  ^^^  statement  line.     Then,  since  20  shillings 

3-"        ■  "^  ■  =  1  jt.",  there  are  A  as   many  pounds   as  shillings, 

which  is  indicated  by  writing  the  20  on  the  division 
side  of  the  statement  line.  The  result  of  the  work  thus  far  gives  the  fractional  equivalent  of  Jd. 
in  the  unit  of  pounds,  which  is  ^£.  This  we  reduce  to  a  decimal  in  the  usual  way  and  obtain 
.003125  of  a  £,  answer. 


266 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  Eeduce  i^  of  a  pint  to  the  decimal  of  a  buslu'l.     Aiis.  .013020SJ  of  a  bu. 

3.  Eeduce  n,-  of  an  inch  to  the  decimal  of  a  rod.        Ans.  .003oiy|  of  a  rod. 
i.     lieduce  j  of  a  grain  to  the  decimal  of  a  i)ound  Troy. 

Ans.  .OOOOTM-JL-  of  a  lb. 
5.     Eeduce  i-j  of  a  square  inch  to  the  decimal  of  a  sq.  yd. 

Ans.  .OOOGai  of  a  sq.  yd. 
0.    Eeduce  {|  of  a  second  to  the  decimal  of  an  hour. 

Ans.  .00026  1-  of  an  hour. 
7.     Eeduce  f  of  a  "  to  the  decimal  of  a  degree.        Ans.  .0001^  of  a  degree. 


TO  REDUCE  A  DECIMAL  DENOMINATE  NUMBER  TO  A  DECIMAL 

DENOMINATION. 


OF    A    HIGHER 


605.     1.    Eeduce  .3.5  of  a  pint  to  the  decimal  of  a  gallon. 

OPERATION. 

.35  of  a  pint.  .35 


.175  of  a  quart.  or  thus  :     4 


.01375  of  a  gallon. 


Explanation  and  Season.^- 
Here  again  we  have  .a  problem 
very  much  like  the  one  uiuler 
Article  602.  We  are  to  liud  the 
equivalent  of  .35  of  a  pint  in 
the  ileciinal  of  a  gallon.  We 
first  write  the  .3.5  of  a  pint  on 
Since  2  pints  =  1  qn.art,  there  are  +  as  many  quarts  as 
175  of  a  quart.     Then,  since  4  quarts  =  1  gallon,  there 


.01375  of  a  gallon. 


the  statement  line,  and  reason  as  follows: 

pints,  which  is,  as  shown  by  the  operation 

are  i  as  many  gallons  as  quarts,  which  is  .04375  of  a  gallon. 

In  the  second  statement,  the  reasoning  is  the  same  as  in  the  first,  hut  the  operation  of  division 
is  not  iierformeil  in  detail,  with  each  separate  divisor. 

2.    Eeduce  .75  of  a  grain  to  the  decimal  of  a  jiound  Troy. 

Ans.  .0001302  i„-  of  a  lb. 
Eeduce  .96  of  a  minim  to  the  decimal  of  a  pint.  Ans.  .000125  of  a  pint. 
Eeduce  .16  of  a  rod  to  the  decimal  of  a  mile.  Ans.  .0005  of  a  mi. 

Eeduce  .30  of  a  foot  to  the  decimal  of  a  cable-length. 

Ans.  .0005  of  a  c.-l. 
Eeduce  .012  of  a  rotl  to  the  decimal  of  a  league.  Ans.  .0000125  of  a  L. 
Eeduce  .072  of  a  farthing  to  the  decimal  of  a  £.      Aus.  .000075  of  a  £, 


3. 
4. 
5. 

C. 


TO  EEDUCE  A  DECIMAL  DENOMINATE  NUMBER  TO  A  FRACTION  OF  A  HIGHER 

DENOMINATION. 


606. 


100 
60 
60 


1.     Eeduce  .35  of  a  second  to  the  fraction  of  an  hour. 

OPERATION. 

Explanation  and  Heason. — In  this  problem,  we  wish 
to  determine  the  equivalent,  in  the  fraction  of  an 
hour,  of  .35  of  a  second  expressed  as  a  common  frac- 
tion. We  therefore  write  the  .35  of  a  second  on  the 
statement  line  as  a  common  fraction,  -pj',;  and  then 
reason  as  follows:  Since  60  seconds  =  1  minute, 
there  .are  jV  as  many  minutes  as  seconds;  then,  since 

60  minutes  =  1  hour,  there  are  -^Sj  as  many  hours  as  :uinutes.     This  statement,  when  worked,  gives 

tjSott  of  an  hour,  answer. 

2.    Eeduce  .36  of  a  dram  to  the  fraction  of 


35  of  a  second. 


720^00  oi  an  hour,  Ans. 


..  a.__  of 

6  4  00    Ol 

sh  of  a  bbl. 


a  pound.  Ans. 

3.  Eeduce  .75  of  a  pint  to  the  fraction  of  a  barrel.  Ans. 

4.  Eeduce  .124  of  a  farthing  to  the  fraction  of  a  £.        Ans.  -s 

5.  Eeduce  .27  of  a  square  inch  to  the  fraction  of  a  rod. 

Aus.  Tr-... 
0.    Eeduce  .0156  of  an  inch  to  the  fraction  of  a  mile. 

A  ji  o    la 

-^"''-     D2SO0OOO 

7.  Eeduce  .324  of  a  grain  to  the  fraction  of  a  Troy  ounce. 

Ans.  TTo^o  of  an  oz, 

8.  What  would  .7875  of  a  hogshead  of  Tvine  cost  at  5  cents  per  gill  ? 

Aus.  $79.38. 


lib. 
bbl. 
40 oFo  ot  a  £. 

of  a  rod. 

of  a  mi. 


I 


Edition  of  Compound  Denominate  Numbers. 


» i»i  < • 

607.  Addition  of  Compouud  Deuomiuate  Numbers  is  tbe  process  of 
uiiitiTic'  two  or  more  coinpouiul  deuomiuate  uuuibers  iuto  oue  equivaleut  number. 

The  process  of  adding  is  the  same  as  adding  simple  numbers,  excei)t  that  tbe 
scale  of  increase  and  decrease,  by  i)assing  from  one  denomination  to  another, 
varies  with  every  system  of  measurement  aud  with  almost  every  denomination  of 
each  system. 

We  shall,  therefore,  not  discuss  the  subject  at  length,  but  shall  proceed  to 
illustrate  the  process  by  which  results  iu  compound  addition  are  determined. 

1.  Add  5  bu.  3  pks.  7  qts.  1  pt.,  8  bu.  3  qts.  1  pt.,  and  3  bu.  3  pks.  1  pint. 

OPEEATION. 

Iju.      pks.     qts.       pt.  Exjilanaiion. — In  all  problems  of   this  kin<l,    we    first 

5  3  7  1  ■write  tlie  imiiil>ers  so  tluit  units  of  tlio  siime   <li'nomiiia- 

g  Q  3  2^  tiou  staiul  iu  tlie  same  coluiuu,  aud  begin  with  the  lowest 

o  .,  ..  ^  denomination  to  add. 

■*■  Accordingly,  we  hero  first  add  the  pints  .and  find  the 

-^        —       —       —  sum  to  be  3  pts.,  which  we  divide  by  2,  (since  2  pts.  =  1 

17  3  3  1,  Alls.  <lt.)  and  obtain  1  qt.  and  1  pt.  remainder.     The  1  pt.  we 

write  under  the  column  of  pts.,  and  carry  or  add  the  1 

q     i        .     Q    XT         f      f      •  qt.  to  the  c<dnnin  of  qts.,  which  added  gives  11  qts.  and 

^  pts.  —  J,  xNO.  Ot  pts.  in  a  -jvhich  we  divide  by  8,  (since  8  qts.  =  1  pk.)  and  obtain  1 

qt.,  =  1  qt.  and  1  J)t.  pk.  and  3  qts.  remainder.     The  3  qts.  we  write  under  the 

column  of  qts.,  and  adil  the  1  pk.  to  the  column  of  pks., 

11  qts.  ^  S,  No.  of  qts.  in  a  T^'''^  f l'?«'i  g'X''?  »  f""!  of  7  pks     This  we  di vide  by  4 

I       ,'    .  1  Q     f  (since  4  pks.  =  1  bu.)  and  obtain  1  bu.  and  3  pks.     The  3 

pii.,  _  1  pj£.  ana  c5  qts.  p^^   ^.g  ,,.,.ifg  u,„ler  the  column  of  pks.  and  add  the  1  bu. 

to  the  column  of  bushcd.s,  which  gives  a  sum  of  17  bu., 

7  pks.  —•  4,  Ifo.  of  pks.  in  a  which  we  write  under  the  column  of  bushels. 

]jU_    _.j^  jjl^   and  3  llks  This  completes  the  operation  and  produces  17  bu.,   3 

'  *'  '  '  -^       *  pks.,  3  qts.,  1  Jit.  as  the  complete  result. 

2.  Add  15£.  10s.  9d.,  S£.  9s.  7d.,  1£.  12s.  lOd.,  and  1£.  18s.  Gd. 

Ans.  27£.  ILs.  8d. 

3.  Add  12  yds.  3  qrs.  2  na.  2  in.,  5  yds.  2  qrs.  1  na.  1.^  in.,  2  qrs.  1  na. 
IJ  ill.,  8  yds.  2  in.  Ans.  27  yds.  0  qr.s.  3  na.  ^  in. 

Note. — In  working  the  above  problem,  the  student  must  remember,  when  dividing  the 
inches,  that  whenever  the  dividend  is  foiirtha,  the  remainder  is  also  fourths.  The  remainder  is 
always  like  the  dividend. 

4.  Add  10  lbs.  1  5.  6  5.  0  9.  10  grs.,  10  5.  2  3.  1  3.  16  grs.,  and  13  lbs.  2  3, 
73.  33.  12  grs.  Ans.  24  lbs.  3  5.  0  3.  2  g.  18  grs. 

5.  Add  3  yds.  2  ft.,  9  in.,  4  rds.  2  yds.  1  ft.  11  in.,  and  5  rds.  4  yds.  2  ft.  8  in. 

Ans.  11  rds.  0  yd.  1  ft.  4  in. 

6.  Add  21  gals.  3  qts.  1  pt.  3  gi.,  32  gals.  1  qt.  0  pt.  2  gi.,  and  47  gals.  2  qts, 
1  pt.  1  gi.  Ans.  101  gals.  3  qts.  1  pt.  2  gi. 

7.  Wliat  is  the  sum  of  the  following  quantities,  Troy  weight  1  25  lbs.  10  oz; 
15  pwt.  18  grs. ;  42  lbs.  9  oz.  20  grs. ;  26  lbs.  i  oz.  7  pwt. ;  11  oz.  19  pwt.  23  grs. ;  100 
lbs.  6  oz.  14  pwt.  16  grs.  Ans.  196  lbs.  6  oz.  18  pwt.  5  grs. 

(267) 


268 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Add  the  following : 


TROY  WEIGHT. 


m. 


Ans.  319 


pwt. 


gr. 


74 

<l 

16 

22 

36 

7 

19 

1 

87 

10 

4 

20 

65 

8 

17 

11 

54 

0 

1(1 

16 

AVOIRDUPOIS  WEIGHT. 

T.  cwt.  qr,  lb.  oz.  dr. 

60  15  3  24  13  12 

75  18  0  16  9   4 

44  11  2  19  11   9 

33   3  3   3  3   3 

Ans.  214   9  2  14  5  12 


APOTHECARIES'  WEIGHT. 


CLOTH  MEASURE. 


LINEAR  MEASURE. 


m. 

oz. 

<lr. 

scr. 

Sr. 

yil. 

qr. 

n;i. 

in. 

L. 

ni. 

i'm- 

rd.  yil. 

ft.  in. 

24 

11 

6 

2 

19 

1200 

2 

1 

1 

7 

2 

6 

35     4 

2    6 

46 

9 

7 

1 

5 

954 

3 

2 

2 

0 

1 

7 

24     5 

1    2 

16 

8 

3 

2 

15 

675 

1 

3 

2 

3 

2 

0 

16    3 

0  11 

22 

2 

2 

2 

2 

544 

3 

2 

0 

6 

1 

5 

39     2 

2    9 

7 

0 

20 

4 

0 

4 

7    4 

1    8 

'' 

Ans 

3375 

3 

2 

i 

Ans.  163 

4 

2 

1 

1 

Ans. 

28 

0 

1 

4    4 

1    6 

SQUARE  MEASURE. 


sq.  m. 
6 

4 

7 

22 


R.    P.  sq.  y.  sq.  ft.  sq.  in. 


A. 

120 

560    2    36    24 


'"il 


639 

0'>2 


3    39    30 

2     ■T2     '~>2 


Ans.  41    264    1      5     11 


100 

44 

143 

92 


57 


SOLID  OR  CUBIC 
MEASURE. 
C.  c.  ft.  cu.  ft.  cu.  in. 


24  7 

36  6 

44  5 

52  7 


Ans.  159    4 


14 
12 
9 
15 


1224 
1111 
1725 
1000 


WINE  MEASURE. 


hbd.    gal. 
74      56 


36 
124 
146 


1604     Ans.  383 


44 
32 
62 


qt.       pt.     gl. 

3      12 

2  0      3 
111 

3  13 


DRY  MEASURE. 


bu. 
476 
581 
436 
324 


Ans.  1822 


pk.  qt. 

3  7 

2  6 

1  5 


pt. 
1 
0 
1 
1 


TIME  MEASURE. 


100 

240 

70 

33 


42 
50 
24 
33 


d.       h. 


Ans.  445      47 


15 

22 

18 

3 


34 

38 
54 
33 


41 


28 
56 
24 
33 


CIRCULAR  MEASURE. 

•leg. 
74 
34 


58 
68 


Ans.  235 


m. 
44 
28 
22 
15 


51 


sec. 
36 
56 
45 
10 


27 


8.     Add  18  cu.   yds.  13  cu.  ft.  1431  cu.  in.,  IG  cu.  yds.  12  cu.  ft.  931  cu.  in., 
'jud  30  cu.  yds.  I'O  cu.  ft.  1240  cu.  in.  Ans.  05  cu.  yds.  20  cu.  ft.  152  cu.  in. 

0.     Add  9  mi.  07  cli.  3  rd.s.  17  1.,  17  mi.  01  ch.  1  id.  12  1.,  and  16  mi.  46  ch.  2 
rd.  14,  l.  Ans.  44  mi.  15  cU.  3  rd.  18  1. 


10.     Add  §  wk.  f  da.  |  lir.  and  §  min. 


Ans.  5  da.  16  lirs.  14  min.  10  sec. 


OPERATION   INDICATED. 


f  wk. 

-_ 

d:i. 
5 

hr. 
0 

mi. 

0 

sec. 

0 

Ida. 

= 

9 

30 

0 

fir. 

^ 

37 

30 

§  mill. 

= 

40 

5 

10 

14 

10   Ans 

I 


I 


I 


fubtraction  of  Compound  Denominate  Numbers. 


£. 

8. 

il. 

f. 

18 

4 

7 

3 

11 

9 

11 

»> 

608.  Subtraction  of  Compound  Denominate  Numbers  is  the  process  of 
decreasing  one  comi)ound  deiiomiiuite  number  hy  another  of  the  same  sj'stem  of 
measurement. 

As  in  addition,  the  scale  of  increase  and  decrease  varies,  otherwise  the  work 
is  the  same  as  iu  subtraction  of  simiile  numbers. 

1.    From  18£.  4s.  7d.  3f.  subtract  11£.  9s.  lid.  2f. 


Explanation. — In  all  problems  of  this  kind,  we  first 
write  the  nnmbers  so  that  units  of  the  same  denomiua- 
tion  stand  in  the  same  colinnn,  and  beyin  with  the 
lowest  denomination  to  subtract.  Accordingly,  we  here 
commence  with  the  farthings  and  say  2  far.  from  3  far. 
6         11  S  1     Ans.  leaves  1  far.  which  we  write  under  the  column  of  far- 

things. We  now  come  to  the  coliimu  of  pence  and 
observe  that  lid.  cannot  bo  taken  from  7d.,  because  the  lid.  is  the  greater  number;  we  therefore, 
according  to  the  law  that  the  difference  between  two  numbers  is  the  same  as  the  difference  between  the  two 
numbers  tcJicn  equally  increased,  as  demonstrated  in  Article  129,  add  12d.  to  the  7d.,  making  19d. 
From  this  we  subtract  the  lid.  and  have  8d.  remainder,  which  we  write  under  the  column  of  pence. 
Then,  as  we  added  12d.  to  the  minuend,  we,  now,  to  compensate  thereftu',  according  to  the  above 
law,  add  Is.,  the  equivalent  of  12d.,  to  the  column  of  shillings  in  the  subtrahend  and  thus  we  have 
10s.  to  subtract  from  4s.  which  we  cannot  do.  Hence,  for  reasons  above  given,  we  add  20s.  to  the 
4s.  which  makes  24s.,  from  which  we  take  10s.  and  have  14s.  remainder,  which  we  write  under  the 
column  of  shillings.  AVe  now  add  1£,  the  equivalent  of  20s.  to  the  11£,  making  12.t;.,  which  we 
take  from  18£  and  have  a  remainder  of  6£,  which  we  write  under  the  column  of  pounds.  This 
completes  the  oi)eration. 

It  will  be  observed  that  when  we  added  the  12d.  to  the  column  of  pence,  and  the  20s.  to  the 
column  of  shillings,  that,  in  each  case,  we  added  such  a  number  of  that  order  as  made  one  of  the 
next  higher  onler.  This  must  always  lie  done  in  simple  numbers  or  in  any  of  the  systems  of  com- 
pound numbers,  when  the  subtrahend  number  or  denomination  exceeds  the  minuend  number  of  like 
denomination. 

2.  From  45  bbls.  19  gals.  2  qts.  take  24  bbls.  24  gals.  3  qts. 

Explanation. — We  proceed  in  this  as  in  the  previous 
example,  to  find  the  several  denominate  remainders; 
but  since  one  of  them,  the  25^  gals.,  is  partly  fraction- 
al, we  reduce  the  i  gallon  to  quarts,  equal  to  2  qts., 
which  we  add  to  the  3  qts.  in  the  remainder,  making 
5  qts.,  equal  to  1  gallon  and  1  quart;  we  set  down  the 
1  qt.  and  add  the  1  gal.  to  the  25  gals.,  giving  26  gals. 
20  -O  1  for  tiie  final  result. 

3.  From  3  mi.  5  fur.  30  rds.  take  1  mi.  7  fur.  32  rds. 

Ans.  1  mi.  5  fur.  38  rds. 

4.  From  7  lbs.  3  oz.  12  pwt.  20  grs.  take  3  lbs.  5  oz.  10  pwt.  15  grs. 

Ans.  3  lbs.  10  oz.  2  jjwt.  5  grs. 

(269) 


OPERATION. 

bbl. 

gal. 

qt. 

45 

19 

^ 

24 

24 

3 

20 

25J 

3 
2 

^2^o 


SOULE  S    PIIILOSOPHIC    PRACTICAL    MATHEMATICS. 


5.     Prom  30°  25'  32"  take  25°  34'  35' 


Alls.  4°  50'  57". 


6.  From  4  lbs.  12  oz.  13  drs.  take  2  lbs.  9  oz.  15  drs. 

Ans.  2  lbs.  2  oz.  14  drs. 

7.  From  13  lirs.  24  iiiiii.  7  sec.  take  10  hrs.  30  miu.  12  sec. 

Ans.  2  lirs.  53  niiii.  55  sec. 

8.  From  3  sq.  rds.  5  sq.  ft.  95  sq.  in,  take  1  sq.  rd.  10  sq.  ft.  100  sq.  in. 

Ans.  1  sq.  rd.  267  sq.  ft.  31  sq.  in. 

SOLUTION. 


sn.  rd. 

3 
1 


sq.  ft. 

5 
10 


sq.  m. 

95 
100 


26G1 
\  -- 


139 
36 


267 


31 


Explanation. — By  proceeding  as  in  the  explanations 
above  given,  we  obtain  1  sq.  rd.  2G6i  sq.  ft.  139  sq.  in. 
as  a  result;  but  it  is  nut  iiropcr  to  have  a  fractional 
expression  in  any  but  llie  lowest  denomination  of  a 
compound  denominate  number,  so  we  proceed  to  reduce 
J  of  a  sq.  ft.  to  sq.  in.;  i  of  144  sq.  in.  =:  36  sq.  in. ; 
139  sq.  in.  -f- 36  sq.  in.  =:  175  eq.  in.,  or  1  sq.  ft.  and 
31  si|.  iu.  So  we  write  31  sq.  in.  and  add  1  sq.  ft.  to 
266  sq.  ft.  and  tiud  the  answer  to  be  1  S(i.  rd.  267  sq.  ft. 
31  sq.  iu. 


9.     Subtract  the  foUowinc: : 


A. 

R. 

40 

3 

16 

3 

Ans.  23 


P.  sq.  yd.  sq.  ft.  sq.  in. 

22  30  4  10 

30  35  5  100 


23 

3 

31 

24i 
i    = 

7 

=  2i 

i   = 

54 
=      36 

31 


90 


10.  From  f  of  a  hogshead,  subtract  §  of  a  gallon. 

Ans,  37  gal.  0  qt,  1  pt.  Oi^  gi. 

OPERATION    INDICATED. 

i  of  a  Mid.     =     37  gal.  3  qt.  0  pt.  If     gi. 
§  of  a  gal.       =  2  qt.  1  pt.  1^     gi, 

37  gal.  0  qt.  1  pt.  0^^.  gi. 

11.  From  2f  bu.  subtract  f  of  a  peck.  Ans.  2  bu.  1  pk.  4  qt.  1^1  pt. 

12.  From  555  A.  3  R,  30  P.  take  ,335  of  a  square  mile, 

Ans,  341  A,  2  E.  6  R 

13.  From  5  in.  4  fur.  24  r.  take  2  m.  G  fur.  37  r.  3  yd.  2  ft.  9  in. 

Ans.  2  m.  5  fur.  26  rd.  li  yd.  0  ft.  3  in, 

14.  From  the  sum  of  f  of  365^  days  and  |  of  2§  hours,  take  I  of  -ja-  of  15J 
minutes.  Ans.  15G  d.  14  h.  47  m.  25-ii^  sec. 

15.  From  the  sum  of  f  of  5^  miles  and  .875  furlongs  take  |  of  2J  yards. 

Ans.  2  m.  2  fur.  34  rd.  3J  yd.  0  ft.  4J  in. 


ultiplication  of  Compound  Denominate  Numbers. 


^ 


609.  Multiplicatiou  of  Compound  Denominate  Numbers  is  the  process 
of  tletermiuing  the  product  of  two  uumbers,  when  the  number  to  be  niultiplied 
is  a  compound  denominate  number. 

I  Compound  multiplication  dififers  from  multiplication  of  simple  numbers,   in 

that  when  the  product  iu  any  denomination  equals  or  exceeds  one  of  the  next  higher 
denomination  in  the  same  system  of  measurement,  said  product  nmst,  by  a  process 
of  division,  be  reduced  to  that  next  higher  denomination,  before  the  number  iu  it 
can  be  multiplied. 

1.    Multiply  -i  gals.  3  qts.  1  pt.  2  gi.  by  9. 


OPERATION. 


gal. 

4 


qts. 

3 


pt. 
1 


g"' 


44 

18  gi. 
13i)ts, 
33  qts, 

8  gals,  anil  1  q 
of  galUiiiS.  L 
we  -(vrite  unde 
product. 

Note.  1 
Note.  2 
factors. 


Explannlinn. — In  all  problems  of  this  l<in(l,  we  write 
the  multiplier  uuder  tlie  lowest  deiioniiuation  of  the 
multiplicand  and  nuilf  iply.  We  first  say  9  times  2  gi. 
are  18  gi.  wliicli,  as  sbowu  iu  the  operation,  ei|ual3  4 
pts.  and  2  gi.  The  2  gi.  we  write  under  the  gills  and 
reserve  the  4  pts.  to  add  to  the  product  of  pints.  We 
theu  say  9  times  1  pt.  are  9  pis.  and  4  pts.  added  make 
13  pts.  which  reduced  to  the  next  higher  denomination 
equals  6  qts.  and  1  ])t.  The  1  pint  we  write  under  the 
unit  or  deuominati(JU  pints,  and  reserve  the  6  qts.  to 
add  to  the  product  of  quarts.  Then  we  say  9  times  3 
qts.  are  27  qts.  and  6  qts.  added  make  o3  qts.  e(|ual  to 
t.  The  1  qt.  we  write  under  the  quarts,  and  reserve  the  8  gals,  to  add  to  the  product 
astly  we  say  9  times  4  gals,  are  36  gals,  plus  the  8  gals,  reserved,  are  44  gals,  which 
r  the  deuomiuatiou  of  gallons.     This  gives  44  gals.  1  qt.  1  jit.  2  gi.  for  the  entire 

— The  multiplier  must  always  he  considered  an  abstract  number,  Art.  149,  page  75. 
— When  the  multiplier  is  large,  and  a  composite  number,  we  may  multiply  by  the 


4  =  4  l^ts.  2  gi. 
r  2  =  6  qts.  1  pt. 
r  4  =  8  gals.  1  qt. 


2.  IMidtiply  7  Cong.  3  O.  15  f  3    by  6. 

3.  iMultiply  27  cords  34  cu.  ft.  by  12. 


Ans.  44  Cong.  7  O.  10  fg 


4.  Multiply  3  mi.  7  fur.  20  rds.  by  7. 

5.  Multiply  7  hrs.  30  nu.  24  sec.  by  5. 
G.  Multiply  3  francs,  7  dec.  3  cen.  4  mil.  by  5. 
7.  Multiply  4£.  3s.  4d.  by  12. 
S.  In  5  bbls.  of  pecans,  each  containing  2  bu.  3  pks.  5  qts.  1  pt.,  how  many 

bushels?  Ans.  14  bii,  2  pks.  3  qts.  1  pt. 

(271) 


Ans.  327  cords  24  cu.  ft. 
Ans.  27  mi.  4  fur.  20  rds. 
Ans.  1  da.  L3  hrs.  32  min. 
Ans.  IS  fr.  G  dec.  7  cen. 
Ans.  £50. 


272 


SOULE  S    rHILOSOPHIC     PRACTICAL    MATHEMATICS. 


9.     Multiply  23  gallons,  3  quarts,  1  pint,  by  3.5. 


OPEKATION. 


25  gals.  3  qts.  1  pt.  =  207  pts. 
4  3.5 


2)  724.5  pt.. 


103 


1035 
6-21 


4)  362  qts.  and  .5  or  i  pt. 


90  J 


^'als.,  2  qts,  i  pt. 


$1585.65. 
llow  many 


i,'als.  2  (jts. 

■ Aus.  'M I 

207  724.5  pt. 

10.  How  many  busbel.s  in  12  barrels  of  grain,  each  barrel  containing  4  bus., 
3pks.,  5  qrs.,  1  pt.  1  Ans.  5!»  bus  ,  2  qts. 

11.  Uow  many  gallons  in  15  casks  of  wine,  each  cask  containing  42  gallons, 
2  quarts,  1  pint,  and  wbat  will  they  cost  at  $2  48  per  gallon  ? 

Ans.  Contents:  039  gals.,  1  qt.,  1  pt.     Cost; 

12.  A  man  has  7  lots  of  land,  each  containing  24  A.,  3  li.,  25  P. 
acres  has  he  altogether,  and  what  are  tliey  worth  at  $15.65  per  acre? 

Ans.  174  A.,  1  li  ,  15  P.     Cost:  $2728.471^. 

13.  If  a  boy  studies  5  hours  25  mmutes  each  day,  how  many  d»ys  of  24  hours 
each  will  he  .study  in  54  days  1  Ans.  12  ds.,  4  hrs.,  30  m. 

14.  What  is  the  distance  around  an  octagonal  or  eight  sided  field,  each  side 
measuring  1  fur.,  25  rds.,  3  yds.,  2  ft;.,  8  in.  ? 

Ans.  1  mi.,  5  fur.,  5  rds.,  3  yds.,  1  ft.,  10  in. 

15.  A  man  bought  25  pieces  of  broad  cloth,  each  containing  32  yds.,  3  qrs. 
2  na.     IIow  many  yards  did  he  buy,  and  what  ilid  they  cost  at  $5.84  per  yard  ?       ' 

Ans.  821  yds.,  3  qrs.,  2  ua.     Cost:  $4799.75. 

SPECIAL  PROBLEM. 

A  farmer  has  a  piece  of  land  that  he  measures  with 
a  pair  of  compasses  whose  points  are  6  feet  apart, 
and  finds  the  following  dimensions  of  his  land  :  A  to 
B  4  acres  28  coinjjasses ;  A  to  C  4  acres  13  compasses; 
C  to  D  2  acres  9  compasses.  How  many  acres  in  the 
plat  of  ground  ?  Ans.  15  acres,  98  sq.  rods, 

16  sq.  yds.,  8  sq.  ft.,  5.52  sq.  in, 

Remauks. — The  compasses  for  measuiin<;  land  is  a  priniative 
instrument  in  the  form  of  compasses,  whose  points  are  6  feet 
apart.  35  spans  or  units  of  the  comjiasses  (210  feet)  is  consid- 
ered an  acre  for  approximate  work  on  a  farm.  This  is  incor- 
rect, as  an  acre  or  sipiare  piece  of  ground  containing  one  acre, 
is  208  7103  ft  on  each  side.  See  Comparative  Tahle  on  page 
254  of  this  hook,  also  tahle  on  page  244.     iSee  page  253  for  arpeut,  -which  is  the  old  French  word 


for  acre. 


SOLUTION. 
4x210  =  840  ft 
13x6      ==78     " 


918  ft.  A  to  C. 


2x210=  4''0  ft   ) 

<.■  ^  474  ft.  C.  to  D  X  918  =  435132  sq.  ft.  in  the  main  body  of  laud. 


9x6 


54 


Theu    4X210  =  840  ft. 
28x6      =  168 
Less     -     - 
Gives  the  base  of 

triangle     -     -     -     -     .534      "   x  to  B. 
Theu  534x918  A  to  C  —  2  for  the  triangle  =  245106  sq.  feet. 


I  1008  feet  A  to  B. 
474     "   C  to  1). 


Total  sq.  feet  in  the  plat  of  laud  -  680238     " 

680238  —  43560  sq.  feet  in  an  acre,  gives 
15.616  +  acres.     .616x160  =  98.560  sq.  rods. 
.56x30i  =  16.94  sq.  yds. 
.94  X  9    =    8.46  sq.  ft. 
.46x12    =    5.52  sq.  in. 
Note.— See  page  254  for  Paris  and  Woodland  Arpeuts. 


ivision  of  Compound  Denominate  Numbers. 


^ 


610.  Division  of  Compound  Denominate  Nnmbers  is  the  process  of  deter- 
mining any  required  part  of  a  compound  denominate  number. 

It  is  the  reverse  of  compound  multiplication,  and  we  first  divide  the  highest 
denomination  in  the  number,  and  if  any  remainder  occurs  in  the  result,  it  must  be 
reduced  to  the  next  lower  denomination  before  the  number  in  that  denomination  is 
divided. 

1.    Divide  32  lbs.  12  oz.  12  drs.  by  5. 


OPEKATION. 


lbs. 
5)32 


oz. 
12 


dr. 
12 


Explanation. — In  all  problem3  of  this 
kiud,  we  write  the  quantity  to  be  di- 
vided iu  tlie  order  of  its  deuominationa 
and  place  the  divisor  on  the  left,  as  iu 
division  of  simple  numbers. 

Having  thus  stated  the  problem,  we 
first  divide  the  32  lbs.  by  5  and  obtain 
a  quotient  of  6  lbs.  with  2  lbs.  remain- 
der. We  write  the  6  ll)s.  in  the  quo- 
tient line  under  the  pounds,  and  re- 
duce the  2  Iba.  remainder  to  ounces  = 
32  oz.  -|-  12  oz.  ^44  oz.  which  we  divide 
by  5  aud  obtain  8  oz.  with  a  remainder 
of  4  oz.  The  8  oz.  we  write  in  the 
quotient  line,  under  the  ounces,  and 
reduce  the  4  oz.  to  drams,  =  64  drs.  + 
12  drs.  =  76  drs.  which  -:-  5  =  15  drs. 
with  1  dr.  remainder,  or  15|^  drs. ;  this 
is  written  in  the  quotient  Hue  under  the  drams,  and  completes  the  operation. 

Note. — When  the  divisor  is  large,  and  a  composite  number,  we  may  divide  by  the  factors. 
2.    A  box  contains  8  bu.   3  pks.   5  qts.    How  many   smaller  boxes,   each 


6  8  15^,  Ans. 

32  lbs.  4-5  =  6  lbs.  and  2  lbs.  remainder. 

2  lbs.  X  16  =  32  oz.  -4-  12  oz.  =  4i  oz. 
44  oz.  ->  5  =  8  oz.  and  4  oz.  remainder. 

4  oz.  X  16  =  64  drs.  +  12  drs.  =  76  drs. 
76  drs.  -^  5  =  15  drs.  and  1  dr.  remainder. 


Idr. 


dr. 


hokling  1  pk.  1  qt.  1  pt.,  can  be  filled  from  the  larger  box  1 


Ans.  30  boxes. 


I 


OPERATION    INDICATED. 

8  bu.  3  pks.  5  qts.  =  570  pts. 
1  pk.  1  qt.  1  pt.  =  19  pts. 

670  pts.  -^-  19  =  30  boxes,  Ans. 

3.  Divide  37  mi.  3  fur.  4  rds.  by  4 

4. 

5. 

6. 

7. 


Explanation. — In  all  problems  of  this  kind, 
we  first  reduce  both  dividend  and  divisor  to  the 
same  denomination,  and  then  divide  as  in  simple 
numbers. 


Ans.  9  mi.  2  fiu-.  31  rds. 
Divide  6°  24'  32"  by  7.  Ans.  54'  66". 

Divide  24  lbs.  3  oz.  12  pwt.  12  grs.  by  12.  Ans.  2  lbs.  0  oz.  6  pwt.  1  gr. 
Divide  25  yds.  3  qrs.  2  nails  by  13.  Ans.  1  yd.  3  qrs.  3  nails,  1^  in. 
Di\ide  3  cu.  yds.  19  cu.  ft.  996  cu.  in.  by  12.     Ans.  8  cu.  ft.  659  cu.  in. 

(273) 


274 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


S.     Divide  3  mi.  75  cli.  21 1.  by  7.  Aiis.  45  cli.  o  1. 

9.     What  is  f  of  2  m.  3  fur.  1  yd.  2  ft.  Aiis.  7  fur.  5  rds.  1  ft.  lOJ  iu. 

10.  If  300  faiuilies  live  iu  one  towusliip,  and  the  huid  be  equally  divided,  how 
large  a  homestead  will  each  have  1  xVus.  7G  A.  3  li.  8  P. 

11.  How  long  will  it  take  4  men  to  excavate  a  cellar  IG  feet  .square  and  13 
feet  deep,  if  1  man  removes  3  cubic  yard.s  and  15  cubic  feet  in  1  day  ?    Ans.  8  days. 

12.  If  a  man  travels  771  miles  in  30  days,  what  distance  does  he  pass  over  in 
each  day  ?  Ans.  25  mi.  5  fur.  24  rd. 

13.  If  one  spoon  weighs  3  oz.  12  pwt.  12  gr.,  how  many  dozen  spoons  can  be 
made  from  38  lbs.  11  oz.  12  pwt.  12  gr.  of  silver?  Ans.  10|  dozen. 

14.  If  a  man  steps  2  ft.  9J  in.,  how  many  steps  will  he  make  in  going  33  J  miles  ? 

Ans.  C33G0  steps. 

15.  If  the  distance  around  the  earth  were  25000  miles,  and  a  man  could  travel 
33J  miles  each  day,  how  long  would  it  take  him  to  make  the  circuit  of  the  globe, 
allowing  365|  days  to  the  year  ?  Ans.  2  years  19  days  12  hours. 


COMPARISON  OP  THEEMOMETERS. 

611.  There  are  three  scales  in  use  for  registering  temperature  : 

1st.  The  Fahrenheit  (marked  F.)  in  which  the  freezing  point 
of  water  is  marked  32°,  and  the  boiling  jwint  212°.  Zero  is  32^ 
below  freezing  in  this  scale. 

Note. — This  F.  scale  is  generally  used  iu  the  United  States  for  ordinary 
hoiiseliold  and  business  purposes. 

2nd.     The  Centigrade  (marked  C.)  in  which  the  freezing  point 

of  water  is  marked  0°,  and  the  boiling  point  100°. 

Note.— This  C.  scale  is  used  in  France  and  several  countries  in  Europe, 
and  by  scientists,  generally. 

3rd.  The  Reaumur  (marked  E.)  in  which  the  freezing  point  of 
water  is  marked  0°,  and  the  boiling  point  80°. 

Note. — This  R.  scale  is  used  generally  in  Germany  and  Spain. 

From  the  above  facts,  we  make  the  following  statement  of 
ratios  by  which  one  scale  may  be  reduced  to  any  other  : 

1st.  Since  the  number  of  degrees  between  the  freezing  and 
boiling  points  in  the  F.  scale  is  (212°  —32°  =)  180°  and  lOOo  in  C, 
therefore,  (a)  1°  F.  =  igno  q.  =  |o  C;  and  (b)  1°  C.  =  i^o 
F.  =  f  o  F. 

2nd.  Since  the  C.  and  E.  scales  both  have  the  freezing  point 
at  0°  or  zero,  and  since  the  boiling  point  is  100°  C.  and  80°  E., 
therefore,  (a)  1°  C.  =  -h%o  =  |°  E. ;  and  (b)  1°  R.  =  W°  =  -1°  C., 

3rd.  Since  the  number  of  degrees  between  the  freezing  and 
boiling  points  in  the  R.  scales  is  80°  and  iu  the  F.  scale  180°,  (212o_  32o  =  180°), 
therefore,  (a)  1°  R.  =  W°  F.  =  F  F;  and  (b)  1°  F.  =  -ss.-o  E.  =  ao  e. 


-^          0         \ 

f       R  Q 

220  1 — 

ST'    •- 

J 

W.B. 

B 

^) 

80  - 

-  00 

310 

= 

200 

- 

-90 

190 

— 

70  " 

180 

- 

-80 

170 

— 

60 

160 

— 

-70 

IBO 

— 

50 

140 

- 

-60 

130 

— 

120 

- 

lo- 

- 50 

110 

- 

-40 

100 

— 

se" 

90 

- 

-30 

80 

— 

20 

70 

- 

-  20 

60 

- 

10 

60 

— 

-  10 

40 

_ 

F-l- 

30 

= 

-0-- 

--0- 

20 

_ 

; 

10- 

1 

\ 

10  " 

-  10 

♦  COMPARISON    OF    THERMOMETERS.  275 

PROBLEMS. 

1.  Convert  60°  F.  to  its  equivalent  in  Centigrade  degrees  or  scale. 

Note. — In  changing  the  F.  scale  to  C.  or  R.  scale,  or  in  changing  the  C.  or  R.  scale  to  F.,  it 
must  be  remembered  that  the  number  of  degrees  that  expresses  the  temperature  on  a  F.  scale 
(which  marks  freeziug  at  32-\)  does  not  express  the  number  of  degrees  above  freezing  point,  as  it 
does  on  the  C.  and  K.  scales.  Thus,  60^  F.  is  not  60^  above  freezing  point,  but  60'^  —  32='  =  28° 
above  it.  Hence  to  reduce  a  Fahreuheit  reading  to  a  Centigrade  reading,  first  subtract  32°  from 
the  F.  reading  or  number,  and  then  multiply  the  remainder  by  |S3  =  5;  ^^^  to  reduce  F.  to  R.  first 
subtract  32*^  and  multiply  by  Yg\  or  |. 

To  reduce  a  C.  to  a  F.  reading  or  scale,  first  multiply  the  C.  number  by  -ffg  = 
I  and  then  add  to  the  product  32° ;  and  to  reduce  li.  to  F.  first  multiply  the  E. 
reading  by  \%^  or  f  and  add  32°  to  the  product. 

OPERATION. 

•^J^^vlOO        28='v'> 

60°  — 320  =  28°,  above  the  freezing  point  of  F.    Then  '-\^^-  -or  ^^=15|oC.  Ans. 

2.  Eeduce  3oO  C.  to  the  F.  scale. 

OPERATION. 
35°X180  35'^x9         „,,-,     ,  -.  .  e  T^  e  ^ 

— 100 —  oi'  ~5 —  =  ^■^    above  freezing  of  F.,  or  zero  of  C. 

Then  03°  +  32©  =  95°  F.  Ans. 

3.  Eeduce  60°  F.  to  Eeaumui-  scale. 

OPERATION. 


60°  —  32°  =  28,  above  freezing  point  of  F.  and  above  zero  of  R. 
4.    Reduce  12°  R.  to  F.  scale. 


_,  28-^x80         28^X4      ioin-D       a 

Then      lao     ^^  ~~d — =12 PR-  Ans. 


OPERATION. 

12^x180  l''^x9 

— yu —  or  -^^  =  27°  above  the  freezing  point  of  F.    Then  27°+  32°  =  59°  F.  Ans. 

5.  Reduce  50°  C.  to  R.  scale. 

OPERATION. 
50°x80  50°  X  4 

-ToQ-o'" 

6.  Reduce  60°  R.  to  C.  scale. 

OPERATION. 

600x100  600x5 


2/6  soule's  philosophic  practical  mathematics.  * 

TO  EXPRESS  TEMPERATURE  WHEN  BELOW  ZERO. 

612.  WLen  the  temperature  is  below  0  or  zero,  in  any  scale,  it  is  indicated 
by  a  minus  sign  jilaced  before  the  number  of  degrees.  Thus:  —  10°  F.  indicates 
10°  below  0,  or  42°  below  freezing  point  iu  F.  scale. 

*     —  15°  C.  indicates  15°  below  0,  or  zero,  according  to  the  C.  scale. 
—  20°  R.  indicates  20°  below  0,  or  zero,  according  to  the  R.  scale. 

7.  Reduce  — 10°  C.  to  its  equivalent,  iu  the  F.  scale. 

OPERATION. 

— 160  X  I  =  —  28.8°  F.     Then,  — 28.8o  +  32°  =  3.2°  F.  Ans. 

Note. — Iu  adding  or  subtracting  the  32°  where  we  have  minus  numbers,  it  should  be  done 
algebraically. 

8.  Reduce  22°  P.  to  its  equivalent  on  the  C.  scale. 

operation. 

10°  y  100  10°  V") 

220  —  320  =  _  10°  ;  ^      or  —  -^  or  —  IQo  x  f  =  —  5|o  C.  Ans. 

9.  Change  —  13o  p.  to  the  0.  scale. 

operation. 
_  130  _  300  =  _  430  below  freezing  of  F.;  —  45o  x  i  =  —  25°  C.  Ans. 

10.  Change  —  12o  C.  to  the  R.  scale. 

operation. 

12°x80 

-^~  or  —  120  X  I  =  _9.Co  R.  Ans. 

11.  Change  —  20o  R.  to  the  F.  scale. 

operation. 

—^  20^ V 180 

•    go       or  —  200  X  f  =  —  450  F. ;     Then,  —  45°  —  32o  =  —  13°  F.  Ans. 

12.  Change  —  20o  R.  to  the  C.  scale. 

operation. 
—  20°  X  J-  =  —  250  C.  Ans. 
Reduce  each  of  the  following  problems : 

13.  660  F.  to  C.  17.  —  320  F.  to  C. 

14.  8O0  C.  to  F.  18.  _  140  C.  to  P. 

15.  2120  F.  to  R.  19.  —  I80  R.  to  C. 

16.  750  C.  to  R.  20.  —  240  p.  to  R. 


i 


i 


iscellaneous  Problems, 


IN  DENOMINATE  AND  IN  COMPOUND  DENOMINATE  NUMBERS. 


613.     1.     Paid  $2  for  sawing  3  cds.  2G  cii.  ft.  of  wood.     How  iiiucli  could  be 
sawed  for  $1,  at  tlie  same  rate  ?  Aus.  1  cd.  77  cu.  ft. 

ed.          cu.  ft. 
OPERATION    INDICATED.  2  |  3 2C 

1  77  Ans. 

2.  nenry  traded  4  rubber  balls  for  3  pks.  4  qts.   1  pt.   of  pecans.     How 
much  did  he  get  for  each  ball  ?  Ans.  7  qts.  ^  pt. 

3.  An  English  weaver  sold  7J  webs  of  cloth  of  an  equal  number  of  yards  in 
each  web  at  30£.  8s.  3d.  2  far.  a  web.     How  much  did  he  receive  for  all  ? 

OPERATION    INDICATED. 


£. 

8. 

d. 

far. 

30 

8 

3 

2 

^4 

228 

i> 

lA 

3 

•> 

228 


1  Ans. 


£. 

s. 

(1. 

far 

30 

8 

3 

2 

or, 

7i  times  30£ 

n 

=225 

0 

0 

0 

7J  times  8s. 

r= 

'60 

0 

0 

7|  times  3d. 

= 

22 

o 

7^  times  2  far. 
An 

= 

15 

s.  228 

i> 

o 

1 

4.    38  lbs.  12  oz.  15  drs.  of  butter  cost  $7  J. 


How  much  can  be  bought  for  $1  ? 
Ans.  5  lbs.  4  oz.  lOJ-J  drs. 


OPERATION    INDICATED, 


m. 

oz. 

<lr. 

38 

12 

15  -H 

3 

22)116 

6 

13 

Explanation  and  Heason. — In    this    problem,    we 
7J  or  -3-.  have  tho  quantity  that  $7^  =  S";f  will  buy,  and  we 

reason  as  follows  :  Since  $'3^  will  buy  38  lbs.  12  oz. 
15  drs.,  i  of  a  dollar  will  buy  the  22(1  part  and  J, 
or$l  will  buy  3  times  as  much.  In  the  operation 
of  all  problems  of  this  kind,  it  is  best  to  first 
multiply  the  dividend  by  the  denominator  of  the 
5  4  IQll     Ans.  divisor,  and  then  divide  the  product  by  the  numer- 

"  "'  '  ator  of  the  divisor,  as  shown  in  the  above  partially 

worked  operation. 

5.  A  farmer  sowed  3  bu.  3  pks.  6  qts.  of  oats  on  each  of  3.5  acres.     How 
much  oats  did  he  sow  on  all  ?  Ans.  13  bu.  3  pks.  1  qt. 

6.  Divide  26  tons,  13  cwt.  3  qrs.  12  lbs.  of  coal  equally  among  12  families, 
and  find  what  each  will  have  to  pay  at  $12  per  ton.  Ans.  $26.6935. 

7.  How  many  days  at  $1.25  per  day  will  it  take  a  man  to  earn  200  lbs.  12  oz. 
of  beef  at  6  cents  a  pound  1  Ans.  9.6  +  days. 

(277) 


278  soule's  philosophic  practical  mathematics.  * 

8.  A  merchant  exchanged  3  yds.  2  qrs.  of  broad  cloth  worth  $4  per  yard, 
for  26  gals.  3  qts.  1  pt.  of  molasses.     What  was  the  molasses  worth  per  gallon  ? 

Ans.  52^3  cents. 

9.  At  a  certain  distance  20  lbs.  of  steam  expanded  the  mercury  in  a  Fahr- 
enheit thermometer  3.5  degrees.  IIow  many  lbs.  at  the  same  distance  would  be 
required  to  expand  it  from  zero  to  the  freezing  point?  Ans.  182"  lbs. 

Note. — There  are  32-^  between  zero  and  the  freezing  point,  F. 

10.  From  17  lbs.  5  oz.  take  2  lbs.  9  oz.  Ans.  14  lbs.  12  oz. 

11.  One  side  of  a  square  field  is  lOSO  ft.  long;  how  many  rods  of  straight 
fence  will  inclose  it?  Ans.  201  rd.  4  yd.  1ft.  Gin. 

12.  The  wheels  of  a  locomotive  are  IS  ft.  3  in.  iu  circumference  and  make  4J 
revolutions  a  second ;  what  time  will  be  required  to  run  99  miles  ? 

Alls.  1  h.  46  mi.     5  sec. 


phactical  phoblems. 


614.  N.  B.  See  pages  96  to  101,  of  this  book,  for  various  contracted  methods 
of  working  business  problems  containing  the  elements  of  per  bushel,  per  hundred 
(c),  per  thousand  (m),  per  ton  and  per  dozen,  and  also  problems  where  one  or  both  of 
the  factors  are  aliquots  or  multiples  of  10,  100,  or  1000. 

13.  What  cost  1521  pounds  operation   indicated.         -Reason.— One  bushel  costs 

Of  corn  at  84/  per  bushel,  and       1521  ^  56  f^,  Hr,  J^^;-  ^-ff  >„^- 

how  many  bushels  are  there  ?         =27  bu.  9  lbs.     56     84      the    56th    part,    and    1521 

Ans.  $22.81  J  ;  27  bu.  9  lbs.  1^21  P^^^^^s^"'"!  <=«st  1^21  times 

14.  What    cost    2842    bu.     16      operation   indicated.         iJcnsoii.— One  bushel  costs 
lbs    of  wbpit  nt  ftl  '>o„^_i„,„r,„i  9  a.  Si. 22.     Since  1  bushel  or  60 

IDS.  01  wueat  at  *i.^..  per  bushel  ?  .»  j^^  ^^^t,  ^i  22 1  pound  will 

Ans.   $3467.56/5 .  60      1.22  cost  the  60th  part,  and  170536 

I  170536  pounds  will  cost  170536  times 

as  much. 

SECOND  operation.  THIRD  OPERATION. 

2842        ®  $1.22  =  $3467.24  2842  bu.  ®  $1  z=  $2842. 

U  =  A  ®  $1.22  =  .32-,^  ®  (20/)  $1  =      568.40 


®  (2c)  ,i  of  $i    56.84 
$3467.56^^  16  lbs.  ®  {2,\^)  .32,^ 


5 


$3467.56/5 

See  pages  96  to  101,  for  work  similar  to  that  of  the  third  operation. 


15.     What  cost  342506  lbs.   of  wlieat 
at  $9.80  per  imperial  quarter  ? 

Alls.  $C992.S3iL. 


16.     Corn  is  60c'  per  bu.     What  is  it 
vrorth  per  cental?  Aiis.  $1,071. 


17.     Wheat  is  $2.90  per  cental.     What 
is  it  worth  jier  bushel  ?         Aus.  $1.74. 


IS.     Cloth  is  $1.S0  per  yard.     What  is 
it  worth  per  metre  ?         Ans.  $1,968+. 


19.  Cloth  is  $3,937  per  metre.  What 
is  it  worth  per  yard?  Aus.  $3.60. 

20.  Cloth  cost  in  Mexico  $2  per  vara. 
What  is  it  worth  per  yard  ? 

Aus.  S2.1S1  +  practically. 

$2,183  +  accurately. 

Note, — A  vara  of  Mexico  is  32.97  in.,  prac- 
tically, 33  inches. 

21.  What  cost  378  pounds  of  hay  at 
$15  per  ton  7  Ans.  $2.82. 

22.  Cloth  is  worth  $1.45  per  yard. 
What  is  it  worth  per  Spanish  vara? 

Ans.  $1,342  +  practically. 
$1,344  +  accurately. 
Note. — A  Spanish  vara  is  33.3864  inches,  prac- 
tically, 33^  inches. 

23.  Sold  5294  pounds  of  hay  at  $23.75 
per  ton.  How  many  tons  were  there,  and 
what  was  the  value  of  it  ? 

Ans.  2  tons,  1294  lbs.  $62.S6§  value. 


24.     What  cost  4  tons  1420  pounds  of 
hay,  at  $16.25  per  ton  ?     Ans.  $76.53^, 


PRACTICAL  PROBLEMS. 

OPERATION  INDIOATKD. 


279 


480 

9.80 
342506 

OPERATION   INDICATED. 

50 

60 
100 

OPERATION   INDICATED 

100 

2.90 
60 

OPERATION 

$ 

36  1 

INDICATED 

1.80 
39.37 

OPERATION 

39.37 

INDICATED 

$ 

3.937 
36 

OPERATION    INDICATED. 

Practically.  Accurately. 


33 


36 


32.97 


36.00 


OPERATION    INDICATED. 


2000 


15. 
376 


or. 


15. 
.376 


OPERATION    INDICATED. 

Practically.  Accurately. 


36 
3 


1.45 
100 


36.0000 


1.45 
33.3864 


OPERATION    INDICATED. 

$  5294  —  2000  =  2  t.  1294  lbs. 
23.75    $23.75  x  2   =  $ 
5.294  °^  $23.75  x  1294  =  $ 


2000       $  62.86  § 

OPERATION    INDICATED. 

16.25         or,  as  in  the  jireced- 
9.420  ing  problem. 


28o 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


25.     What  cost  1205  pounds  of  bran  at 
80/percwt,?  Aiis.  $10.12. 


26.     What  is  the  value  of   5790  hoop 
poles  at  $18  per  M.  ?         Ans.  $104.22. 


27.    What  is  the  value  of  1304  piue 
apples  at  $11 J  per  C.  ?      Ans.  $156.86. 


28.  What  is  the  cost  of  31845  feet  of 
lumber  at  $22.25  per  M.  1 

Ans.  $70S.55J. 

29.  Bought  3  coops  of  chickens  cou- 
taining  2  dozen  and  7  chickens  each,  at 
$4.35  per  dozen.     What  did  they  cost? 

Ans.  $33.71  J. 

30.  Butter  is  worth  35/  per  pound. 
How  much  can  you  buy  for  lOf  ? 

Ans.  4f  ounces. 

31.  Sell  4J  inches  of  silk  at  $2.75  per 
yard,  and  state  the  amount. 

Ans.  $.34|. 

32.  A  lady  wishes  to  buy  40,0"  worth 
of  silk  which  is  $3.00  per  yard.  How 
much  will  you  sell  her  ?        Ans.  44  in. 


OPERATION    INDICATED. 

.80  12.05 

100     1205   "•'         80/ 

OPERATION     INDICATED. 


1000 


IS. 
5790 


or, 


5.790 
18 


OPERATION    INDICATED. 


100 


11.50 
1304 


or, 


13.64 


OPERATION     INDICATED. 


I  22  25 
1000    31845     *^''' 


31.845 


02  i 


OPERATION    INDICATED. 


4.35 
93 


OPERATION    INDICATED. 


OZ. 


35 


16 
10 


OPERATION    INDICATED. 


36 


i.ti> 


OPERATION    INDICATED. 

in. 


300 

36 
40 

OPERATION    INDICATED. 

( 

30 

65 
74    «■ 

$65 

'           oil 

—  a  0 

33.  A  clerk  commenced  work  on  the 
17th  of  January,  and  discontinued  April 
1st.  He  received  $05  per  month.  IIow 
much  was  due  him  counting  January 
1 7th,  but  not  April  1st  ?    Ans.  $160.33^. 

Note. — In  computing  salaries  and  rents,   all  months  are  considered  as  containing  30  days. 

34.  What  will  4  bu.  3  pks.  1  qt.  1  pt.  cost  at  10/  i)er  pint  1        Ans.  $30.70, 

35.  What  cost  4  bu.  3  pks.  1  qt.  1  pt.  at  $6  per  bushel? 

36.  What  cost  4  bu.  3  pks.  1  qt.  1  pt.  at  25/  per  quart? 

37.  What  cost  4  bu.  3  pks.  1  qt.  1  pt.  at  $1.75  per  iieck  ? 

38.  A  grocer  has  8  jars  of  butter,  each  weighing  14  lbs 


pounds  in  all,  and  what  is  it  worth  at  32J/  per  pound? 


Ans.  $28.78J. 

Ans.  $38.37*. 
Ans.  $33.57}!. 
OZ.     How  many 

Ans.  $37,533. 


♦  PRACTICAL    PROBLEMS.  201 

39.  Bought  3  bales  of  hay  weighing  as  follows :  (1)  421  pounds,  (2)  394  pounds, 
(3)  487  pounds,  at  $22.50  per  ton.     What  did  they  cost?  Ans.  $14.C4§. 

40.  What  is  the  cost  of  2417  cocoanuts  at  $8.25  per  C.  ?  Ans.  $199.40^. 

41.  What  is  the  value  of  8750  shingles  at  $8.75  per  M.  ?  Ans.  $7G.5f)J. 

42.  What  is  the  value  of  11428  fence  pickets  at  $9  per  M.  1      Ans.  $102,852. 

43.  Maple  syrup  is  worth  $1.92  a  gallon.     IIow  much  can  be  bought  for  25/*  ? 

Ans.  41  gills,  or  1  pt.  i  gi. 

44.  Tea  is  worth  $.75  per  jiound.     How  inucli  will  you  sell  for  20<?'  T 

Ans.  4-,\-  oz. 

45.  Bought  4692  pounds  of  barley  at  $.88  per  bushel.  How  many  bushels 
-were  there  and  what  was  the  cost?  Ans.  97  bu.  36  lbs.     Cost  $86.02. 

46.  Bought  2765  pounds  of  oats  at  76/  per  bushel.  What  was  the  cost,  and 
how  many  bushels  were  there  ?  Ans.  $65. 66 J  cost.     86  bu.  13  lbs. 

47.  What  is  the  cost  of  4S7S  pounds  of  wheat  at  $2.45  per  cental  f 

Ans.  $119,511. 

48.  What  is  the  cost  of  200  sacks  of  guano  each  weighing  162  pounds,  at 
$52i  per  ton  ?  Ans.  $846.45. 

49.  A  planter  shi^iped  6  dozen  dozen  boxes  of  peaches  to  market,  but  being 
delayed  on  the  way  ^  a  dozen  dozen  boxes  spoiled ;  the  remainder  were  sold  at  70 
cents  per  box.     What  did  they  amount  to  1  Ans.  $554.40. 

50.  A  grocer  bought  a  cask  of  rum  containing  84  gals.,  at  $1.25  per  gal.;  he 
lost  f  of  it  by  leakage ;  at  what  rate  must  he  retail  the  remainder  per  quart  to  gain 
$27  on  his  purchase  ?  Ans.  55c. 

51.  A  box,  16  in.  long,  12  in.  wide,  and  11.2  in.  deep,  contains  1  bushel;  how 
many  bushels  will  a  proportionally  shaped  box  contain,  each  dimension  of  which  is 
as  many  feet  as  the  first  is  inches,  that  is,  12  times  as  great  1  Ans.  1728  bu. 

52.  A  box  16  in.  square  and  8.4  in.  deep,  contains  one  bushel,  how  much  will 
a  box  contain,  whose  dimensions  are  J  times  as  great? 

Ans.  0.669921875  bu.,  or  2  pks.  5  qts.  |  pt. 

53.  A  can  or  vessel,  11  in.  deep,  7  in.  long,  3  in.  wide,  contains  1  gallon ;  how 
many  gallons  will  a  similar  vessel  contain,  each  of  whose  dimensions  is  10  times 
as  great  ?  Ans.  1000  gals. 

The  last  three  questions  give  convenient  dimensions  for  constructing  the  bushel 
and  gallon  measures ;  and  the  given  problems  are  easily  solved  by  cubing  the  mul- 
tiple of  the  dimensions;  thus,  the  3d  problem  is  equal  to  12^,  the  4th  to  (5)%  and 
the  5th  to  10^ 

54.  If  cloth  costs  $6.40  per  yard,  and  it  takes  6  yds.  3  qrs.  3  na.  to  make  a  suit, 
and  a  tailor  charges  $15  a  suit  for  the  making,  what  will  be  the  entire  cost  of 
making  up  173  yds.  1  qr.  3  na.,  and  how  many  suits  can  be  made  ? 

Ans.  $1485.     25  suits. 

55.  How  many  bottles,  holding  1  qt.  1  gi.,  will  it  require  to  hold  the  contents 
of  1  hhd.  of  wine  ?  Ans.  224  bottles. 


282 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


56.  A  hardware  merchant  received  from  Liverpool  an  invoice  of  iron  weighing 
2  T.  2  cwt.  3  qrs.  20  lbs.,  long  ton  weight  j  the  invoice  price  was  £12.  17s.  Gd.  per 
ton  ;  what  did  the  whole  cost  in  sterling  money  ? 


FIRST  OPERATION. 

2  T.  2  cwt.  3  qrs.  20  lbs.  =  4808  lbs., 
and  £12  17s.  Gd.  =  3090d. 


1  T.  =  2240  lbs.  )  1485G720d. 


12  )  C632i|d. 


20  )  552s.     SMd. 
Ans.    £27    12s.  8i|d. 


SECOND   OPERATION. 

2  T.  2  cwt.  3  qrs.  20  lbs.  =  T.  2.1464  + 
and  £12  17s.  Gd.  =  £  12.875 


£27.6349000 
20 


s.  12.6980 
12  " 


Ans.  £27    12s. 


d.  8.376 
8.376d. 


Explanation. — In  this  and  similar 
questions,  we  iirst  reduce  the  inte- 
gers of  the  commodity  to  the  lowest 
denominutiou  given,  =  4808  lbs., 
and  thcu  reduce  the  price  to  its 
lowest  term  or  denomination.  We 
multiply  these  two  amounts  togeth- 
er, giving  14856720d.,  as  the  cost  of 
4808  lbs.  at  3090d.  per  pound;  but 
since  this  last  number  is  the  cost  of 
one  ton  instead  of  one  pound,  and 
as  there  are  2240  lbs.  in  one  ton  long 
"weight  we  must  divide  by  2240  to 
reduce  it  to  the  rate  per  ton,  which 
gives  us  6632J|d.  as  the  total  cost, 
and  this,  when  reduced,  gives  £27. 
12s.  8J|d.  for  the  answer. 


Explanation. — Here  we  reduce  each 
term  to  the  decimal  of  its  own 
highest  denomination,  then  multi- 
ply them  together,  and  reduce  the 
result  according  to  the  principles  of 
reduction  of  denominate  decimals. 


57.  A  cargo  of  castings  was  received  from  Wales,  amounting  to  225  T.  17 
cwt.  2  qrs.  long  ton  weight;  the  invoice  price  was  £10. 12s.  6d.  per  ton ;  the  freight 
was  4s.  per  cwt.  Considering  the  pound  sterling  to  be  equal  to  $4.84,  what  would 
be  the  cost  in  Federal  money,  of  the  invoice  and  the  freight,  and  also  the  entire 
cost  of  the  whole  cargo  ?  Ans.  Cost  of  invoice,      $11615.62-1%. 

Cost  of  freight,       $  4372.94. 

Cost  of  the  whole,  $15988.56^. 


58.     What  will  be  the  cost  of  1450  tons,  2  cwt.  3  qrs.  of  coal,  at  $4  per  long 
ton  ?  Ans.  $5800.55 


OPERATION   BV   OUR   NKW   MK.TKOD. 
T.      CWT.      QRS. 
1450        2  3 

.05      .OU  =  .0125 


Explanation. — 1  cwt.  =  112  lbs.  =  ^^^^i^  ton,  = 
112  -r-  2240  =  .05  ton.  1  qr.  =  i  of  a  cwt.  or  i 
of  .05  of  a  ton,  which  is  .05  r-  4  =  .0125  of  a 
ton. 


.10 +  .0375  =  .1375 
1450.1375  X  $4  =  $5800.55.     Ans. 

Note.— In  calculating  the  wholesale  cost  of  coal,  it  is  the  custom  to  consider  the  pounds  only  to  the  nearest  quarter. 


SECOND   NEW   METHOD. 

Multiply  total  number  of  pounds  by  the  price,  divide  by  4,  4 

8  and  7,  and  point  off  1  place  in  the  dollars.     Or  make  the  8 

Btatement  thus,  and  cancel :  7 


=  lbs. 

=  price  per  ton. 


iills  and  Invoices. 


^ 


615.  Bills,  in  a  general  sense,  embrace  all  written  statements  of  accounts  and 
many  legal  instruments  of  wiiting ;  but  in  a  more  common  and  limited  sense  tliey 
are  statements  of  goods  sold  or  delivered,  services  rendered  or  work  done,  with  tlie 
price  or  value,  quality  or  grade  of  each  article  or  item.  Bills  or  invoices  of  mer- 
chandise should  state  the  place  and  date  of  each  sale,  the  names  of  the  buyer  and 
seller,  the  price,  the  extra  charges  or  the  discount  to  be  allowed,  the  marks  and 
numbers  on  the  goods,  and  the  terms  of  the  sale. 

When  goods  are  bought  to  be  sold,  or  when  bills  are  rendered  to  the  jobber 
or  retailer,  or  consigned  to  an  agent,  the  bill  is  then  called  an  iiwoice. 

It  is  the  custom  of  accountants  and  merchants,  wheu  making  bills,  to  com- 
mence the  name  of  each  article  with  a  capital. 

When  a  charge  is  made  for  the  box,  barrel,  jar,  etc.,  containing  goods,  it  is 
customary  to  write  its  price  above  and  to  the  right  of  it,  and  add  the  same  to  the 
cost  of  the  goods  it  contains. 

In  making  extensions,  fractions  of  cents  are  not  used  in  the  product;  when 

they  are  ^  or  more,  they  are  counted  cents,  when  they  are  less  than  i,  they  are  not 

counted. 

Note. — Some  bouses  count  only  every  other  ic.  as  a  ■n'liole  cent,  thus  equalizing  tlie  matter 
with  their  customers. 

In  making  the  following  bills,  students  shoidd  use  pen  and  ink,  and  give 
earnest  attention  to  the  proper  form  and  spacing,  to  plain,  neat  and  rapid  penman- 
ship of  both  words  and  figures ;  and  above  all,  to  the  accuracy  of  extensions  and 
additions. 

Note. — AVhen  notes  or  drafts  are  given  in  payment,  the  student  should  draw  the  same  and 
correctly  mature  the  notes. 


[No.  1.     Groceries]. 


H.  A.  &  R.  C.  SPENCER, 

Terms— Cash. 


New  Orleans,  Jau'y  2,  1894. 
Bot.  of  A.  L.  &  E.  E.  Soule. 


1893 
Dec. 


16 


2  bags  Rio  Coffee,  325  lbs. 

50c. 
1  bbl.  Sugar,  234*  lbs. 
i  Chest  Black  Tea,  35  lbs. 
Ibbl.  Rice,  243-16=^227 
40  gals.  N.  O.  Molasses 
6  doz.  Brooms 

3  bbls.  XXX  Family  Flour 
25  lbs.  Cream  Crackers 

.50  lbs.  Graham      do. 
20  lbs.  W.  Butter 


-S  $    23ic. 

"         9  c. 

"        87ic. 

8  c. 

75  0. 
"    4.15 
"     8.12} 

16  c. 
"        15  c. 

30  c. 


Rec'd  payment, 
(283) , 


$76 

21 

30 

18 

30 

24 

24 

4 

7 

6 


38 

61 
63 
16 
00 
90 
38 
00 
50 
00 


$    243  56 
A.  L.  &  E.  E.  SOULfi 

Per  F.  Richardson, 


jj84 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 
[No  .2.     Groceries]. 


MAYER  ifc  BARRIERE, 


New  Orleans,  Apr.  1,  1894. 
Bot.  of  P.  ff.  Thoele  <&  Co. 


523J  lbs.  Butter              ®  25  c. 

13(;j   "     La.  Pecans               "  12Jc. 

56  doz.  C.  Eggs "  18Jc. 

249  boxes  Lobsters "  37Je. 

140  jars  Pickles              "  62Jc. 

48ibs.  Rice "  6|c. 

83  bu.  AV.  Corn            "  87  c. 

64:   "     W.  Oats              "  44  c. 

92  bags  Salt                  "  93  c. 

84  lbs.  B.  Tea               "  97  c. 

108  gals.  R.  Whiskey "  $1.06 

113  doz.  I'ine  Apples              --....  "  1.15 

119  lbs.  Y.  H.  Tea "  96  c. 

324  boxes  Shrimp           .......  "  35  c. 

240  gals.  Fire  Proof  Coal  Oil               .        .        .        .  "  15  c. 

164  lbs.  Lard "  TJc. 

286    "    Green  Peas "  2|e. 

320    "    N.  Y.  C.  Cheese "  ITic. 

112  gals.  La.  Molasses "  22ic. 

124     "      B.  Whiskey "  $1.25' 

72  bbls.  La.  Oranges "  3.75 

61     "      N.  Potatoes "  2.50 

16ilbs.  C.  Soap "  8ic. 

9J    "     Persian  Dates "  6ic. 

Ibbl.  50c.  Sugar,  241-24  =  217  lbs.           ..."  5ic. 

43  lbs.  L.ard  Oil "  58'c. 

45     "     Elgin  C.  Butter                "  35  c. 

34  bn.  W.  Oats             "  48  c. 

425  bbls.  Flour  "  $8.75 


Note. — All  extensions  of  the  above  bill  should  be  made  mentally. 


5620 


70 


[No.  3.  Drugs.] 
Henry  L.  Landfried,  bought  of  E.  S.  Winn,  of  Bayou  Tunica  Post  Office, 
La.,  November  2,  1894,  the  following  Drugs:  60  oz.  Morphiue  at  $3.15;  20  lb.s. 
Insect  Powder  at  33  c' ;  28  lbs.  Gum  Camphor,  at  55/?" ;  15  lbs.  Gum  Opium,  at 
GOc';  2  lbs.  Blue  Mass,  English,  at  90/;  46  lbs.  Potassium  Bromide,  at  60/;  8 
lbs.  Senna  Leaves,  at  55/;  1  Carboy  Nitric  Acid,  74  pounds,  at  11/.  What  was 
the  amount  of  his  bill  1     (Make  and  receipt  the  bill).  Ans.  $262.34. 


[No.  4.     Books]. 


H.  C.  SPENCER  &  CO., 

TERMS— Cash. 


New  York,  Dec.  8,  1894. 
Bot.  of  B.  I>.  Roiclee  &  Co. 


20  doz.  Missionary  Bibles, 
108    "     small  New  Testaments, 

65    "     Prayer  Books, 

65    "     Hymn  Books, 
3    "     Bible  Dictionaries, 
i    "    Webster's  Dictionary, 


®     $15.25 
2.50 
2.25 
3. 
4. 
50. 


Rec'd  payment,  $    953  25 

B.  D.  ROWLEE  &  CO., 

Per  E.  Conrad. 


BILLS    AND    INVOICES. 


285 


[No.  5.     Tobacco]. 


WM.  MELCHERT  &  CO., 
TERMS— Cash. 


New  Oi-.LEAXS,  Jaii'y  31,  1894. 
Bot.  of  L.  L.    WiUtams  &  Co. 


321  lbs.  Tobacco,  Low  LugR, 

1140  "  "  Med.  Lugs, 

509  "  "  Low  Leaf, 

965  "  "  Med.  Leaf, 

398  "  "  Good  Leaf, 

2416  "  "  Fine  Leaf, 

713  "  "  Selections 


® 


6  c. 

7ic. 

9ic. 
lUc 
13ic. 
15  c. 
16|c. 


Rec'd  payment, 


L.  L. 


$    798  49 
WILLIAMS  &  CO. 


[No.  6.     Sugar]. 


F.  L.  RICHARDSON,  Jr., 
TERMS— Dft.  30  days. 


New  Obleaxs,  Dec.  17,  1894. 
Bot.  of  C.  J.   Sinnott. 


1420  lb: 

1927 

2810 

902 

813 
2741 


s.  Sugar,  Common, 


Good, 

Fair, 

Prime, 

Choice, 

Yellow  Centrifugal, 


@    5Jc. 

"    7ic. 

"    7|c. 

"    8ic. 

"    9i^c. 

"  lOJc. 

Rec'd  payment  by  dft.  ®  30  days'  sight. 


I 


$    878  78 
C.  J.  SINNOTT. 


[No.  7.     Tobacco]. 


MONTGOMERY  &  TREPAGNIER, 

TEEMS— 3  moa. 


New  Orleans,  Dec.  23,  1894. 
Bot.  of  Sadler  <&  Smith, 


7  Gross  Chewing  Tobacco,    .-------®  $13. 

180  lbs.  Smoking  do., "        1.40 

6  M.  Havana  Cigars "70. 

2  M.  N.  O.  Manufacture  do "30. 

Rec'd  payment,  1 


[No.  8.     Pork  and  Beef]. 

Jacob  Broders,  bought  of  G.  F.  Sauter,  of  St.  Louis,  Mo.,  October  15, 
1894,  the  following  articles:  15  bbls.  of  Prime  Mes.s  Pork,  $13.50;  10  bbls.  Extra 
Mess  Beef,  $8.75;  152  lbs.  Bacon,  Shoulders,  at  7^/;  75  lbs.  Long  Clear  Sides,  7 J/; 
146  lbs.  Sugar  Cured  Hams,  at  12J/;  78  lbs.  Breakfast  Bacon,  canvassed,  at  HJ/. 
Make  out  a  receipted  bill.  Total,    $333.58. 


286 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


[No.  9.     Candles]. 


i\  h.  &.  W.  p.  KICIIARDSON, 
TEEMS— 1  mo. 


New  Orleans,  Feb.  4,  1894. 
Bot.  of  P.  W.  Sherwood  &  Co. 


30  liox.  Sperm  Candles,  596  lbs. 

24  do.    Adam  Extra  Candles,  483  lbs. 

15  do.    Silver  Gloss  Starch  360  lbs. 


® 


.35t 
.28 


Rec'd  payment, 
[No.  10.     Shoes]. 


$    385  53 


M.  G.  WUERPEL, 
TEEMS— Cash. 


New  Orleans,  August  8,  1894. 
Bot.  of  Albert  Bohnet. 


6 
4 
5 
12 
1 

2i 
2 

35 

6 

7 
20 


dozen  Men's  Boston,  17  In.  Kip  Boots 

"  Ladies'  Calf,  Piuk  Lined,  Balmorals 

"  Youths' Was  Brogans        ... 

"  Youths'  Buff  Congre.s8,  Balmorals 

"  Misses'  Lasting  Congress,  No.  1,  sewed 

"  Misses'  Goad  Polish  No.  1,  sewed 

"  Boys'  Buff  Balmorals,  Cloth  Top, 

''  Men's  17  in.  French  Top  Calf  Boots 

"  Misses'  Calf  Balmorals  No.  1,  pegged 

"  Misses'  Lasting  Congress,  No.  2,  sewed 

"  Boys'  Split  Wax  Brogans 


® 


$.50.00 
25.00 
15.50 
17.45 
23.75 
32.00 
25.75 
35.90 
17.00 
19.00 
12.00 


2573  65 


[No.  11.    Domestic]. 

Eujiene  Ellis,  bought  of  Wm.  Dirker,  of  New  Orleans,  La.,  November 
4,  1894,  the  following  Dry  Goods:  370i  yds.  of  Toledo  TlaidsatlSc^;  42yds.  Cotton 
Check  at  ]2it' ;  81  yds.  Cotton  Duck  at  10^ c^;  90  yds.  of  Hickory  Stripes  at  li^f  ; 
C7  yds. Tickings  at  13(.'';  86  yds.  Lowell  Sheetings,  10-4,  bleached,  2Gff;  82  yds. 
Sheeting,  Pembroke  10-4,  at  29.J^;  3SJ  yds.  Cotton  Drills,  Pride  of  the  South,  at 
7J/;  143  yds.  Lonsdale  Cambric  at  124/ ;  45  yds.  Wamsette  at  14('.  Make  and 
receipt  the  bill.  "  Total,  $164.72. 


[No.  12.     Drugs]. 


HENRY  HIRSCH, 
TEEMS— Note  60  days. 


New  Orleans,  August  7,  1894. 
Bot.  of  Henry  Palmer. 


55  lbs.  Acetic  Acid,  No.  8.,         ... 
123     "     Carbolic  Acid,  Cryst. 
98     "     Sulphuric  Acid,  Carboy,  per  jiound 
67     "    Citric  Acid  -     "  . 

34  oz.  Antipyrine  ..... 
78  lbs.  Aqua  Ammonia  .... 
88     "    Bismuth 

44  "    Cream  Tartar,  pure        ... 

35  "    Castile  Soaji,  white         ... 
2  cans  Glycerine,  50  pounds      ... 

24  lbs.  Gum  Arabic,  French,  powdered 
67     "     Oil,  Berganiot  .        .        .        - 

45  "  Potassium  Chlorate  ... 
34    "    Potassium  Iodide  ... 


® 


$l.i 

3.; 


25c. 

60c. 

6c. 

,53c. 

50 

15c. 

50 

.S5c. 

25c. 

35c. 

70c. 

00 

35c. 

50 

"" 

"1 

825l84 


^ 


BILLS    AND    INVOICES. 
[No.  13.     Butter  and  Cheese]. 


=  87 


J.  A.  SABATIER, 

TERMS— Due  bill  1  mo. 


New  Orleans,  August  4,  1894. 
Bot.  of  W.  E.  Slaughter. 


167    lbs.  Extra  Fancy  Creamery,  Bntter 

112  "  Faucy  Creamery  " 

75  "  Choice  Creamery  ■•        .        -        . 

55  "  Prime  Creamery  "... 

85  "  Fancy  Dairy  "... 

62J  "  Choice  Dairy  "... 

Ill  "  Prime  Dairy  "        -        -        - 

48  "  Common  "... 

75  "  New  York  Fancy  Cream  Cheese 

88 J  ''  Western  Chedtlers,  Full  Cream  Fancy  Cheese 

74  "  America  Full  Cream  Fancy  Cheese 
90  "  Twins  Full  Cream  F.ancy  " 

75  "  Flats  Full  Cream  Fancv  " 

38  "  Best  Skims  "  "  -        - 

92  "  Skims 


® 


6ic. 
25  c. 
20  c. 
17  c. 
16  c. 
14  c. 
12  c. 

8|c. 
12ic. 
llic. 
12  c. 
11  c. 
lOic. 

8"c. 

5ic. 


157,40 


[No.  14.     Wood  and  Coal]. 


H.  J.  CALVERT, 


I 


New  Orleans,  Jau'y  29,  1894. 
Ill  acc't  with  L.  B.  Keiffer. 


Jan. 


29 


To  old  balance  as  per  bill  rendered, 

12  cords  Ash  Wood         - ®  $7. 

4  cords  Oak  Wood        .........  ''  6.50 

50  bbls.  Pittsburg  Coal           ........  "          60c. 

Cr. 

By  Cash "    $50 

•''  6  days'  Labor,  at  $4, "24 

Balance  due  Jan'y  29,  1894,  $    157  10 

Settled  by  note  at  60  days. 

L.  B.  KEIFFER. 


[No.  15.     Poultry  and  Eggs]. 


J.  H.  McNEELY, 

TEEMS— Note  30  days. 


New  Orleans,  August  9,  1894. 
Bought  of  S.  Delerno. 


6i  doz.  Western  Grown  Chickens, 

7J     "     Louisiana  Young  Chickens, 

4Ji  "     Ducks, 

6i     "    Western  Geese, 

2t     "     Western  Y'oung  Turkeys, 

H    "    Western  Old  Turkeys,    - 

76      "    Texas  Eggs, 


® 


$  5.00 

3.00 

2.5.1 

6.25 

15.00 

18.00 

.18i 


188  63 


i'88 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


D.  S.  DRENNAN, 
TEEMS— Dne  Bill,  30  days. 


[No.  16.     Rice]. 

New  Orleans,  August  13,  1894. 

Bought  of  A.   J.   Duclos. 


Aug. 


13 


160  lbs.  Rice,  No.  2, 
136         "         Common 
240  "  Ordinary      - 

110         "         Fair     - 
273  "         Good    - 

140  "  Prime 

310  "  Choice 

150         "         Head, 
285  "  Fancy 

25  bWs.  of  162  lbs.  Eougb  Rice 


® 


lie 

25c. 
3ic. 
3Jc. 
4ic. 
4ic. 
5ic. 
5ic. 
5Jc. 


$4.50 


189  22 


FRANK  DUMAS, 


[No.  17.     Plantation  Sugar,  Open  Kettle]. 

New  Orleans,  August  15,  1894. 


Bought  of  If.   P.   Knobloch. 


July 


10 


280  lbs.  Choice  Sugar 

197    "  Strict  Prime 

230   "  Prime 

2.54   "  Good  Fair 

290    "  Good  Prime 

260   "  Common 

425   "  Inferior 


®      4Jc. 

"       5  c. 

"       4|c. 

"       4ic. 

"       4|c. 

"       4ic. 

"      3ic. 

85 

S.  FORET, 
TERMS— Cash. 


[No.  18.    Cotton]. 


New  Orleans,  August  9,  1894. 
To  J.  J.  Foley,  Dr. 


2  bales,  452,  461  =  913  lbs.  Low  Ordinary  Cotton 

1  "  4.58  lbs.  Ordinary  Cotton  .... 

3  "  450,  455,  460  =  1365  lbs.  Good  Ordinary  Cotton 
1  "  459  lljs.  Low  Middling,  Cotton         .        .        - 

1  "  462  lbs.  Middling  Cotton  .... 

1  "  455  lbs.  Good  Middling  Cotton         ... 

2  "  4.52,  460  =  912  lbs.  Middling  Fair  Cotton 

1  "  458  lbs.  Fair  Cotton  .        .        .        •        - 


® 


5Jc. 
6  c. 

6Jc. 
6jc. 
lie 
Tic. 
He 
8  c. 


372  56 


BILLS    AND    INVOICES. 


289 


[No.  19.     Refiueil  Sugar]. 


I 


EDWARD  LeBLAXC, 
TEEMS— 60  days. 


New  Orleans,  August  10,  1894. 
Bot.  of  D.  y.  Moore. 


252    lbs.  Cut  Lnaf  Su<rar 


ITGi 

240 

11'} 

2.52 
144 
208 
197 


Powdered  Sugar 

Standard  Granulated  Sugar 

Off  Granulated  Sugar 

Candy  A.  Sugar 

Confectioners'  A.  Sugar 

Extra  C.  Sugar 

C.  Sugar  ... 


® 


61  c. 

6f  c. 

6i  c. 

6f  c. 

6i  c. 

oi  c. 

5  c. 


99  95 


[No.  20.     Lumber]. 


A.  &  S.  H.  SOULfi, 


New  Orleans,  Jau'y  12,  1894. 
To  Jas.  Mc.  Hugh.  Dr. 


To  4378  feet  Common  Boards      -         -         - 
"  1760  feet  Dressed  Flooring 
"  5125  Brick  .        .         .        .        . 

"  9250  Cypress  Shingles 
"  Cartage  aud  Labor, 

Received  payment, 


21.00  per  M. 

28.50   " 

14.25   " 

6.50   " 

14 

25 


I      289  51 


GEO.  F.  BARTLEY  &  CO., 


[No.  21.     Freight  Bill]. 

New  Orleans,  Jan'y  3,  1894. 

To  Steamship  Knickerhocker  and  Owners,  Dr. 


For  Freight  on  439f  cubic  ft.  ®  25c.     The  same  being  contents  of  8  boxes 
measuring  as  follows : 

Nos.  1,  2  &  3  :  5  ft.  4  in.  X  4  ft.  6  in.  x  2  ft.  8  in.  = 
"  4,  5  &  6 :  6  ft.  2  in.  X  3  ft.  0  in.  x  2  ft.  11  in.  = 
"    7  &  8 :  12  ft.  3  in.  X  2  ft.  4  in.  x  1  ft.  6  in.        — 

Received  payment, 


$      109  91 


I 


[No.  22.     Services]. 

A  carpenter  receives  $3.25  per  day  of  9  Lours.     How  mncli  will  lie  earn  work- 
ing each  week  day  iu  the  mouth  of  January,  1894^,  which  commenced  on  Monday  1 
(Make  the  bill).  Ans.  $87.75. 


[No.  23.     Services]. 

A  laborer  receives  SOe'  per  hour,  and  he  works  lOJ  hours  per  day.    How  much 
wUl  he  receive  for  6  days  work  ?     (Make  the  bill).        '  Ans.  $18.90. 


290 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


[No.  24.     Services]. 

A  clerk  commenced  work  June  20tli,  and  worked  till  Sej)tember  the  15tli,  at 

$65  per  month.    He  paid  $15  per  month  for  board  and  $3  per  mouth  for  washing, 

including  June  20th,  but  not  September  15th,  how  much  money  has  he,  not  allowing 

for  other  expenditures  ?     (Make  tlic  bill).  Ans.  $1.33.16§. 

Note. — In  calculating  salaries  ami  rents  by  the  raontli,  30  days  are  allowed  for  a  month;  i.  e. 
the  same  salary  or  rent  is  jiaid  for  each  mouth  whether  of  2(J,  29,  30  or  31  days. 


[No.  25.     Rent  and  Services]. 


J.  T.  O'QUINN, 


New  Orleans,  Jan'y  4,  1894. 
To   W.  Hermann,  Dr. 


For  Rent  of  honse  No.  386  Dryades  Street,  from  Oct.  7,  1894,  to  Jan.  1,  1895, 

Ist  date  included,  2|;^  months  at  $35  

"    Services  as  collector  from  Sept.    19,   1894,  to   Jan.   4,    1895,    both   dates 
included,  3^^  mouths,  at  %lb         ........ 

Received  payment,  \ 


363  00 


[No.  26.     Services]. 
MISSISSIPPI  VALLEY  TRANSPORTATION  CO., 


New  Orleans,  Feb.  1,  1894. 
To  Buck  &  Richardson,  Dr. 


For  services  rendered  in  cause  No.  55472. 
vs.  Miss.  V.  T.  Co." 


'  Steamer  R.  E.  Lee,  and  Owners 


Received  payment. 


[No.  27.     Services]. 

A  mechanic  receives  $4  per  day  of  9  hours,  and  60/  an  hour  for  "overtime." 
During  the  first  three  days  of  the  week  be  worked  30  hours,  and  the  last  three 
days  of  the  week  he  commenced  at  6:30  A.  M.  and  worked  till  10  P.  M.,  less  two- 
hours  for  dinner  and  sui)pei'.    What  are  his  wages  for  the  week's  work  ? 

(Make  the  bill).  Ans.  $33.90. 


[No.  28.     Carpenters'  Time  Sheet]. 
M.      T.      W.        T.      F.      S.      da. 


Rate 
per  (lay. 


R.  Hirsh 

F.  Smith 

O.   Wood 

P.  Murphy  . . . 
H.  Brown . . . . 
J.  Fitzpatricl< 
J.  S.    Smith. . 


Foreman  . . 
Carpenter . 

do. 

do. 

do. 

do. 
Apprentice 


Oct. 

oo 

10 

10 

10 

10 

15 

10 

6i 

$4 

00 

22 

10 

10 

10 

10 

15 

10 

6i 

3 

50 

- 

oo 

10 

10 

10 

10 

X 

10 

5 

3 

00 

•>') 

10 

JO 

10 

10 

10 

10 

fi 

3 

00 

oo 

10 

5 

5 

5 

10 

10 

4i 

3 

00 

26 

X 

X 

X 

X 

10 

10 

2 

2 

50 

22 

10 

10 

X 

10 

10 

10 

5 

2 

00 

— 

114 

00 

^    Ans. 


In  this  time  slieet,  what  is  due  to  each  mechanic,  and  what  is  the  total 
amount  at  the  close  of  the  week ;  10  hours  being  a  day's  work,  and  double  price 
being  paid  for  overtime  ?  Ans.  $111.    Total  amount. 

Note. — The  numbers  written  after  the   names   represent   the  number  of  hours   that  each 
worked  on  the  different  days  of  the  week.     The  x  signifies  no  hours. 


BILLS    AND    INVOICES. 


291 


[No.  29.     Factory  Time  Sheet]. 


M. 

T. 

■ft'. 

T. 

F. 

s. 

d3. 

n.ate 
per  day. 

J  c  Bell 

Oct. 

8 
8 
8 
8 
8 
8 

10 
10 
5 
10 
10 
10 

10 
10 
10 
X 
10 
10 

10 
10 
10 
10 
10 
10 

10 
X 
10 
10 
10 
10 

10 
10 
10 
10 
10 
10 

10 
10 
10 
10 
10 
10 

6 
5 
5^ 
5 
6 
6 

4 
3 
3 
3 
3 
2 

no 

Thos.  A.  Clark 

50 
nO 

T.  J.  Fatjo  

T-f  A  Hornor  .   . .     

50 
50 

H  C  Lowrv 

00 

In  this  time  sheet,  what  is  due  each  workman  Saturday  night,  10  hours  being 


a  day's  work,  and  what  is  the  total  amount  ? 


Ans.  To  the  last,  $111.25. 


[No.  30.     Paints,  Oils,  etc]. 


FLYNN  &   KERWIN, 
TERMS— 30  (lays. 


New  Orleans,  Mar.  11,  1894. 
Bought  of  Donelon  tfc   Haight. 


2  ca.ses,  12  gals.  Green  Paint 
i      "        3     "     Vermilion 
12  gals.  Roofing  Paints 
9      "     Carriage     "      -        - 
6       "     Copper        "       -         - 
.50  lbs.    White  Lead 
75    "      Pntty 
12  gals.  Coach  Varnish  - 
18      "     West.    Linseed  Oil  - 
45  lbs.    Eng.  Vermilion 
20     "     Vandyke  Brown 


® 


$1.80 

2.40 

.65 

1.85 
.07 
.02 

4.25 
.51 
.90 
.081 


163  01 


I 


[No.  31.     Building  Materials]. 


M.  A.  TERRY  &  CO., 
\ 

TERMS— Cash. 


Nkw  Orleans,  Mar.  9,  1894. 
Bought  of   H.   M.    Thompson   i£-   Son. 


750  ft.  1}X3}  Clear  Pine  Flooring       -        -        -  per  M.  |25.00 

1200  "    No.  1.  Weatherboards          ....            "  13.00 

5.50  "     Ceiling         --....-            "  15.50 

3500  Laths         -        - "  1.90 

45  bbls.   Lime         .--.---             ®  .92 

18      "     Domestic  Cement    -        .        .        -        .            "  1.65 

800  Paving  Brick             ......  per  M.  11.25 

25  bbls.  Sand         .......             ®  .22i 

12    "      Small  Shells "  .33^ 

2000  Am.  Fire  Brick per  M.  26.00 

750  ft.    Sewer  Pipe          - @  .21 

600    "    Cliimney  Flue  Lining        ....              "  .Igf 


461 


26 


2g: 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 
[No.  32.     Hardware]. 


New  Oulkans,  Bee.  22,  1893. 
E.  WEIDIG, 

Bot.  of  Reynes  tO  Carman. 

TEEMS.— Cash.  Days X  Discount Days %  Interest  after  Maturity  and  subject  to  sight  Draft  without  notice. 


49i  doz.  Claw  Hammers 

ISJ  "  Lathiug  Hatchets 

16  "  Chain  Holts 

75  lbs.  Nails,  ea.  ed.-8d.-10d. 

14i  doz.  Measuring;  Tapes 

92  "  Marking  Brushes 

lOi  "  Butoher  Saws 

104  "  Gem  Wrenches 

18  "  Chisels,  ea.  |  aud.  |  in. 

98  "  File  Handles    - 

67i  "  Curry  Combs    - 

85J  "  Gas  Flyers 

89  '•  10"/14"  Shelf  Brackets 

142^  "  Ivory  Rules 

65  "  Pointing  Trowels     - 

106i  '•  Mortise  Locks 

55^  "  Kim  " 

165  "  Shutter  Bolts 

18i  "  P.adlocks 

46  "  Drawer  Pulls    - 

26  "  Dashboard  Lanterns 


© 


$  4.95 
8.49 
1.75i 

.04* 
4.50 

.95 
5.65 
2.35 
6.80 

.88 
1.76i 
8.46 
4.98 
10.20 
4.75 
3.30 
2.85 
1.60 
2.38 
1.86i 
4.25 


[No.  33.     Grain,  Hay,  etc]. 


McDANIEL  &  KELLER, 


TERMS— Note  at  CO  ilays. 


New  Okleans,  Jan'y  25,  1894. 


Bot.  of  Eedfield  d;  Halsmith. 


2144  llis.- 

1242  "  - 

8.52  "  - 

792  '•  - 

1427  '•  - 

3745    "  - 

1701  "  - 


1)11.  Yellow  Corn,         .... 

-  "     Texas  Wheat,  -         -  »      - 

-  "     Wliite  Oats,  ..... 

•  "     Barley,  ...... 

Cwt.  Bran,  ..... 

■  Tons  Timothy  Hay,  ... 

•  Tons  Clover  Hay,         .... 

Bec'd  pay't  by  note  at  60  days, 


® 


63c. 

1.70 

56c. 

83c. 

75c. 

18.50 

20. 

Ji    150  27 


REDFIELD  &  HALSMITH. 


BILLS    AND    INVOICES. 


293 


[No.  34.     Sugar— Centrifugals]. 


J.  HEINTZELMAN, 


TEEMS-Cash. 


New  Orleans,  August  4,  1894. 
Bot.  of  Win.  O'Brien. 


244  11 

>s.  Plantation         S 

355 

'    Granulated 

197 

'     Off  Granulated 

280 

'     Choice  White 

175 

'    Off    White 

290 

'     Gray  Off  WTiite 

178 

'     Fancy  Yellow 

230 

'     Choice  Yellow 

312 

'    Prime  Y'ellow 

230 

'    Off  Yellow 

234 

'     Yellow  Fair 

226 

'     Seconda 

Sugar 


®     oi  c. 


6 
6J- 


61 
6J- 
6'i 
6 

5|Sc 
5i  c 
4J  0 
4f  c, 


174  34 


[No.  35.     Produce]. 


J.  P.  POWELL, 


TEEMS— 1  mo. 


New  Orleans,  August  7,  1894. 


Bot.  of  W.  E.  Richards. 


27  hbls.  Neshannock  Potatoes 

15     "       Onions  .         .         _         _ 

6  Crates  Cabbage  ... 

12  Strings  La.  Garlic         ... 

30  bbls.  Sweet  Potatoes,  Yams 
6     "      Western  Cucumbers 

40  doz.  Egg  Plants  ... 

140  lbs.  Choice  Northern  White  Beans 
150     •'    Northern  Red  Kidney        " 

75     "    Northern  AVhite  Kidney     " 

70     "    Green  Peas  ... 

67     "    Dried  Peaches 
128    "    Evaporated  Peaches 


® 


$  3.25 
4.50 
4.25 
1.25 
2.25 
2.55 

.20 

4ic. 

7Ac. 

8  c. 

3ic. 

9ic. 
12  c. 


332 


26 


294 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


ETHICAL  AMBITION, 
TERMS— Fidelity  to  Duty. 


[No.  36.     Character,  Virtue,  Knowledge,  etc]. 

New  Orleans,  Oct.  1,  1894. 

Bot.  of  Soule  College. 


116|  lbs.  Energy  .and  Perseverance 

45  "  Self  Keliance  .  .  .  .  . 
100      "     Honesty 

17  "  Business  Sagacity  -  .  -  . 
150      "     Punctuality  -         .         .        .         . 

49  "  Cheerfulness  ..... 
56  "  Faithfulness  ..... 
73  boxes  Accuracy  ..... 
38      "      Thoughtfulness  .... 

i  yd.  Talkativeness         ..... 

25    yds.  Economy  

2^     "      Recreation  ..... 

50  bu.  of  Politeness  ..... 
75  "  "  Neatness  and  Order  ... 
75      "      "  Obedience  and  Good  Deportment 

i    "      "  Attention  to  other  Peoples'  Business 
50      "      "   Attention  to  my  own  Business 
500    bbls.    Practical  English  Education 
500      "        Practical  Business  Education 
500    facts  of  Physiology  and  Hygiene 
500      "      "   Phrenology  and  Sociology 


® 


.87i 
.95 

jiioo.'oo 

1.15 

1.16ff 

.81 

.94 

.75 

.661 

.01 

.83 

.33i 

2.00 

1.00 

2.50 

1.00 

5.00 

4.00 

4.00 

5.00 

5.00 


WEHRT  &  GEORGE, 
TERMS-30  days. 


[No.  37.     Dry  Goods]. 

New  Orleans,  Feb'y  14,  1894. 

Bot.  of  Morris  tfc  Childress. 


25    yds.  Serge               ®  .22J 

3~i  "  Hopsacking             -  "  A2i 

181  "  Diagonal  Serge               .....  "  .39" 

15|  "  French  Siiteens                "  .27f 

12  "  Brocaded  Silk -  "  .62i 

47  "  Colored  Satins "  .43' 

22  "  Table  Damask        .        -        -        .        .  "  .47i 

310  "  Crash               .......  "  .06 

82.}  "  Canton  Flannel                ......  «  .09i 

64  "  Storm  Serge              '•  .35 

9  "  Brocaded  Cashmere                  ....  "  .141 

.53  "  Ladies'  Cloth    ' "  .87} 

87  "  Figured  Satinet               .        .         .         _        .  "  .9,5' 

20  "  French  Velours                "  1.47;V 

35  "  Surah  Silk               .......  .<  .75" 

33J  "  .Scotch  Plaida          - "  .65 

28  "  Whipcords                ....  u  ,33^ 

17}  "  Henriettas                .....  "  .35' 

142}  "  Silk  Ribbon             ....  <<  .12^ 

75  "  Satin     " "  .16| 

IJ  doz.  Initial  Linen  Handkerchiefs         ....  ''  4.75 


382  16 


*  BILLS    AND    INVOICES.  295 

[No.  38.     Invoice  of  Goods  Shipped]. 

New  Orleans,  Jau'y  16,  1894. 

Invoice  of  Sundries,  irarcliased  by  H.  Strong  &  Co.,  and  shipped  per  Steamer 
La  Belle,  for  account  and  risk  of  Clias.  B.  Thorn,  Shi-eveport,  La. 


87  l)l)ls.  Molasses,  3498  fjalloiis 
■20  liliils.  Sngi-r.  23780  lbs. 
10  bbl,s.  Eice,  2ir>0  lbs. 


Dr.iy.age, 

Insurance  on  $4800.40 

Commissiou  ou  $4364.00 


-Charges.- 


®  60c. 
"  9c. 
"      5c. 


4346 

17 

30 

109 


50 
50 
00 
10 


$    4503  10 


[No.  39.     Slaters' Bill]. 


C.  H.  REYNOLDS, 


New  Orleans,  Jan'y  14,  1894. 
To  H.  Marsden,  Dr. 


For  slating  a  roof  measuring  72  ft.  4  in.  by  49  ft.  10  in.,  and  containing 

36.04}i8qs.         -.■-.-■ ®     $14.50 

For  239  It.  Guttering  "  .90 

Eeceived  payment, 


$       737  77 


[No.  40.     Cistern  Makers' Bill]. 


NEW  ORLEANS,  ST.  LOUIS,  AND  CHICAGO  E.  E., 


New  Orleans,  Nov.  4,  1894. 


To   W.  L.  &  H.  Jackson,  Dr. 


For  150  Cisterns  holding  766782.45  gals.  ®  2ic.  per  gal.  The  inside  meas- 
urement of  each  cistern  is  as  follows:  11  it.  3  in.  perpendicular  height, 
lower  base  9  ft.  2  in.  in  diameter  and  upper  base  8  ft.  5  in.  in  diameter. 

Eeceived  payment, 


$  19169  56 


Note.— For  other  Ijills  involving  discount,  rebate,  etc  ,  and  embracing  different  lines  of  busi- 
ness, see  other  pages  of  this  book  as  per  index. 


0  Find  the  Interval  of  Time  Between  Two  Dates, 


^ 


616.  1.  How  many  years,  months,  days,  hours,  and  minutes  from  4:20 
o'clock  p.  M.  June  10,  1889,  to  9:15  o'clock  A.  M.  August  14,  1894,  not  allowing  for 
leap  years,  and  counting  30  days  to  the  month  ? 

OPERATION.  Explanalioit. — In  all  prob. 

J  ,  •  leius  of  this  kind,  we  write 

y-  "^°-  ^-  ^'"-  '"'°-  the  later  date  first,  since  it 

1894  8  14  9  15  expresses  the  greater  period 

1889  G  10  IG  20  of  time,  and  the  earlier  date 

— ■ ■      beneath.    Then  subtract  as  iu 

5  yrs.  2  mOS.  3  ds.  IG  hrs.  55  min.      Ans.    compound  denomiiuite  num- 

bers. 
Note. — The  months  are  numbered  from  January,  and  the  hours  are  counted  from  12  o'clock 
at  night. 

SECOND   OPERATION 

Thetimefrom  June  10,  1889,  to  June  10,  1894,     -         -        -        -  =  5  yrs. 

"  "      June  10,  1894,  to  Aug.  10,  1894,     -         -         -        -  =  2  mos. 

"  "      4:20  P.  M.  Aug.  10, 1894,  to  4:20  P.  M.  Aug,  13,  1894,  =  3  ds. 

"  "      4:20  p.m.  Aug.  13, 1894,  to  8:20  A.M.  Aug.  14, 1894,  =  IGhrs. 

'<  "      8:20  A.  M.  Aug.  14,  1894,  to  9:15  A.  M.  Aug.  14,  1894,  =  55  min. 

Note. — While  both  of  the  above  operations  are  correct  according  to  the  conditions  of  the 
problem,  and  conform  to  the  usual  method  of  finding  the  time  between  dates,  neither  operation 
gives  an  accurate  result,  for  the  reason  that  some  years  and  some  months  contain  more  days  than 
other  years  and  other  months.  The  only  accurate  way  is  when  the  time  is  less  than  one  year,  to 
count  the  exact  number  of  days  in  each  month  of  intervening  time,  or  refer  to  a  time  table.  When 
the  time  is  more  than  one  year,  fiud  the  days  for  the  mouths,  as  above,  and  allow  3G5  days  for  com- 
mon, and  366  for  leap  years. 

2.  The  Declaration  of  Independence  was  ratified  July  4,  1776  ;  the  battle  of 
New  Orleans  was  fought  Jan'y  8,  1815.  What  is  the  time  between  these  two  dates, 
by  the  usual  method  ?  Ans.   38  yrs.  6  mos.  4  ds. 

3.  Washington  was  born  February  22, 1732.  How  old  was  he  at  the  Declar- 
ation of  Independence,  Jnly  4,  1770?  How  long  did  he  live  after  that  event,  his 
death  occurring  December  14, 1799,  and  what  was  his  age  f 

Ans.  44  yrs.  4  ms.  12  ds. 
23    "      5    "     10  " 
G7    "      9    "     22  " 

Jamestown,  Va.,  was  first  settled  May  23,  1007,  and  the  Pilgrims  lauded 
at  Plymouth,  December  22,  1G20.     What  time  intervened,  by  the  usual  method  ? 

Ans.  13  yrs.  6  mos.  29  ds. 

5.  A  note  was  dated  January  10, 1892,  and  was  made  payable  2  years  after 
date.  When  did  it  become  due;  how  many  days  did  it  run,  by  the  usual  method; 
and  how  many  days  counting  actual  time;  no  allowance  to  be  made  for  days  of 
grace  ?  Ans.  January  10,  1894,  it  matured ; 

720  ds.  by  the  usual  method ; 


731  ds.  actual  time. 


(296) 


bk  V- 


*  TO  FIND  THE  LENGTH   OF  THE    DAY  AND  THE  NIGHT.  297 

fi.  A  note  is  drawn  Oct.  15,  1894,  and  made  payable  3  months  after  date. 
When  does  it  mature,  allowing  3  days  of  grace  according  to  business  custom,  and 
how  many  days  does  it  run,  actual  time  ?       Ans.  Jlatures  Jan.  18, 1895 ;  runs  95  ds. 

7.  A  note  is  drawn  Oct.  15,  1894,  and  made  payable  90  days  after  date. 
When  does  it  mature,  allowing  the  customary  3  days  of  grace,  and  how  many  days 
does  it  ruu,  actual  time  ?  Ans.  Matures  Jan'y  10,  1895 ;  runs  93  days. 

Note. — It  is  the  custom,  Trhen  matnrin<j  commercial  instruments,  to  be  governed  strictly  by 
the  terms  expressed  therein.  AVhen  they  are  nKiilo  payable  in  years  or  mouths,  they  are  matured 
in  years  or  months,  counting  them  as  they  run  from  the  date  of  the  instrument.  When  they  are 
made  payable  iu  days,  they  are  matured  in  days,  the  actual  number  of  days  in  each  month  of  the 
intervening  time  being  counted. 

The  day  that  a  note  or  other  instrument  is  dated,  is  not  counted  as  one  of  the  days  which  it 
has  to  run.     The  day  it  matures  is  however  counted. 

When  discounting  notes,  bills  of  exchange,  etc.,  the  actual  number  of  days  -n-hich  the  instru- 
ment has  to  run,  is  counted,  to  compute  the  interest,  whether  it  is  drawn  iu  years,  months  or  days. 

See  Banker's  Discount  in  this  book  for  extended  work  of  jnaturing  and  discounting  commei> 
cial  paper. 

HOW   TO   FIND    THE   LENGTH    OF   THE   DAY   OR   NIGHT   WHEN    WE 
HAVE   THE   TIME   THE   SUN   RISES    OR   SETS. 


617.     1.    If  the  Sun  sets  at  7h.  5m.  10s.,  what  is  the  length  of  the  day  and 

i  Sun  rise  ? 

OrERATION. 


the  night,  and  at  what  time  does  the  Sun  rise  ? 


Sun  sets 
add 

7h.     5m. 
12h. 

10.S.  X  2 

=  14. 

10.    20. 

Time  after  midnight  that  the  Sun  set 

=       19h.     5m. 
24h. 

10s. 

Sun  rises 

4h.  54m. 

50s. 

19h.     5m. 

10s. 

Length  of  day 

14h.  10m. 
24h. 

20s. 

Length  of  night  9h.  49m.     40s. 

Explanation. — In  all  problems  like  this,  first  add  12h.  to  the  time  the  Sun  sets,  -which  gives 
the  time  past  midnight  that  the  Sun  sets.  Then  subtract  this  from  24h.  and  in  the  remainder  we 
have  the  time  the  Sun  rises.  Then,  since  the  sun  rises  4h.  54m.  50s.  A.  M.  and  sets  19h.  5m. 
10s.  past  midnight  the  ditierence  will  bo  the  length  of  the  day.  Then  the  length  of  the  day 
subtracted  from  24h.  will  give  the  length  of  the  night. 

2.  If  the  Sun  rises  at  4h.  54m.  50s.,  what  is  the  length  of  the  night  and 
the  day  ? 

OPERATION. 

Sun  rises  4h.  54m.  50s.  X  2  =  91i.  49m.  40s.  length  of  night. 

24h. 

14h.  10m.  20s.  length  of  day. 

Explanation. — In  all  problems  of  this  kind,  multiply  the  time  the  Sun  rises  by  2  and  in  the 
product  we  have  the  length  of  the  night;  Then,  to  liud  the  length  of  the  day  we  subtract  the 
length  of  the  night  from  24h. 

JJOXE. — The  above  methods  of  finding  the  length  of  the  day  and  the  night,  and  the  rising  and 
the  setting  of  the  Snn  are  not  absolutely  correct,  ou  account  of  the  obliquity  of  the  earth's  orbit, 
and  other  astronomical  reasons. 

3.  When  the  Sun  sets  6h.  34m.  15s.,  what  is  the  length  of  the  day  and  what 
time  does  the  Sun  rise?  Ans.  Length  of  day,  ]3h.  8m.  30s. 

Sun  rises  5h.  25m.  45s. 

4.  When  the  Sun  rises  at  5h.  14m.  15s.  past  12  midnight,  what  is  the  length 
of  the  night?  Ans.  lOh.  28m.  30s. 


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*  DIRECTIONS    FOR   THE    OPERATION.  299 

TO  DETERMESTE   THE   DAT  OF  THE   WEEK  ON    WHICH    AN    EVENT 

DID  OR  MAT  OCCUR. 

619.  A  convenient  method  of  determining  immediately  what  day  of  the 
week  any  date  transpired,  or  will  transpire,  from  the  commencement  of  the  Christian 
era,  for  the  term  of  three  thousand  years. 

620.  The  following  table  shows  the  ratio  to  be  added  for  each  month : 

TABLE  OF  MONTHS. 

September,  ratio  is  -  1 

October,          "  -  -  3 

November,      "  -  -  6 

December,      "  -  -  1 

Note. — In  leap  years,  tbe  ratio  of  January  13  2  and  the  ratio  of  February  is  5.     There  is  no 
change  in  the  ratios  of  the  other  mouths. 

621.  The  following  table  shows  the  ratios  to  be  added  for  each  centmy  of 
the  Christian  era : 

TABLE  OF  CENTURIES. 

200,  900,  ISOO,  2200,  3000,  the  ratio  is         -     -  0 

300,  1000,  the  ratio  is 6 

400,  1100,  1900,  2300,  2700,  the  ratio  is     -     -     -  5 

500,  1200,  1600,  2000,  2400,  2800,  the  ratio  is     -  4 

600,  1300,  the  ratio  is  3 

700,  1400,  1700,  2100,  2500,  2900,  the  ratio  is 


January,  ratio  is 

-     3 

May,      ratio  is 

-     -     4 

February,     " 

-     G 

June,         '■ 

-     -     0 

March,          " 

-     G 

July,          " 

-     -     2 

April,            " 

'> 

August,    " 

-     -    5 

o 


100,     800,  1500,  the  ratio  is 1 


DIRECTIONS  FOR  THE  OPERATION. 

622.     1.    Add  to  the  given  year,  {omitting  the  century  figures)  one-fourth  part 
of  itself,  rejecting  the  fractions,  if  any. 

2.  To  this  sum  add,  1°,  the  day  of  the  given  month;  2°,  add  the  ratio  of  the 
month,  as  per  table  of  months  ;  3°,  add  the  ratio  of  the  century  as  per  table  of  centuries. 

3.  Divide  this  sum  hy  7.     The   remainder   is   the  day  of  the  week  counting 
Sunday  as  the  first  day,  Monday  as  the  second,  etc. 


300  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

PROBLEMS. 

623,     i      The  battle  of  Sliiloli  was  commenced  on  tlie  6th  of  April,  1S62. 
•^Iiat  VIS  eke  tiay  of  the  week  ?  Aus.  Sunday. 

OPEKATION. 

The  given  j'ear,  omitting  the  centuries,  is     -  62 

One-fourth  of  same,  rejecting  fractions,  is     -  15 

The  day  of  the  month  was  the        .        .        -  6 

The  ratio  of  April  is 2 

The  ratio  of  ISOO  is 0 

This  sum,  divided  by  7,  gives  -        -        7)  85 

12  and  1  remainder  -        -        -  12  +  1 

The  1  remainder  signifies  the  first  day  of  the  week — Sunday. 

2.    The  Declaration  of  Independence  was  signed  July  i,  1776.     Wliat  was  the 
day  of  the  week  ?  Ans.  Thursday. 

OPERATION. 

Given  year  is      -     -    -  -  76 

One-fourth  of  same  is  -  19 

Day  of  month  is     -    -  -  4 

Eatio  of  July  is  -    -    -  -  2 

Eatio  of  1700  is  -    -    -  -  2 


Divide  by  7)  103 

ll:+5  remainder — Thursday, 

3.  Gen.   E.   E.   Lee  was  born  June  19,  1807.     What  was  the  day  of  the 
week?  Ans.  Friday. 

4.  Martin  Luther  was  born  Nov.  10,  1483.    What  was  tlie  day  of  the  week  ? 

Ans.  Monday. 

5.  What  day  of  the  week  will  January  1  occur  in  2000 1      Ans.  Saturday. 

Note. — When  there  is  no  remainder  .after  dividing  by  7,  tlie  day  of  the  week  is  Saturday. 

6.  A  note  is  dated  Wednesday,  October  24th,  1894,  payable  in  120  days. 
In  what  year  and  month,  and  on  what  day  of  the  month  and  day  of  the  week,  will 
it  become  due,  allowing  3  days  of  grace  ?  Ans.     Sunday,  Feb.  24,  1895. 

7.  On  what  day  of  the  week  were  you  born,  and  if  you  live  to  be  150  years 
old,  as  we  wish  you  may,  on  what  day  of  the  week  will  you  die  ? 


ifference  of  Latitude. 


621.     Latitude  is  the  distance  iu   ilej^'recs,   minutes,    and   seconds,   of  any- 
place on  tlie  globe,  Xorth  or  South  of  the  equator. 

Latitude  is  reckoned  from  the  e(iuator  to  eacli  pole  of  tlie  earth,  and  like  arcs  of  all  circles, 
is  measured  iu  degrees,  minutes  and  seconds,  and  can  never  be  greater  tliau  a  quailrant,  or  90 
degrees.  The  earth  not  being  a  perfect  siihere,  hut  ohlate,  or  flattened  at  the  poles,  the  degrees 
vary  sliglitly  iu  length  toward  the  poles. 

The  dilVereuce  of  latitude  hetween  two  places  is  found  by  subtraction  or  addition,  as  iu 
compound  denominate  numbers, 

1.     The  latitude  of  Washington  City  is  38°  53'  39"  Xorth,  and  that  of  Kew 
Orleans  is  29°  5(5'  59"  Xorth.     Required  the  ditfeience  of  hititude. 


OPERATION. 

Lat,  of  Washington    =  38°  53'  39"  IS". 
"      "  Xew  Orleans    =  29°  5G'  59"  X. 


Dif.  of  latitude  =    S°  5G'  40"  Ans. 

Explanation. — Since  both  places  are  on  the  same  side  of  the  equator,  that  is  iu  North  lati- 
tude, we  subtract  the  lesser  latitude  from  the  greater.  When  the  latitude  of  one  place  is  North, 
and  that  of  the  other  South,  of  the  equator,  we  add  the  two  latitudes  together,  and  tlie  sum  will 
be  their  ditVerence  of  latitude. 

2.  The  latitude  of  :\robile  is  30°  41'  26"  I^orth,  and  that  of  Quebec  is  4Co 
48'17"Xorth.     AMiat  is  their  difference?  Aus.  10°  C  51". 

3.  Philadelphia  is  in  latitude  39°  56'  53"  iforth,  and  Eio  de  Jtmeiro  22°  54' 
24"  South.     What  is  their  difference  ?  Ans.  62°  51'  17". 

4.  Xe\Y  York  is  iu  hititude  40  deg.  42  iniu.  Xorth,  and  Cape  Horn  is  in  lati- 
tude 55  deg.  58  min.  South.    What  difference  intervenes  ?    Ans.  96  deg.  40  miu. 

5.  Havana  is  in  latitude  23  deg.  9  min.  North,  and  San  Francisco  lies  11 
deg.  39  min.  further  Xorth.    What  is  the  latitude  of  Sau  Francisco  ? 

Ans.  37  deg.  48  min. 
(301) 


^Jfference  of  Longitude. 


@\N 


625.  The  Lougitude  of  ;i  place  is  its  distance  iu  degrees,  minutes,  and 
seconds,  East  or  AVest,  from  a  given  meridian. 

A  degree  of  longitude  ou  the  equator  ia  69.1638  Btaiute  miles,  Lut  since  all  meridians  ar"? 
drawn  through  the  polua  and  meet  in  a  point,  tlioy  gradually  converge  as  we  advance  from  the 
equator,  and  vary  in  length  with  each  degree  of  latitude,  uiitil  they  meet  in  the  poles,  and  the 
longitude  becomes  nothing. 

626.  A  Meridian  is  an  imaginary  circular  line  ou  the  surface  of  tlie  earth, 
passing  through  the  poles  and  any  given  place. 

The  meridian  from  -tt-hich  longitude  is  reckoned  is  called  the  first  meridian,  or 
Siandard  lueridiau,  and  is  marked  0°.  All  places  East  of  the  iirst  or  standard 
meridian,  within  180°,  are  in  East  longitude ;  and  all  places  West  of  the  first  or 
standard  meridian,  within  180°,  are  in  West  longitude. 

The  English  reckon  longitude  from  the  meridian  of  Greenwich ;  the  French  from  that  of 
Paris.  The  government  of  the  United  .States  usually  reckons  longitude  from  the  Knglish  standard 
meridian,  Greenwich.  In  American  maps,  the  meridian  of  Greenwich  is  printed  at  the  top  and  the 
meridian  of  Washington,  the  capital  of  the  U.  S.,  is  printed  at  the  bottom. 

The  difference  of  longitude  between  two  places  is  found  like  the  difference  of  latitude,  by 
subtraction  or  addition,  as  iu  compound  denominate  numbers. 

1.  The  longitude  of  Galveston  is  94°  47'  20"  West;  of  Liverpool,  3°  4'  16" 
West.     What  is  their  difference  ?  Ans.  01°  43'  10". 

OPERATION.  Explanation. — Since  both  places  are 

T        _       r  r\    1        i                t\ir^    fi   cr'ti  -rtr  11  West    louijitude,   we   subtract  the 

Long,  of  Galve-Ston    =    04°  4<'   20"   W.  i^sser   from   tl,e  gr^^ater.      When    one 

u        a    Liverpool     =      3°     4'   10"  W.  place    is   ia    West    and  the   other  iu 

East    longitude,    we    add    the    longi- 

Dif.  of  longitude       =  91°  43'  10"    Ans.        !"'}''^  °^  *''?  ^'^^  I'^-'f,''^  ^°f^},^V'  ""^ 

*'  if  the  .sum  IS   more    than    180  degrees 

we  subtract  it  from  360  degrees,  because  the  shortest  distance  between  the  places  would  then 
be  the  other  way  around  the  world. 

2.  Cape  Flattery,  Wash.,  is  iu  longitude  124°  44'  30"  W.  Hong  Kong,  Ch., 
is  in  longitude  114°  9'  32"  East.     \\Tiat  is  their  difference  ?       Ans.  121°  5'  58". 

OPERATION. 

Long,  of  Cape  Flattery     =     124°  44'  30"  W. 
"       "    Hong  Kong         =     114°     9'  32"  E. 


238°  54' 


300°     0'     0" 


238°  54' 


nil 


121°     5'  58' 


3.  The  longitude  of  Calcutta  is  88°  20'  11"  East,  of  Trieste,  13°  40'  East. 
What  is  their  difference  ?  Ans.  74°  34'  11". 

4.  The  longitude  of  Havana  is  82°  21'  17"  West,  and  of  Antwerp  4°  24' 
44"  East.     What  is  their  difference  1  Ans.  80°  40'  1". 

5.  St.  Petersburg  is  in  longitude  30°  19'  22"  East.     Cedar  Keys  is  in  longi- 
tude 83°  1'  57"  West.    Find  their  difference.  Ans.  113°  21'  19". 

(302) 


longitude  and  Time. 


^ 


627.  The  circumfereuce  of  the  earth,  being  a  great  circle,  is  divided  into 
360  equal  parts  called  Degrees  of  Longitude. 

The  earth  revolves  on  its  axis,  from  West  to  East,  once  iu  24  hours — which 
gives  the  sun  the  appearance  of  passing  around  the  earth  from  Hast  to  West. 

Now,  since  the  earth  revolves  once  in  24  hours,  all  parts  of  its  surface  or 
circumference — the  300°  of  longitude — jjass  under  the  sun  during  that  space  of 
time.  Hence,  since  360°  are  passed  under  the  sun  in  24  hours,  -}^  part  of  360°,  or 
15°  of  longitude  are  passed  in  1  hour.  Since  15°  are  passed  in  1  hour  or  00  minutes, 
fo  part  of  15°,  or  15'  of  longitude  are  passed  in  1  minute.  Since  15'  of  longitude 
are  passed  in  1  minute,  or  GO  seconds,  eV  part  of  15',  or  15"  of  longitude  are  passed 
in  1  second. 

Then  since  15°  of  longitude  =  1  hour  of  time, 

"  =  -,V  of  1  hour  ((»()  min.)  =  4  min.  of  time. 

''  =:  1  minute  of  time, 

"  =  -jij  of  1  minute  (00  sec.)  =  4  sec.  of  time. 

"  =1  second  of  time, 

"  =  'i\'  of  a  second  of  time  ; 

"  =4  seconds  of  time, 

'<  =  -ig  of  4  sec.  =  tjV)  o'"  'At  *>f  a  second  of  time. 

628.  From  the  foregoing,  we  deduce  the  following : 

COMPARATIVE  TABLE  OF  LONGITUTE  AND  TIME. 

360°  long,  make  a  difference  of  24  hours  of  time. 
15°     •• 
15'      " 

15"     "         " 

V      "         '' 

2"       U  it 

629.  Longitude  and  Time  give  rise  to  two  classes  of  problems,  as  follows : 

1.  To  reduce  time  t(»  longitude,  or  to  find  the  difference  of  longitude  between 
two  places,  when  the  difference  of  time  is  known. 

2.  To  reduce  longitude  to  time,  or  to  find  the  difference  of  time  between  two 
places,  when  the  difference  of  longitude  is  known. 

{303) 


1° 

Then 

since 

15' 

1' 

Then 

since 

15" 

1" 

Or,  since 

1' 

1" 

1       " 

1  minute 

1  second 

4  minutes 

4  seconds 

"iV  second 

304  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

PROBLEMS  UNDER  THE  FIRST  CLASS, 

OR,  TO  FIND  THE  DIFFERENCE  OF  LONGITUDE  BETWEEN  TWO  PLACES,  WHEN  THE 

DIFFERENCE  OF  TIME  IS  KNOWN. 

630.     1.     Reduce  11  lirs.  20  min.  40  sec.  of  time  to  longitude. 

OPERATION. 

11  hrs.     20  min.     40  sec.  Explanation. — Since  aceordiiig  to   the   foregoing  eluci- 

-i  -  (latioiis,  1  hr.  ^  15^  of  longitude,  1  min.  =  15'  of  lougi- 

tilde,  and  1  second  =  15"  of  longitude  there  are  15  times 

~  as  many  °,  ',  and  "  of  longitude  as  there  are  hrs.,  min., 

170°  10'  0"   Alls.  an<l  sec.  of  time.     Hence  in  all  jiroblems  of  this  kind,  we 

multijily  the  different  units  of  time  by  15,  as  in  multipli- 
cation of  comiiound  dcnomin.ate  numbers,  and  thus  convert  them  into  units  of  longitude. 

2.  Tlie  diftereiice  of  time  between  two  places  is  3  hrs.  15  min.  24  sec.    Wliat 
is  their  difference  of  longitude  ?  Ans.  48°  51'  0". 

3.  The  difference  of  time  between  New  York  and  Chicago  is  54  min.  19  sec. 
What  is  the  difference  of  longitude  1  Ans.  13°  34'  45". 

4.  When  it  is  11  o'clock  16  minutes  1,V  seconds  A.  m.  at  Boston,  it  is  10 
o'clock  A,  M.  at  New  Orleans.     Find  the  difference  of  longitude. 

OPERATION    INDICATED. 


Time  at  Boston 

11  hrs. 

10  min. 

1 1\  sec. 

"     "  New  Orleans 

10  hrs. 

0  mill. 

0  sec. 

Difference  of  time  1  hr.      16  min.    1 1^  sec. 

15 


19°  0'  19".       Ans. 

5.  The  longitude  of  Rome,  Italy,  is  12°  27'  E.;  the  difference  of  time 
between  Rome  and  Mobile,  Ala.,  is  6  hrs.  41  min.  57|f  sec.  What  is  the  longitude 
of  Mobile,  West  ?  Ans.  88°  2'  28"  W. 

OPERATION  INDICATED. 

6  hrs.     41  min.     57yi5-  sec.  =  dif.  of  time. 
15^ 

lOOO          29^                 28^'  =  dif.  of  longitude. 
120  27' 0^  =  longitude  of  l^ome,  E. 

"88°  2'  28"  =  longitude  of  Mobile,  W. 

Note. — ^^^len  the  Dif.  of  Long,  given  is  that  of  two  places  which  are  in  opposite  longi- 
tudes, and  the  Long,  of  one  of  them  is  given  to  find  that  of  the  other,  we  subtract  the  lougitudo 
of  the  given  place  from  the  Dif.  of  Long,  and  thereby  find  the  longitude  of  the  other. 

This  is  done  liecanse  the  Dif.  of  Long,  between  the  two  places  is  equal  to  their  sum  (one 
being  E.  and  the  other  W.)  and  cousefiuently,  when  the  Long,  of  one  is  given,  that  of  the  other  is 
foun<l  by  subtraction.  Shouhl  the  Dif.  have  been  that  of  two  ])laces  in  the  same  kind  of  Long., 
we  would  then  have  added  the  dift'erfuce  of  lougituilc.  if  the  lesser  bmgitude  were  given ;  and  sub- 
tracted, if  the  greater  longitude  were  given,  to  find  the  longitu<le  of  the  other. 

6.  When  it  is  12  m.  in  New  York,  it  is  18  min.  2|  sec.  past  11  o'clock  at 
Cincinnati.     What  is  their  difference  of  longitude  ?  Ans.  10°  i>9'  21". 

OPERATION    INDICATED. 

12  hrs.     0  min.     0    sec. 
11  hrs.  18  mill.     2}  sec. 

41  min.  57f  sec.  x  15  =  10°  29'  21". 

7.  The  difference  of  time  between  Jerusalem  and  Baltimore  is  7  hr.  28  min. 
36  sec. ;  the  longitude  of  Baltimore  is  76  deg.  37  min.  West;  what  is  the  longitude 
of  Jerusalem,  East  ?  Ans.  35  deg.  32  min.  E, 


*  LONGITUDE    AND    TIME.  3o5 

PROBLEMS  OF  THE  SECOND  CLASS. 

OR,  TO  FIND  THE  DIFFERENCE  OF  TIME  BETWEEN  TWO   PLACES,   WHEN  THE  DIF- 
FERENCE OF  LONGITUDE  IS  GIVEN. 


» 


631.     1.     Eeduce  30°  42'  50"  of  longitude  to  time. 

OPERATION.  Explanaiion. — As  shown  by  the  foregoinj;  elu- 

15^  QQO  4'>'  5()//  cidated  table  of  lougitude  and  time  equivalents, 

'  ^  15^  of  longitude  =  1  br.  of  time,  15'  of  lougi- 

tude  =  1  uiin.  of  time,  and  15  "  of  longitude  =: 
2  lir«.  2  mill.     51 J  sec.   Ans.  l  sec.  of  time;   therefore,   there  will  be   ^^^  as 

m.any  hours,  minutes,  and  seconds  of  time  as  there 
are  degrees,  minutes,  and  seconds  of  longitude,  and  hence  we  divide  the  °,  ',  and  ",  by  15,  as  in  divis- 
ion of  compound  denominate  numbers,  and  thus  reduce  longitude  to  time. 


¥ 


2.  The  longitude  of  Paris,  France,  is  2°  20'  15"  E.,  and  of  Berlin,  Germany, 
13°  23'  Ai"  E.     What  is  the  difference  of  time  ?  Ans.  44  min.  13|4.  sec. 

OPEEATION. 

130  23'  44"  =  longitude  of  Berlin,  E. 

■  20  20'  15"  =         "  "    Paris,  E. 

15)  11°  3'  21)"  =  difference  of  longitude. 

44  min.     13|*  sec.  =  difference  of  time. 

3.  The  longitude  of  Bombay,  India,  is  72°  54'  East ;  of  St.  Louis,  Mo.,  90° 
15'  15"  West.  What  is  the  difference  of  time,  and  when  it  is  10  A.  M.  in  St.  Louis, 
what  is  the  time  in  Bombay  1 

OPERATION. 

720  54'  0"  =  longitude  of  Bombay,  E. 

90O  15'  15"  =         "         "     St.  Louis,  W. 

15)  103°  9'  15"  =  difference  of  longitude. 

10  hrs.      52  min.    37  sec.  =  difference  of  time,  or  tlie  length  of  time  it  is  10 

o'clock  in  Bombay,  before  it  is  10  a.  m.  in  St.  Louis. 
10  lirs.  =  St.  Louis  time. 

20  hrs.      52  min.    37  sec,  or  8  hrs.  52  min.  37  sec.  p.  M.  in  Bombay. 

4.  The  difference  of  longitude  between  St.  Paul  and  Cincinnati  is  10°  35' 
24".    What  is  the  difference  of  time  1  Ans.  42  min.  213-  sec. 

5.  The  longitude  of  New  Orleans  is  90°  4'  9".  The  longitude  of  San  Fran- 
cisco is  122°  26'  45".  What  is  the  difference  in  time,  and  when  it  is  12  m.  in  New 
Orleans,  what  is  the  time  in  San  Francisco  ? 

OPERATION    INDICATED. 

122°  20'  45"  =  longitude  of  San  Francisco. 

90°  4'  9"  =         "         "     New  Orleans. 

15)  320  22'  "sF'  =  difference  of  longitude. 

2  hrs.        9  min.    30f  sec.  =  dif.  of  time. — 1st.  Ans. 

12  hrs.        0  min.      0    sec.  =  New  Orleans  time. 
2  hrs.        9  min.    30f  sec.  =  dif.  of  time,  or  time  before  12  m.  in  San  Francisco. 

9  hrs.      50  min.    29|  sec.  =  50  minutes  29f  seconds  past  9  A.  M.,  San  Franciscp 
time. — 3d  Ans. 


3o6  soule's  philosophic  practical  mathematics.  * 

6.    The  lougitude  of  Boston  is  71°  3'  30",  aud  the  longitude  of  Chicago  is 
87°  37'  45".     What  is  the  time  in  Boston  when  it  is  10  o'clocli  A.  M.  in  Chicago? 

Ans.  11  o'clock  G  min.  and  17  sec.  A.  M. 

OPERATION. 

87°  37'  45"  =  longitude  of  Chicago. 

71°  3'  30"  =  "         "     Boston. 


15)  KJO  34'  15"  =  difference  of  longitude. 

1  hr.         6  min.     17  sec.  =difference  of  time,  or  the  length  of  time  it  is  10  A. 

M.  in  Boston  before  it  is  10  A.  M.  in  Chicago. 
10  hrs.  —  Chicago  time  added. 


11  hrs,        6  min.     17  sec.  A.  M.  Ans, 

7.  In  traveling  from  Washington,  longitude  77°  0'  15"  W.,  to  New  Orleans, 
longitude  90°  4'  9"  W.,  how  much  time  will  an  exactly  running  watch  appear  to 
gain  ?  Ans.  52  min.  15f  sec, 

8.  The  longitude  of  Greenwich  is  0,  of  Astoria,  Oregon,  123°  49'  42".  How 
much  earlier  does  the  sun  rise  in  Greenwich  than  in  Astoria,  Oreg.  1 

Ans.  8  hrs.  15  min.  18f  sec. 

9.  When  it  is  noon  at  Pekin,  longitude  118  deg.  East,  what  time  is  it  at 
Paris,  longitude  2  deg.  20  min,  15  sec.  East?  Ans.  4h.  17  m.  21  sec.  A.  M. 

OPERATION    INDICATED, 
118. 


2,     20,     15. 


15)  115.    39.     45. 

7.     42.     39. 


12, 
7.     42.     39. 
4.     17,     21. 


10.    When  it  is  midnight  at  Halifax,  longitude  63  deg.  36  min.  West,  what 

de? 
Ans,  11  h.  47  m.  20  sec.  A.  M. 


time  is  it  at  Canton,  113  deg.  14  min.  East  longitude  ? 


THE  LOSS  OR  GAIN  OF  A  DAY  BY  TRAVELING  AROUND  THE  WORLD. 

632.  Two  very  curious  facts  would  be  noticed,  could  we  start  on  any  given 
meridian  with  the  sun  on  a  certain  date  and  journey  with  it  around  the  earth,  so  to 
speak :  The  first  fact  to  be  noticed  would  be  that  in  going  around  the  earth  from 
Uast  to  West,  and  upon  arriving  at  our  original  meridian,  we  would  find  that  we 
had  lost  one  day's  time,  i.  e.  that  we  were  one  day  behind  the  correct  date.    The 


*  LOSS  OR  GAIN  OF  A  DAY  BY  TRAVELING  AROUND  THE  WORLD.  30/ 

second  fact  noticed  would  be  that  if  we  should  reverse  our  first  direction  and  go 
around  from  West  to  East,  we  would  gain  one  day's  time,  i.  e.  we  would  be  one  day 
ahead  of  the  correct  date. 

The  meridian  at  Greenwich,  Eng.,  is  one  of  the  standards  of  longitude  for 
the  world.     When  it  is  midnight  on  Sunday  there,  it  is  Sunday  from  there  to  180° 

West,  and  Monday  from  Greenwich  to  180°  East,  or  to  the  same  meridian.  Hence 
ship  captains  change  their  time  at  this  180th  meridian.  Thus  in  traveling  from  the 
West  towards  the  Uast,  and  upon  crossing  the  ISOth  meridian  on,  say  Sunday,  the 
time  on  shipboard  woidd  be  changed  to  Monday.    And  in  traveling  ft-om  East  to 

■  West  (in  the  ojiposite  direction)  and  arriving  at  the  180th  meridian,  on  Sunday,  the 
time  would  be  changed  to  Saturday.  These  changes  are  always  made  by  nautical 
men  in  order  that  they  may  have  the  correct  time  of  all  localities  East  or  West  of 
the  180th  meridian. 

There  is  also  another  line  which  may  be  named  an  International  Bate  Line. 
This  line  differs  slightly  from  the  180th  meridian.  The  direction  of  this  line,  as 
indicated  by  Mr.  Olney,  the  famous  Scientist  and  Mathematician,  is  that  it  starts 
from  the  Chatham  Islands,  latitude  44°  S.,  longitude  177°  W.,  and  runs  north  afld 
west,  to  the  East  of  Xew  Zealand,  New  Guinea,  and  Borneo,  thence  through  the 
Philippines,  approaches  China  near  Canton,  and  goes  to  the  northwest  of  the  Japan 
Isles,  through  Behring  Sea,  and  terminates  in  the  North  Pole.  Tlie  date  on  the 
east  of  this  line  is  generally  one  day  behind  what  it  is  on  the  west  of  it. 


ilAXIO, 

^(Sl'Mm^-s. ^^^^- 


>  •  * 

633.  Ratio,  in  the  mathematical  sense  iu  which  it  is  here  used,  means  the 
measure  of  the  relation  which  one  quantity  bears  to  another  of  the  same  kind,  as 
expressed  by  the  quotient  of  the  first  divided  by  the  second;  that  is,  it  is  the  num- 
ber of  times  that  one  quantity  (the  first)  is  equal  to  another,  (the  second)  which  is 
used  as  a  unit  of  measure.  Thus,  the  ratio  of  G  to  2,  is  G  4-  2,  and  is  equal  to  3- 
If  we  ask  what  is  the  relation  of  G  to  2,  the  correct  answer  would  be  6  is  3  times 
2.  We  thus  see  that  the  ratio  three  is  the  number  which  measures  the  relation  of  6 
compared  with  2,  and  therefore,  that  ratio  is  not  merely  the  relation  of  two  similar 
numbers,  but  the  measure  of  this  relation. 

Note  1. — Mathematical  Tvriters  are  not  agreed  in  their  tleflnitions  of  the  ratio  of  one  num- 
ber or  quantity  to  another.  Many  contend  that  it  is  the  quotient  of  the  second  quantity  divided 
by  the  first.  Thus,  the  ratio  of  6  to  ii  is,  2  -H  6  is  |  — =  i^.  .  This  is  the  preferred  detinition  of  Webster 
and  of  the  Dictionary  of  Mathematics.  But  nearly  all  of  the  Later  writers  as  well  as  the  German, 
English  and  luost  of  the  French,  prefer  the  definition  first  above  given,  and  we  are  in  full  accord 
with  thera. 

The  result  of  the  two  systems  is  the  s.ame.  .although  differently  obtained.  By  the  first  definition 
we  saw  thiit  6  is  equal  to  2,  3  times ;  and  by  the  second  detinition  we  see  that  2  is  equal  to  6,  4-  of  a 
time,  which  expresses  precisely  the  same  ratio  between  the  numbers.  The  question  is,  which 
number,  the  frst  or  the  second,  shall  be  used  as  the  unit  of  measure,  or  the  divisor?  The  prexjon- 
derance  of  authority  is  largely  in  favor  of  the  second. 

Note  2. — Numbers  or  magnitudes  can  have  no  ratio  to  each  other  unless  they  are  of  the 
same  kind.  Thus,  there  can  be  no  ratio  between  6  dollars  and  2  yards,  for  6  dollars  are  not  equal 
to  2  yards  any  number  of  times. 

634.  The  Terms  of  a  Ratio  are  the  numbers  compared.  The  first  term  of 
a  ratio  is  the  antecedent,  which  means  going  before  ;  the  second  term  is  the  con- 
sequent, which  means  foUoicing. 

635.  The  Sign  of  Ratio  is  the  colon  (  :  ),  which  is  the  sign  of  division,  with 
the  horizontal  line  omitted.  Thus,  the  ratio  of  G  to  2,  is  written,  G  :  2.  Eatio  is 
also  indicated  by  writing  the  consequent  under  the  antecedent  iu  the  form  of  a  frac- 
tion.    Thus,  the  ratio  of  G  :  2  is  ofteu  written  f. 

636.  An  Inverse  Ratio  is  the  quotient  of  the  consequent  divided  by  the 
antecedent.     Thus,  the  inverse  ratio  of  8  :  4,  is  ^. 

637.  The  Valne  ot  a  ratio  is  the  quotient  of  the  antecedent  divided  by  the 
cousequeiit,  and  is  always  an  abstract  number. 

638.  A  Simple  Ratio  is  the  ratio  of  two  numbers,  as  8  :  4. 

639.  A  Componnd  Ratio  is  the  ratio  of  the  products  of  the  corresponding 
terms  of  two  or  more  simple  ratios,  as  follows : 

G  •  2  ) 

e|.>  =  G  X  8:2  X  4;  or  f  x  |isa  compound  ratio,  =  C. 

640.  The  Reciprocal  of  a  ratio  is  the  quotient  of  1  divided  by  the  ratio,  or 
it  is  the  quotient  of  the  consequent  divided  by  the  antecedent.  Thus,  the  ratio  of 
6  to  2  is  G  :  2  or  f ,  and  its  reciprocal  is,  1  -h  J  =  f ,  or  ^. 

641.  The  Ratio  of  two  fractions  is  obtained  by  reducing  them  to  a  common 
denominator  and  then  comparing  their  numerators.  Thus,  the  ratio  of  f  :  f  is  the 
same  as  5  :  2. 

642.  The  Ratio  of  Compound  Denominate  numbers  is  found  by  reducing 
them  to  the  same  denomination  and  then  making  the  comparison. 

(308) 


*  GENERAL    PRINCIPLES.  309 

From  the  foregoing  deflnitious  and  elucidations,  the  following  formulas  and 
general  princijiles  are  deduced: 

FOEMULAS. 

643.     1.     The  Ratio  =  the  Antecedent  -4-  Consequent. 

2.  The  Consequent  =  Antecedent  -H  Ratio. 

3.  Ike  Antecedent  =  Consequent  x  Ratio. 

GENEEAL  PRINCIPLES. 
64:4.    1.  Multiplying  the  antecedent  or  dividing  the  consequent  multiplies  the  ratio. 

2.  Diriding  the  antecedent  or  multiplying  the  consequent  divides  the  ratio. 

3.  Multiplying  or  dividing  both  terms  by  the  same  number  does  not  change  the 
value  of  the  ratio. 

645.  These  general  principles  may  be  formulated  into  one  General  Law,  as 
follows:  Any  change  in  the  antecedent  produces  a  LIKE  change  in  the  ratio;  but  any 
change  in  the  consequent  produces  an  opposite  change  in  the  ratio. 

PROBLEMS. 

646.  What  is  the  ratio  of  the  following  numbers : 

1.  10  to  5  3.     4  to    8  5.     54  to    8  7.     if  to  -,% 

2.  18  to  6  4.     7  to  42  6.     11  to  60  8.     ff  to  ^h 

9.    What  is  the  ratio  of  5  gallons  to  3  quarts  ?  Aus.  6§. 

10.  What  is  the  ratio  of  20/  to  $3  ?  Ans.  h- 

11.  The  antecedent  is  12  and  the  consequent  4.    What  is  the  ratio  ?    Ans.  3. 

12.  The  antecedent  is  12  and  the  ratio  is  2.     What  is  the  consequent  1 

Ans.  6. 

13.  The  consequent  is  16  and  the  ratio  is  8.    What  is  the  antecedent  ? 

Ans.  128. 

14.  The  antecedent  is  5J  and  the  ratio  is  4.     What  is  the  consequent  ? 

Ans.  1^. 

15.  What  is  the  reciprocal  ratio  of  16  to  48  ?  Ans.  3. 

16.  The  reciprocal  of  the  ratio  of  two  numbers  is  3  and  the  antecedent  is  16. 
What  is  the  consequent  ?  Ans.  48. 

17.  What  is  the  ratio  of  a  pound  Troy  to  a  pound  Avoirdupois  t    Ans.  jf |. 

18.  What  is  the  ratio  in  grains  of  a  pound  of  iron  to  a  pound  of  gold  ? 

A  n  «     ITS 

-a.ns.  YJ4. 
10.    The  ratio  of  two  numbers  is  3,  the  antecedent  is  15.    What  is  the  con- 
sequent? Ans.  5. 

20.  The  ratio  of  two  numbers  is  4  the  consequent  is  IJ.  What  is  the  ante- 
cedent? Ans.  5J. 

21.  The  inverse  ratio  of  two  numbers  is  3J  and  the  antecedent  is  20.  What 
is  the  consequent  ?  Ans.  70. 


00 


(    8*1'') 
What  is  the  value  of    ]  1  a  !    k\i       Ans.  IJ. 

23.    What  is  the  value  of    ]  qI  .'  1 1  [  « 


OPERATION  INDICATED. 
3 


30 

24.    What  is  the  difference  between  the  direct  3 

and  the  inverse  ratio  of  2|  and  7^  ?  Ans.  2-2^.  8 

Ans. 


7 

10 

5 


i-rf- 


(4^'roportion. 

647.  Proportion  arises  from  tlic  coiniiaiisou  of  ratios.  It  is  a  comparison 
of  tbe  results  of  two  previous  coiiiparisoiis.  l^vciy  i)ro])ortiou  involves  three  eoiu- 
parisous :  tlie  tirst  two  were  tliose  wliicli  produced  tlie  ratios,  and  the  third,  that 
■which  compares  or  ecprates  the  ratios. 

Proportion  is  the  expression  of  the  equality  of  equal  ratios,  or,  it  is  the 
comparison  of  two  equal  ratios.  Thus,  G  :  13  :  :  15  :  5  is  a  projiortion,  and  is  read 
(5  is  to  2  as  15  is  to  5,  or  the  ratios  of  C  to  2  equals  the  ratio  of  15  to  5. 

Tlius,  the  ratio  of  G  :  2  as  15  :  5  is  a  proportion,  i.  e.  four  quantities  are  in 
proportion,  -when  the  Jirnt  is  the  same  imdtlple  or  i)art  of  the  second,  that  the  third 
is  of  the  fourth. 

648.  The  Sign  of  Proportion  is  a  double  colon  ( :  : ),  or  the  sign  of  equality 
(  =  ).  Thus,  the  above  proportion  is  expressed  G  :  2  :  :  15  :  5,  or  G  :  2  =  15  :  5.  The 
first  is  read,  G  is  to  2  as  15  is  to  5.  The  second  is  read,  the  ratio  of  G  to  2  equals  the 
ratio  of  15  to  5. 

649.  The  Terms  of  a  proportion  are  the  numbers  compared. 

650.  The  Antecedents  of  a  proportion  are  the  first  and  third  terms. 

651.  The  Consequents  are  the  secow«Z  and  y'o?<r^/i  terms. 

652.  Tlie  Extremes  are  the  first  and  fonrth  terms. 

653.  The  Means  ai'e  the  second  and  third  terms. 

654.  In  the  proportion,  3  :  G  :  :  4  :  8,  all  the  numbers  are  the  proportionals ; 
3  and  4  are  the  antecedents ;  G  and  8  are  the  consequents  ;  3  and  S  are  the  extremes; 
and  C  and  4  are  the  means. 

655.  Three  numbers  are  proportional,  when  the  ratio  of  the  first  to  the  second 
is  equal  to  the  ratio  of  the  second  to  the  third.  Thus,  4,  8,  and  IC  are  proportional, 
since  4  :  8  :  :  8  :  IG,  each  ratio  being  4. 

In  this  kind  of  proportion,  the  second  term  is  called  a  mean  proportional 
between  the  other  two. 

656.  A  Simple  Proportion  is  an  expression  of  the  equality  between  two 
equal  ratios,  as  above  elucidated. 

657.  A  Compound  Projjortion  is  an  expression  of  the  equality  between 
two  ratios,  when  one  or  both  of  which  are  compound. 

658.  There  are  several  other  classes  or  divisions  of  proportion  ;  viz :  Con- 
joined, Partitive,  IJeciprocal,  JMi'dial,  etc.,  but  as  we  shall  iicrlorm  all  kinds  of 
jjroportional  problems  by  our  pliilosoi)hic  system,  indc])(Mideiitly  of  the  various 
terms  and  the  ingenious  classilications,  or  mechanical  formulas  used  by  most  authors, 
we  will  not  now  occupy  space  with  them.  As  we  ])resent  the  problems,  which  are 
generally  classed  under  these  various  heads  or  divisions  of  proportion,  we  shall 
then  define  each,  with  our  iihilosophic  solution. 

659.  The  following  General  Principles  are  derived  from  the  preceding 
elucidations : 

1.  In  every  proportion,  the  jyroduct  of  the  e-xtremes  is  equal  to  the  product  ofthemeans. 

2.  The  product  of  the  extremes  divided  hy  either  of  the  means,  gives  the  other  mean. 

3.  Thi}  product  of  the  means  divided  by  either  extreme,  gives  the  other  extreme. 

(310) 


Relationship  and  Equivalency  of  Numbers. 


^ 


660.  Ill  defining  ratio,  we  showed  that  there  is  a  difference  between  the 
relatioushiii  and  the  ratio  of  two  numbers;  we  shall  now  show  in  the  following 
work,  other  differences  which  were  not  then  indicated. 

Eelationship  of  numbers  is  a  more  general  term  with  a  wider  significance 
than  ratio.  And  used  as  we  shall  employ  it,  it  includes  far  more  than  the  numerical 
measure,  ratio,  of  one  number  by  another. 

Tlirough  the  relationship  and  the  equivalency  of  numbers,  we  shall,  by 
comparison,  analysis,  and  synthesis,  solve  all  forms  of  ratio  and  proportional 
questions,  simple,  compound,  reciprocal,  etc.,  without  regard  to  any  of  the  preced- 
ing mathematical  conventional  terms,  ratios,  jtroportious,  general  principles,  etc. 

Different  kinds  of  relationship  and  equivalency  of  value  may  exist  between 
things  of  the  same,  and  things  of  different  kind.  Thus,  the  relationship  and 
equivalency  of  monetarj'  value,  exist  when  5  yards  cost  50c',  1  hat  cost  $4,  etc.;  the 
relationshii)  and  equivalency  of  numerical  value  exist  when  8  =  5,  2  =  3,  etc. ;  the 
relationship  and  equivalency  of  exchange,  or  barter  value  exist  when  10  pounds  of 
rice  =  24  oranges,  6  days  labor  =  18  bushels  of  corii,  etc.;  the  relationship  and 
equivalency  of  service  or  force  and  results  exist  when  30  men  make  uO  yards  of 
levee  in  20  days  ;  or  when  a  steamboat  carries  1000  bales  of  cotton  500  miles  for 
$2000,  etc. 

And  thus  some  kind  of  relationship  and  equivalency  exists,  like  the  law  of 
compensation  in  nature,  in  every  conceivable  statement  of  numbers. 

The  following  problems  and  their  solutions  will  more  clearly  show  the  rela- 
tionship and  equivalency  of  numbers,  and  at  the  same  time  elucidate  the  philosophic 
system  of  work. 

SIMPLE  PEOPOETIOK 

PROBLEMS. 

1.     9  pounds  cost  45/.     "What  will  3  pounds  cost  at  the  same  rate  ? 


PR0BLK3I     CLASSIFIED, 
lbs.  c. 

9  45 

3 

OPERATION. 


45 
3 


Explanation  and  Heason. — This  is  a  problcui  in  simple  pro. 
portidii,  ami  by  our  system  of  reasoning,  it  is  as  easily 
solveil  as  a  qnesticm  in  division. 

In  all  problems  of  all  classes  of  proportion,  there  is  one 
number  wliich  is  of  the  same  name,  nature,  or  denomination, 
as  the  number  or  answer  required.  This  number  we  use  as  the 
nature  of  the  ansirer,  and  in  the  solution  of  jirolilems,  we  first 
write  it  at  the  top  of  t  lie  inereasingside  of  tlie  statement  line. 

In  this  problem,  tlie  nature  of  the  answer  is  45c.  which 
we  write  on  the  statement  lino  and  reason  as  follows : 

Since  9  pounds  cost  450.  1  pound  will  cost  the  9th  part, 
which  we  indicate  by  writing  the  9  on  the  decreasing  side 
of  the  statement  line ;  then  since  1  pouud  costs  the  result  of 
this  statement,  3  pounds  will  cost  3  times  as  much,  which 
worked,  gives  15c.  the  cost  of  3  pounds. 

It  will  be  observed  that  we  classified  the  ])roblem  before  stating  the  operation. 
This  the  learner  or  calculator  should  always  do,  in  order  to  see  more  clearly  the 
relationship  and  the  equivalency  of  the  numbers  which  are  compared  with  the 
nature  of  the  answer. 

(311) 


loc,  Ans. 


312 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


PKOBLEM 

CLASSIFIED, 

boys. 

$ 

tls. 

3 

72 

12 

2 

15 

OPERATION. 


3 
12 


15 


$G0,  Aus. 


•  COMPOUND  PROPORTION. 

2.  Three  boys  earned  $72  in  12  days.  How  many  dollars  can  2  boys  earn  in 
15  days  T 

Explanation  and  Reason. — This  is  a  jiroljlcm  in  compound 
proportion,  and  by  the  philosophic  system  of  work  it  is  as 
easily  solved  as  a  problem  in  simple  projiortion.  By  inspec- 
tion and  reason,  we  see  that  $72  is  the  nature  of  the  answer. 
We  therefore  write  the  same  on  the  increasing  side  of  the 
statement  line  and  reason  thus:  3  boys  earned  $72.  Since  3 
boys  earned  $72,  1  boy  earned  the  3d  part,  and  2  boys  can 
earn  twice  as  much;  then,  since  $72  are  earned  in  12  days, 
in  1  day  the  12th  part  will  be  earned,  and  in  15  days,  15 
times  as  much  will  be  earned.  This  completes  the  reasoning. 
The  operation  statement  is  worked  out  by  cancellation,  as 
explained  in  cancellation,  and  in  other  subjects  treated. 

It  will  be  observed  that  in  giving  the  reason  for  writing 
the  terms,  boys  and  days,  on  the  statement  line,  we  did  not 
state  all  of  the  conditions  or  relationships  of  the  syllogistic 
premises  from  which  we  deduced  our  conclusions  of  increase  and  decrease  of  the  dollars.  Thus, 
per  example,  with  the  term  boys,  we  said  :  Since  3  boys  earned  $72,  1  boy  earned  the  3d  part,  etc. 

To  consider  the  full  conditions,  or  to  state  the  full  premises  of  this  term,  we  would  reason 
thus  :  Since  3  boys  earned  $72  in  12  days,  1  boy  earned  the  3d  part,  etc. 

To  consider  the  full  conditions  of  the  term  days,  we  would  reason  thus :  Since  $72  are 
earned  by  3  boys  in  12  days,  in  1  day  the  12th  part  will  be  earned,  etc. 

To  name  all  the  conditions  at  each  step  of  the  analysis  would  require  much  time,  and  would 
rather  confuse  than  aid  the  learner.  For  the.se  reasons,  and  because  the  analysis  is  clear,  correct, 
and  logical,  without  naming  all  the  conditions,  we  omit  their  useless  repetition. 

CONJOINED  PROPORTION. 

661.  Conjoined  proportionals  a  proportion  in  ^vllicll  there  are  several  terms 
of  different  kinds  of  which  each  preceding  term  or  antecedent  is  of  an  equivalent 
value  to  a  succeeding  term  or  consequent. 

3.  If  4  apples  are  worth  3  oranges,  and  2  oranges  are  worth  100  grapes,  and 
200  grapes  are  worth  6  peaches,  and  9  iieaches  are  worth  30  cents,  then  how  many 
apples  can  we  buy  for  90  cents  1 

PROBLEM   CLASSIFIED. 


r- 

Or. 

4 

=     3 

Gr. 

2     - 

=      100 

Pe. 

200 

=     6 
9 

c. 
30 
90 

Explanation  and  Reason. — This  is  a  problem  in  conjoined 
proportion,  and  by  the  philosophic  method  of  work  it  is 
solved  in  the  same  manner  as  the  two  preceding  problems. 
Writing  on  the  statement  line  the  4,  which  represents  apples 
and  which  is  the  nature  of  the  answer,  we  reason  thus: 
Since  4  apples  are  worth  3  oranges,  conversely  3  oranges  are 
worth  4  apples;  and  since  3  oranges  are  worth  4  apples,  1 
orange  is  worth  the  J  part,  and  2  oranges  are  worth  2  times 
as  many;  then  since  2  oranges  are  worth  100  grapes,  con- 
versely 100  grapes  are  worth  2  oranges ;  and  since  100  grapes 
are  worth  two  oranges,  1  grajie  is  worth  the  -yfnj  part,  and 
200  grapes  are  worth  200  times  as  many.  Then  since  200  grapes  are  worth  6  peaches,  conversely 
6  peaches  are  worth  200  grapes  ;  and  since  6  peaches  are  worth  200  grapes,  1  peach  is  worth  the  \ 
part,  and  9  peaches  are  worth  9  times  as  many.  Then  since  9  peaches  are  worth  30c.,  conversely 
30c.  are  worth  9  peaches  ;  and  since  30e.  are  worth  9  peaches,  1  cent  is  worth  the  ^\j  part,  and  90c. 
are  worth  90  times  as  many. 

The  above  is  the  full  reasoning  for  the  problem,  and  it  gives  healthful  exercise  to  the  mind  of 
the  student — exercise  which  sharpens  and  strengthens  and  logicalizes  the  mind  for  service  not  only 
in  the  domain  of  mathematics,  but  upon  all  rjuestions  of  life. 

The  reasoning  may  be  abbreviated  thus  :  Since  3  oranges  =  4  apples,  1  orange  =  J  part,  and 


OPERATION 

APPLES. 

4 

3 

2 

100 

200 

6 

9 

30 

90 

24  apjiles,  Ans. 


■*■  PROPORTION,  313 

2  oranges  =  2  times  as  many;  then,  since  100  grapes  =  2  oranges,  1  grape  =  fi'ju  part,  and  200 
grapes  =  200  times  as  many ;  then,  since  6  peaches  =  200  grapes,  1  peach  =  ^  jiart,  and  9  peaches 
=  9  times  as  many  ;  then,  since  30c.  =  9  peaches,  Ic.  =  /„  part,  and  90c.  =  90  times  as  many. 

NoTK.  1. — The  student  should  repeat  tlie  reasoning  of  the  foregoiug  prohlems  until  it  is 
perfectly  understood,  before  advancing  to  other  problems. 

NoTK.  2. — By  this  philosophic  process  of  solution, -we  consider  and  compare  with  the  nature 
of  the  answer  but  one  number  at  a  time,  and  the  conclusion  of  all  comparisons  cither  increases  or 
decreases  the  nature  of  the  answer  at  the  head  of  the  sUilemnit  line.  Should  the  conclusion  increase  or 
decrease  anythiug  else,  it  is  wrong,  and  the  reasoning  must  be  revised. 

XoTE.  3. — In  all  the  reasoning  for  each  number  of  every  term,  it  -n-ill  be  observed  that  wo 
make  1,  which  is  the  basis  of  all  numbers,  the  standard  unit  of  comparison  or  measurement;  and 
that  the  reasoning  is  always  from  1  to  a  collective  number,  ur  from  a  collective  number  to  1. 


GENERAL  DIEECTIOXS  FOR  PROPORTION 

662.  From  the  foregoiug  i)roblems,  explanatious,  ami  reasoniug,  we  deduce 
tlie  following  general  directious  for  all  proj)ortioual  work : 

1.  Classify  the  prohlem. 

2.  Write  the  nature  of  the  answer  at  the  top  of  the  increasing  side  of  the  state- 
ment line. 

3.  Commence  tvith  such  number  of  any  term  in  the  classified  statement,  that  is 
the  equivalent  of  the  nature  of  the  answer,  and  comparing  one  by  one,  each  number  of 
all  the  terms  with  the  nature  of  the  answer,  consider  and  determine  the  equivalency 
or  relationship  existing  between  them,  and  lohether  from  the  conditions  and  relationship 
the  nature  of  the  answer  is  to  be  increased  or  decreased,  and  then  write  the  number  on 
the  increasing  or  decreasing  side  of  the  statement  line,  according  to  the  conclusion 
reached  by  the  comparison,  the  relationship  and  the  conditions  of  the  case. 

4.  When  each  number  of  all  the  different  terms  of  the  classified  statement  has 
been  compared,  considered,  and  written  on  the  statement  line,  then  worlc  out  the  problem, 
canceling  as  much  as  possible. 

REMARKS. 

663.  Before  presenting  other  problems,  we  desire  to  place  before  tlie  student 
a  few  more  facts  pertaining  to  proportion.  We  wish  to  compare  true  proportion, 
the  analytic  system  of  proportion,  and  the  philosophic  system  of  jiroportion,  and  to 
make  a  few  pertinent  remarks  regarding  the  illogical  and  mechanical  systems  of 
proportion  taught  by  nearly  all  the  text  books  now  before  the  imblic. 

The  philosoi)hic  system  of  solving  proportional  iiroblems  is  not  proportion, 
nor  is  it  analysis.  To  elucidate  this,  let  us  take  the  following  problem  and  solve  it, 
1°,  by  proportion ;  2°,  by  analysis ;  3°,  by  the  philosophic  method : 

5  yards  of  cloth  cost  $14.    What  cost  15  yards  of  cloth  at  the  same  rate  ? 


314.  soule's  philosophic  practical  mathematics.  * 

661.  riEST  SOLUTION  BY  PEOPORTIOK. 

Cost  of  15  yds.  :  $11  :  :  lo  yards  :  5  yds.  Explanation.— This  is  the  state 

11  y^  J-,  nicnt  by  true  proportion,  and  by 

Cost  of  15  yds.  = p — '-  =  $42,   Ans.  it  we  soo  that  the  cost  of  15  yards 

/■«      1  -      -    o  J.-                 "^   1  r,  i-            ^11          Ajn  «                   bears  the  same  relation  to  $14,  the 

Or,  lo  :  o,  3  times;  and  3  times  $14  =  $42,  Ans,         ^.g^t  of  5  yards,  that  15  yards 

bear  to  5  yards. 

By  tliG  arbitrary  rtilcs  of  nearly  all  the  scliool  arithmetics  in  the  land,  the 
principles  of  proiiortion  are  ignored,  and  in  place  thereof  are  substituted  illogical 
and  absurd  mechanical  operations  by  Avhich  answers  are  obtained. 

More  than  one  hundred  authors  say  in  effect:  "Write  that  number  which  is 
like  the  answer  sought  as  the  third  term;  then  if  the  answer  is  to  be  greater  than 
the  third,  make  the  greater  of  the  two  remaining  numbers  the  second  term,  and  the 
smaller  the  first  term ;  "  etc. 

This  process  will  produce  the  answer,  but  it  is  illogical,  arbitrary,  and  absurd. 
It  robs  proportiou  of  all  claims  to  a  scientific  process,  and  the  student  who  thus 
learns  to  solve  proportional  problems  knows  no  more  of  true  proportion  than  the 
Hottentot  knows  of  the  Christian  religion. 

In  the  language  of  I'rof.  Brooks,  Avlieii  thus  treated :  "  The  whole  subject 
becomes  a  piece  of  charlatanism,  utterly  devoid  of  all  claims  to  science." 

If  proportion  is  used  at  all.  it  should  be  used  as  a  logical  process  of  reasoning, 
and  the  solution  statements  should  be  made  by  a  truthful  proportional  comparison 
of  the  elements  of  the  problem. 

For  the  higher  mathematics,  proportion  is  indispensable ;  but  in  arithmetic 
proportional  problems  are  more  easily  solved  by  analysis  than  by  iJroportion,  and 
still  more  easily  by  the  i^hilosophic  method  than  by  analysis. 

665.  SECOND  SOLUTION  BY  ANALYSIS. 

OPERATION. 

$14-i-5  =  $2|;  Analysis. — 5  yards  cost  $14.     AVhat  ■wi]115  yards  cost  ?    Since 

5  yards  cost  $14,  1  yard  frill  cost  the  i  part  of  $14  =  $2^.     Thea 

$04    v^   1-         «lo      A„o  since  1  yard  cost  I'JJ,  15  yards  will   cost   15   times   $2|  =  $42. 

-■1   X   -li>  —  ^■i-',   ^ns.  Therefore,  if  5  yards  cost  |14,  15  yards  will  cost  $42. 

By  the  analytic  system  no  proportional  statement  is  made,  and  no  false  claim 
thereto  is  assumed.  The  process  of  reasoning  is  less  scientific  than  in  proportion, 
but  much  easier,  aiul  fiir  more  practical. 

The  only  serious  objection  that  can  be  urged  against  the  analytic  system  is 
the  labor  to  produce  each  intermediate  result  of  the  alternate  operations  of  division 
and  multiplication. 

This  objection  is  entirely  obviated  by  the  philosophic  system,  and  all  the  merit 
of  the  axiomatical  reasoning  is  retained. 

666.  The  analytic  system  originated  with  Pestalozzi,  a  Swiss  teacher  and 
educational  reformer,  who  was  born  in  1746,  and  who  wrote  and  taught  from  1781  to 
1827.  His  first  methods  were  quite  crude,  but  yet  they  served  as  the  germs  from 
which  have  grown  a  large  part  of  all  modern  improved  methods  of  teaching. 


*  SOLUTION    BY    THE    PHILOSOPHIC    SYSTEM.  3l5 

He  aimed  to  take  nature's  method.  He  first  tauglit  Lis  pupils  to  tJiinlc,  then 
to  icrife,  and  then  to  read.  He  claimed  that  "  the  first  one  to  read  must  have  written 
before  he  had  anything  to  read." 

In  arithmetic  he  eommeuced  with  the  simplest  problems  and  taught  his  pupils 
to  tliink,  to  reasou,  and  to  analyze  every  problem  without  regard  to  any  rules. 

lu  1805,  his  disciple,  Jo.seph  Xeef,  came  to  this  country  and  opened  a  school 
on  the  Pestalozzian  systeni  near  riiiladelpliia.  But  the  people  were  not  then  pre- 
pared for  the  improved  methods,  and  iu  a  few  years  the  school  was  abandoned. 

The  spirit  of  the  new  system  had  however  taken  root  in  the  minds  of  pro- 
gressive teachers  who  were  pleased  to  find  something  better  adapted  to  the  wants 
of  the  educational  world  than  the  old  arbitrary  system ;  and  from  that  time  to  the 
present,  the  Pestalozzian  system  has  been  growing  in  iwpular  favor,  and  it  now  con. 
stitutes  the  basis  of  progressive  educational  systems  throughout  the  civilized  world. 

For  more  than  GO  years  past,  most  of  the  authors  of  arithmetics  have 
utilized,  to  some  extent,  in  various  modified  forms,  the  system  of  Pestalozzi.  There 
was  no  sj-stem  of  mental  arithmetic  before  his  day. 

Iu  1820,  Warreu  Colburu  published  what  he  called  an  "Introduction  to 
Arithmetic  on  the  Inductive  or  Intellectual  System."  In  this  work  he  presented 
many  of  Pestalozzi's  ideas  and  methods. 

From  Pestalozzi's  Analytic  System  of  Arithmetic  has  been  evolved  the 
Philosophic  System — a  sy.stem  without  a  peer  in  the  annals  of  mathematics,  and  to 
extend,  to  practicalize,  to  logicalize,  and  to  perfect  which,  the  author  of  this  book 
has  labored  with  tongue  and  i)en  for  more  than  a  third  of  a  century. 

He  has  taught  the  system  to  over  9000  students  and  has  by  his  lectures  and 
publications  advanced  the  philosphic  system  to  loftier  plains  and  to  more  rational 
and  logical  methods  than  were  ever  before  achieved.  In  no  ancient  or  modern  work 
on  numbers,  have  comparison,  analysis,  and  synthesis  been  woven  into  such  a  chain 
of  logical  and  jihilosophical  reasoning  as  is  presented  iu  this  and  the  author's  other 
works  on  numbers. 

The  day  is  not  far  distant,  when  the  banner  of  the  philosophic  system  will 
wave  in  triumph  from  every  spire,  pinnacle,  and  dome  of  the  Temple  of  Practical 
Mathematics. 


667.  THIRD  SOLUTION  BY  THE  PHILOSOPHIC  SYSTEM. 

OPERATION. 

..  ,  Premise. — 5  yards  co.st  .$14.     Season. — Since  5  yards  cost  $14,  1 

^.^  3  yard  -n-ill  cost  the  i  part,  and  15  yards  -frill  cost  15   times   a^ 

much,  which  is  $42. 


12, 


Ans. 


By  this  philosophic  system,  all  the  difficult  scientific  statements  of  true  pro- 
portion and  all  the  objections  to  the  analytic  method  are  obviated,  and  solutions 
are  performed  with  the  least  possible  number  of  figures,  and  by  the  easiest  possible 
process  of  logical  and  axiomatical  reasoning. 


3i6  soule's  philosophic  practical  mathematics.  * 

668.  Practical  arithmetic  has  been  sadly  neglected  by  niatlieinaticians, 
ancient  and  modern.  Pythagoras,  Plato,  Enclid,  and  Archimedes,  of  ancient  fame, 
Newton,  La  Place,  Leibnitz,  and  La  Grange,  of  modern  renown,  have  enriched  the 
literature  of  arithmetic,  but  they  did  not  give  to  it  the  penetrating  thouglit  which 
they  gave  to  the  higher  branches  of  mathematics.  And  our  more  recent  authors  of 
note,  Davies,  Peck,  Kay,  Greenleef,  Perkins,  Dodd,  Brooks,  liobinson,  Thompson, 
Wentworth,  Hill,  and  others,  have  adopted  largely  the  arbitrary  non-reasoning 
methods  of  their  jircdecessors.  They  have  failed  to  keep  pace  with  the  progress  of 
an  ingenious  age. 

The  world  of  thought  in  other  departments  of  applied  science  has  moved 
grandly  forward  and  left  the  mathematicians  of  the  19th  century  contemplating' 
and  approving  many  of  tlie  monstrous  absurdities  of  the  "  dark  ages"  of  arithmetic. 

With  the  ancients,  magic  squares  and  the  properties  of  numbers  seem  to  have 
been  deemed  more  important  than  the  science  or  the  practical  application  of  arith- 
metic. With  the  moderns,  mechanical  methods  to  produce  answers,  without  regard 
to  reason,  science,  or  brevity,  seem  to  be  the  leading  characteristic. 

And  thus  arithmetic  has  been  slighted.  Only  feeble  efforts  have  been  made 
to  unfold  and  evolve  its  logical  relations  as  a  science,  and  to  elucidate  the  iiliilo- 
sophic  princij)les  which  extend  throughout  all  its  parts. 

The  brain-entangling  cause  and  effect  method  of  Prof.  Robinson  and  some 
others,  the  complicated  and  confounding  methods  of  Prof.  Davies  and  a  hundred 
others,  seem  to  us  a  disgrace  to  the  science  of  numbers,  a  discredit  to  the  intelli- 
gence and  genius  of  arithmeticians,  and  an  insult  to  the  common  sense  of  mankiiul. 

The  damaging  effect  of  the  arbitrary  rules  of  these  authors,  on  the  minds  of 
the  rising  generation,  cannot  be  estimated.  They  fail  to  present  the  full  meaning- 
of  problems,  prevent  the  exercise  of  the  reasoning  faculties  of  the  student,  and 
produce  a  distaste  for  the  science  of  numbers. 

TO  THE  STUDENT. 

The  philosophic  system,  or  the  reasoning  we  employ  in  solving  problems  in 
proportion,  is  applicable  to  the  solution  of  all  arithmetical  questions,  and  hence 
efficiency  herein  is  a  condition  precedent  to  every  other  consideration.  The  reason- 
ing combines  that  of  multiplication  and  division  of  whole  and  fractional  numbers, 
in  alternation,  and  there  is  no  other  exercise  that  can  be  given  in  numbers,  that  will 
give  so  much  strength,  acuteuess,  flexibility  and  comprehensibility  to  the  reasoning 
organs  of  the  brain.  Thus  preparing  the  mind  not  only  to  solve  easily  all  practical 
pi'oblems  in  the  most  rapid  manner,  but  also  capacitating  the  person  for  efiicient 
service  in  all  the  positions  and  relationships  of  business  life  where  conclusions  must 
be  deduced  from  the  facts  or  premises,  or  from  the  circumstances  and  conditions  of 
things  relating  to  the  subject  under  consideration.  Therefore,  we  have  presented  a 
large  number  of  jiroblems  in  proportion  and  given  the  full  reasoning  in  the  solution 
of  many  of  them. 

Some  of  the  more  intricate  problems  are  presented  and  solved  more  as  logical 
drills  for  the  mind  than  for  the  practical  value  of  the  problems.     The  student  should 


PROBLEMS  IN  PROPORTION. 


317 


critically  read  and  re-read  tlie  reasoning  of  such  solutions  and  then  solve  the  prob- 
lems and  write  the  reasoning  without  reference  to  that  in  the  book.  By  this  means, 
a  high  degree  of  capacity  to  reason,  to  conclude  aud  to  express  thought  with  the 
pen,  logically  and  grammatically  may  be  attained. 


669. 


PROBLEMS  IX  PEOrOETIOX. 


1.  If  8  pounds  of  sugar 
cost  oG/,  what  will  27 
pounds  cost  at  the  same 
rate?  Ans.  $1.89. 


2,     If  8  iiounds  of  sugar 
cost  50/,  how  many  pounds 
can  be  bought  for  $1.89  1 
Ans.  27  pounds. 


3.  If  $1.89  will  buy  27 
pounds  of  sugar,  how  many 
pounds  can  be  bought  for 
5C^  ?        Ans.  8  pounds. 

4.  If  f  of  a  pound  cost 
$1J,  what  will  2|  pounds 
cost?  Ans.  $5^. 


CLASSIFICATION, 
lbs.  c. 

8  56 

27 

OPERATION. 


2< 


.89,  Ans. 


CLASSIFICATION, 
lbs.  c. 

8  56 

189 

OPERATION. 

lbs. 


^(5 


X?9 


27 


27  lbs.  Ans. 


CLASSIFICATION. 

lbs. 

$1.89  27 

.56 

OPERATION, 

lbs. 


m 


?^ 


8  lbs.  Ans. 


CLASSIFICATION. 

lbs.  $ 


OPERATION. 


21 


3 
i 
21 


$5J,  Ans. 


Note. — Tbe  student  should  work  the  above  problem,  and 
and  all  other  problems  that  are  solved. 


Heason. — 8  pounds  cost  56c. 
Since  8  pounds  cost  56c.,  1  pound 
will  cost  the  J  part,  and  27 
pounds  will  cost  27  times  as 
much. 

Note. — The  student  should 
work  this  problem  and  write 
the  reasoning. 

Jieasoti. — 8  pounds  cost  56c. 
Since  8  pounds  cost  56c.  con- 
versely 56c.  cost  or  bought  S 
pounds,  and  since  56c.  bought 
8  pounds,  le.  will  buy  the  jV 
part,  and  189c.  will  buy  189 
times  as  many. 

Note. — The  student  should 
work  this  problem  and  write 
the  reasoning. 

iieason.— $1.89  bought  27 
pounds.  Since  189c.  bought  27 
pounds,  Ic.  will  buy  j^j  part, 
and  56e.  will  buy  56  times  as 
many. 

Note. — The  student  should 
work  this  problem  aud  write 
the  reasoning. 

Eeason.  J  of  a  pound  cost 
$li.  Since  f  of  a  pound  cost 
$li,  or  81,  i  of  a  pound  will  cost 
the  i  part,  and  J,  or  1  pound, 
will  cost  4  times  as  much.  Then 
since  1  pound  cost  this  result,  ^ 
of  a  pound  will  cost  the  I  part, 
and  V  pounds  will  cost  21  times 
as  much. 

We  leave  this  operation  un- 
cancelled in  order  that  th« 
statement  may  the  better  assist 
the  student  in  understanding 
the  reasoning. 

should  write  the  reasoning  for  it 


3i8  soule's  riiiLosoPHic  practical  mathematics.  * 

5.  A  man  walks  320  miles  in  25  days,  when  he  walks  6  hours  per  (lay.  How 
many  miles  can  lie  walk  in  20  days  liy  walking  8  hours  per  day  ?     Ans.  341 J  miles. 

G.  J  of  a  steamboat  is  valued  at  $35700.  What  is  the  value  of  the  whole 
boat  at  the  same  rate  T  Ans.  $40800. 

7.  A  merchant  borrowed  $4000  for  04  days  at  6  pr.  ct.  He  now  desires  to 
comi)eusate  the  party  from  wliom  he  borrowed  the  money  by  loaning  him  $1500  at  8 
pr.  ct.    How  many  days  must  he  loan  it  1  Ans.  128  days. 


Note. — The  student  should  classify  the  problem  aud 
write  the  reasoninir. 


8.  A  merchant  invested  $15000  for  4  months  and  gained  $800.  How  much 
Investment  would  be  required  to  gain  $1400  in  3  months,  at  the  same  rate  ? 

Ans.  $35000. 

9.  A  note  broker  loaned  $5000  for  30  days  at  8%.  The  party  to  whom  he 
loaned  the  money  wishes  to  reciprocate  the  favor  by  loaning  the  note  broker  $2000 
at  6%.     How  long  a  time  should  lie  lend  it  ?  Ans.  100  ds. 

10.  A  farmer  owned  J  of  a  farm  and  sold  J  of  his  share  for  $12400.     What  is 
the  value  of  the  whole  farm  at  the  same  rate  ?  Ans.  $49000. 

11.  Tea  costs  80  cents  per  pound.    How  much  can  you  buy  for  15/,  at  the 
same  rate  7  Ans.  3  ounces. 

CLASSIFICATION.  OPERATION  INDICATED, 

c.  lb.         oz. 


OPERATION 

INDICATED. 

ds 

1500 

8 

4000 
C 

80  1  or  16 

15  80 


16  oz.         Write  the 
15  reasoning. 


12.  A  baker  charges  5/  for  a  loaf  of  bread  when  he  pays  $7.25  per  barrel  for 
flour.     What  ought  he  to  charge  when  he  pays  $8.70  per  barrel  for  flour  ?    Ans.  6/. 

13.  If  a  5/  loaf  of  bread  weighs  15  ounces  when  wheat  is  worth  $1.40  per 
bushel,  what  ought  to  be  the  weight  of  a  12J/  loaf  when  wheat  is  worth  $1.00  jier 
bushel  ?  Ans.  52  J  ounces. 

CLASSIFICATION.  OPERATION. 

The  student  should  write  the 
reasoning  by  which  the  figures 
are  placed    on    the    statement 
line. 
52^  OZ.,  Ans. 

14.  A  baker  charges  ten  cents  for  a  loaf  of  bread  weighing  8  oz.,  when  he 
pays  $0.30  jier  barrel  for  flour.  How  many  ounces  should  the  same  price  loaf  weigh 
■when  he  pays  $7.20  per  barrel?  Ans.  7  oz. 

15.  If  the  fore-wheel  of  a  carriage  is  7  ft.  6  in.  in  circumference,  and  turns 
round  616  times,  how  often  will  the  hind  wheel,  which  is  12  ft.  10  in.  in  circumfer- 
ence, turn  round  in  going  the  same  distance  ?  Ans.  360  times. 


c. 
5 

oz. 
15 

1.40 

o 

z. 

15 

m 

1.00 

5 

2 

1.00 

25 
1.40 

PROBLEMS    IN    PROPORTION. 


319 


OPERATION 

10.     13i  gallons  cost  $4J. 

$ 

What  will   ^   t)f  a  gallon 
cost  at  the  same  rate  f 

0 

27 
4 

U 
0 

3 

Alls.  $1. 

Note. — The  student  should 
classify  the  problem  aud  write 
the  reasoninjr. 


17.  If  it  requires  12  yds.  of  cloth  to  make  a  diess  -n-heii  the  cloth  is  38  inches 
■wide,  what  must  be  the  width  of  a  piece  20  yds.  long,  to  aiiswer  the  same  purpose  ? 

Alls.  224  inches. 

18.  A  grocer  has  coffee  at  12^/  per  pound,  and  a  coal  dealer  has  coal  at  40/ 
per  barrel.  If,  in  exchanging,  the  grocer  puts  his  cofl'ee  at  14/,  what  should  the 
coal  dealer  charge  for  his  coal  ? 


Ans.  44-i/  iier  barrel. 


CLASSIFICATION. 


OPERATION. 


12J 
40 


14 


25 


14 

o 

40 


Write  the 
reasoning. 


44|/,  Ans. 


or  thus: 


CLASSIFICATION. 


14/  —  12J^  =  IJ/  gain 
40/ 

Note. — The  student  should   write  the  rea- 
soning for  the  operations. 


OPERATION. 

/  gain. 
"     3 

2 

40 


4i/  gain. 
40/  value  of  coal. 


444/  price  of  coal. 


19.  A.  has  rice  at  6/  per  pound,  and  B.  has  sugar  at  7/  per  pound,  which  they 
wish  to  barter.  Before  bartering  it  is  agreed  that  B.  shall  value  his  sugar  at  7i/, 
and  that  A.  shall  advance  his  rice  accordingly.  What  should  be  the  exchange  price 
of  A's  rice  ?  Ans.  6f /. 

NoTi:. — See  Miscellaneous  Prohlems  in  Percentage  for  full  elucidation  of  this  class  of  problems. 

20.  If,  with  $8.10,  yoii  can  buy  9  yards  of  cloth,  bow  many  yards  can  you  buy 
for  $5.40  ?  Ans.  6  yds. 

21.  If  C  yds.  cost  $7.20,  how  much  will  12  yds.  cost  ?  Ans.  $14.40. 

22.  If,  for  12  cents,  you  can  buy  §  of  a  yard,  how  many  yards  can  you  buy  for 
36  cents  ?  Ans.  2  yds. 

23.  If  J  of  a  ton  of  hay  cost  $15,  how  much  will  3000  pounds  cost  1 

Ans.  $30. 

24.  If  20  pounds  of  sugar  will  supply  a  family  of  six  persons  for  7  days,  how 
many  pounds  will  be  required  to  supply  the  same  family  30  days  ?     Ans.  85f  lbs. 


320 


SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


L'5.    If  3  men  eat  3  dozen  oysters  in  3  minutes,  liow  many  men  will  it  require 
to  eat  100  dozen  oysters  in  100  minutes  ? 


CLASSIFICATION. 
M.        D.  Oys.        Mill. 


OPERATION, 


3 
100 


Men. 


3 
100 


3 
100 


100        AYiite  tbe 
3  reasoning. 


3  men,  Ans. 

20.     If  3  men  eat  3  dozen  oysters  in  3  minutes,  in  how  many  miiuites  can  100 
men  eat  100  dozen  oysters  ? 

CLASSIFICATION.  OPERATION. 

M.        T>.  Oys.     Mill.  Min. 


3 
100 


3 

100 


100 

o 

o 


3 
3 

100 


Write  the 
reasoning. 


3  mill.  Ans. 
If  3  men  eat  3  dozen  oysters  in  3  minutes,  how  many  dozen  oysters  can 


100  men  eat  in  100  minutes? 

CLASSIFICATION, 
M.       r>.  Oys.         Min. 

3  33 

100  100 


D.  Oys. 
'3 


100 
100 


OPERATION. 

Write  the 
reasoning. 


3333J  doz.  oysters,  Ans. 

28.  If  2.J  acres  of  land   will   produce   550   bushels   of  potatoes,   how   many 
bu.shels  will  8 J  acres  produce  at  the  same  rate  ?  Ans.  1925  bus. 

29.  If  IJ  hens  lay  1 J  eggs  in  IJ  days,  how  many  egjrs  will  6  hens  lay  in  7 
days  ?  Proportional  Answer,  28  eggs,     Logical  Answer,  2i  eggs. 


OPERATION   FOR    PROPORTIONAL   ANSWER, 

2 
3 


3  =  eggs, 


Write  the 
reasoning. 


lieason  for  Iht,  Logical  Answer. — Since  IJ 
Lous  lay  l.J  eg{;s,  in  1|  days,  the  period 
for  laying  is  hence  36  hours.  And  as  7 
days  =  lC8  hours  there  will  be  as  many 
periods  fur  laying  as  168  is  e(|ual  to  36, 
■nbich  is  168  —  36  =  4  periods  and  24 
hours  oyer.  Therefore  the  number  of  eggs 
laid  will  equal  the  product  of  the  4  pe- 
riods  by  the  6  hens,  which  is  4  X  6  =  24 
28  eggs,   Ans,  eggs  answer. 

Note. — Since  36  hours  is  the  laying  period,  it  is  presumed  in  this  solution  that  the  hens  do 
not  lay  any  eggs  during  the  21  hours. 

30.    If  1.J  cats  catch  1 J  rats  in  IJ  minutes,  how  many  cats  will  it  take  to  catch 

200  rats  in  50  minutes  ?  Ans,     G  cats, 

OPERATION. 

Cats. 

3 


CLASSIFICATION. 

C.  E.  M, 

H  IJ  ij 

200  50 


o 
3 

o 

60 


o 

200 
3 


PROBLEMS    IN    PROPORTION. 


321 


I 


31.  A  railroad  contractor  agreed  to  grade  450000  cubic  yards  in  90  days.  He 
employed  on  tlie  work  560  meu  for  30  days ;  then,  measuring  the  grading  completed, 
lie  finds  that  they  have  made  120000  cubic  yards.  Hovr  many  more  men  must  he 
employ  at  the  expiration  of  the  30  days,  in  order  to  complete  the  work  in  the  con- 
tract time  ?  Ans.  210. 


PROBLEM 

Yds. 
450000 
120000 


330000 


CLASSIFIED  : 
D.  M. 

90 

30  560 

00 


I 


2  00 

X^oppp 

OPERATION. 
M. 

330p0p 
11 

770  men 
560  men 

required, 
emjilojed 

210  men  needed. 

Explanation. — In  all  problems  of  this  character,  we  must  first  fiud  the  number  of  men  that 
■will  be  required  to  perform  the  unfinished  work  in  the  unexpired  time.  We  therefore  phice  560, 
the  number  representing  men,  on  the  increasing  side  of  our  statement  line;  and  then  to  find  the 
proportional  numbers,  ■vre  deduct  from  the  whole  work  that  was  to  be  performed,  and  the  time  in 
which  it  was  to  be  finished,  the  work  that  has  been  completed  and  the  time  consumed  in  its  com- 
pletion, and  in  the  remainder,  as  shown  in  the  problem  classified,  we  have  the  numbers 
containing  the  direct  relationship  with  the  numbers  representing  the  work  performed  and 
the  time  consumed.  Having  these  proportional  numbers,  we  reason  as  follows:  If  it  requires 
560  men  30  days  to  grade  120000  yards,  to  do  the  work  in  1  day  instead  of  30,  it  will 
require  30  times  as  many  men,  and  to  do  the  work  in  60  days  instead  of  1,  it  will  require 
the  00th  part  of  the  number  of  men  ;  then,  if  it  requires  500  meu  to  grade  120000  yards,  to  grade  1 
yard  instead  of  120000,  it  will  require  the  120000th  part  of  the  number  of  men,  and  to  grade  330000 
yards  instead  of  1,  it  will  require  330000  times  as  many  men.  This  completes  the  reasoning  and 
statement,  the  result  of  which  is  770  men,  as  the  number  required  to  perform  the  unfinished  work 
in  the  xinexpired  time,  and  as  560  men  are  now  employed,  it  is  clear,  by  the  exercise  of  our  reason, 
that  as  many  additional  men  must  be  employed  as  770  is  greater  than  5G0,  which  is  210. 

32.  If  125  men,  in  45  days,  working  8  hours  per  day,  can  build  15  walls,  each 
150  feet  long,  3  feet  thick,  and  12  feet  high,  how  many  men  will  be  required  to  build 
21  walls,  each  200  feet  long,  5  feet  thick,  and  9  feet  high,  in  14  days,  working  10 
hours  per  day  ?  Ans.  750. 

PROBLEM   CLASSIFIED  : 

Student  should  make  the 
statement  and  write  the 
reasoning. 

33.  A  merchant  owning  f  of  a  steamboat,  sold  |  of  his  share  for  $45000. 
What  was  the  steamer  worth  at  that  rate  ?  Ans.  890000. 

34.  A  besieged  city,  containing  a  population  of  45000,  has  provisions  for  4 
•weeks.  How  many  citizens  must  be  sent  away,  that  the  provisions  may  support  the 
remaining  population  40  days  ?  Ans.  13500. 

35.  If  a  man  walks  50  yards  in  f  of  a  minute,  how  far  will  he  walk  in  16f 
hours  ?  Ans.  66400  yds. 

36.  Allowing  the  number  of  pulsations  in  a  person  to  be  72  per  minute,  and 


PROBLEM    classified: 

M. 
125 

D. 
45 
14 

H.               W.               F.  L. 

8            15            150 
10            21            200 

F.  T. 
3 
5 

F.  H. 

12 

9 

322 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


that  the  velocity  of  sound  is  1118  feet  per  second,  what  will  be  the  distance  of  the 
electric  Ihiid  when  10  pulsations  are  counted  between  seeing  the  flash  of  lightning 
and  hearing  the  sound  of  the  thunder  1  Ans.  9.'51G§  ft. 


PKOBLEM    CLASSIFIED  : 

STATEMENT 

P.             M.               Ft. 

Sec.         Ft. 

P. 

Ft. 

72     =     1            1118 

or, 

1        1118 

72 

1118 

10 

72 

60 

1  m.  or  CO  sec. 

10 

10 

37.  A  builder  contracts  to  erect  a  wall  140  feet  long,  11  feet  high,  and  2J  feet 
thick,  in  10  days.  lie  employed  8  men  who  labored  on  the  wall  G  days,  during  which 
time  they  built  the  wall  G  feet  high,  the  required  length  and  thickness.  How  many 
additional  men  must  be  employed  to  complete  the  work  in  the  required  time  1 

Ans.  2. 

38.  In  a  piece  of  machinery  there  are  two  wheels  running  into  each  other: 
one  contains  42  cogs  and  the  other  33  cogs ;  in  how  many  revolutions  of  the  larger 
wheel  will  the  smaller  gain  10  revolutions  ?  Ans.  3G§. 


PEOBLEM    classified: 


c. 

R. 

42 

10 

33 

9 


statement : 
R. 
1 

33 
10 


39.  A  contractor  engaged  to  build  480000  cubic  yards  of  levee  in  90  days. 
He  employed  500  men  who  worked  60  days  and  built  300000  cubic  yards.  How 
many  additional  men  must  be  employed  at  the  end  of  60  days  to  complete  the  work 
in  the  contract  time  ?  Ans.  100  additional  men. 

40.  A  passenger  train  leaves  the  depot  and  runs  at  the  rate  of  30  miles  per 
hour.  5  hours  after  the  departure  of  said  train,  the  fast  mail  train  leaves  and  runs 
at  the  rate  of  40  miles  an  hour.  How  far  will  the  fast  mail  train  run  before  it  over- 
takes the  passenger  train  ?  Ans.  600  miles. 

OPERATION    INDICATED. 

30  X  5  =  150  miles  =  distance  of  passenger  train  when  fast  express  starts. 
40  miles,  speed  of  fast  express  train. 
30  miles,       "    "    passenger  train. 

10  miles,  gain  in  a  run  of  40  miles. 

M.  Fast  Ex. 
40 
250  Note. — The    etndeut    should    write    the 


10 


reasoning  for  the  operation. 


41.  A  thief  steals  an  article,  and  in  ninning  oif  with  it  he  takes  steps  6  feet 
long ;  after  he  has  gone  150  feet,  a  police  officer  starts  after  him  and  takes  steps  Gi 
feet  long;  allowing  that  they  both  step  equally  fast,  how  far  will  the  officer  run 
before  he  catches  the  thief  ?  Ans.  1950  feet. 


i 


PROBLEMS  IN  PROPORTION. 


323 


42.  The  steamer  Katie  leaves  the  wharf  at  New  Orleans,  and  runs  at  the 
average  speed  of  15  miles  an  hour.  When  the  Katie  had  run  25  miles,  the  steamer 
E.  E.  Lee  leaves  the  wharf  and  runs  at  the  average  speed  of  18  miles  an  hour.  How 
far  will  the  Lee  run  before  she  overtakes  the  Katie  ?  Ans.  150  miles. 

43.  A  man  has  six  hands  hired,  four  of  them  at  75/  a  day  and  two  of  them 
at  $1.00  a  day.  He  wants  to  advance  them  $G.00  on  account,  each  one  in  the  propor- 
tion of  his  daily  i^ay.    How  much  will  each  receive  1 

"'600 
3 


OPERATION. 

4  men 
2  men 

at  75/ 
at  $1. 

per 
per 

day  = 
day  = 

$3.00 
2.00 

$5.00 

5 

"coo 

0 

.60  -^  4  =  90/  to  each  of  4 
men  at  75/  per  day. 


or  thus: 
12.40  -4-  2  =  $1.20  to  each  of  2 
men  at  $1.  i^er  day. 

44.    Mr.  A.  leaves  New  Orleans  on  a  journey, 


6 
75 


90/. 


6 
1.00 

$1.20 


and  travels  at  the  rate  of  25 
miles  a  day ;  6  days  after  Mr.  B.  leaves  New  Orleans,  and  travels  the  same  road  at 
the  rate  of  30  miles  a  day.    How  many  days  will  it  be  before  Mr.  E.  will  overtake 

Mr.  A.  and  how  far  does  each  travel  1  Ans.  30  ds. 

A.  travels  750  mi. 

B.  travels  900  mi. 

45.  The  interest  on  $5000,  for  135  days,  at  8  per  cent,  is  $150.    WTiat  will  be 
the  interest  on  $4500,  for  96  days,  at  6  pr.  ct.  1  Ans.  $72. 

46.  If  a  grocer  uses  a  gallon  measure  that  is  J  a  pint  too  small,  what  will  be 
the  true  measure  for  250  gallons  of  the  false?  Ans.  234|. 


OPERATION   INDICATED. 


1  gal.  =  8  pts.  —  i  pt.  1=  7i  pts.  sold  for  8  pts.  =  -^  of  250  gals.  =  234|  gals, 
true  measure.  qj.  « 


J  pt.  short  weight  on  each  of  250  gals.  =  125  pts.  =  15|  gals,  short  weight. 
250  —  15|  =  234|  gals,  true  measure. 


47.  A.  paid  B.  $50  for  goods,  bought  by  the  pound.  When  B.  weighed  the 
goods,  he  used  a  false  weight  of  14  ounces!  What  were  A's  loss  and  B's  gain,  and 
what  was  the  true  amount  of  goods  received  by  A.  ? 

Ans.  A.  loses  ^  of  weight  bought. 
B.  gains  \  of  weight  sold. 
A.  received  $43.75  true  value  of  goods. 


OPERATION. 

=  -h  =  i  of  the 


A.  lost  2  oz.  on   each   16   bought 

weight  bought. 
A.  having  lost  2  oz.  on  each  10,  he  therefore  received 

-Jl  of  $50  worth  of  goods  =  $43.75  —  true  value 

of  goods  received  by  A. 


B.  gained  2  oz.  on  each  14 
sold  =  ^  =  I  of  weight 
sold. 


324  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

48.  A  grocer  used  a  false,  weis'lit  of  i~>\  ounces,  instead  of  10  oimces,  for  a 
jionud  ;  liow  many  pounds  did  one  of  liis  customers  lose  wlien  be  bouglit  22o  pounds 
just  weight!  Ans.  II-/1  lbs. 

OPERATION    INDICATED. 


oz.  oz. 

15.^  =  false  weight.  4 


1(5    =  true  weight.  15^ 

5  oz.  loss  on  each  pound  of  15^  oz. 


o  =  oz.  loss  on  15J  oz. 
IG 

225 


177e\oz.  =  lllbs.l,f,-oz.orlli;Mbs. 


49.  In  the  above  problem,  what  would  the  customer  have  lost  had  he  bought 
225  lbs.  by  the  false  weight  ?  Ans.  lOfl  lbs. 

OPEKATION     INDICATED, 
oz.  oz. 

10  4        3  =  oz.  loss  on  16  oz.  or 

15^  IG     225  1  pound. 

f  oz.  loss  on  a  pound  of  16  oz. 

or, 
I  oz.  X  225  =  16Sf-  oz.  -^  16  =  lOtf  lbs. 

50.  If  a  raixture  of  80  gallons  of  brandy  and  water,  consisting  of  |  of  brandy 
and  ^  of  water,  be  worth  $400,  what  would  be  the  value  of  100  gallons  of  pure 
brandy  of  the  same  quality?  Ans.  $66G.6G§. 

51.  If  a  perpendicular  post  that  is  6  feet  high  produces  a  shadow,  on  level 
ground,  of  8  feet,  what  is  the  height  of  a  tree,  the  shadow  of  which  at  the  same 
time  is  170  feet  ?  Ans.  127J. 

52.  If  $5  will  buy  12  peaches,  how  many  peaches  will  $2J  buy  ?       Ans.  40. 

53.  If  it  requires  12J  yards  of  cloth  to  make  a  suit,  when  the  cloth  is  1^  yards 
wide,  bow  many  yards  will  it  require  when  it  is  1 J  yards  wide  ?  Ans.  15. 

PROBLEM    CLASSIFIED  :  STATEMENT  : 

Yds.  Vd.x.  W.  "  Yds. 


12  J  li 


H 


25 

3 

4 


54.  If  a  boy  can  run  320  rods  in  8i  minutes,  how  many  minutes  will  it  take 
him  to  run  2J  miles  ?  Ans.  19J  min. 

55.  4  men  can  pick  1500  pounds  of  cotton  iu  2^  days,  how  many  men  will  it 
require  to  pick  8^  bales  of  450  pounds  each,  in  1 J  days  ?  Ans.  20  men. 

56.  If  5  men,  in  6i  days  of  5?-  hours  each,  can  build  a  wall  149J  feet  long,  3^ 
feet  thick,  and  5J  feet  high,  how  many  men  will  it  require  to  build  2  walls,  each  90g 
feet  long,  4  feet  thick,  and  5:^  feet  high,  in  lOJ  days  of  8J  hours  each  ?        Ans.  3. 

CLASSIFICATION. 


M. 

w. 

D. 

II. 

F.  L. 

F.  T. 

F.  H 

5 

1 

6i 

5i} 

149:\ 

3k 

5i 

2 

10^ 

8i 

90§ 

4 

5i 

PROBLEMS    IN    PROPORTION. 


!25 


M. 

5 

2 

o 

13 

21 

O 

5 

28 

17 

2 

448 

3 

3 

272 

13 

4 

4 

10 

3 

4 

21 

Heason. — Since  it  requires  5  men  to  Imilil  1  Arall,  to  linilil  2  ■walls  it  'will  re- 
quire 'J  times  as  many  men;  then  since  it  recjnires  5  men  when  they  ■work  6J 
or  -y-  days,  if  they  worii  but  i  day,  it  will  require  13  times  as  many  men,  and 
if  they  work  |  or  1  whole  day,  it  ■will  require  i  as  many  men  ;  tljen  since  it 
requires  the  number  of  men  shown  l)y  the  result  of  this  statement  when  they 
■work  1  day,  to  work  |-  day,  it  ■will  re(|uiro  2  times  as  many  men,  and  to  work 
h^  (10}  reduced)  days,  it  will  require  the  21st  part  of  the  nnml>erof  men  ;  then, 
since  it  requires  5  men  ■^■hcn  they  work  5*  or  %^  hours  per  day,  if  they  work 
but  +  of  an  hour  it  will  recjnire  28  times  as  many  men,  and  if  tliey  work  f  or  1 
whole  hour  it  will  require  i  as  many  men  ;  then,  since  it  requires  the  number 
of  men  shown  by  the  result  of  this  statement  when  they  work  1  hour  per  day, 
to  work  i  an  Imur,  it  will  require  2  times  as  many  men,  and  to  work  \}  (8J- 
reduced)  lumrs,  it  will  require  the  17th  partof  thenumberof  men  ;  tlicn,  since 
it  requires  5  men  to  build  a  wall  ^^5'^  (UOi  reduced)  feet  louR,  to  build  it  4-  of  a 
foot  long  will  require  the  44Srh  part  of  thenumberof  men,  and  to  build  it  !|  or  1 
whole  foot  loni;,  will  require  3  times  as  many  men  ;  then,  since  it  requires  the 
number  of  nu'ii  shown  by  the  result  of  this  statement  to  build  it  1  foot  Ion;;, 
to  build  it  J  of  a  foot  long,  will  require  J  part  of  the  number  of  men,  and  to  build  it  *p  feet  long 
will  reiinire  272  times  as  many  men;  then,  since  it  requires  5  men  to  build  a  wall  3J-  or  '^  feet 
thick,  to  build  it  i  of  a  foot  thick  will  require  the  l3th  part  of  the  number  of  men,  and  to  build  it 
}  or  1  whole  foot  thick,  will  require  4  times  as  many  men:  then,  since  it  requires  the  number  of 
men  shown  by  the  result  of  this  statement  to  build  it  1  toot  thick,  to  build  it  4  feet  thick  will 
require  4  times  as  many  men ;  then,  since  it  requires  5  men  to  build  a  wall  5J-  or  ^-f  feet  high,  to 
build  it  I  of  a  foot  high  will  recjuire  the  IGth  part  of  the  number  of  men,  and  to  build  it  |  or  1 
whole  foot  high,  will  require  3  times  as  nuiny  men;  then,  since  it  requires  the  number  of  men 
shown  by  the  result  of  this  statement  to  build  it  1  foot  high,  to  build  it  J  of  a  foot  high  will  re- 
quire J  part  of  the  number  of  men,  and  to  build  it  ^^  feet  high  will  retiuire  21  times  as  many  men. 
This  completes  the  reasoning  and  the  statement. 

OPERATION. 


57.  If  4  boys  can  work  GO  prob- 
lems in  3  hours,  how  many  boys 
will  be  required  to  work  40  problems 
in  8  hours!  Ans.  1  boy. 


58.  In  what  time  can  50  men  do 
a  piece  of  work  that  2  men  caTi  do 
in  25  days  ?  Ans.  1  day. 


Boys. 
4 


60 

8 


40 
3 


OPERATION. 
Ds. 

25 


50 


25 
o 


Note.— The  student  should 
classify  the  problem  and  write 
the  reasoning. 


Note. — The  student  should 
classify  the  problem  and  ■write 
the  reasoning. 


59.  If  it  costs  $3750  to  supply  250  laborers  for  30  days,  when  they  are  fur- 
nished with  28  ounces  of  provisions  per  day,  what  will  it  cost  to  supply  175  laborers 
for  40  days,  when  they  are  furnished  with  24  ounces  of  provisions  j)er  day  ? 

Ans.  $3000, 

CO.  If  it  requires  126  pieces  of  paper  to  cover  the  walls  of  a  room  when  the 
paper  is  18  inches  wide,  how  many  pieces  will  it  require  when  it  is  21  inches  wide? 

Ans.  108. 

61.  If  the  freight  on  200  bales  of  cotton  for  450  miles  is  -worth  2400  pounds  of 
sugar,  when  sug.ar  is  worth  12^/  per  lb.,  how  many  pounds  of  sugar  would  the 
freight  on  350  bales  of  cotton  for  300  miles  be  worth,  when  the  sugar  is  valued  Sge' 
per  lb  1  ^         .  -  -  ■»' 


B. 
200 
350 


PROBLEM 
M. 
450 

300 


CLASSIFIED  : 
P. 

2400 


c. 
8f 


Ans.  4000 

STATE3IENT  : 

K. 

2400 

200 

350          Write  the 

450 

300          reasoning. 

2 

25 

35 

4 

326 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


02.  If  24  men,  in  124  days  of  9|  hours  each,  dig  a  ditch,  when  the  earth  is 
estimated  at  4  degrees  of  hardness,  542J  feet  long,  4f  feet  wide,  and  4.}  feet  deep,  in 
how  many  days  of  7J  hours  each,  will  248  men  dig  a  ditch,  wlien  the  earth  is  esti- 
mated at  7  J  degrees  of  hardness,  387 J  feet  long,  3f  feet  wide,  and  li  feet  deep, 
allowing  that  time  is  required  in  proportion  to  the  degrees  of  hardness  of  the  earth  ? 

Ans.  45  days. 

63.     How  many  planks  13  ft.   9  in.   long  and  3Jin.   thick  are   equivalent   to 


5000  planks,  9§  ft. 


long  and  25  in.  thick  1 


Ans.  27o0  planks. 


CLASSIFICATION. 

P.       rt.  L.           In.  T 

)00       9|              2f 

1  "?    !>                             11 

OPERATION. 

P. 

5000 

8 

77 

165 

12 

4 

11 

7 

2 

Reason. — Placing  the  5000  planka  on  the  statement  line,  we 
reason  as  follows :  Since  there  are  5000  planks  ''^^  ft.  long,  if 
they  were  but  ^  of  a  foot  long  there  would  be  77  times  as 
many,  and  if  j  or  a  whole  foot  long  the  8th  part;  then,  since 
there  is  the  result  shown  by  this  statement,  when  the  ])lank8 
are  1  foot  long,  if  they  were  but  i^of  a  foot  long  there  would 
be  12  times  as  many,  and  if  ^fi^  ft.  long,  there  would  be  the 
165th  part;  then,  since  there  is  this  indicated  result  when 
the  i)lanks  are  -V  in.  thick,  if  they  were  but  J  in.  thick  there 
would  be  11  times  as  many,  and  if  J  or  1  in.  thick  the  4th 
part ;  then,  since  there  is  this  result  when  they  are  1  in.  thick, 
if  they  were  |  in.  thick  there  would  be  2  times  as  many,  and 
if  I  in.  thick  the  7th  part. 

64.  If  it  costs  $59.75  to  carpet  a  room  32  ft.  by  12  ft.  how  much  will  it  cost  to 
carpet  a  room  48  ft.  by  16  ft.  with  the  same  kind  of  material  ?  Ans.  $119.50. 

65.  If  8  men  in  6  days  of  10  hours  each  cut  50  cords  of  wood,  how  many 
cords  will  15  men  cut  in  15  days  of  8  hours  each  ?  Ans.  187J. 

66.  If  3  compositors,  in  2J  days  of  9.J  hours  each,  set  up  28^  pages  of  type, 
how  many  compositors  will  be  required  to  set  680  pages  in  4|  days  of  lOf  hours 
each  ?  Ans.  33^  men. 

PROBLEM    CLASSIFIED: 
C.  D.  H.  P. 

3  2i  9J  2SJ 


101 


680 


STATEJVEENT : 

4 
21 

C. 

3 

9 

5            Write  the 

0 

70 
85 

19          reasoning. 

7 

3 

680 

67. 

68, 
09. 


70. 


If  5J  yards  cost  $8.20,  what  will  22J  yards  cost  ?  Ans.  $36. 

If  J  of  a  pound  costs  $i,  what  will  ly^.  pounds  cost?  Ans.  $2.12J. 

If  10  men  in  6  days  earn  $240,  how  much  can  6  men  earn  in  10  days  1 

Ans.  $240. 
How  many  chickens  at  70/  a  piece  must  be  given  for  2  barrels  of  flour 
at  $7  per  barrel  ?  Ans.  20. 

71.  If  it  costs  a  family  of  8  jjersons  $4000  a  year  to  live  in  New  Orleans,  what 
will  it  cost  a  family  of  9  persons  to  live  in  St.  Louis  in  the  same  style  for  10  months, 
supposing  the  prices  to  be  in  St.  Louis  J  less  than  in  New  Orleans  ? 

Ans.  $2812.50. 


PROBLEMS    IN    PROPORTION. 


1^-1 


72.    If  %  of  a  pint  of  wine  cost  37J/,  what  -will 
same  rate  % 

PROBLEM    CLASSIFIED  : 


H. 

4. 
5 


c. 
37J 


i 

of  a 

liogshead  cost  at  the 

Aus.  $201.00. 

STATEMENT : 

o 

C. 

75 

3 

4 

2 

Write  the 

4 

reasoning. 

03 

5 

4 

Ans.  §  of  a  day. 

STATEMENT : 

4 
3 

Ds. 

3 

2          Write  the 

2 
9 

3          reasoning. 

8 

73.  A  garrison  consisting  of  4500  men,  has  provisions  for  9  weeks,  allowing 
each  man  24  ounces  per  day.  How  many  reinforcements  may  be  sent  into  the 
garrison,  so  that  the  provisions  will  last  but  5  weeks  when  the  daily  allowance  ia 
reduced  to  10  ounces  ?  Ans.  7050. 

74.  If  it  requires  2  men  ^  of  a  day  to  perform  §  of  a  piece  of  work,  how  many 
days  will  it  require  3  men  to  perform  f  of  it  ? 

PROBLEM    classified: 
M.  D.  \V. 

a       I      § 

3  t 


75.  If  it  requires  5  men  |  of  a  day  to  perform  f  of  a  piece  of  work,  how  long 
will  it  require  3  men  to  perform  f  of  it  ?  Ans.  2 J  d. 

76.  If  the  freight  on  85  bales  of  cotton  for  300  miles  is  worth  1000  pounds  of 
sugar,  when  the  sugar  is  worth  8Jf  per  iiound,  how  many  pounds  of  sugar  must  be 
given  for  the  freight  on  110  bales  of  cotton  for  200  miles,  when  the  price  of  sugar  is 
6^/ per  pound?  Ans.  1173J  pounds. 

OPERATION   INDICATED, 
lbs. 

1000 

85     110  '      Note. — The  student  should  classify  the  prob- 

300     .-00  lem  and  write  the  reasoning. 


25 


17 
4 


77.  If  a  loaf  of  bread  worth  5/  weighs  Gi  ounces,  when  wheat  is  worth  $1.80 
per  bushel,  what  should  be  the  weight  of  a  Vl\f  loaf,  when  wheat  is  worth  $1.02J 
per  bushel?  Ans.  18  oz. 

78.  If  it  requires  5  men  f  of  f  of  a  day  to  do  f  of  §  of  J  of  a  piece  of  work, 
what  time  will  it  require  4  men  to  do  §  of  f  of  the  work  ?  Aus.  1^  d. 


PROBLEM    CLASSIFIED  : 


M. 
5 


Ds. 

5    '^     8    —    4 


w. 


STATEMENT : 


4 
4 

15 


Ds. 


Write  the 
reasoning. 


328 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


79.  If  12  pounds  of  rice  are  worth  9  pounds  of  sugar,  liow  many  pounds  of 
rice  can  be  bartered  for  217  pounds  of  sugar?  Ans.  289J  lbs. 

SO.  If  25  oranges  are  worth  48  peaches,  how  many  oranges  can  be  bought 
for  240  peaches  ?  Ans.  125  oranges. 

81.  If  it  requires  1800  brick,  8  inches  long,  and  4  inches  wide,  to  pave  a  side- 
walk 40  ft.  long  and  10  ft.  wide,  how  many  brick  10  inches  long  and  5  inches  wide, 
will  be  required  to  pave  a  yard  50  feet  long  and  30  feet  wide  I       Ans.  4320  brick. 

82.  If  4  of  a  gill  of  wine  costs  12J/,  what  will  423  gallons  cost,  at  the  same 
rate?  Ans.  $342. 

83.  If  4  boys  can  make  100  hats  in  6  days  of  10  hours  each,  how  many  hats 
■would  0  girls  make  in  10  days  of  8  hours  each,  allowing  that  4  girls  can  do  as  much 
■work  as  5  boys  in  the  same  time  !  Ans.  250  hats. 

CLASSIFICATION. 


Boys.         Hats.             Ds.            Hr. 

4             100             (J             10 
10                8 
Girls. 

(j     =     7  J  boys. 

FIRST    OPEEATION. 

4 

o 

6 
10 

Hats. 
100 

15 
10 

8 

250  hats,  Ans. 

SECOND    OPERATION. 

4 

4 

6 

10 

Hats. 

100 

5 

6 

10 

8 

250  hats,  Ans. 

Explanation. — Since  the  working  capacity  of 
each  girl  is  greater  than  that  of  each  boy,  w© 
must  first  find  the  equivalent  -svorking  capacity 
of  the  6  girls  in  the  working  capacity  of  boys. 
And  since  4  girls  can  do  the  work  of  5  boys,  1 
girl  can  do  i  part  =  IJ  or  f  times  as  much  as  1 
boy,  and  6  girls  can  do  6  times  as  much,  =  J  x 
6  =  'V'  =  7i  boys ;  ».  e.  the  work  done  by  6 
girls  is  equal  to  the  work  done  by  7^  boys. 


Explanation  to  the  second  operation. — Since  4 
boys  make  100  h.ats,  1  boy  will  juako  the  i  part, 
and  5  boys  or  4  girls  will  make  5  times  as  many. 
And  since  4  girls  make  the  result  of  this  state- 
ment, 1  girl  will  make  the  i  part,  and  6  girls 
will  make  6  times  as  many.  >  Then,  since  100  hats 
are  made  in  6  days,  in  1  day  ^  part  would  bo 
made,  and  in  10  days  10  times  as  many.  Then, 
since  100  hats  are  made  when  the  hatters  work 
10  hours  a  day,  to  work  1  hour  a  day  they  woulif 
make  -,'„  part,  and  8  hours  8  times  as  many. 

84.  If  20  girls  can  make  200  shirts  in  10  days,  how  many  days  would  it  re- 
quire 15  boys  to  make  300  shirts,  estimating  that  5  girls  can  do  as  much  work  as  6 
boys  in  the  same  time  1  Ans.  24  ds. 

PROBLEM    CLASSIFIED  : 
G.  li.  S.  D. 

20  0  200  10 

5  15  300 


5 

OPERi 

D. 
10 

20 

ltion: 
Write  the 

15 
200 

6 
300 

reasoning 

PROBLEMS    IX    PROPORTION. 


329 


I 


85.  If  it  takes  7  men  J  of  a  day  to  do  |  of  a  piece  of  work,  how  long  will  it 
require  5  men  to  do  |  of  the  work  T  An.s.  J  day. 

86.  A  garrison  of  3000  men  has  provisions  for  42  days,  allowing  each  man  2 
lbs.  4  oz.  per  day.  How  many  reinforcement*  may  be  sent  in,  so  tliat  the  provisions 
will  la.st  10  weeks,  when  the  allowance  is  reduced  to  1  lb.  per  day  ? 

Ans.  1050  reinforcements. 

87.  If  200  iwnnds  of  first  grade  rice  cost  ?12,  what  will  be  the  cost  of  600 
pounds  of  second  grade,  estimating  that  5  ix)unds  of  the  first  grade  are  equal  to  six 
pounds  of  the  second  ?  Ans.  $30. 


CLASSmCATIOy, 

lbs. 

200  1st  grade  $12. 
600  2d  grade, 
X  f  = 
500  1st  grade. 

FrE.ST    OPEBATIO". 


200 


12 
5<XJ 


§30.  Ans. 


SECOSD   OPEEATIOX. 


200 
6 


12 
5 

GOO 

?30,  Ans. 


Explanation. — Since  1  ponndof  the  gecond  grade  ^  |  of  the 
first  grade,  600  pounds  of  the  second  grade  =  600  X  f  =  300 
pounds  of  the  first  grade.  Then,  since  200  poonds  cost  $12, 
1  pound  will  cost  fi-Q  part,  and  500  pounds  500  times  as  much. 


Explanation  of  the  second  operation. — Since  200  pounds  of 
first  grade  cost  $12,  1  ponnd  wiU  cost  ;-  -  part,  and  5  poonds 
oi  fint  grade  or  6  pounds  of  gecond  grade  will  cost  5  limes  as 
much.  And  since  6  pounds  of  second  grade  cost  this  result, 
1  pound  of  second  grade  will  cost  i  part,  and  600  pounds  of 
second  grade  ■will  cost  600  times  as  much. 


88,  If  4.J  pounds  of  coffee  cost  $9,  what  wiU  be  the  cost  of  90  pounds  of  a 
different  quality,  estimating  that  4  pounds  of  the  first  quality  are  equal  to  5  pounds 
of  the  second!  Ans.  $14.40. 

89.  If  5  oxen  or  7  cows  eat  3  A  tons  of  hay  in  87  days,  in  what  time  will  2  oxen 
and  3  cows  eat  6^  tons  ?  Ans.  210  days. 


CLASSIPXCATIOX. 

Ox.        C.  T.  Ds. 

5  or  7        3^        87 

'*TT 


2  and  3 


Explanation. — Since  by  the  first  condition  of  the  problem, 
5  oxen  or  7  cows  performed  the  eating,  and  by  the  second 
condition  2  oxen  and  3  cows  performed  the  eating,  it  is  there- 
fore necessary  to  first  find  the  eqnivalent  eating  capacity  of 
1  ox  compared  with  that  of  1  cow  ;  or  the  ef|nivalent  eating 
capacity  of  1  cow  compared  with  that  of  1  ox.  Hence, 
since  5  oxen  eat  as  much  as  7  cows,  1  ox  will  eat  f  part  ^=  1?,  or  J  times  as  much  as  1  cow  ;  or  since 
7  cows  eat  as  much  as  5  oxen  1  cow  will  eat  A  part  =  *  as  much  as  1  ox. 

Having  now  the  eqnivalent  eating  capacity  of  1  ox  and  of  1  cow  each  measured  by  the 
capacity  of  the  other,  we  nest  tind  the  eating  capacity  of  the  2  oxen  in  that  of  cows;  or  the  eating 
capacity  of  3  cows  in  that  of  oxen,  and  since  the  eating  capacity  of  1  ox  is  Ij,  or  i  that  of  1  cow, 
the  eating  capacity  of  2  oxen  is  2  times  as  much  :=  t  x2  =  -^  =  2*  cows.  This  added  to  the  3 
cows  ^  3  +  2*  ^  5f  cows. 

Or,  since  the  eating  capacity  of  1  cow  is  ^  that  of  1  ox,  the  earing  capacity  of  3  cows  is  3 
times  as  much  ^  i  x  3  =  ^  ^  2*  oxen.     This  added  to  the  2  oxen  =  2  -j-  2*  ^  4*  oxen. 

Xow,  from  these  reductions  of  the  capacity  of  oxen  to  cows,  or  cows  to  oxen,  and  the  other 
nmditions  of  the  problem,  we  may  solve  the  problem  from  either  of  the  following  classifications : 


330 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


CLASSIFICATION. 


With  the  eating  capacity  of  oxen  ex 

pressed  in  that  of  cows. 

C.               T.               Ds. 

7            3-A-         ^  87 

5*          6-fi 

OPERATION. 
Ds. 

87 

7 

29 
37 
11 

5 

11 

71 

CLASSIFICATION. 

With  the  eating  capacity  of  the  cows 
reduced  to  that  of  oxen. 


Ox. 
6 

*7 


29 
37 
11 


T. 
3-A- 

CA 

OPERATION. 
Ds. 

87 

5 

7 

11 

74 


Ds. 

87 


210  ds.,  Ans.  210  ds.,  Ans. 

90.  If  4  men  or  5  women  can  perform  J  of  a  piece  of  work  in  27  days,  how 
long  will  it  require  1  man  and  1  woman,  working  together,  to  perform  the  whole 
work  ?  Ans.  80  ds. 


CLASSIFICATION. 
Men.       Wo.      Ds.        W. 


4  or  5 
1  audi 


whole  or 


2i 


1st  OPERATION. 
Ds. 

27 
4 
5 
4 


9 


80  ds.,  Ans. 


2d  OPERATION. 

Ds. 

27 

5 

4 

4 


80  ds.,  Ans. 


Explanation. — Since 
the  work  of  4  men  is 
=  to  th.at  of  5  women, 
the  work  of  1  man  is 
=  to  that  of  1 J  women  ; 
and  the  work  of  1  man 
and  1  woman  is  =  to 
li  +  1  i.  2i  women's 
work. 


Or,  since  the  work  of  5  women  is  =  to  that  of  4  men,  the  work  of  1  woman  is  =  to  that  of  4 
of  a  man,  or  to  ^  of  the  work  of  1  man ;  and  the  work  of  1  woman  and  1  man  is  =  to  ^  -f-  1  ^  1| 
men's  work. 

From  these  reductions  and  the  conditions  of  the  prohlem,  the  classification  and  operation 
statements  are  made. 


Note. — The  student  should  write  the  reasoning  for  each  operation. 


6  X 

108 


of  the  work. 

flOSI        135      


=  -A  which  is 


This  problem  may  also  be  solved  as  follows  ; 

4  X  27  =  108  =  days  for  1  man  to  do  f  of  the  work. 
27  =  135  =  days  for  1  woman  to  do 
work  done  by  1  man  iu  1  day. 
-j^g  =  work  done  by  1  woman  iu  1  day. 
the  work  done  by  1  man  and  1  woman  in  1  day.  Then,  since  -^g  of  the  f  of  work 
is  done  in  1  day,  to  do  f|f  of  the  ^  work,  will  require  60  days.  And  since  it 
requires  00  days  to  do  f  of  the  work,  to  do  ^  will  require  J  part  of  the  time,  which 
is  20  days  ;  and  to  do  ^  or  the  whole  work  will  require  4  times  as  many  days,  which 
is  80. 

91.  If  ^  of  a  pound  costs  50/,  what  will  4  pounds  and  11  ounces  cost,  at  the 
same  rate  ?  Ans.  $9.37J. 

92.  A  grocer  sold  40  gallons  of  a  compound,  consisting  of  J  whiskey  and  J 
water,  for  $60.  What  would  be  the  value  of  50  gallons  of  pure  whiskey  at  the  .same 
rate?  Ans.  $150. 


PROBLEMS  IN  PROPORTION. 


331 


93.  If  3  meu  aud  5  women  can  do  a  piece  of  work  in  l-t  days,  0  women  and  8 
boys  can  do  the  same  work  in  lOJ^  days,  and  4  men  and  7  boys  can  do  it  in  V2^  days. 
How  long  will  it  take  2  men,  10  boys  and  3  women  to  do  it  ?  Aus.  9j  J-  ds. 


OPERATION. 


First  Stej). 

Men.  Women.  Boys.  Days.  AVork  in  1  day. 

3  +         5  ■         =         14         =         -i^ 

6         -+  8         =         lOi       =         #1- 

4  +  7         =         ll'l       =         -A. 


Classification. 


M.         W.                                   W.           AV. 

(3  +  5  =  J-J  X  6  =  18  +  30 

AVorl 

=       f 

W.        B.                                           W. 

(6  +  8  =   ,?,)  X  5  =               30  + 

B. 
40 

=     if 

Subtract   upper  Equa- 
tion from  lower. 


Second  Step. 
M.  B. 

IS  +     40        =    J^i-    ]Midti])ly  by  2. 

4  +  7        =    -A-    Mnltipl'y  b'y  9. 


M. 

B. 

36 

+ 

80 

=       "2T 

36 

+ 

03 

—  34 

6  1 

==    -2^    Add  the  two  Equations. 


143        —  3.S.S. 

-"■*"  3  5  7 

1  B.    =   afr 


Work  iu  1  day. 


M. 
4 

B. 

+    7    : 
(I' 

4 

—  51" 

=     3t7) 
_    _ii_ 

—  3  5  7 

X       7 
subtract. 

M. 
4 

1 

M. 

^     5^r 
=  T02 

Work  in  ] 

-  day 

Fourth 

Step. 

M. 

AV. 

B. 

1 

1 

1 

ro2 
M. 

r+  3t-7- 
B. 

2 

3 

10 

2 
I02 

+  -xh 

1     20    

r  -f-  3-5-7  — 

,J2  2 

12  13 

Third  Ste2). 


a 


M.      AA'. 
3  +  5  = 
(1  =  -i,)  X  3 

:  3  =  Tirir    subtract. 


5  W. 
1  W. 


Work  in  1  day. 


■^,%i^  =  work  in  1  day ;  12138  -^  1224  =  9^4  days,  Ans. 


I 


332  SOULES    PHILOSOPHIC    PKACTICAI.    MATHEMATICS.  * 

PA'RTITIVE  PROrORTION. 

670.     Partitive  Proportion  is  a  i)roporti(iii  wliere  a  iiuniber  is  divided  into 
parts  which  bear  certain  quantitative  relations  or  ratio  to  each  other. 

OPERATION.  Exiilnnafinv. —Hince    A.    re- 

1.      Divide     35      oranges  a.  li.         A.  &  H.  ceivcH  3  to  ir.s  4.  they  together, 

,  ,  ,    -r-.     •      7i  •:>        I         I  7  n^ceivo   3  -f  4  =  7.     Heuoe,    A. 

between  A.  and  B.  in  the  ^     +     -i     —     i  ^^.m  receive  }  and  B.  f  =  J  of 

ratioof3to4.      Ilowmany  ^-^^  J'„_  the  oranges      And  since    f  are 

.  .       ™  L)'J  '>ii  ***   receive  35   oranges,   j-    will 


will  each  receive  1 

Ans.  A.  15;  B.  20. 


3 


15,  A. 


^  receive  f  part  and  'i,  3  times  as 

many.     This  gives    A's    sluire. 
]  Similar  reasoning   with   the    f 

1!0,   B.        gives  B's  share. 


2.  A  father  divided  $GC  as  a  Christmas  i)resent  among  his  six  children  in 
proportion  to  their  ages,  which  were  as  follows:  Albert  16,  Eddie  14,  Mamie  12, 
Frank  10,  Robert  8,  and  Lillie  G.    How  many  dollars  did  he  give  to  each  ? 

Aus.  Albert  $10,  Eddie  $11,  Mamie  $12, 
Frank  $10,  Robert  $8,  and  Lillie  $G. 

3.  Divide  $2222  into  4  parts,  which  shall  be  to  each  other  as  1,  2,  3,  and  5. 
How  many  dollars  will  there  be  in  each  part  ? 

Ans.  1st,  $202 ;  2d,  $404 ;  3d,  $006 ;  4th,  $1010. 

4.  A  business  man  has  $42000  invested  in  three  kinds  of  stock — bank,  insur- 
ance, and  manufacturing.  He  has  three  times  as  much  bank  stock  as  insurance 
stock,  and  2J  times  as  much  manufacturing  stock  as  bank  and  insurance  stock 
together.    What  is  his  investment  in  each  kind  of  stock  ? 

Aus.  Insurance  stock,  $3000 ;  bank  stock,  $9000 ; 
jnaniifacturing  stock,  $30000. 

Memorandrim. — Insurance  stock  1  +  bank  stock  3  =  4  x  2|  =  10  =  manu- 
facturing stock ;  then,  1  -f  3  -f  10  =  14  =  $42000. 

5.  An  Egyptian  army  commander  had  1300  camels,  which  he  distributed 
among  three  separate  general  ofdcers,  in  proportion  to  ^,  ^,  and  J.  How  many  did 
each  receive  ?  Ans.  The  1st,  600 ;  the  2d,  400 ;  and  the  3d,  300. 

6.  A  benevolent  man  has  $600,  of  which  he  wishes  to  give  to  A.  J,  to  B.  J, 
to  C  ^,  and  to  D.  ^.     What  amount  will  each  receive  ? 

Ans.  A.,  $200;  B.,  $150;  C,  $120 ;  and  D.,  $100. 

7.  Another  benevolent  man  has  $600  which  he  wishes  to  give  to  W.,  X.,  T., 
and  Z.,  in  proportion  to  J,  ^,  |,  and  |.     What  amount  will  each  receive? 

Ans.  W.,  $210.53;  X.,  $157.89;  Y.,  $126.32;  Z.,  $105.26. 

NoTK. — See  Partnership  Average  for  the  solution  of  the  two  preceding  problems. 

8.  James  and  Louis,  together,  have  $124,  and  James  has  three  times  as  much 
as  Louis.    How  many  dollars  has  each  ?  Ans.  James  $93.,  Louis  $31. 

Solution. — Let  $1  represent  what  Louis  has;  then,  since  James  has  3  times  as  many,  he  has 
$3,  and  together  they  have  |1  -j-  $3  =  $4 ;  then,  $124  divided  in  the  ratio  of  1  to  3,  gives  Louis  $3i. 
and  James  $93. 


*  PROBLEMS  IX  PROPORTION.  333 

9.     Henry  and  William,  tojjetlier,  liave  $296,  and  three  times  Henry's  sliare 
equals  5  times  William's  sliare.     How  many  dollars  lias  each  1 

Ans.  Heury  $185,  William  $111. 

Solution. — Since  3  times  Henry's  share  =  5  times  William's  share,  1  time  Henry's  share  =  i 
of  5  times  William's  share,  =  5  —  3  =  IJ  or  ^  of  William's  share;  and  this  added  to  William's 
share,  which  isl,  =  5  +  5  =  §  =  what  both  have,  or  $296.  Then,  $296  divided  iu  the  ratio  of  ^ 
to  J  gives  Henry  .f  185  and  William  $111. 

Or,  thus:  Since  5  times  William's  share  =  3  times  Henry's  share,  1  time  William's  share  =  | 
of  Henry's  share  which  added  to  f  (Henry's  share  iu  fifths)  =  »  the  sum  of  both  shares.  Then 
dividing  $2d6  in  the  ratio  of  f  to  |,  gives  William  JIU  and  Henry  $185. 

Or,  thus:  To  avoid  fractions,  give  Henry  5  shares  and  William  3,  making  8  shares.  Then 
since  8  shares  equal  $296,  1  share  equals  the  i  part,  which  is  $37.  Then  5  times  $37  ecpials  $185 
equals  Henry's  share ;  and  3  times  $37  equals  $111  equals  William's  sh.are. 

10.  A.  and  B.  have  $1020.  §  of  A's  is  equal  to  a  of  B'.s.  How  nuiL-h  has 
each  ?  Ans.  A.  $540.     B.  $480. 

Sohtfioii.  J  of  A's  =  i  of  B's,  then  i  of  A's  =  i  of  B's,  which  is  i  of  J  =  |,  and  3  thirds  or 
the  whole  of  A's  =  3  times  |  or  g  of  B's.  Hence  t  +  I,  B's  share  =  \^  which  is  the  sum  of  A's  and 
B's  shares,  or  $1020.  And  -,V  of  $1020  is  $60  ;  and  9  times  $60  =  $540,  A's  share  ;  and  S  times  $60  = 
$480,  B's  share. 

11.  X.  and  Y.  own  a  plantation  containing  3500  acres,  ^  of  X's  interest  is 
equal  to  f  of  Y's.     How  many  acres  has  each  ?  Ans.  X.  1500 ;  Y.  2000  acres. 

12.  A  pomologist  has  an  orchard  containing  1744  trees,  apple,  peach  and 
pear;  there  are  3  times  as  many  apple  as  peach  trees,  plus  150;  and  4  times  as 
many  peach  as  pear  trees,  minus  120.    How  many  trees  are  there  of  each  kind  ? 

Ans.  122  pear,  368  peach,  1254  apple. 

XoTK. — Problems  of  this  character  belong  properly  to  Algebra;  but  as  they  are  frequently 
met  with  in  arithmetics,  we  have  given  it  jdace,  and  fur  those  who  have  not  studied  Algebra,  we 
give  a  brief  solution. 

Solution.— l.et  pears  =  1 ;  then  peaches  =  4  —  120,  (which  is  4  times  the  pears  —  120);  then 
apples  =  12  —  360  +  150,  (which  is  3  times  the  peaches  +  150).  Then,  since  the  pears,  peaches 
and  apples  =  1744,  it  follows  that  1  +  (4  —  120)  +  (12  —  360  +  150)  =  1744.  Then,  transposing 
or  classifying  we  have  17  =  1744  +  360  +  120  —  150  =  2074.  Then,  17  =  2074,  1  =  122  pear  trees. 
(122  X  4)  —  120  =  368  peach  trees.     (368  X  3)  +  150  =  1254  apple  trees. 

MEDIAL    TEOPOETIOIf. 

671.  Medial  Proportion  is  a  proportion  where  two  or  more  quantities  of 
different  values  are  combined  to  find  the  mean  or  average  value  of  the  combination. 
This  division  of  proportion  is  generally,  and  will  be  in  this  book  treated  under  the 
subjects  of  Alligation  Medial  and  Alligation  xVlternate. 

COXJOINED    rROPORTIOX. 

672.  For  a  definition  of  Conjoined  Proportion,  see  page  312. 

1.  If  40  oranges  are  worth  60  apples,  and  75  apples  are  worth  7  dozen 
.peaches,  and  100  peaches  are  worth  1  box  of  grapes,  and  3  boxes  of  grapes  are 
■worth  40  pounds  of  pecans,  how  many  pounds  of  pecans  can  be  bought  for  100 
oranges  ?  ^iis.  22f  lbs. 


334 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  If  40  lbs.  of  sugar  are  worth  GO  lbs.  of  rice,  and  75  lbs.  of  rice  are  worth 
600  oraugcs,  and  150  oranges  are  worth  200  apples,  and  250  apples  are  worth  16 
doz.  bananas,  and  10  doz.  bananas  are  worth  8  doz.  peaches,  and  IG  doz.  peaches 
are  worth  50  lbs.  of  pecans,  and  40  lbs.  of  pecans  are  worth  8  boxes  of  grapes,  and 
3  boxes  of  grapes  are  worth  2  boxes  of  raisins,  and  7  boxes  of  raisins  are  worth  2 
bbls.  of  flour,  and  3  bbls.  of  flour  are  worth  6  bbls.  of  potatoes,  and  2  bbls.  of 
potatoes  are  worth  $7,  then  how  many  pounds  of  sugar  can  we  buy  fur  $40  ? 

Ans.  70Jvlbs. 


EECIPEOCAL  PROPOETION. 


673.  Eeciprocal  Proportion  is  a  proportion  where  a  number  is  divided  into 
jjarts  which  bear  certain  reciprocal  quantitative  relations  to  each  other. 

1.  Four  citizens  of  a  neighborhood,  A.,  B.,  C,  and  D.,  agreed  to  build  a 
schoolhouse  and  to  pay  for  the  same  in  reciprocal  proportion  to  the  distance  that 
each  resides  from  the  selected  location.  A.  resides  2  miles;  B.,  1  mile ;  C,  J  mile ; 
and  D.,  ^  mile  from  the  schoolhouse.  The  cost  of  the  house  was  $1500.  What  does 
each  owe  ?  Ans.  A.  $100 ;  B.  $200 ;  C.  $400 ;  D. 


OPERATION. 


A.  2  miles. 

B.  1  mile. 

C.  J  mile. 

D.  I  mile. 


The  reciprocal  of  2  is  J 

"  "  "   1  isl 

"  "  "    i  is  2 

iis4 


u 


u 


The  sum  of  the  reciprocals  is  7i 


A. 


B. 


15 


1500 

o 


$100 


15 


1500 


$200 


C. 


D. 


15 


1500 


$400 


15 


1500 


$S00 


Explanation. — In  solving  pro])lem.s  of 
this  kind,  we  first  find  the  reciprocal  of 
each  number  which  represents  the  dis- 
tance that  the  several  parties  named 
reside  from  the  school,  or  other  object, 
for  which  the  money  is  contribnted.  In 
this  problem,  the  reciprocal  numl>ers  are 
shown  in  the  operation  of  the  problem, 
and  were  obtained  in  accordance  with 
Article  250,  page  147. 

These  reciprocals  show  the  respective 
proportional  parts  that  each  party  i3 
liable  for,  and  the  sum  of  the  same,  (7|) 
represents  tlie  whole  cost  of  the  school- 
house,  $;1")00.  Hence  to  determine  each 
party's  liability,  or  the  amount  that  each 
has  to  pay,  we  write  the  $1500  on  the 
statement  line,  commence  with  A.  and  reason  thus:  Since  \^  (7i)  parts  are  equal  to,  or  have  to 
pay  §1500,  i  part  equals  or  will  pay  the  i^^  part  and  f,  or  a  whole  part,  twice  as  much  :  and  since 
1  ]lart  equals  or  will  pay  this  indicated  result,  i  part  (which  is  A's)  is  equal  to,  or  will  pay,  i  as 
much.     This  shows  A's  amount  to  be  $100. 

Similar  reasoning  is  given  for  the  statements  of  B.,  C,  and  D.,  which  are  exhibited   in  the 
operation. 

2.     A  father  left  an  estate  of  $81G00,  to  be  divided  among  his  5  children  in 
reciprocal  proportion  to  their  ages,  which  are  3,  8,  10,  12,  and  15  years.     How  many 

Ans.  3  years,  $38400.     10  years,  $11520. 

8       "         14400.     12       "  9G00. 

15       "  7GS0. 


dollars  will  each  receive  ? 


*  PROBLEMS  IN  PROPORTION.  335 

3.  Suppose,  in  the  above  problem,  tbe  father  had  ordered  that  the  money 
should  be  divided  in  proportion  to  their  ages,  how  much  would  each  have  received  ? 

Aus.  3  years,  $  5100.    10  years,  $17000. 

8       "      $13000.     12       "      $20100. 

15       "      $25500. 

4.  A.,  B.,  C,  and  D.,  agree  to  contribute  the  necessary  funds  for  the  erec- 
tion of  a  mill  at  a  certain  specified  place,  in  reciprocal  iiroportion  to  the  distance 
that  they  respectively  reside  from  the  location  selected  for  the  mill,  on  condition 
that  they  should  be  joint  owners  in  the  mill  in  proportion  to  the  amount  contributed 
by  each  in  its  erection.  The  distance  that  each  resides  from  the  selected  location 
of  the  mill,  is  as  follows :  A.  ^  of  a  mile,  B.  |  of  a  mile,  C.  1  mile,  and  D.  2.f  miles. 
The  mill  cost  $3500 ;  what  is  the  interest  of  each.  A.,  B.,  C,  and  D.,  in  the  mill  ? 

Alls;     A    2-0  •   T?   -fi- •  (^,    S^'  nnd  T)     ^ 


iiPensuration. 

#^ 


671.  To  the  Reader.  Mensuration  of  surfaces  and  solids  is,  by  most 
autliors,  located  in  the  latter  part  of  arithmetics.  But,  because  of  the  general 
importance  of  the  subject  in  business  affairs,  we  present  mensuration  in  advance 
of  square  root  and  percentage  and  the  several  subjects  involving  per  cent.  The 
subject  is  herein  treated  far  more  extendedly,  practically  and  philosophically  than 
in  any  other  treatise. 

If  the  student  has  any  special  reason  for  acquiring  a  knowledge  of  percent- 
age, interest,  exchange,  etc.,  before  he  learns  mensuration,  he  may  omit  this  work 
until  he  has  studied  the  iierceutage  subjects. 

The  few  problems  involving  the  principles  of  square  root,  may  be  omitted  by 
those  unacquainted  with  that  subject,  or  if  preferred  they  may  turn  to  the  subject 
of  square  root,  and  in  a  few  minutes  acquire  the  knowledge  thereof  necessary  to 
work  the  problems. 

It  is  presvxmed  that  those  who  possess  this  book  are  already  familiar  with  the 
first  principles  of  square  root. 

675.  Mensuration  is  the  process  of  finding  the  length  of  lines,  the  area 
of  surfaces,  and  the  volume  or  solidity  of  solids.  The  jirinciples  that  govern 
the  process  of  work  are  derived  from  Geometry,  a  very  important  and  inter- 
esting branch  of  mathematics,  but  which  cannot  be  fully  explained  in  a  treatise  of 
this  character.  In  order,  however,  to  explain  some  of  the  most  simple  and  fre- 
quently occurring  geometrical  terms  and  figures,  we  present  the  following 

GEOMETRICAL  DEFINITIONS  AND  FIGURES. 

676.  A  Point  is  that  which   has   position    with- 
•                              out  measurable  length,  width,  or  thickness. 

677.  A  Line  is  that  which  has  length  without 
measurable  width  or  thickness. 

■ —        678.    A  Straight  Line  is  one  that  does  not  change 

direction  at  any  point. 

679.  A  Curved  Line  is  one  that  changes  direction 
at  every  point. 

680.  A  Croolied  or  Broken  Line    is  one  that 
changes  direction  at  some  of  its  points. 

(336) 


I 


MENSURATION. 


337 


681.  A  Surface  is  that  which  has  length  and  width,  without  measurable 
thickness. 

682.  A  Plane  Surface  is  a  surface  which  lies  even  throughout  its  whole 
extent. 

683.  A  Curved  Surface  is  a  surface  that  has  length  and  width,   without 
measurable  thickness,  and  like  a  curved  line,  changes  its  direction  at  every  point. 

684.  A  Volume,  Solid  or  Body,   is  that  which  has  length,  width  and 
thickness. 


ANGLES. 


685.  An  Angle  is  the 
opening  or  divergence  of 
two  lines  from  a  certain 
point. 

687.  An  Acute  Angle  is 
less  than  a  right  angle. 


X.. 


686.    A  Right  Angle 

is  an  angle  formed  by 
one  straight  line  meeting 
another,  so  as  to  make 
equal  angles. 

688.  An  Obtuse  Angle 
is    greater  than  a    right 
le. 


.«     an 


FIGUEES. 


689.  A  Plane  Figure  in  geometry  is  a  portion  of  a  plane  bounded  by  lines, 
straight  or  curved,  or  by  both  combined.  When  the  lines  which  bound  a  figure  are 
straight,  the  space  that  they  inclose  is  called  a  polygon,  or  rectilineal  figure. 


690.  A  Polygon  is  a  plane 
figure,  or  portion  of  a  surface 
bounded  by  straight  lines. 


692.    A  Rectangle   is  a 

quadrilateral  polygon  which 
has  its  opposite  sides  equal 
and  parallel,  and  all  its  an- 
gles right  angles. 


691.    A  Square  is  an  equi- 


lateral rectangle. 


693.  ARight-An- 
gled  Triangle  is  a  tri- 
angle that  has  one  of 


g-its  angles  a  right  an- 


In  the  right-angled  triangle,  ABC,  the  side  A  C,  opposite  the  right  angle 

B,  is  called  the  hypotenuse  ;  the  side  A  B,  the  hase^  and  the  side  B  C,  the  perpendic- 

■ulat 

Upwanl  of  2400  year.s  ago,  PythaRoras,  a  celebrated  Grecian  geometer,  discovered  that  the 
square  described  on  the  hypotenuse  of  a  right-angled  triangle  is  equal  to  the  sum  of  the  squares  described  on 
the  other  two  sides.  And  on  making  this  discovery,  to  express  his  joy  and  gratitude,  we  are  told  that 
he  sacrificed  one  hundred  oxen  to  the  Muses.  Pythagoras  also  discovered  that  a  circle  is  the  great- 
est of  all  figures  of  equal  perimeter. 

^  691.  A  Diagonal  is  a  straight  line  connecting  the  vertices 
of  two  opposite  angles  of  a  iwlygon.  As  the  line  A  B,  in  this 
rectangle.  The  sum  of  all  the  lines  bounding  a  polygon  constitute^ 
its  perimeter. 


ioS 


soule's  philosophic  practical  mathematics. 


'     695.  A  Parallelogram 

is  a  quadrilateral  liaviiij? 
its  opposite  sides  parallel. 


696.    A  Khombus  is  an 

equilateral     oblique-angled 
parallelogram. 


697.    A  Trapezoid  is  a 

quadrilateral  having  only 
two  of  its  opposite  sides 
parallel. 


698.    A  Trapezium  is 

a  quadrilateral  liaviug  no 
two  sides  parallel. 


699.  An  Equi- 
lateral Triangle 

is  one  having  its 
three  sides  equal. 


700.  An  Isos- 
celes  Triangle 

is  one  having 
only  two  of  its 
sides  equal. 


701.  A  Sea- 
lene  Triangle 

is  a  trian  gl  e 
I  which  has  all  its 
-'  sides  unequal. 


Peutagon.  Hexagon.  Heptagoo.  Octagon.  Nonagon.  Decagon. 

702.  A  Circle  is  a  plane  figure  bounded  by  a  regular  curved 
line,  every  point  of  which  is  equally  distant  from  a  point  within 
called  the  center. 

For  the  definition  of  tlie  Diameter,  the  Circumference,  the  Radius,  a  Chord,  an 
Arc,  a  Sector,  an  Angle,  anil  a  Segment  of  a  Circle,  see  Circular  Measure,  page  252. 

703.  The  Ratio  between  the  diameter  and  the  circumference  of  a  circle 
Las  been  demonstrated  in  geometry  to  be  as  1  to  3.1416 ;  i.  e.,  when  the  diameter  of 
a  circle  is  1,  the  circumference  is  3.141G,  practically. 

Note. — This  ratio  cannot  be  exactly  expressed  in  figures.  The  decimal  has  been  carried  by 
mathematicians  to  two  hundred  and  titty  places,  and  yet  the  exact  ratio  was  not  obtained.  It  is 
imijossible  to  square  the  circle. 

Van  Cenlen,  a  German  mathematician,  first  extended  the  approximation  to  36  places  by 
continually  bisecting  the  arc  of  a  circle.  In  honor  of  this  achievement,  the  36  numbers  expressing 
the  ratio  of  the  diameter  of  a  circle  1o  its  circumference  were  engraved  on  his  tomb-stone.  The 
following  are  the  numbers :     3. 1415926535«9793238462643383279502884. 

704.  The  Area  of  a  figure,  or  of  a  described  object,  is  the  measure  of  its 
surface  in  some  square  lonit,  as  the  inch,  the  foot,  the  yard,  the  mile,  etc. 

705.  The  Ratio  between  the  area  of  a  circle  and 
of  a  square,  one  side  of  which  is  equal  to  the  diameter 
of  the  circle,  has  been  demcmstratcd  in  geometry  to  be 
as  1  to  .7854: ;  i.  e.,  when  the  area  of  a  square  is  1,  the 
area  of  the  circle  is  .7854.  Or,  if  we  divide  a  square 
surface  into  10000  smaller  squares  and  then  cut  otf 
the  four  corners  so  as  to  make  a  circle,  we  will  thus 
cut  away  2146  of  the  smaller  squares  and  have 
remaining:  in  the  circle  7854  of  the  smaller  squares. 


Area  10000  J^rea  "854. 

10000—2146=7854. 


MENSURATION. 


339 


706.     An  Ellipse  is  a  figure  bounded  by  an  oval  curved  line. 

Tho  transcerse  diameter,  or  axis  of  an  ellipse   is  a,  line  passing  tlirough  its 
i   center  in  tho  direction  of  its  length,  as  the  line  A   B. 

The  conjugate  diameter,  or  axis,  is  a  line  passing  throngh  the  center  of  the 
elliiise  in  the  direction  of  its  width,  as  the  line  E  D. 


TABLE  OF  MULTIPLES. 

707.    The  following  table  may  be  often  used  to  advantage  in  the  computation 
of.  mechanical  and  geometrical  problems : 

TABLE. 

Diameter  of  a  circle  x  3.1416  =  Circumference. 
Radius  of  a  circle  X  G.283185  =  Circumference. 
Square  of  the  radius  of  a  circle  x  3.1416  =  Area. 
Square  of  the  diameter  of  a  circle  X  0.7854  =  Area. 
Square  of  the  circumference  of  a  circle  X  0.07958  =  Area. 
Half  the  circumference  of  a  circle  X  hy  half  its  diameter  =  Area. 
Circumference  of  a  circle  X  0.159155  =  Radius. 
Square  root  of  the  area  of  a  circle  x  0.56419  =  Radius. 
Circumference  of  a  circle  X  0.31831  =  Diameter. 
Square  root  of  the  area  of  a  circle  X  1.12838  =  Diameter. 
Diameter  of  a  circle  X  0.866  =  Side  of  inscribed  equilateral  triangle. 
Diameter  of  a  circle  x  0.7071  =  Side  of  an  inscribed  square. 
Circumference  of  a  circle  X  0.2251  =  Side  of  an  inscribed  square. 
Circumference  of  a  circle  X  0.2821  =  Side  of  an  equal  square. 
Diameter  of  a  circle  X  0.8862  =  Side  of  an  equal  square. 
Base  of  a  triangle  x  by  i  the  altitude  =  Area. 

Multiplying  both  diameters  and  .7854  together  =  Area  of  an  ellipse. 
Surface  of  a  sphere  X  by  ^  of  its  diameter  =  Solidity. 
Circumference  of  a  sphere  x  by  its  diameter  =  Surface. 
Square  of  the  diameter  of  a  sphere  x  3.1416  =  Surface. 
Square  of  the  circumference  of  a  sphere  x  0.3183  =  Surface. 
Cube  of  the  diameter  of  a  sphere  x  0.5236  =  Solidity. 
Cube  of  the  radius  of  a  sphere  x  4.1888  =  Solidity. 
Cube  of  the  circumference  of  a  sphere  x  0.016887  =  Solidity. 
Square  root  of  tho  surface  of  a  sphere  X  0.56419  =  Diameter. 
Square  root  of  the  surface  of  a  sphere  X  1.772454  =  Circumference. 
Cube  root  of  the  solidity  of  a  sphere  X  1.2407  ==  Diameter. 
Cube  root  of  the  solidity  of  a  sphere  X  3.8978  ^  Circumference. 
Radius  of  a  sphere  X  1.1547  =  Side  of  inscribed  cube. 

Square  root  of  (J  of  the  square  of  )  the  diameter  of  a  sphere  =  Side  of  inscribed  cube. 
Area  of  its  base  X  by  J  of  its  altitude  =i  Solidity  of  a  cone  or  pyramid,    whether  round, 
square,  or  triangular. 

Area  of  one  of  its  sides  x  6  =  Surface  of  a  cube. 

Altitude  of  trapezoid  X  i  the  sum  of  its  parallel  sides  =  Area. 


340  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


LINEAR  MEASURE 


708.     In  the  mensuration  of  lines,  surfaces,  and  solids,  we  have  three  Icinds 
of  measure — line  or  linear,  surface  and  solid. 

The  U7iit  of  measure  for  lines  is  an  inch,  a  foot,  a  yard,  a  rod,  etc.,  or  some 
subdivision  thereof. 

PROBLEMS. 

•*  ■  1.    The  diameter  of  a  circle  is  50  feet.    What  is  the  circumference? 

Ans.  157.08  feet. 

Explanation. — In  all  problems 
OPERATION  "^  *^'^   kind,    we   multiply   the 

diameter  by  3.1416,  which  is  the 
50  X  3.141G  =  157.0S00  ft.,    Ans.  ratio  between  the  diameter  and 

the  circumference. 


Note. — In  all  problems  In  Linear,  Surface,  or  Solid  Measure,  the  student  should  draw  on  his 
paper  the  outline  of  the  figure  to  be  measured,  before  performing  the  operation. 

2.  The  circumference  of  a  circle  is  40  feet.    What  is  the  diameter  ? 

Ans.  12.73  +  ft. 

OPERATION.  Explanation. — In  all  problems  of  this  kind, 

we  divide  the  circumference  by  3.1416,  which  is 
,„     „„  „^  .    „         _      „„  „         .  the  ratio  between  the  diameter  and  the  circum- 

40.0000  -4-  3.1416  =  12.73  +  ft.,    Ans.         ference  of  a  circle. 

3.  What  is  the  circumference  of  a  circular  garden,  the  diameter  being  25 
yards  and  2  feet  ?  Ans.  241.9032  ft. 

4.  What  is  the  diameter  of  a  circle  whose  circumference  is  OS  feet  9  inches  f 

Ans.  262.005  in. 

5.  The  radius  of  a  circle  is  40  feet.    What  is  the  circumference  ? 

Ans.  251.328  ft. 

6.  What  is  the  perimeter  of  a  nonagon,  each  side  of  which  is  61  inches  ? 

Ans.  549  inches  =  45  ft.  9  in. 

Note. — The  perimeter  of  a  body  or  figure  is  its  outer  boundary  or  the  sum  of  all  its  sides. 

7.  What  is  the  perimeter  of  a  hexagon  whoso  sides  are  each  4  ft.  3  in.? 

Ans.  25  ft.  6  in, 

8.  The  perimeter  of  a  duodecagon  is  27  feet.    What  is  the  length  of  one 
£ide  ?  Ans.  2^  ft.  =  2  feet  3  inches. 


I 


*  LINEAR    MEASURE.  34I 

TO  FIND  THE  HYPOTENUSE  OF  A  EIGHT-ANGLED  TRIANGLE.  WHEN 
THE  BASE  AND  THE  ALTITUDE  ARE  GIVEN. 

PROBLEMS. 

709.     1.     The  base  of  a  right-angled  triangle  is  60  feet  and  the  altitude  25 
feet,  what  is  the  hypotenuse  ?  Ans.  65  feet. 

Note. — Students  who  have  not  studied  square  root  may  omit  this   and  the  two  following 
problems,  until  they  have  learned  square  root. 

OPERATION. 
60^      :=      3600  Explanation. — In  all  problems  of  this  kind,  we  extract 

the  iquare  root  of  tlie  sum  of  the  squares  of  the  hase   and 


^25  altitude,  and  the  result  will  be  the  hypotenuse. 

•/I225  =  65  feet,  Ans. 

2.    If  one  side  of  a  square  park  is  one  mile,  what  is  the  distance  from  the 
center  to  either  corner  ?  Ans.  3733.52  +  feet. 


TO  FIND  THE  BASE  OR  THE   ALTITUDE   OF  A    RIGHT-ANGLED  TRI- 
ANGLE, WHEN  THE    HYPOTENUSE    AND    ONE    OF    THE 
OTHER    SIDES  ARE   GIVEN. 

710.  1.    The  hypotenuse  of  a  right-angled  triangle  is  50  feet  and  the  base 
40  feet,  what  is  the  altitude  I  Aus.  30  feet. 

OPERATION. 

50=     =     2500 

^Qa      _.      1000  Explanation. — lu  all  problems  of  this  kind,  we  eiiract 

.  the  square  root  of  the  difference  between   the   squares   of  the 

Dinerence,       JOO  hypotenuse  and  the  given  side,  and  the  result  will  be  the 

, required  side. 

V  000  =  30  feet,  Ans. 

2.    The  hj'poteuuse  of  a  right-angled  ti-iangle  is  410  feet,  and  the  base  is 
90  feet.    What  is  the  perpendicular  or  altitude  ?  Ans.  400  feet. 

TO  FIND   THE  DIAMETER   OR  THE   CIRCTJMEERENCE   OF  A  CIRCLE, 
WHEN  THE  AREA   IS  GIVEN. 

NoTE.^The  area  of  a  figure  is  the   quantity   of  its  surface   expressed  in   units  of  square 
measure,  as  an  inch,  foot,  yard,  acre,  etc. 

PROBLEMS. 

711.  1-     The  area  of  a  circle  is  47124  square  feet,  what  is  the  diameter? 

Ans.  244.94+  feet. 


342 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS, 


OPERAXrOX. 

.7S54  )  471i!4.0000  (  GOOOO 


V  (iOWK)  =  244.94+  ft.,    Ans. 


Ezplanaiion. — In  all  problems  of 
this  kind,  we  first  diride  the  area  bi/ 
.7Sr>4.  and  tkeit  ejctraci  the  square  root  of 
the  quotient.  To  find  the  circnm/erencr. 
■we  would  mnltiply  the  diameter  by 
3.1416,  or  divide  the  area  by  .07958, 
and  extract  the  square  root  of  the 
quotient. 


WTiat  is  the  diameter  of  a  circle  \rliose  area  is  5026.56  feet  T 

Ans,  SO  feet. 


TO  FDvD  THE  SIDE  OF  A  SQUARE  THAT   IS  EQUAL  IX  AREA  TO  A 

GIVEN  CIRCLE. 

PKOBLEMS. 

712.     1.    "Wliat  is  the  side  of  a  square  equal  in  area  to  a  circle  whose  diame- 
ter is  40  feet  f  Ans,  35.44S  feet. 

OPEKATION, 

,SSG2  ^Tplanaiion. — In  all  problems  of  this  kind,  we  multi- 

40  diameter.  j>Jy  the  diameter  by  .8S62,  or  where  the  circumference   is 

given,  multiply  it  by  .2821;  the  result  in  each  case  is  the 

33.4480  fe^t,   Ans.  side  of  an  equivalent  square 

Note. — The  decimal  .SS62,  is  the  square  root  of  .7^4  and  is  the  side  of  a  square  equal  in 
area  to  a  circle  whose  diameter  is  1.  The  decimal  .2821  is  the  square  root  of  .07958,  and  is  the  side 
of  a  square  equal  in  area  to  a  circle  whose  circumference  is  1 

2.  The  circumference  of  a  race  track  is  one  mile  or  52S0  feet.    "VThat  is  the 
side  of  a  square  field  of  equal  area  !  Ans.  14S9.48S  feet. 

3.  The  diameter  of  a  circle  is  60  feet,  what  is  the  side  of  a  square  of  equal 
areaT  Ans.  53.172  feet. 


TO  FIND  THE  SIDE  OF  A  SQUARE  LS'SCEIBED  IN  A  GIVEN  CIRCLE. 

PROBLEilS. 

713.     1.     The  diameter  of  a  circle  is  40  feet,  what  is  the  side  of  the  inscribed 
square  t  Ans,  2S.2J»4  feet. 


OPEKATION, 

.7071 
40 


2S.2S40  feet,  Ans. 


Explanation. — In  all  problems  of  this 
kind,  we  multiply  the  diameter  by  .7071,  or 
where  the  circumference  is  given,  multiply  it 
by  .2251 ;  the  result  in  each  case  is  the  sida 
of  the  inscribed  sqnai«. 


2.     What  is  the  width  of  one  side  of  the  largest  square  piece  of  timber  that 
can  be  sawed  from  a  log  30  inches  in  diameter  t  Ans.  2L213  inches. 


\. 


LINEAR    MEASURE. 


343 


TO  FIXD  OR  MEASURE  THE  HEIGHT  OF  A   TREE,   OR  THE   HEIGHT 

TO  A  CERTAIN  POINT  OF  THE  SAME  WITHOUT  CUTTING 

OR  CLIMBING  THE  TREE. 

PROBLEMS. 

714.     The  figure  A  B  C  T),   represents   a 

tree.     What  is  the  distance  from  C  to  B  T 

Aus.  36  feet. 

Note. — The  point  C  is  placed  3  feet  from  the  ground 
to  facilitate  the  observation  liy  the  eye. 

Erplanation. — First  measure  from  the  base  of  the  tree 
a  distance  of,  say  10  feet,  and  at  this  point,  vertically 
erect  a  pole  E  F,  that  h.-vs  feet  and  inches  marked 
thereon.  Then,  place  a  second  pole  G 11,  say  2  feet  outside 
of  the  first,  and  at  a  point  I,  3  feet  from  the  ground, 
pl.ice  the  eye  and  prolong  the  sight  by  the  side  of  the 
first  pole  to  the  point  B,  on  the  tree  and  note  at  what 
height  the  line  of  eight  strikes  the  first  polo.  Then, 
suppose  in  this  example,  that  the  line  of  sight  touches 
the  ]i()le  at  the  height  of  9  feet,  we  have  the  following 
proportion  of  triangles: 

As  2  feet  is  to  6  feet,  so  is  12  feet  to  the  distance  from 
C  to  B,  or,  thus:  If  a  triangle  of  2  feet  base  gives  6 
feet  height,  1  ft.  b.i8e  will  give  i  part,  and  a  triangle  of 
12  ft.  base,  (the  distance  from  C  to  I )  will  give  12  times 
aa  many  feet  height. 

STATEJIENT. 


6 
12 


ft.  high. 


3G  ft.,  Ans. 

Note  1. — Tlio  6  ft.  height  is  found  by  subtracting  3  feet  from  9;  9  being  the  pole  mark  of 
the  line  of  sight  and  3  feet  being  the  distance  that  the  line  C  I  is  from  the  ground. 

Note  2. — The  difference  in  the  diameter  of  the  tree  at  the  ground  and  at  the  point  B,  is  not 
considered  in  the  ])roblem.     For  jiractical  work  of  this  kind  this  element  need  not  be  considered. 

This  problem  is  of  value  to  lumbermen,  for  by  it  they  may  determine  the  height  of  trees  up 
to  a  r(i|uired  diameter,  or  to  a  limb  or  crook  before  cutting  the  tree  and  in  case  the  height  is  not 
what  is  required,  the  tree  need  not  bo  felled. 

2.  In  the  above  problem,  suppose  the  line  of  sight  on  the  pole  E  F  had 
passed  the  figures  or  iiulieated  a  height  of  say  10  ft.  5  in.;  what  woiihl  have  been 
the  height  of  the  tree  from  the  ground  to  B  T  Ans.  47i  feet. 


STATEMENT. 
Ft. 


12 
o 


89 

12 


10  ft.  5  in.  —  3  ft.  =  7  ft. 


=  7j§jft.  =  f5  ft. 


•i4.J  ft.  from  C  to  B. 
3     ft.  =  distance  from  C  to  the  ground. 


47j  ft.,   Au3. 


344 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


3.  TVliat  is  the  lieigbt  of  a  tree,  from  the  ground  to  the  top,  when  the  pole  E 
F  is  placed  40  ft.  from  the  tree,  the  pole  H  G  is  placed  14  feet  beyond  the  pole 
E  F,  and  the  line  of  sight  strikes  the  pole  E  F  at  a  height  of  18  ft.  4^  inches,  and 
the  eye  point  being  3  feet  above  the  ground  1  Ans.  42S|  ft. 


OPERATION. 

18  ft,  4J  in.  —  3  ft.  =  15  ft.  4 J  in.  =  1S4J  in. 
40  ft. +  1J  ft.  =  41 J  ft. 

12    in.  in  1  ft. 

498    in. 


=  ^P-  in. 


Ft. 


Ft. 


2 

309 

2 

369 

12 

or 

12 

498 

3 

2 

18 

2 

83 

425| 
3 

ft. 
ft. 

428| 

ft, 

Ans 

4.  If  in  triangles  like  the  above,  the  pole  F  E  is  24  ft.  from  the  tree, 
the  pole  n  G  is  3  feet  beyond  pole  F  E,  the  eye  point  or  line  C  I  being  3  feet 
from  the  ground,  and  the  line  of  sight  on  pole  F  E  crosses  the  figure  16  ft.  3  in., 
■what  is  the  length  of  the  tree  from  G  to  B  ?  Ans.  119J  ft. 


TO  FIND  THE  LENGTH  OF  A  PART  OF  A  TREE  BROKEN  OFF  AT 

THE  TOP  WHEN  ONE  END  OF  THE  BROKEN  PART  RESTS  ON 

THE  STUMP  AND  THE  OTHER  END  IS  ON  THE  GROUND. 

PROBLEM. 

715.     1.     A  tree  100  feet  high  broke  in  a  storm,  and  the  top  touched  the 
'  ground  40  feet  from  the  foot  of  the  tree.     What  was   the   length   of  the  portion 
broken  off  ?  Ans.  58  feet. 


Explanation. — In  all  problems  of  this  kiiifl,  we  .add  the  square  of  the  base 
to  the  square  of  tho  height  of  the  tree,  and  divide  the  snm  by  twice  the 
height  of  tlio  tree,  tho  qnotient  will  he  the  length  broken  off.  Or,  divide 
the  square  of  the  base  by  twice  the  height  of  the  tree,  and  to  the  quotient 
add  one-half  the  height  of  the  tree. 


OPERATION. 


40"  =  1600  1600  +  10000 
100=  =  10000 


11600  4-  200  =  58  feet. 


LINEAR    MEASURE. 


345 


I 


TO  FIND  OR  MEASURE  THE  DISTANCE  BETWEEN  TWO  POINTS   ONE 
OP  WHICH  IS  INACCESSIBLE. 

I  A. 

PROBLEMS. 

716.     1.     What  is  the  di.staiice  between  A  and  B 
measured  from  the  point  B  -witliout  access  to  A? 

Ans.  90  feet. 

Explanation. — To  determine  this  distance,  we  resort  to  the 
princiiiles  of  Geometry  and  proceed  thus:  First  prolong  an  indef- 
inite straight  line  drawn  through  the  points  A  and  B  say  to  C. 
Then  from  the  point  B  measure  at  right  angles  to  the  line  ABC, 
.any  convenient  distance,  say  30  feet  to  the  point  D.  Then  project  an 
indefinitelinedrawu  from  Athrough  Dmakingline  A  D.  Then  from 
point  D  and  perpendicular  to  line  A  V>  project  a  line  to  the  line  A 
C  until  it  intersects  the  line  A  B  C  at  the  point,  say  O;  then 
measure  the  distance  from  B  to  O  which,  s.ay  is  10  feet.     Then  the 

10 


BD2 


distance  from  A  to  B  =  ^.     Thus :  30  x  30  =  900  ft.  900  ft 

BO 
=  90   feet. 

This  principle  of  Geometry  may  often  be  applied  to  advantage 

Ly  any  one,  to  measure  the  distance  across  rivers  or  other  inaccess. 

ible  places. 

2.  Considering  the  above  diagram,  what  is  the  dis- 
tance from  A  to  B,  when  the  distance  from  B  to  D  is 
60  feet  and  the  distance  from  B  to  O  is  6  ft.  4  in.  t 

Ans.  394|i  ft. 

3.  Wliat  is  the  width  of  a  river  in  paces  and  feet, 
when  the  distance  from  B  to  D  in  a  diagram  like  tlie 
above,  is  60  paces  and  the  distance  from  B  to  O  is 
9  J  paces,  supposing  each  pace  to  be  3  feet  T 

Ans.  385f  paces,  1157^^  feet. 

TO  FIND   THE   DISTANCE  BETWEEN  TWO   INACCESSIBLE    OBJECTS, 
AS  TWO  TREES  ON  THE  OPPOSITE  SIDE  OF  A  RIVER. 


PROBLEMS. 

717.  1.  WTiat  is  the  distance  between  A  and  B 
in  the  diagram,  which  are  located  on  the  opposite  side 
of  a  river!  Ans.  700  feet. 

Explanation. — In  all  problems  of  this  kind,  first  find  the  dis- 
tances AC  and  BC  by  the  preceding  proposition,  which  are,  say, 
500  and  600  feet.  2d.  Divide  these  distances  by  10  or  100,  by  point- 
ing off  one  or  two  places  and  thus  obtain  proportional  distances; 
in  this  problem,  wedivide  by  100  and  thus  obtain  6  and  ;"  as  the 
proportionals.  3rd.  Lay  off  on  the  prolongation  of  AC  and  BC 
their  respective  proportionals.  4th.  Connect  and  measure  ab, 
which  is  the  proportional  for  AB,  and  multiply  the  s.ame  by  100, 
the  product  will  equal  the  distance  AB.  In  this  problem,  we 
assume  that  ab  is  7  feet,  which  multiplied  by  100  equals  700  feet 
the  distance  from  A  to  B. 


346 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  Standiug  on  a  plateau,  I  can  see  two  mountain  peaks  to  the  westward.  I 
ascertain  the  distance  to  the  one  on  the  left  to  be  25  miles,  and  to  the  one  on  the 
right  to  be  30  miles.  Laying  off  the  proportionals  25  feet  and  30  feet,  I  find  the 
distance  between  them  to  be  iO  feet.     What  is  the  distance  between  the  peaks  T 

Ans.  40  miles. 


MENSURATION  OF  SURFACES. 


718.  The  unit  o(  mensnre  for  surfaces  is  a  square  whose  side  is  some  fixed 
length,  such  as  an  inch,  foot,  yard,  rod,  etc. 

In  the  solution  of  problems,  in  mensuration  of  surfaces,  much  time  may  often 
be  saved  by  applying  the  following  principle  which  has  been  demonstrated  in 
geometry : 

That  all  similar  surfaces  or  areas  are  to  each  other  as  the  squares  of  their  like 
dimensions.  •  •  .  , 


TO  FIND   THE   AREA    OF  PARALLELOGRAMS,   EITHER  RECTANGLE, 
RHOMBOID,  RHOMBUS,  OR  SQUARE. 

PEOBLEMS. 


719.     1.    What  is  the  area  of  a  rectangular  garden  240  ft.  long  and  120  ft. 
wide?  Ans.  28800  sq.  ft. 

OPERATION. 

240  ft.   long.  Mxplanation. — In   all   squares   or   rectangular    figures,     we 

""          '              '  multiply  the  length   by   the   width,    in   the   same   units   of 

28800   sq.   ft.,    Ans.  measure,  and  in  the  product  wo  have  the  required  area. 


2.  How  many  square  yards  are  there  in  the  surface  of  a  wall  212  feet  long 
and  32  feet  high  ?  Ans.  753|  sq.  yards. 

3.  What  is  the  area  of  a  rectangular  floor  20  feet  long  and  14  feet  wide  ? 

Ans.  280  sq.  feet. 


I 


MENSURATION  OF  SURFACES. 


347 


4.    How  many  square  feet  in  a  right-angled  triangle  wLose  base  is  12  feet,  and 


perpendicular  height  is  8  feet  ? 

OPEEATIOK. 

12  ft.  long. 
8  ft.  high. 


96^2=48  sq.  ft.,  Ans. 


Ans.  48  sq.  ft. 

Explanation. — lu  all  ^iroblems  of  this  kind, 
■we  multiply  the  leiifjth  by  the  height  which 
gives  the  sq.  ft.  of  a  rectangle  of  equal 
length  and  height.  This  result  is  then  di- 
vided liy  2,  since  a  right-angled  triangle  is 
equal  to  hut  oiie-hal/of  a  rectangle  of  equal 
length  and  height. 

5.    What  is  tlie  area  of  a  triangle  wliose  base  is  40  feet  and  altitude  or  per- 
pendicular height  18  feet?  Aus.  360  sq.  feet. 

TO  FIND  THE  AIIEA  OF  A  TItAPEZOID. 

PKOBLEMS. 

720.     1.     IIow  many  square  feet  in  the  surface  of  a  trapezoid  whose  parallel 
Bides  are  respectively  108  aud  130,  aud  the  perpendicular  distance  between  them 


32  feet? 


OPERATION. 


108 
130 


238  X  32  =  7G16 
7616  -^    2  =  3808  sq.  ft.,  Ans. 


Ans.  3808  sq.  feet. 

Explanation. — In  all  problems  of 
this  kind,  we  multiphj  the  sum  of  the 
parallel  sides  hij  the  })erpendicular 
distance  hetxeeen  them,  and  take  i  of 
the  product. 


2.    What  is  the  area  of  a  trapezoid  whose  parallel  sides  are  30  and  40  feet 
and  whose  i^erjiendicular  is  15  feet?  Ans.  525  sq.  feet. 


TO  FIND  THE  AEEA  OF  A  TRAPEZIUM. 

PROBLEMS. 

721.     1.     How  many  square  feet  in  a  trapezium  whose  diagonal  is  42  feet, 
and  the  perpendiculars  18  feet  aud  16  feet?  Ans.  714  sq.  feet. 


OPERATION. 

42  ft.  length  of  diagonal. 

34  ft.  sum  of  tlie  perpendiculars. 

2)1428  product. 


714  sq.  feet.,  Ans. 


Explanation. — In  all 
prohlenis  of  this  kind, 
\ve  multiply  the  diagonal 
by  the  sum  of  tlie  per- 
pendiculars which  in- 
tersect it  from  opposite 
angles,  aud  take  i  of  the 
product  for  the  area. 


2.    What  is  the  area  of  a  trapezium  whose  diagonal  is  30  feet,   and  the 

perpendiculars  are  respectively  7  feet  6  inches,  aud  5  feet  8  inches? 

Ans.  197J  sq.  feet. 

Note. — To  find  the  area  of  an  irregular  polygon,  divide  the  tigure  into  triangles  and  tra- 
peziums, and  find  the  area  of  each  separately.  The  stun  of  these  areas  will  be  the  area  of  the 
whole  polygon. 


548  soule's  philosophic  practical  mathematics.  * 

TO  FIND  THE  AEEA  OF  A  TRIANGLE. 

EXAMPLE. 

722.    "What  is  the  area  of  a  triaiijfle  whose  base  is  40   feet  and  altitude 
or  perpendicular  heiglit  18  feet  ">.  Aus.  3G0  sq.  feet. 

OPERATION.  Explanation. — In  all  problems  of  this  kind, 

40  ft    leno'th  of  base  ^®  multiply  the  iase  hy  the  altitude,  and  take 

18  ft,  altitude.  one-half  of  the  product,   and  the  result  will 

o  ,   rjf)n  be  the  area. 

360  sq.  ft.,  Ans. 


10  FIND  THE  AEEA  OF  A  SCALENE  TRIANGLE,   WHEN   THE   THREE 

SIDES   ARE   GIVEN. 

723.  1.  How  many  square  feet  are  there  in  a  triangle  whose  sides  are  30, 
40  and  50  feet  ?  Ans.  600  sq.  feet. 

OPERATION. 

60 
50 

20  2d  remainder.        10  3d  remainder. 

60 

60  X  30  X  20  X  10  =  3C0000 

V  360000  =  600  feet.,  Ans. 

Explanation. — In  all  problems  of  this  kind,  we  first  add  the  three  sides  together  and  take 
half  their  sum  ;  then,  from  this  half  sum  we  deduct  each  side  separately ;  then  multiply  together 
the  half  sum  and  each  of  the  three  remainders,  and  extract  the  square  root  of  the  product;  the 
result  will  bo  the  required  area. 

2.  Suppose  the  base  of  a  triangle  to  be  200  feet  and  each  of  the  other  two 
Bides  is  50  feet ;  what  would  be  the  area  ?  Ai.s.  Nothing. 

NoTK.— This  problem  has  puzzled  thousands  of  students.     Does  it  mystify  you  ? 

3.  What  is  the  area  of  aa  equilateral  triangular  figure  each  side  of  which  iS 
40  feet?  Ans.  692.8+  ft. 

OPERATION    INDICATED.  Explanation. — In  all  problems  of  this  kind, 

square  one  side  and  multiply  the  product  by 
40  X   40  =  ICOO  ;  .433.     The  .433  beinR  the  r.atio  between  the 

are.a  of  a  square  and  the  area  of  a,  triangle, 

IGOO  X  .433  =  692.8  +  ft.  =  Area.  ^^^  '"'''  "J^  ^.li'^'^ '"  *"!"=''  *"  """  ^''^'^  "^  ^'^^ 

^  n  square.     See  Table  on  page   3oO. 

To  find  the  area  of  the  face  or  base  of  different  kinds  of  regular  polygons, 
multiply  the  square  of  one  side  by  the  tabular  area  set  opposite  the  polygon  of  the 
same  number  of  sides,  as  shown  in  the  Table  on  page  350. 


30 

60 

60 

40 

30 

40 

50 



— 

30  1st  remainder. 

20 

2  )  120 

*  MENSURATION  OF  SURFACES.  349 

TO    FIND    THE    AREA    OF    A    CIRCLE,    WHEN    THE    DIAMETER    OB 
THE  CIRCUMFERENCE  IS  GIVEN. 

PROBLEMS. 

724.     1.     TVliat  is  the  number  of  square  yards  iu  a  circular  piece  of  ground 
wliicli  is  -0  yards  in  diameter?  Ans.  314. IG  sq.  yds. 

OPERATION.  Kxplartnlion. — In  all  prob- 

lems of  tbis  kind,   we  uiul- 
tiplv  the  (liamettT  by  itself 
20  yds.  diameter  =  lennjtb.  as  is  showuiu  the  operation, 

20  yds.  diameter  =  ■width.  and  theninnltiply  this  pro- 

, duct  by   .7854,    which    has 

.nr,  .1  1       •  1  •   X,  •     nr\      ^      1  been  demonstrated  In  KPoiu- 

400  =  the  sq.  yds.  in  a  square  which  is  20  yds.  long        ^,,y  ^^  ,,^  praHically  the 

and  20  yards  wide.  ratio   between  the   area   of 

400  X  .7854  =  314.1600  sq.  yds.,  Ans.  a   square  and   the  area   of 

a  circle,  the  diameter  of 
which  is  eqnal  to  one  Bide 
of  the  siinare. 

2.  What  is  the  area  of  a  circle  whose  circumference  is  157.08  feet  1 

Ans.  1903.5  sq.  feet. 

OPERATION    INDICATED. 

157.08  -^  3.  1416  =  50  ft.  =  diameter. 

Then  50  x  50  =  2500;  2500  x  .7854  =  1903.5  square  feet  area. 
Or,  if  it  is  preferred,  the  square  of  the  circumference  may  be  imdtiplied  by 
,07958. 

Note. — The  decimal  .7854  is  i  of  3.1416,  and  is  the  area  of  a  circle  whose  diameter  is  1 ;  and 
.07958  is  i  of  (1  —  3.1416);  or,  .07958  is  ^  of  tlie  diameter  of  a  circle  whose  circumference  is  1. 

3.  The  diameter  of  a  circular  garden  is  56.4188  +  feet.     How  many  square 
feet  does  it  contain?  Ans.  2499.99  +  =  practically  2500  sq.  ft. 


TO  FIND  THE  AREA  OF  A  CIRCULAR  RING,  OR  THE  AREA  INCLUDED 

BETWEEN  THE  CIRCUMFERENCES  OF  TWO  CIRCLES 

HAVING  A  COMMON  CENTER. 

PROBLEM. 

725.     What  is  the  area  included  between  two  concentric  circles,  the  diam- 
eters of  which  are  15  and  8  feet?  Ans.  126.4494  sq.  feet. 

OPERATION. 
-,_„         t,r)_  Explanation. — In   all  problems   of 

g2   ^     (J J.  this   kind,    we    multiply    the    difference 

hetKeen  the  squares  of  the  diameters  by 

Difference,  101  t/ie  decimal  .7854,  and  the  product  will 

.7854  ,    J.. 

be  the  area. 


126.4494,    Ans. 


35o 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TO  FIND  THE  AREA  OE  AN  ELLIPSE, 


PROBLEM. 


726.    "WTiat  is  tlie  area  of  an  ellipse  whose  diameters  or  axes  are  20  and 
15  feet?  Ans.  235.62  sq.  feet. 


OPERATION. 

20 
15 

300 

.7854 


Ans. 


Explanation. — In  all  prob- 
lems of  this  kind,  we  multiply 
the  two  diameters  together, 
and  their  product  hy  the  deci- 
mal .7854,  and  the  result  will 
be  the  area. 


235.0200  sq.  feet., 

For  the  operations  to  find  the  entire  surface  of  cylindrical,  triangular,  octag- 
onal, and  other  solids,  see  Mensuration  of  Solids. 

TABLE. 

727.  The  following  Table  gives  the  names  of  the  ten  regular  polygons,  and 
shows  their  areas  when  the  side  of  each  is  equal  to  1.  It  also  shows  the  length  of 
the  radius  of  the  inscribed  circle. 


Xo.  of  Biiles, 

Names. 

Areas. 

Radius  of  inscribed  Circle, 

3 

Triangle, 

0.4330127 

0.2886751 

4 

Square, 

1.0000000 

0.5000000 

5 

Pentagon, 

1.7204774 

0.6881910 

6 

Hexagon, 

2.5980762 

0.8660254 

7 

Heptagon, 

3.6339124 

1.0382617 

8 

Octagou, 

4.8284271 

1.2071068 

9 

Nonagon, 

6.1818242 

1.3737387 

10 

Decagon, 

7.0942088 

1.5388418 

11 

Undecagon, 

9.3656404 

1.2028437 

12 

Duodecagon, 

11.1961524 

1.8660254 

To  illustrate  the  use  of  this  table,  we  present  the  following  problems : 

1.  What  is  the  area  of  an  equilateral  triangle,  whose  sides  are  40  inches  ? 

Ans.  692.82033  sq.  in. 

OPERATION.  Explanation. — In  all  problems  of  this 

kind,  square  one  of  the   sides   of  tho 

40  X  40  =  1600;  1600  x  .4330127  =  692.8203200     {^S.lnr'^^eSlliJr.l^t'^qll;^ 

the  named  polygon  in  the  above  table. 
Note. — This  operation  is  based  ujion  the  principle  that  the  areas  of  similar  polygons  are  to 
each  other  as  tho  squares  of  their  like  sides. 

2.  What  is  the  area  of  an  octagonal  polygon  one  side  of  which  is  30  inchest 

Ans.  4345.58439  sq.  in. 
Operation  indicated.     ZO^  X  4.8284271. 

3.  The  side  of  a  decagon  is  20  J  feet.    What  is  its  area  I 

Ans.  3233.4912  -f  sq.  feet 
Operation  indicated.     (20 J)  =  x  7.6942088. 


MENSURATION    OF    SURFACES, 


35i 


TO  FIND  THE   AEEA   OF  AX    IRREGULAR    POLYGON    OF    FOUR    OR 

MORE  SIDES. 

728.  1.  Divide  tlie  figure  into  triangles  by  diagonals  connecting  some  one 
angular  point  with  each  of  the  others.  2.  Compute  the  area  of  each  triangle.  3. 
Add  their  respective  areas  and  the  sum  will  be  the  area  of  the  polygon. 


» 


TO  FIND  THE  CO]SrVEX  SURFACE  OF  A  PRISM  OR  A  CYLINDER. 


PROBLEMS. 

729.     1.    What  is  the  area  of  the  convex  surface  of  a 
pentagonal  jjrism  whose  altitude  is  8  ft.  and  each  side  4  ft.  ? 

Ans.  IGO  square  feet. 

OPERATION. 

4  ft.  X  5  =  20  ft.  perimeter,  20  ft.  x  8  =  IGO  sq.  ft. 

ETpJnnation. — In  all  jiroblems  of  this  kiuil,  multiply  the  perimeter  of 
the  base  by  the  altitude. 


I 


2.  "WTiat  is  the  area  of  the  convex  surface  of  a  trian- 
gular prism  whose  altitude  is  6J  feet,  and  the  sides  of  its  base 
are  3,  5,  and  7  feet  respectively  'I  Ans.  97^  sq.  ft. 

OPERATION. 

3  ft.  +  5  ft.  +  7  ft.  =  15  ft.  perimeter. 

15  ft.  X  C.J  ft.  =  97J  sq.  ft.  Ans. 

3.  What  is  the  area  of  the  convex  surface  of  a  cylinder, 
whose  altitude  is  i  feet  3  inches,  and  the  circumference  of  its 
base  is  5  ft.  6  in. 

OPERATION. 

4  ft.  3  in.  =  51  in. ;  5  f t  0  in.  =  GG  in. ;  G6  in.  x  51  =  33G6 

sq.  in.  equals  23  sq.  ft.  44  sq.  in.  convex  surface. 


Note. — To  find  the  entire  surface,  add  the  area  of  the  bases  or  ends  to  the  couvex  surface. 


^52  soule's  philosophic  practical  mathematics.  * 

TO  FIND  THE  AREA    OF    A    SEGMENT   OF    A    CIRCLE    WHEK    THE 
DIAMETER  AND  THE  HEIGHT  OK  VERSED  SINE  ARE  GIVEN. 

PROBLEM. 

730.    What  is  the  area  of  a  segmeut  of  a  circle  whose  diameter  is  72  inches 
and  the  versed  sine  is  12  inches?  Aus.  440.0313  +  sq.  in.,  area. 


OPERATION 
BY  THE   VEUSED   SINE   METHOD. 

Versed  sine  12  in.  ■-■  the  diameter,  72  in.  = 
.166G|  =  quotient  of  versed  sine. 


D 

Find  the  quotient  of  .106  in  the  cohimn  of  versed  sines,  in  tlie  following 
table ;  then  take  the  area  number  which  is  in  the  next  column  to  the  right,  and 
multiply  it  by  the  square  of  the  diameter,  the  result  will  be  the  area  of  the  segment 
ADB. 

The  tabular  number  is  .08554  x  72-  =  443.4393  +  sq.  in.,  area  of  segment. 
As  the  quotient  of  the  versed  sine  by  the  diameter  was  .lOCg  instead  of  .100,  the 
area  would  be  nearer  accurate  to  use  as  a  multiplier  the  number  opposite  .107  in 
the  table.  This  number  is  .08029;  and  this  x  72=  =  447.3273  +  sq.  in.,  area  of 
segment. 

For  greater  accuracy  we  would  proceed  as  follows : 

.100  =  .08554 
.107  =  .08029 


Difierence,  .00075  x  .Og  =  .00050  +  .08554  =  .08004,  the  sum  by  which  the 
square  of  the  diameter  is  to  be  midtiplied. 

Then.  .08004  x  72^  =  440.0313  +  sq.  in.,  area  of  segment. 

Note  1. — To  find  the  area  of  a  segment  greater  than  a  semi-circle,  find  the  area  of  the  lesser 
segment  and  subtract  the  same  from  the  area  of  the  whole  circle. 

Note  2. — To  find  the  area  of  a  zone  of  a  circle,  find  the  areas  of  the   two  segments   and 
subtract  the  sum  from  the  area  of  the  whole  circle. 

Note  3. — To  find  the  chord  of  a  segment  of  a  circle  when  the  diameter  and  versed  sine  are 
given,  see  Article  734,  page  356,  also  Article  807,  page  398. 


*  MENSURATION    OF    SURFACES.  353 

731.      TABLE  OF  THE  AREAS  OF  THE  SEGMENTS  OF  A  CIRCLE. 


THE    DIAMETER    OF    A    CIRCLE  ASSUMED    TO    BE    UNITY,  AND    DIVIDED   INTO   ONE 

THOUSAND  EQUAL  PARTS. 


Versed 

aiue. 

Seg.  Area. 

1  Versed 
1  Sine. 

Seg.  Area. 

Versed 
Sine. 

Seg.  Area. 

.001 

.00004 

.061 

.01972 

.121 

.05404 

.002 

.00012 

.062 

.0202 

.122 

.05ni9 

.003 

.00022 

.063 

.02U08 

.123 

.05.534 

.004 

.00034 

.064 

.02117 

.124 

.056 

.005 

.00047 

.005 

.02165 

.125 

.05606 

.006 

.00002 

.066 

.02215 

.126 

.05733 

.007 

.000/8 

.067 

.02265 

.127 

.05799 

.008 

.00095 

.068 

.02315 

.128 

.05.S66 

.009 

.00113 

.069 

.02336 

.129 

.05933 

.01 

.00133 

.07 

.02417 

.13 

.06 

.011 

.00153 

.071 

.02408 

.131 

.06067 

.012 

.00175 

.072 

.Oi'519 

.132 

.06135 

.013 

.00197 

.073 

.112571 

.133 

.00203 

.014 

.0022 

.074 

.02624 

.134 

.00271 

.015 

.00244 

.075 

.02076 

.135 

.06339 

.016 

.00268 

.076 

.02729 

.136 

.06407 

.017 

.00294 

.077 

,02782 

.137 

.0()476 

.018 

.0032 

.078 

.(12835 

.138 

.06545 

.019 

.00347 

.079 

.02889 

.139 

.00614 

.02 

.00375 

.08 

.02943 

.14 

.061)83 

.021 

.00403 

.081 

.02997 

.141 

.0()7.53 

.022 

.00432 

.082 

.03052 

.112 

.0(W22 

.023 

.00462 

.083 

.03107 

.143 

.Oi;,S92 

.024 

.00492 

.084 

.03162 

.144 

.()09i;2 

.025 

.00523 

.085 

.03218 

.145 

.07(133 

-026 

.00555 

.086 

.03274 

.146 

.07103 

.027 

.005^7 

.087 

.0333 

.147 

.07174 

.028 

.00019 

.088 

.033S7 

.148 

.07245 

.029 

.00053 

.089 

.03144 

.149 

.07316 

-03 

.00086 

.09 

.03501 

.15 

.07387 

.031 

.00721 

.091 

.03558 

.151 

.07459 

.032 

.011/56 

.092 

.03016 

.1.52 

.07531 

.033 

.■11)791 

.093 

.O3071 

.1.53 

.070(13 

.034 

.1)0.^27 

.094 

.03732 

.154 

.(17675 

.035 

.0I1S64 

.095 

.0379 

.1.55 

.07747 

.036 

.00901 

.096 

.(I3S19 

.1.56 

.0782 

.037 

.00938 

.097 

.(13908 

.157 

.07892 

.038 

.00976 

.098 

. 03908 

.1.58 

.07965 

.039 

.01015 

.099 

.(J4(i27 

.1.59 

.0S(I3.S 

.04 

.01054 

.1 

.0I0S7 

.16 

.OSIU 

.041 

.0l0!t3 

.101 

.04118 

.161 

.081K5 

.042 

.01133 

.102 

.04208 

.102 

.0S258 

.043 

.01173 

.103 

.042{i9 

.163 

.0S332 

.044 

.01214 

.104 

.0431 

.164 

.0S406 

.045 

.01255 

.105 

.04391 

.165 

.0848 

.046 

.01297 

.106 

.044.52 

.166 

,().X554 

-047 

.01339 

.107 

.04514 

.167 

.(iN(;_':i 

-048 

.013,s2 

.108 

.04575 

.108 

.087(11 

-049 

.01125 

.109 

.04638 

.169 

.(I,S779 

-05 

.01408 

.11 

,047 

.17 

.0S,X53 

.051 

.01512 

.111 

.04763 

.171 

.0s92il 

-052 

.01.5.56 

.112 

.04826 

.172 

.0: 1(1(14 

.053 

.01001 

.113 

.04889 

.173 

.0908 

-054 

.OIOIG 

.114 

.04953 

.174 

.091.55 

-055 

.01091 

.115 

.0.5016 

.175 

.09231 

.056 

.01737 

.116 

.0508 

.176 

.09307 

.057 

.01783 

.117 

.05145 

.177 

.09384 

-.058 

.0183 

.118 

.05209 

.178 

.0946 

.059 

.01877 

.119 

.05274 

.179 

.09537 

.06 

.01924 

.12 

.05338 

.18 

.09613 

Versed 
Siuo. 

.181 

Seg.  Area. 
.091)9 

Versed 
Siue. 

.241 

Seg.  Area. 

.1458 

.182 

.09707 

.242 

.14665 

.1X3 

.09S15 

.243 

.14751 

.184 

.09922 

.244 

.14837 

.i;<5 

.1 

.245 

.14923 

.1^6 

.10077 

.246 

.15009 

.187 

.1(11.55 

.247 

.15095 

.188 

.10233 

.248 

.15182 

.1X9 

.10312 

.249 

. 1 5268 

.19 

.1039 

.25 

.15355 

.191 

.10468 

.251 

.15441 

.192 

.10547 

.252 

.15528 

.193 

.10020 

.2.53 

.15615 

.194 

.10705 

.254 

.15702 

.195 

.10784 

.255 

.  1 5789 

.196 

.1(1864 

.2.56 

.15876 

.197 

.10943 

.257 

.15964 

.198 

.11023 

.258 

.16051 

.199 

.11102 

.259 

.16139 

2 

.11182 

.26 

.  1 6226 

.201 

.11262 

.261 

,16314 

.202 

.11343 

.262 

,  1 6102 

.203 

.11423 

.263 

.1649 

.204 

.11.503 

.264 

.16578 

.205 

.11584 

,265 

.  1  6666 

.206 

.11005 

.206 

.  1  6755 

.207 

.11746 

.207 

,  1 6844 

.208 

.11X27 

.21)8 

. 1 6931 

.209 

.11908 

.269 

.1702 

.21 

.1199 

.27 

.17109 

.211 

.12071 

.271 

.17197 

.212 

.12153 

.272 

. 1 7287 

.213 

.12235 

.273 

.17376 

.214 

.12317 

.274 

.17465 

.215 

.  1 23;i9 

.275 

.175.54 

.216 

.12-1X1 

.276 

.17643 

.217 

.12.5»;3 

.277 

,17733 

.218 

.12046 

.278  , 

.17822 

.219 

.12728 

.279 

.17912 

'>j 

.12811 

.28 

.18002 

.221 

.12894 

.281 

.18092 

.222 

.12977 

.282 

.18182 

!223 

.1306 

.283 

.18272 

.224 

.13144 

.284 

.18361 

.225 

.13227 

.285 

.18452 

.226 

.13311 

.286 

.18542 

.227 

.133!  14 

.287 

.18633 

.22.S 

.13178 

.288 

.18723 

.229 

.13502 

.289 

.18814 

.23 

.13046 

.29 

.18905 

.231 

.13731 

.291 

.18995 

.232 

.13815 

.292 

.19086 

.233 

.139 

!293 

.19177 

.234 

.13984 

.294 

.19268 

.235 

.14069 

.295 

.1936 

.236 

.14154 

.296 

.19451 

.237 

.14239 

.297 

.19543 

.238 

.14324 

.298 

.19634 

.239 

.14409 

.299 

.19725 

.24 

.14494 

.3 

.19817 

3^4 


souLE  s  rniLosoinnc  practical  mathematics. 


TABLE  OF  THE  AREAS  OF  THE  SEGMENTS  OP  A  ClVlQUSs— Continued. 


Vorsed 
Sine. 

Seg.  Area. 

Versed 
Siue. 

Seg.  Area. 

Versed 
Siue. 

Seg.  Area. 

Versed 
Siue. 

Seg.  Area. 

Versed 
Sine. 

Seg.  Area. 

.301 

.19908 

.341 

.23642 

.381 

.27483 

.421 

.31403 

.461 

.35374 

.802 

.2 

.342 

.23737 

.382 

.27580 

.422 

.31502 

.462 

.35474 

.:«)n 

.20092 

.343 

.23832 

.383 

.27677 

.423 

.316 

.463 

.35573 

.304 

.20184 

.344 

.23927 

.384 

.27775 

.424 

.31699 

.464 

.35673 

.305 

.20276 

.345 

.24022 

.385 

.27872 

.425 

.31798 

.465 

.35773 

.306 

.20368 

.346 

.24117 

.386 

.27969 

.426 

.31897 

.466 

.35873 

.307 

.2046 

.347 

.24212 

.387 

.28067 

.427 

.31996 

.467 

.3.5973 

.308 

.20553 

.348 

!24307 

.388 

.28164 

.428 

.32095 

.468 

.36073 

.309 

.20645 

.349 

.24403 

.389 

.28262 

.429 

.32194 

.469 

.36173. 

.31 

.20738 

.35 

.24498 

.,39 

.28359 

.43 

.32293 

.47 

.36273 

.311 

.2083 

.351 

.24593 

..391 

.28457 

.431 

.32391 

.471 

.36371 

.312 

.20923 

.352 

.24689 

.392 

.28554 

.432 

.3249 

.472 

.36471 

.313 

.21015 

.353 

.24784 

.393 

.28652 

.433 

.3259 

.473 

.36571 

.314 

.21108 

.354 

.2488 

.394 

.2875 

.434 

.32689 

.474 

.36671 

.315 

.21201 

.355 

.24976 

.395 

.28848 

.435 

.32788 

.475 

.36771 

.316 

.21294 

.356 

.25071 

.,396 

.28945 

.436 

.32887 

.476 

.36871 

.317 

.21387 

.357 

.25167 

.397 

.29043 

.437 

.32987 

.477 

.36971 

.318 

.2148 

.358 

.25263 

.398 

.29141 

.438 

.33086 

.478 

.37071 

.319 

.21573 

.359 

.25359 

.399 

.29239 

.439 

.33185 

.479 

.3717 

.32 

.21667 

.36 

.25455 

.4 

.29337 

.44 

.33284 

.48 

.3727 

.321 

.2176 

.361 

.25551 

.401 

.29435 

.441 

.33384 

.481 

.3737 

.322 

.21853 

.362 

.25047 

.402 

.29533 

.442 

.33483 

.482 

.3747 

.323 

.21947 

.363 

.25743 

.403 

.29631 

.443 

.33582 

.483 

.3757 

.324 

.2204 

.364 

.25839 

.404 

.29729 

.444 

.33682 

.484 

.3767- 

.325 

.22134 

.365 

.25936 

.405 

.29827 

.445 

.33781 

.485 

.3777 

.326 

.22228 

.366 

.26032 

.406 

.29925 

.446 

.3388 

.486 

.3787 

.327 

.22321 

.367 

.26128 

.407 

.30024 

.447 

.3398 

.487 

.3797 

.328 

.22415 

.368 

.26225 

.408 

.30122 

.448 

.34079 

.488 

.3807 

.329 

.22509 

.369 

.26321 

.409 

.3022 

.449 

.34179 

.489 

.3817 

.33 

.22603 

.37 

.26418 

.41 

.30319 

.45 

.34278 

.49 

.3827 

.331 

.22697 

.371 

.26514 

.411 

.30417 

.451 

.34378 

.491 

.3837 

.332 

.22791 

.372 

.26611 

.412 

.30515 

.452 

.34477 

.492 

.3847 

.333 

.22886 

.373 

.26708 

.413 

.30614 

.453 

.34557 

.493 

.3857 

.334 

.2298 

.374 

.26804 

.414 

.30712 

.454 

-.34676 

.494 

.3867 

.335 

.23074 

.375 

.26901 

.415 

.30811 

.455 

.34776 

.495 

.3877 

.336 

.23169 

.376 

.26998 

.416 

.30909 

.456 

.34875 

.496 

.3887 

.337 

.23-63 

.377 

.27095 

.417 

.31008 

.457 

.34975 

.497 

.3897 

.338 

.23358 

.378 

.27192 

.418 

.31107 

.458 

.35075 

.498 

.3907 

.339 

.23453 

.379 

.27289 

.419 

.31205 

.459 

.35174 

.499 

.3917 

.34 

.23547 

.38 

.27386 

.42 

.31304 

.46 

.35274 

.5 

.3927 

MENSURATION    OF    SURFACES. 


t55 


TO   FIND   THE    AREA    OF    A    SEGMENT    OF    A    CIRCLE    WHEN    THE 
CHORD  AND  THE  HEIGHT  OR   VERSED  SINE  ARE  GIVEN. 

PROBLEM. 

732.     What  is  the  area  of  a  segment,  the  chord  of  which  is  9  ft.  3  in.  and 
the  height  or  versed  sine  is  4  ft.  ? 

Aus.  •±036.9828  +  sq.  iu.,  area  of  the  segment  A  D  B. 

OPERATION. 

111'" 


D  D 

First  Step  to  find  the  diameter  of  the  circle  of  zchich  the  given  segment  is  a  part. 


A  B  =  111    in.  =  whole  chord. 
A  P  =    5r»i  in.  =  J  chord. 
P  D  =    4S'  iu. 


(55J)-  -4-  48  =  64.17  +  in.  difference  between 

diameter  and  versed  sine. 
64.17  +  48  =  112.17  in.  diameter  of  circle. 
112.17  -^  2  =  oG.085  in.  radius  A  C. 


Second  Step  to  find  the  quotient  of  the  versed  sine  divided  by  the  diameter. 

48  ill.  versed  sine  4-  the  diameter,  112.17  =  .4270,  quotient. 
The  tabular  number  fur        .427  =  .M'Mii 
"       "  "         "  .428  =  .32095 


Difference  is,  .00009 

Then,  .00099  x  .9,  the  difference  between  .4279  and  .427,  =  .000891.  Hence 
.31990  +  .000891  =  .320851,  the  sum  by  wliich  the  square  of  the  diameter  is  to  be 
multiplied. 

.320851  X  112.17=  =  403C.9828  +  sq.  in.,  area  of  the  segment  A  D  B. 


A  SHORT  APPROXIMATE    METHOD  TO  FIND  THE   AREA  OF  A   SEG- 
MENT OF  A  CIRCLE,  WITHOUT  THE  VERSED  SINE  TABLE. 

PROBLEM. 

733.     VThat  is  the  area  of  a  segment  of  a  circle,  the  chord  of  the  segment 
being  24  feet  and  the  height  or  versed  sine  of  the  segment  6  feet? 

Ans.  lOOi  sq.  feet. 


21  feet. 


OPERATION. 


4  X  24  X  6 

„ii    3x6' 
add,  -^j— 

divide  by 


area 


=    576 

G)  003 

100.5  sq.  ft 


E.rplanniion. — In  all  problems 
of  this  kind,  in  •n-hich  the  seg- 
meut  is  lesa  than  a  Bemi-circle, 
■vre  proceed  as  follows :  To  4 
times  the  product  of  the  chord 
by  the  versed  siue.  add  the  quo- 
tieiit  of  3  times  tlio  cube  of  the 
versed  sine  divided  l)y  tlio  chord, 
and  then  divide  the  sum  by  6. 
Or,  if  preferred,  -we  may  perform 
the  operation  as  follows  :    To  1 


356 


soule's  philosophic  practical  mathematics. 


times  the  square  of  the  chord,  add  3  times  the  square  of  the  versed  sine;  then  multiply  the  sum 
by  the  versed  sine  and  divide  the  product  by  6  times  the  chord.     Thus  : 

24=  X  4  =  2304 
6-  X  3  =    106  2412  X  6    =  14472. 

14472  -H  (6  X  24)  =  lOOJ 

2412 

ADVANTAGE  OF  THIS  METHOD. 

The  advantage  of  this  method  is  that  it  can  be  used  without  the  table  of 
versed  sines.  The  result  produced  is  slightly  different  from  that  ])roduced  by 
the  use  of  the  versed  sine  tabuhir  method  as  shown  above,  but  for  most  practical 
purposes,  it  may  be  used  when  the  versed  sine  table  is  not  at  hand. 


OPEEATION   OF     THE 


24  feet. 


ABOVE     PROBLEM    BY 
TABLE. 


THE    USE     OF     THE     VERSED     SINE 


24  ft.  length  of  chord  -f-  2  =  12  ft.,  J  of  chord. 
12^  -1.  6  ft.  versed  sine  =  24  ft.,  diiference  be- 
tween the  diameter  and  versed  sine. 
24  +  C  =  30  ft.,  diameter,  6  4-  30  =  .2  quotient. 
Tabuhir  number  for  .2  is  .111823. 
.111823  X  30==  100.G407  sq.  ft.,  area  of  segment. 


TO  FIND   THE    CHORD   OF   A    SEGMENT   OF   A   CIECLE    WHEN   THE 
DIAiMETEK   AND  VERSED  SINE  ARE  GIVEN. 

PROBLEM. 

734.    What  is  the  chord  of  a  segment,  of  a  circle,  the  diameter  of  which  is 
72  inches  and  the  versed  sine  12  inches  ?  Ans.  53.CG4  in.,  length  of  chord. 

OPERATION. 

First  Step. — Diameter  72  in.  —  versed  sine  12  in.  =  CO  x  versed  sine,  12  in.  =  720, 

Second  Stej). 

■/720  =  26.832  iu.  =  J  of  chord. 
4  2 


4")  320  =   53.6G4  in.  chord  of  segment. 
276 


52^)  4400 
4224 


636=) 17600 
160S9 


5366=) 151100 
107324 
Explanation. — In  all  operations  of  this  kind,  first  subtract  the  versed  sine  from  the  diameter 
and  multiply  the  remainder  by  the  versed  sine  ;  then  extract  the  square  root  of  the  product  and 
multiply  the  quotient  by  2;  the  result  will  be  the  length  of  the  chord  of  the  segment. 


Ji^ 


MENSURATION    OF    SURFACES. 


}57 


TO  FIND  THE  RADIUS  OF  THE   INSCRIBED  CIRCLE   OF   A  REGULAR 

rOLYGOX. 

TROBLEMS. 

735.     1.    What  is  tlie  radius  of  an  inscribed  circle  on   a  hexagon,   whose 
eides  are  each  50  inches?  Aus.  43.301L'7  radius  of  circle. 


OPEKATION. 

50  X  .8660254  =  43.3012700 


Explanation. — In  all  problems  of  this  kiiul, 
multiply  one  side  of  the  pol.vf;oii  by  the  tabular 
radius  number  set  opposite  the  named  polygon 
in  the  radius  column  in  the  table  of  areas,  on 
page   350. 


2.    The  respective  sides  of  a  nonagon  are  3  feet.    "What  is  the  diameter  of 
the  inscribed  circle  ?  Aus.  8.2424  +  feet.    Diameter  of  circle. 

OPEEATION. 

3  X  1.3737387  =  4.1212161  =  radius. 
4.1212161  X  2  =  8.2424322  diameter. 


I 


PRACTICAL,  PROBLEMS  IN  MENSURATION  OF  SURFACES. 

PROBLEMS. 

736.  1.  How  many  square  rods  in  a  plat  of  ground,  the  length  of  -which  is 
200  rods  and  the  width  60 1  Ans.  12000  sq.  rods. 

2.  How  many  square  feet  in  a  floor  16  feet  8  inches  long,  and  10  feet  3  inches 
wide  ?  Aus.  170  J  sq.  feet. 

OPERATION. 


50 

41 


Or, 


12 
12 


200 

123 


170 1  sq.  feet.,  Aus. 


170|  sq.  feet.,  Ans. 

3.  How  many  square  yards  in  a  floor  60  feet  long  and  40  feet  10  inches  wide  ? 

Ans.  272|  sq.  yards. 

OPERATION.         60  X  40|  =  2450  sq.  ft. ;  2450  -^  9  =  272|  sq.  yds. 

4.  A  water  pipe  is  50  feet  9  inches  long,  and  its  diameter  is  30  inches.    What 
is  its  concave  surface  ?  Ans.  57397.032  sq.  inches. 

OPERATION. 

30  X  3.1416  =  94.248  circumference,  or  linear  width  of  pipe. 
94.248  X  009  in.,  length  =  57397.032  sq.  in.,  Ans. 

5.  How  many  square  yards  of  plasterins  in  one  wall  of  a  house  32  feet  4 
inches  long  and  15  feet* 3  inches  high  ?  Ans.  M-^^^  sq.  yds. 


358  soule's  philosophic  practical  mathematics.  * 

6.  How  many  square  feet  in  the  bottom  of  a  cistern  •nhose  diameter  is  11 
feiet  8  inches?  Aus.  10G.901G§  sq.  ft. 

7.  Wlaat  is  tlie  area  of  the  base  of  a  cylinder  whose  circumference  is  62.832 
inches?  Ans.  3U.1G  sq.  in. 

S.     A  lumberman  has  25  boards,  each  14  feet  long  and  15  inches  wide.    How 
many  square  feet  iu  all  ?  Ans.  437^  sq.  ft. 

9.    Find  the  area  in  square  feet  of  a  plank  22  feet  3  inches  long,  2J  feet  wide 
at  one  end  and  14  inches  wide  at  the  other  1  Aus.  38  ij  sq.  ft. 

10.  Find  the  area  of  a  piece  of  ground  which  is  406  feet  long  and  is  of  the 
following  width  at  different  points,  equally  distant  from  each  other:  at  the  wider 
end,  210  feet;  at  the  narrower  end,  165  feet;  near  the  wider  end,  ISO  feet;  near  the 
narrower  end,  142  feet;  in  the  middle,  300  feet  ?  Ans.  82164J  sq.  ft. 

Note. — Draw  a  diagram  of  the  ground  before  working  the  iiroblem. 

OPERATION   indicated 
TO  FIND   THE  AVERAGE   WIDTH. 

210  +  165  =  187J,  +  180  +  142  +  300  =  202|  ft.  average  width. 


11.  A  steamboat  boiler  is  45  feet  long  and  50  inches  in  diameter ;  it  has  3 
flues,  each  10  inches,  iu  its  exterior  diameter;  how  many  square  inches  does  the 
boiler  contain  ou  its  entire  interior  surface  1  Ans.  88278.96  sq.  in. 


OPERATION 
TO   FIND   THE   CONCAVE   SURFACE   OF  THE    BOILER. 


50  X  3.1416  =  157.08  in.,  circumference. 
540  in.,  length. 


84823.20  sq.  in.,  in  concave  surface. 
3455.76  area  of  ends. 


88278.90  sq.  in.,  the  entire  interior  surface  of  the  boiler. 


OPERATION 
TO  FIND  THE   SURFACE   OF    THE   2   ENDS. 

50=    =   2500 

2500  X  .7854  =  1963.50 


Area  of  2  ends,  ....  3927.00 

Area  of  the  6  ends  of  3  flues,  471.24 


OPERATION 

TO  FIND  THE  SURFACE  OF  THE  6  ENDS  OP  THE 
3  FLUES. 

10=    =   100 

100  X  .7854  =  78.54 
6 


Area  of  the  6  ends,       .      .      -       471.24 


Area  of  2  ends,  ....  3455.76 

12.     If  a  man  occupies  a  space  20  inches  by  30  inches,  how  many  men  may 
stand  in  a  semicircular  space  which  is  200  feet  in  diameter  1 

Ans.  3769  men  with  552  sq.  in.,  unoccupied. 


MENSURATION    OF    SURFACES. 


,^■'9 


4 

3  3 

4    '  [       4 

3  3 

4 


13.  A  piece  of  board  is  IC  inches  long,  9  iuclies  vide.  Ilo-«'  can  it  be  cut  into 
two  pieces  so  that  by  placing  tliem  togetlier  iu  auotlier  foria  tliey  will  make  a  square 
12  iuclies  on  each  side  ? 

Answer.  1.  Commence 
as  shown  in  tlie  first  fig- 
ure 4  inches  from  either 
end,  and  saw  dowu  3  in. 
2.  Then  at  a  right  an- 
gle from  this  point,  saw 
toward  the  longer  end 
4  inches.  3.  Then  at  a 
right  angle  from  this 
point,  saw  down  3  in- 
ches. 4.  At  a  right  angle  from  this  point  saw  4  inches  iu  the  same  direction  as  the 
first  4  inches  were  sawed.  5.  Lastly  from  this  point  and  at  right  angles  with  the 
last  line,  saw  3  inches,  through  the  board,  and  arrauge  the  pieces  as  shown  iu  the 
second  figure. 

Note. — la  order  to  saw  the  board  as  above  described,  it  would  be  necessary  to  bore  holes  at 
the  four  angles,  -nhere  the  changes  of  direction  were  made.  These  holes  would  be  jjlugged  when 
the  pieces  were  placed  iu  the  square  form. 

14.     A  circular  jiiece  of  ground  contains  10  acres ;  what  is 
its  diameter  in  rods,  and  feet? 

Ans.  45.135  +  diameter  iu  rods.     744.7275  +  feet. 

OPERATION    INDICATED. 

10  X  IGO  =  1000  =  the  sq.  rods  iu  10  acres;  1000  -4-  .7854  =  2037.1785  = 
the  sq.  rods  iu  a  square  oue  side  of  which  is  equal  to  the  diameter  of  the  10  acre 
circle ;  V  :2037.17S5  =  45.135  +  rods  =  one  side  of  the  square  aud  the  diameter  of 
the  circle ;  45.135  x  10^  =  744.7275  +  feet. 


15.  A  gardener  wishes  to  lay  out  a  circidar  gardeu  that 
will  contain  oue  acre  of  ground ;  how  many  feet  long  must  be  a 
rope  that,  fastened  by  oue  end  to  a  stake  iu  the  ceuter  of  the 
garden,  will  enable  him  to  describe  the  oue  acre  ? 

Ans.  117.75  +  feet. 


OPERATION    INDICATED. 


1  acre  contains  43560  sq.  feet;  435G0 -4-  .7854  =  55402.1848+  ;  v'554C2.1848-l- 
=235.50  ft.,  diameter;  235.50+  -^  2  =  117.75  +  ft.  =  radius  or  length  of  rope. 

16.    Cheops,  the  largest  of  the  Egyptian  pyramids,  is  square  at  its  base  and 
measures  764  feet  on  each  side;  how  many  acres  of  ground  does  it  cover? 

Ans.  13  acres  63  sq.  rds.  29J  sq.  yds.  1  sq.  ft. 


360  soule's  philosophic  practical  mathematics.  * 

operation  indicated. 

764  X  764  =  583696  sq.  ft. ;  583696  sq.  ft.  -^  9  =  64S55  sq.  yds.  and  1  sq.  ft. ; 
64855  sq.  yds.  ~  30^  =  -'143  sq.  rods  and  29^  sq.  yds. ;  2143  sq.  rods  ■-  160  =  13A. 
and  63  sq.  rods. 

17.  A  modern  stock  raiser  lias  a  cow  that  lie  wislies  to  feed  npon  2500  sq.  ft. 
of  pasture  i)er  day.  If  lie  ties  tlie  cow  by  tlie  tail  witli  a  rope,  and  if  tlie  head  and 
body  of  the  cow  be  7J  ft.  long  and  the  tail  to  the  point  where  it  is  tied  be  3|  feet 
long,  what  must  be  the  length  of  the  rope  to  give  the  cow  the  desired  feet  of  past- 
urage, no  allowance  being  made  for  the  curve  in  the  line  of  measure,  ou  account  of 
the  height  of  the  cow  ?  Ans.  17.2094  +  feet. 

OPERATION    INDICATED. 


2500  -^  .7854  =  3183.0914+  j  V  31S3.0914  =  50.4188  -^  2  =  28.2094  —  (7J  + 
3J)  11  =  17.2094. 


Artificers'  or  Mechanics'  Work,  in  Surfaces. 


PAVING  YAEDS  AND  WALKS. 

737.    1.     How  many  bricks  will  be  required  to  pave  a  sidewalk  64  feet  long 
and  11  feet  8  inches  wide,  each  brick  being  8  inches  long  and  4  inches  wide? 

Ans.  3360  bricks, 

OPERATION    INDICATED. 


Length  of  brick  =    8 
Width    "       »     =    4 


708    =  length  of  sidewalk  in  inches. 
140    =  width  of  sidewalk  in  inches. 


3360  =  number  of  bricks  to  pave  the  sidewalk. 


Explanation. — In  all  problems  of  this  kind,  -we  first  make  the  statement  to  ascertain  the 
number  of  square  inches  in  the  sidewalk,  by  multiplying  the  length  by  the  width  in  the  unit  of 
inches ;  and  then  we  divide  this  result  by  the  product  of  the  length  and  the  width  of  a  brick. 

2.    Ill  the  above  problem,  how  many  square  yards  in  the  sidewalk,  and  M'hat 
would  be  the  cost  of  the  paving  at  OOf  per  square  yard  ? 

Ans,  825f  square  yards 


$74.66  +  cost. 


OPERATION   TO   FIND   SQUARE    YARDS. 

64 


35 


S2f  s  square  yards  at  90/  =  $74.06§. 


*  artificers'  or  mechanics    work,  in  surfaces.  361 

3.  How  many  Geriiiau  flags,  each  IG  in.  by  16  in.,  will  it  take  to  pave  a 
yard  45  feet  square?  Ans.  llSQ-jii  flags. 

4.  A  yard  is  24  feet  3  inches  long  by  11  feet  5  inches  wide ;  how  many  bricks, 
4  by  8  inches,  will  it  take  to  pave  it,  no  allowance  to  be  made  for  the  openings 
between  the  bricks  ?  Aus.  12453  J  bricks. 

5.  A  circular  court  is  75  feet  in  diameter;  in  the  center  there  is  a  fountain 
15  feet  in  diameter;  how  many  tiles,  each  G  inches  square,  will  it  require  to  tile  the 
court?  Ans.  109G4.G4  tiles. 

OPERATION. 

75=  =  5025  sq.  of  the  diameter  of  court.  Explanaiion.—yVe  first  findtlio  dif- 

152  ^     225    "       "      "  "  "     fountaiu.  ference   between   the    squares    of    the 

"^  *  diameters  of  the  court   and   fountain ; 

■we   then   malti]>ly   this   difference    jy 


5400  difference.  .7854,    and    produce    tlie    number    of 

.7854  square  feet  in  the  area  of  the  ring;  we 

j^j.  then  reduce  this  to  inches  and  divide 

the  same  by  the  number  of  inches  in  a 


169G4.64,  Aus. 


tile. 


Note. — In  actual  practice,  by  reason  of  the  waste  in  making  the  circle,  many  more   tiles 
would  be  required  than  are  produced  by  this  operation. 

6.  A  circular  court  is  30  feet  in  diameter.  How  many  tiles,  each  G  inches 
square,  will  it  require  to  cover  the  court,  making  110  allowance  for  waste? 

Ans.  2827.44  tiles. 

7.  What  will  it  cost  to  pave  a  yard  that  is  140  ft.  long,  34  feet  4  inches 
wide,  with  shillinger  pavement,  at  $3.25  per  square  yard  ?  Ans.  $1735.74/.;. 

8.  A  contract  is  made  with  a  company  to  pave  a  street  with  rosetta  pave- 
ment and  to  curb  the  same  with  4x18  inch  granite.  The  price  is  fixed  at  $14  per 
square  for  the  pavement,  and  $1.10  per  running  foot  for  the  curbing.  "What  will  be 
the  cost  for  paving  one  square  or  block  which  is  320  feet  long,  the  street  being  42 
feet  6  inches  wide  1  Ans.  $1904  for  paving.     $704  for  curbing. 

9.  How  many  sods,  12  by  16  inches  will  it  require  to  sod  a  yard  40  ft.  long 
and  28  ft.  4  inches  wide  1  Ans.  850  sods. 


FEAMING,  BUILDING,  LAYING  FLOOES,   WAINSCOTING,  ETC. 

738.  The  work  of  carpenters  and  joiners  is  done  by  the  lineal  foot,  the 
square  foot,  the  square  yard,  or  by  the  square  of  100  square  feet. 

Framing  the  large  timbers  used  in  building,  such  as  sills,  posts,  principal 
rafters,  etc.,  the  framing  of  hips  and  valleys,  and  the  work  of  chamfering  posts  and 
girders,  raising  plates  and  bond  timbers,  bridging  joists,  making  mouldings,  archi- 
traves, etc.,  are  counted  by  the  lineal  foot. 

The  charge  for  dressing  posts  and  girders,  making  window  shutters,  doors, 
wainscoting,  shelving,  ceilings,  lattice-work,  cornice,  etc.,  is  by  the  square  foot; 
Bometimes  some  of  this  work  is  charged  by  the  liueal  foot. 


o 


62  soule's  philosophic  practical  mathematics. 


The  cliarge  for  framing  joists,  small  rafters,  studding,  putting  on  weatber- 
boarding,  sbingliiig,  sheathing,  laying  floors,  making  i)artitioiis,  board  fences,  etc.,  is 
by  the  square  yard  or  the  square. 

KoTE. — lu  measuring  weather-boarding,  all  openings  over  4  foet  -n-ido  are  to  be  deducted. 

PROBLEMS. 

1.  A  carpenter  in  repairing  a  house,  framed  3  sills  8x6  inches,  each  42  ft. 
long,  and  C  sills  8x8  inches,  each  18  feet  G  inches  long;  what  was  the  number  of 
lineal  feet,  and  what  was  the  cost  at  7/  per  lineal  foot?  Ans.  $16.59. 

OPERATION. 


42     X  3  = 


=  126  ) 
=  111} 


18J  X  6  =  111  i  237  feet  at  7/  =  $16.59. 

2.  A  room  is  22  feet  3  inches  long,  and  IS  feet  6  inches  wide ;  it  has  two 
doors  each  4  feet  wide.  What  will  be  the  cost  to  wainscot  this  room  3  feet  4  inches 
high  at  60/  per  square  yard  ?  Aus.  $16.33J. 


OPERATION. 

22^  X  2  =  44J  2 

18J  X  2  =  37  3 

9 

81J 
4x2=8 


147  =:  distance  around  room. 
10  =  height  of  wainscoting. 
=  feet  in  sq.  yd. 


!7j  sq.  yds.  at  60/.  =  $16.33J. 


73J 

3.  What  will  be  the  cost  to  floor  a  room  34  feet  6  inches  long  and  22  ft.  4 
inches  wide,  at  $4  per  square  1  Ans.  $30.82. 

4.  What  will  it  cost  to  make  twelve  8  panel  doors  8  J  ft.  x  3J  ft.  1 J  inches  thick 
at  25/  per  square  foot  of  1  inch  thick  1  Ans.  $127.50. 

5.  What  will  it  cost  to  make  24  shutters  each  22  inches  wide  1^  inches  thick 
and  7J  feet  long  at  20/  i)er  square  foot  of  1  inch  thick  ?  Ans.  $82.50. 

0.  How  many  squares  in  the  floors  of  a  house  containing  8  rooms  measuring 
as  follows :  2  rooms  are  16  by  20  feet,  2  are  15J  by  20  feet,  2  are  14  by  18  feet,  and 
2  are  12  feet  4  inches  by  18  feet?  Aus.  22/5. 

7.  A  sub-contractor  jiuts  up  the  frame  work  of  a  building  at  90/  per  square. 
The  building  is  85  feet  long,  GO  feet  wide,  and  from  the  bottom  of  the  sill  to  the  top 
of  the  rafter  plate  is  31  feet  8  inches.  From  the  end  of  the  rafters  on  one  side, 
measured  over  the  peak  or  apex  of  the  roof,  to  the  end  of  the  rafters  on  the  oppo- 
site side  is  69  feet  2  inches.  The  studding  in  the  partitions  on  the  first  story  meas- 
ure 270  feet  long  and  15  feet  6  inches  high  ;  and  on  the  second  story  it  is  310  feet 
4  inches  long  and  14  feet  3  inches  high.  The  sleepers  in  all  the  rooms  on  the  lower 
floor  measure  84  feet  6  inches  long  by  59^  feet  wide;  and  the  joists  in  all 
the  rooms  on  the  second  floor  measure  the  same  as  the  sleepers  on  the  first  floor. 
The  two  gables,  or  the  vertical  triangular  ends  of  the  building  from  the  eaves  or 
cornice  to  the  top  are  GO  feet  at  the  base,  and  as  the  rafters  are  J  pitch,  they  have 
an  altitude  of  15  feet.    What  is  the  cost  of  erecting  the  frame  work  1 

Ans.  $311.37+. 


ARTIFICERS      OR    MECHANICS      WORK,    (N    SURFACES. 


363 


270 

310  J 
84i 


X  31^ 
X  COi 


85  +  85  + 
290 

85 

X  15^ 
X  Hi 
X  59J 

84i  X  59i 

60  X   15  X 


OPEKATION    INDICATED. 

GO  +  00  ==  200  ft.  the  moasurement  arouud  the  building. 

=  9183^  sq.  ft.  iu  the  4  .sides  or  walls. 

=  5879^  sq.  ft.  in  the  roof. 

=  4185    sq.  ft.  in  the  partitions  on  the  first  floor. 

=  4422^  sq.  ft.  in  the  partitions  on  the  second  floor. 

r=:  r>013|  sq.  ft.  iu  the  sleepers  on  the  first  floor. 

=  501. '>!  sq.  ft.  iu  the  joists  on  the  second  floor. 

2  =  900    sq.  ft.  in  the  two  gables. 


34597-1I3  sq.  ft.  in  the  entire  frame  work. 
34597,1-  sq.  ft.  ■-  100  =  345.97-ii-  squares,  at  90/  =  $311.37|. 

8.  What  is  the  cost  to  slate  a  roof  127  ft.  4  inches  long  and  66  feet  wide, 
at  $11.50  per  square,  and  what  will  be  the  weight  of  slate  on  the  roof,  allowing  a 
square  of  slate  to  weigh  600  pounds  ?  Ans.  $820,022. 

42784  lbs.  or,  21  tons  and  784  lbs, 

PLUMBEES'    WORK. 


739.  Plumbers'  work  is  done  by  the  pound,  hundred-weight,  or  by  special 
agreement. 

The  following  Table  shows  the  weight  of  a  square  foot  of  sheet  lead,  accord- 
ing to  its  thickness;  and  also  the  common  weight  of  a  yard  of  leaden  pipe,  accord- 
ing to  the  diameter  of  tlie  bore  : 


Thickness  of  Lead. 

Pounds  to  a  square  foot. 

Bore  of  Lead  Pipes. 

Pounds  jier  yard. 

Inch. 

1 
1  0 

5.899 

Inih. 

0-i 

10 

i 

6.554 

1 

12 

i 

7.373 

li 

16 

+ 

8.427 

1-i 

18 

1 

6 

9.831 

If 

21 

X 

5 

11.797               1 

'> 

24 

1.     What  is  tlie  cost  of  261  feet  of  leaden  pipe,  of  li  inch  bore,  at  11  cents 
per  pound,  allowing  as  jier  the  table  that  each  yard  M'eighs  18  pounds? 

Ans.  $172.26. 

OPERATION    INDICATED. 


2G1  4-3     =87  yards. 
87  X  18     =  1566  lbs. 
1566  X  11/  =  $172.26 


or, 


261 

18 

11 


$172.26,  Ans. 

2.  What  is  the  weight  of  lead,  J-  of  an  inch  thick,  that  covers  a  surface  33 
feet  long  and  8  feet  6  inches  wide,  estimating  the  weight  at  7|  jwunds  per  square 
foot  ?  Ans.  2006  pounds. 


364 


SOULE  S    rillLOSOl'HIC    PRACTICAL    MATHEMATICS. 


SLATING  AND  SHINGLING  KOOFS. 


74-0.  1.  The  roof  of  a  building  is  72  feet  G  inches  long  and  measures  46 
feet  9  inclies  from  cave  to  eave.  How  many  slates  and  how  many  squares  of  slating 
are  there  in  the  roof,  allowing  a  slate  to  cover  a  space  4J  by  8  inches  and  not  allow- 


ing for  the  double  course  at  the  eaves  ? 


OPERATION 


TO   FIND   TIIH 

Width  of  slate  =  8 


Length  of  slate  9 
exposed  on  the  — 
roof. 


Nl'MBEU   01'   SLATES. 

870  in.  =  length  of  roof. 

5C1  in.  =  width  "  " 
o 


13557  i  slates,  Ans. 


Ans.  13557^  slates. 

33Ho  sq.  of  slating. 


Explanation. — In  all  ]irol)leiiia  of  this 
kind,  we  Hist  tiiul  the  iiiinibur  of  Bi)nare 
inches  in  theroof  by  mnltiplying togeth- 
er the  length  and  tlio  width  of  the  roof 
in  the  unit  of  inches;  and  then  divide 
the  same  by  the  iinnilier  of  square  inches 
that  each  slate  covers. 


TO  FIND  THR 

11; 

12 
100 


OPERATION 

NTMBEIl   OK   SQUARES   OF   SLATING. 

870 
5C1 


33 Ho  squares,  Ans. 


Explanation. — Here  we  first  make  the 
statement  to  find  the  number  of  square 
feet  and  then  divide  by  100,  which  is  the 
number  of  square  feet  in  a  square,  as  per 
Article  531,  page  244. 


2.  How  many  shingles  will  it  require  to  shingle  a  hoitse  that  is  54  feet  long 
and  33  feet  10  inches  from  eave  to  eave,  estimating  tliat  5  inches  of  each  shingle 
will  be  laid  to  the  weather,  and  allowing  for  the  double  course  at  the  eaves  on  each 
side  1  Aus.  14256  shingles. 

Note  1. — The  unit  width  of  a  shingle  is  4  inches. 

Note  2. — Add  10  inches  to  the  width  for  the  double  courses. 

3.  A  flat  roof  is  208  feet  2  inches  long  and  28  feet  5  inches  wide.  How  many 
square  yards  of  tin  will  be  required  to  cover  it,  and  what  will  be  the  cost  at  $1.15 
per  square  yard  ?  Ans.  657^-Jf  sq.  yds.     $755.85+ cost. 

4.  A  building  is  04  feet  long  and  40  feet  wide;  the  rafters  have  one-quarter 
pitch,  allowing  a  projection  of  G  inches  on  each  end  of  the  roof,  and  1  foot  on  each 
side  for  the  eaves,  how  many  slates  and  how  many  squares  of  slating  will  it  require 
to  slate  the  building,  if  the. slates  are  6  inches  wide  and  5  inches  of  their  length 
are  exposed  to  tlie  weather,  no  allowance  to  be  made  for  the  ridge  of  the  roof  or  for 
the  double  course  at  the  eaves  ?  Ans.  14577  J  slates. 


30j|o  squares. 


OPERATION 


TO  FIND  the   length   OF   THE   KAFTERS. 


10  = 
202 


100 
400 

500 


ARTIFICERS     AND    MECHANICS     WORK,     IN    SURFACES. 


365 


yf  500  =  22.36  +  feet  =  length  of  rafter. 

Practically,  the  length  of  rafter  is 
Projection  of  lafter  is  -        -         -        - 


Length  of  rafter, 


22  ft.  4i  in. 
1  ft.  0    in. 

23  ft.  4  J  in. 


Length  of  two  opposite  rafters,  or  width  of  roof. 


40  ft.  82  ill. 


OPERATION 
TO  FIND  THE  NUMBEU   OK   SLATES. 

64  ft.  +  6  ill.  +  6  ill.  =  Co  ft.  =  780  in. 
46  ft.  8§  in.  ^=  56O5  inches. 


7S0 
1682 


OPERATION 
TO   KIND  THE  Nl'MBEK  OE   SQUARES. 


3 

12 

100 


65 

1682 


oOifJ  squares. 


145774  slates,  Ans. 

Note. — T\y  o^^p-quarli'r  -[iWch  is  meant,  in  merli.nniral  l:mfrna};p,  tliat  tlie  apex  of  tlie  roof  is 
to  lie  J  the  width  of  the  building  higher  than  the  plane  of  the  base  of  the  rafters. 


PLASTEEEES'  WOEK. 


7-1:1.  The  work  of  plasterers  is  of  two  kinds,  viz. :  ceiling,  which  is  plaster- 
ing on  laths,  and  renderhuj,  which  is  jilastering  on  walls. 

In  measuring  i^lasterers'  work,  the  square  foot,  square  yard,  orsrj'jjareisused. 

The  windows,  doors,  fire  places,  and  base  board  are  generally  deducted.  lu 
some  states,  one-half  of  all  openings  over  two  feet  wide  are  deducted. 

Plasterers  have  also  adopted  the  following  regulations  regarding  the  measur- 
ing of  their  work  : 

1.  To  measure  all  circular  walls  and  ceilings,  twice. 

2.  To  charge  extra  for  corners  of  chimneys  and  other  external  angles  when 
made  with  "gauged"  mortar. 

3.  To  add  one  foot  in  length  of  the  cornice  for  each  mitre  or  "return. " 

4.  Stucco  work,  when  more  than  12  inches  wide,  is  done  by  the  square  foot. 

PROBLEMS. 

1.  How  many  square  yards  of  plastering  in  the  walls  and  ceiling  of  a  room 
34J  feet  long,  21  feet  2  inches  wide,  and  14  feet  10  inches  high,  deducting  6  windows, 
7  by  4  feet,  and  2  doors,  9  by  3i  feet,  and  the  base  board  8  inches  wide,  and  what 
\rould  it  cost  at  24/  per  sq.  yard  T  Ans.  ^Zl^^f  sq.  yards. 

$55.50  practically. 


366  soule's  puiLosoriiic  practical  mathematics. 


OPERATION    INDICATED. 


34J  +  34J  +  21^  +  211  3=  111  J  ft.  distance  around  tlie  room. 

14'l't.  10  in.  lieiji'lit  of  room,  —  8  in.  ^vi(ltll  of  base  =  14  ft.  2  in.  height 

of  room  plastered. 
Ill  it  X  1-1  ,f  =  lo77f    sq.  ft.  in  the  walls  of  room. 
34^  X  21i  =    700|    so.  ft.  in  the  ceiling. 


2307^J  sq.  ft.  in  the  walls  and  ceiling. 

DEDUCTIONS. 

9  ft.  —  8  in.  =  8,^  ft.  length  of  doors,  allowing  for  width  of  base  board 

not  measured  in  the  area  of  the  walls. 
8J  X  3i  X  2  =    58 J  sq.  ft.  in  the  2  doors. 
7"    X  4"   X  C  =  1G8    sq.  ft.  in  the  0  windows. 


22CJ  sq.  ft.  deductions. 
2307ii  —  226J  =  2081-3%  sq.  ft.  2081^=8  -^  9  =  231-3^,  sq.  yds.  at  24/  = 
$55.50,  practically. 

2.  A  room  20  feet  C  inches  long,  17  feet  4  inches  wide,  has  two  chimneys 
extending  14  inches  from  the  wall;  what  would  be  the  cost  of  a  cornice  16  inches 
wide  around  this  room,  allowing  for  12  miters,  4  at  the  corners  of  the  room  and  8  at 
the  two  chimneys,  at  18/  per  square  foot?  Aiis.  $22.16. 

OPERATION. 

20J  +  20J  +  17J  +  17i      =  75g  ft.  around  the  room,  not  counting  chimneys. 
14    X    4    =  56    inches      =    43  ft.  width  of  the  chimney  walls. 
1  ft.  allowed  for  each  miter      12  ft. 

92.J  =  length  of  cornice. 
-     92J  X  IJ  =  123i  sq.  feet,  at  18/  =  $22.16. 

3.  What  will  it  cost  to  plaster  the  walls  and  ceiling  of  a  house  of  10  rooms, 
4  of  which  are  20xlCA  feet  and  14i  feet  high;  3  are  18x16  feet  and  13  feet  liigh 
and  3  are  16^xl2J  feet  and  13  feet  high;  allowance  being  made  for  20  windows 
7^X3^  feet,  and  18  doors  9^x3J  feet,  at  23/  per  sq.  yard  ?  Ans.  $275.54-|-. 

Note, — The  measurements  in  this  problem  are  exclusive  of  the  base  board. 

4.  "What  will  it  cost  to  plaster  the  walls  and  ceiling  of  2  rooms  at  37J  cents 
per  square  yard,  each  room  24  feet  3  inches  long,  12  feet  9  inches  wide,  and  11  feet 
6  inches  high,  deducting  5  windows,  0  by  3  feet  4  inches,  and  2  doors,  8  feet  8  inches 
by  3  feet  6  inches,  in  each  room,  and  the  base  board  8  inches  wide  ? 

Alls.  $79.57+. 

TAINTING   AND    GLAZING. 

742.  Painting  is  done  either  by  the  square  yard,  or  the  square,  or  by  esti- 
mate.    Glazing  is  done  by  the  light  or  pane. 

Painters  and  glaziers  have  adopted  the  following  rules  for  measuring  their 
Work: 

1.    In  weathei'-boarding,  measure  not  only  the  width  of  the  board,  but  also 


*  artificers'  and  mecuanics'  work,  in  surfaces.  367 

the  width  of  the  exposed  edge.    For  partition  boards,  when  beaded,  add  to  the  width 
of  the  boards  ^  inrh  for  each  bead. 

2.  For  palin<?  or  picket  fences,  measvire  the  height,  and  add  thereto  the 
■width  of  each  rail,  wliicli  gives  the  width  or  height. 

3.  To  find  the  length,  add  to  the  length  of  the  fence  twice  the  width  of  each 
post,  if  rectangular,  or  one-half  the  circumference,  if  cylindrical. 

4.  In  measuring  the  length  and  width  of  jjanel  doors,  shutters,  etc.,  press 
the  measuring  line  into  all  the  cjuirks  and  mouldings. 

5.  In  measuring  weather-boarding  and  wall  painting,  deduct  half  the  open- 
ings. 

6.  In  measuring  shingle  or  board  roofs,  measure  the  butts  or  edges;  for 
trellis  or  lattice  work,  double  the  measure,  and  for  Venetian  shutters  add  one-half. 

7.  In  measuring  the  width  of  pilasters,  commence  at  the  wall  and  measure 
round  to  the  bead  at  the  end  of  the  jamb  casing,  pressing  the  line  into  all  the 
quirks. 

8.  For  plain  rosettes  in  pilasters,  add  1  foot  to  the  length ;  for  carved 
rosettes,  add  2  feet  to  the  length. 

9.  In  measuring  window  jambs,  cornice,  etc.,  press  the  line  or  tape  into  all 
indentations  and  concave  surfaces. 

10.     When  the  measurement  is  made  by  the  srpiare  foot,  it  is  customary  to 
include  the  whole  sash,  and  circular  or  oval  windows  as  square. 

PROBLEMS. 

1.  What  will  the  glazing  cost  for  a  house  containing  40  windows  of  8  lights 
each,  and  7  windows  of  IS  lights,  at  25^  per  light  ?  Ans.  $123.50 

2.  What  will  be  the  cost  of  the  painting  of  4i  ft.  high  wainscoting  of  4  inch 
•wide  beaded  boards,  of  a  room  18  feet  6  inches  by  13  feet  4  inches,  at  24c'  per  sq. 
yard?  Ans.  $8.59|. 

OPERATION    INDICATED 

18i  4-  ISi  +  13^  +  13J  =  G3§  ft.  measurement  around  the  room. 
C3|  =  704  in.  4-  4  in.,  the  width  of  a  beaded  board,  =  191   beads,    or   half 
inches,  to  add  to  the  measurement  around  the  room;  191  half  inches 
=  95A  inches  =  7ftf  ft. 
G3g  +  7ft J  =  71 J  ft.  length  of  wainscoting. 
7ia.  X  4^  =  322-1%  sq.  ft,  =  351^  sq.  yds.  at  24/  =  $8.59.J. 

3.  What  will  it  cost  to  paint  both  sides  of  a  paling  or  picket  fence  120  feet 
long,  5  ft.  high,  the  posts  being  8  feet  apart  from  center  to  center  and  4x0  in.  with 
the  narrow  side  set  to  the  fence;  the  rails  being  2x4  inches  with  the  wide  sid^ 
facing  the  fence,  at  25/  per  sq.  yard  ?  Ans.  $42.81if. 

OPERATION    INDICATED. 

5  ft.  high  +  4  in.  +  4  in.  =  5|  ft.  =  height  of  fence  including 

the  width  of  the  two  rails. 
10  posts  6  in.  wide,  2  sides  eacli  =  1  ft.  for  each  post,  to  be 

added  to  the  length  of  the  fence. 
120  -f  10  =  130  ft.  length  of  fence,  one  side. 
130  x  2  =  272  ft.  length  of  fence,  both  sides. 
272  X  5§  =  1541i  sq.  ft.  1541^  -f-  9  =  171.7,-  sq.  yds. 
171A-  X  25/  =  $42.81if . 


368 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


4.    How  many  panes  of  glass,  12  by  10  inclies,  in  a  box  of  100  feet  7 

Alls.  120  panes. 

OPERATION    INDICATED. 

Pan  e. 


1  pane  10x12  =  120  inch.     1  sq.  ft.  =  144  sq  in. 
144x100  =  14400  sq.  in.  in  tlie  box.     14400 
4-  120  =  120  panes. 


or  thus:    120 


1 

12 

144 

144  or, 

10 

100 

5.    How  many  panes  of  glass  in  44  boxes,  each  containing  50  sq.  ft.,  the  panes 
being  22x24  inches,  and  what  will  be  the  cost  at  $7  per  100  sq.  feet  1 

Alls.  600  panes.     $154.00  cost. 
G.    "What  will  it  cost  to  glaze  13  windows,  3  feet  by  7  feet  4  in.,  and  22  win- 


dows, 3  feet  by  CJ  feet,  at  25  cents  per  square  foot? 


Ans.  $178.75. 


CAKPETING  FLOORS. 


PROBLEMS. 


743.  1.  How  many  yards  of  carpeting  that  is  27  inclies  wide,  will  be 
required  to  cover  the  floor  of  a  parlor  that  is  32  feet  4  inches  long  and  25  feet  6 
inches  wide,  making  no  allowance  for  waste  in  matching  or  turning  under  ? 

Ans.  122,/,- yds. 


FIRST   OPERATION. 


3G 

27 


388 
306 


122/7  yds.,  Ans. 

SECOND   OPERATION   INDICATED 


or. 


Explanation. — In  this  solution,  ■we  first  find  the  nnmher 
of  square  inches  ill  the   lloor,    by  nmltiplyinj;   together   the 
h-ngth  and  width  in  the  nnit  of  inches.     We  then  divide  by  ~ 
the  jiioduct  of  tho  length  and  ■width  of  a  yard  of  the  carpet, 
■which  is  the  number  of  square  inches  in  one  yard  of  it. 


12 

388 

12 

306 

9 

27 

36 

3 

97 

0 

51 

9 

7 

36 

122/»-yds.,  Ans. 


122/,-  yds.,    Ans. 


the  width  of  tlie 


Explanation. — In  this  solution, 
■we  first  find  the  iiumlier  of  niiiare 
yards  in  the  floor,  -winch  would  be 
the  number  of  yards  required  if 
the  carpeting  -was  1  yard,  or  36 
inches  wide.  But  since  the  carpet- 
ing is  not  I  yard  wide,  we  multiply 
by  SG,  which  j;ive8  the  number  of 
yards  required  if  the  cnriieting  was 
carpeting,  and  produce  the  correct 


but  1  inch  wide,  and  we  tlicu  divide  by 
result. 

We  reason  as  fidlows  :  Since  it  requires  this  expressed  number  of  yards  ■^vhen  it  is  3(1  inches 
wide,  if  it  wciebut  1  inch  wide,  it  would  require  36  times  as  many  yards,  and  if  'Jl  inches  wide,  the 
27th  part. 

KoTE. — In  this  connection,  we  will  remark  that  .although  the  result  of  this  jiroblem  is  math- 
emntically  correi't,  it  is  not  correct  in  a  practical  sense;  i.  c,  in  practice  it  would  reipiiro  more 
yards  of  carpeting  than  the  uuitliematical  measurement  of  the  room  gives.  This  is  occasioned  by 
the  matching  of  tlio  floweis  or  figures  of  the  cari)et,  and  by  the  unavoidable  waste  by  reason  of 
the  room  being  a  little  w  ider  or  longer  than  au  eveii  uumber  of  w  idths  of  the  carpetiug,  which 
necessitates  the  use  of  another  widtJi  of  carpet. 

The  increased  number  of  yards  over  the  mathematical  result,  by  the  foregoing  causes,  will 
depend  upon  the  size  of  the  figure  in  the  carjiet,  and  whether  the  carpet  is  laid  lengthwise  or 
crosswise  of  the  room;  and  the  flower  or  figure  of  the  carpet,  and  the  entrance  to  the  room, 
whetlicr  from  the  side  or  end,  often  determines  which  way  the  carpet  .shall  be  laid.  Tliese  are 
important  facts  for  ]).irties  who  buy  carpets  on  their  own  measurement  of  rooms,  and  if  observed, 
will  often  save  expense  aud«nuch  vexation. 

The  two  following  problems  elucidate  these  jioiuts  and  make  clear  the  practical  manner  of 
working  problems  of  this  kind  ; 


*  artificers'  and  mechanics'  work,  in  surfaces.  369 

2.  How  many  yards  of  carpet,  27  inches  wide,  will  lie  required  to  cover  a  floor 
21  feet  10  inches  long  and  18  feet  6  inches  wide,  laid  lengthwise,  and  allowing  6 
inches  on  each  strip  for  matching?  Aus.  67  yards. 


OPERATION   indicated. 

18  ft.  6  in.  =  222  inches.  222  inches  -i-  27  =  8/,,  practically,  9  widths  of 
carpet,  -HI  of  one  width  will  be  waste,  either  cut  off  or  folded  under. 

21  ft.  10  in.  +  6  in.,  allowed  for  matching,  =  22  ft.  4  in.  =  length  of  each 
strip.     22  ft.  4  in.  x  9  =  201  ft.  -^  3  =  67  yds. 

3.  If  in  the  above  problem,  the  carpet  had  been  laid  crosswise  with  the  same 
allowance,  how  many  yards  would  have  been  required  ?  Aus.  63J  yards. 

operation  indicated. 

21  ft.  10  in.  =  262  inches.  262  in.  -4-  27  =  9^1,  practically,  10  widths  of 
carjiet. 

18  ft.  6  in.  +  6  in.  allowed  for  matching,  =  19  ft.  =  length  of  each  strip. 

19  ft.  X  10  =  190  -^  3  =  63J  yds. 

4.  now  many  yards  of  carpet  36  inches  wide,  and  what  will  be  the  cost  at 
$1.35  per  yard  laid,  to  cover  a  floor  44  ft.  7  inches  long,  31  ft.  6  in.  wide,  laid  cross- 
wise of  the  room,  allowing  4  inches  for  matching?  Aus.  159^  yds. 

$214.S7J  cost. 

5.  A  parlor  is  29  feet  long  by  22  feet  wide.  It  is  carpeted  with  36  inch 
carpet  at  $1.75  a  yard,  surrounded  by  a  carpet  border  18  inches  wide,  at  $1.25  per 
linear  yard.  If  the  border  is  laid  without  being  mitered  and  the  carpet  is  laid 
crosswise  of  the  room,  and  an  allowance  of  4  inches  on  each  strip  is  made  for 
matching  the  carpet,  what  will  be  the  total  cost?  Ans.  $141.50. 

OPERATION    indicated. 

29  ft.  —  3  =  26  ft.  -^  3  =  8|,  practically,  9  widths.    22  ft.  —  3  =  19  ft.  + 
4  in.  (matching)  =  19^  ft.  length  of  strip.    9  x  19J  =  174  ft.  —  3 
=  58  yds.  carpet. 
58  X  $13  =  $101.50  cost  of  carpet. 
Then,  29  x  2  =  58  ft.  length  of  carpet  strips  on  two  sides  of  room.    22  —  3 
t  19  X  2  =  38  ft.  length  of  carpet  strips  on  two  ends  of  room.     58  +  38  =  96  ft. 
•i-  3  =  32  yds.  length  of  car]}et  strips. 

32  X  $1.25  =  $40.00  +  $101.50  =  $141.50,  Ans. 

Note. — When   tbc   liorder  is  mitered,  there  -nill  be  as  many  yards  of  border  as  there  are 
yardi)  distance  in  the  four  sides  of  the  room. 


370 


SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS. 


MENSURATION  OF  SOLIDS. 


744.  The  mensuration  of  solids  is  divided  into  two  parts,  viz.:  1st,  the 
mensuration  of  their  solidities;  and,  2d,  the  mensuration  of  their  surfaces. 

The  Unit  of  Measure  for  solids  is  a  cube  whose  side  is  some  fixed  lengthy 
such  as  an  inch,  foot,  yard,  rod,  etc. 

The  Area  of  a  figure  is  the  measure  of  its  surface. 

The  Solidity  or  Contents  of  a  solid  is  the  number  of  cubes  that  it  is  equal 
to,  or  contains;  or  its  cajiacity  expressed  in  cubic  measure  of  any  fixed  unit. 

In  the  solution  of  problems  in  mensuration  of  solids,  nuich  time  may  ofter» 
be  saved  by  applying  the  following  principles  which  have  been  demonstrated  in 
geometry : 

That  all  similar  solids  are  to  each  other  as  the  cubes  of  their  lihe  dimensions. 

DEFINITIONS, 

745.    A  Eectan  gul  ar  or 


Quadrilateral  Solid  is  a 
solid  which  has  length, 
width,  and  thickness,  and 
is  bounded  by  six  sides  or 
faces. 

747.  A  Cube  is  a  solid 
whose  sides  or  faces  are 
all  equal  squares. 

749.    A  Cylinder  is  a 

solid  having  two  faces  or 
bases,  which  are  equal 
])arallel  circles,  and  which 
has  an  equal  diameter  in 
any  parallel  iilaue  between 
them. 

751.    A    Frustum    of 

a  cone  is  the  jiart  which 
:  remains  after  the  top  is 
cut  off  by  a  jjlane parallel 
to  the  base. 


746.    A  Pyramid  is  a 

solid  whose  base  is  any 
kind  of  a  polygon,  audits 
other  faces,  triangles  unit- 
ed at  a  common  point 
called  the  vertex. 

748.  A  Frustum  of  a 

Pyramid    is   the  part 

_  which  remains  after  the 

,^^  top  is  cut  off  by  a  plane 

^^    parallel  to  the  base. 

750.  A  Cone  is  a  solid 
having  one  face  or  base 
which  is  a  circle,  and  a 
convex  or  curved  surface 
terminating  in  a  point, 
called  the  vertex. 

752.    A   Sphere    is   a. 

solid  bounded  byacurved 
surface,  all  tlie  points  of 
which  are  equally  distant 
^-—-,  fro  I II  a  certain  pointwith- 
in  called  the  center. 


The  Radius  of  a  sphere  is  a  line  flrawn  from  the  center  to  any  part  of  the  circnmferenoe. 
The  axis,  or  diameter  of  a  sphere  is  a  line  passiug  through  the  center  and  termiuateiL  by  the  cir- 
cumfereuce. 

754.  An  Oblate  Sphe- 
roid is  a  solid  fiattened 
or  compressed  at  the 
poles;  or, itis a  solid  gen- 
erated by  the  revolution 
of  an  ellipse  about  its 
shorter  axis. 


753.  A  Prolate  Sphe- 
roid is  a  solid,  elongated 
ill  tlie  direction  of  a  line 
joining  the  jioles  ;  o)',  it  is 
a  solid  generated  by  the 
revolution  of  an  ellipse 
about  its  longer  axis. 


MENSURATIOX    OF    SOLIDS. 


!7I 


i 


TO  FIND  THE  SOLIDITY  OF  A  PRISM,  PARALLELOPIPEDON,  OR  CUBE. 

PROBLEMS. 

755.    1.    A  rectangular  box  is  4  feet  long,  3  feet  4  inches  wide,  and  2  feet 
bigli.     now  many  solid,  or  cubic  feet  does  it  contain  ?  Ans.  26§  cu.  ft. 

Eiplanafion.—1a  all  cu- 
Ijical  or  rectangular  solids, 
Tve  multiply  together  the 
length,  width,  and  height, 
in  the  same  units  of  meas- 
ure, and  in  the  product  we 
have  the  required  solidity, 

2.    "What  are  the  solid  contents  of  a  prism,  the  sides  of  whose  base  are  3  and 
4  feet,  and  whose  altitude  is  6  feet  ?  Ans.  72  cu.  ft. 


OPERATION. 

4 

4 

10           or. 

12 

40 

*> 

2 

26§  cu.  ft., 

2G§  cu.  ft.. 

Ans 

Ans 

TO  FIND  THE  SURFACE  OF  A  PRISM,  PARALLELOPIPEDON,  OR  CUBE. 

PROBLEMS. 


75G.    1. 

above,  contain  1 


How  many  square  feet  of  surface  does  the  box  in  Problem  !No.  1 

Ans.  56  sq.  ft. 

OPERATION. 

4    X  3J  X  2  =  202  sq.  ft.  on  two  sides. 

4    X  2    X  2  =  16    sq.  fr.  on  tlie  other  two  sides. 

3J  X  2     X  2  =  13J  sq.  ft.  on  the  ends. 

56    sq.  ft.  in  the  entire  surface. 
What  is  the  surface  of  a  rectangular  prism,  the  sides  of  whose  base  are 


3  and  4  feet,  and  whose  altitude  is  6  feet? 


Ans.  108  sq.  feet. 


TO  FIND  THE  SOLIDITY  OF  A  SQUARE  OR  RECTANGULAR  PYRAMID. 

PROBLEM. 

757.    A  sqiiare  or   rectangular  pjTamid  is  8  feet  high  and  each  side  of 
the  base  is  4  feet  6  inches.    How  many  cubic  feet  does  it  contain  ?    Ans.  54  cu.  ft. 

Explanation. — In  all  problems 
of  this  kind,  we  multiply  th© 
height  by  the  area  of  the  'jase, 
and  then  divide  by  3,  because  a- 
«(/Ma/-e  jiyramid  hr;  been  demon- 
strated in  gei'metry  to  je  J  of  a. 
rectangular  soliu  of  equal  height 
and  area  of  base. 

Note.— To  find  the  solidity  or  volume  of  a  regular  pyramid  that  is  not  sqnare  or  rectangular, 
nmltiply  the  square  of  one  side  i>y  the  tabular  area  set  opposite  the  polygon  of  the  .ame  umber 
of  sides"  as  shown  in  the  Table  on  page  350,  which  will  give  the  area  of  the  base.  Then  proceed  as 
in  the  above  problem,  by  multiplying  by  the  height  aud  dividing  by  3. 


8 
9 
9              or 

12 

12 

3 

8 

54 

54 

54  cu.  ft., 
Ans. 

54  cu.  ft., 

Ans 

372  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

TO  FIND  THE  SUEFACE  OF  A  SQUARE  OR  RECTANGULAR  PYRAMID. 

PROBLEM. 

758.  How  many  square  feet  of  surface  in  a  rectangular  pyramid  ■whose 
slant  height  is  8  feet  and  whose  base  is  4J  feet  ?  Ans.  92J  sq.  ft. 

OPERATION. 

4J  X  8  X  4  =  144.     144  4-  2  =  72  sq.  ft.  area  of  four  sides. 

4 1  X  4.J  =  2(>|  sq.  ft.  area  of  base. 

73  +  20|  =  92^  sq.  ft.  area,  or  surface  of  pyramid. 

TO    FIND    THE    SOLIDITY    OF    HEXAGONAL    OR    OTHER    REGULAR 
PYRAMIDS  NOT  SQUARE  OR  RECTANGULAR. 

problem:. 

759.  What  are  the  solid  contents  of  a  hexagonal  pyramid  whose  altitude  is 
11  feet,  and  each  side  of  whose  base  is  2  feet  ?  Ans.  38.1051176  cu.  ft. 

OPERATION 

2-  =  4  feet  =  square  of  one  side.  Explanation. — In  all  proWems  of 

this    character,    we    multiply  the 
4  X  2.5980762  =  10.3923048  sq.  ft.,  area  of  base.      square  of  cue  side  l.y  the  tabular 

11  altitude.  area  set  opposite  the  polygon  of 
the  same  iiumher  of  sides,  as  shown 


3)  114.315.3528  in  the  Table,  page  350,  and   this 

product  is  multiplied  by  the  alti- 

38.1051176  solid  feet,  Ans.  tude  and  the  result  is  divided  by  3, 

TO  FIND  THE  ENTIRE  SURFACE  OF  AN  OCTAGONAL  OR  OTHER 
REGULAR  PYRAMID,  NOT  SQUARE  OR  RECTANGULAR. 

PROBLEMS. 

760.     1.     What  is  the  entire  surface  of  au  octagonal  pyramid  whose  slant 
height  is  11  feet  and  each  side  of  whose  base  is  2  feet  ? 

Ans.  107.3130  sq.  ft.  surface. 

OPERATION. 

2  X  8  =  16  ft.,  perimeter  of  the  base. 
16  X  5}  (one  linlf  of  slant  heiglit)  =  88  sq.  ft.,  convex  surface. 
2^  =  4ft.,  square  of  one  side.     4  x  4.8284  =  19.3136  sq.  ft.,  area  of  base. 
88  +  19.3136  =  107.3136  sq.  ft.,  entire  surface. 

ExpJavatioti. — In  all  problems  of  tins  character,  we  multiply  the  perimeter  of  the  base  liy 
Half  the  slant  height;  tlie  result  is  the  convex  surface.  Then  to  find  the  area  of  the  base,  multiply 
tlie  square  of  the  side  by  4.8284,  which  is  tlie  tabular  area  sot  opposite  the  polygon  of  the  same 
uumbcr  of  sides,  as  shown  in  the  Table  on  page  350. 

2.    What  is  the  surface  of  a  regular  octagonal  pyramid,  whose  slant  height 
is  12  feet,  and  each  side  of  the  base  5  feet  ?  Ans.  360.7106775  sq.  ft. 


*  MENSURATION    OF    SOLIDS.  373 

TO  riND  THE  SOLIDITY  OF  THE  FRUSTUM  OF  A  SQUARE  PYRAMID. 

PROBLEM. 

7G1.     A  frustum  of  a  square  or  four-sided  pyramid  is  8  feet  bigb,  lower  base 
7  feet,  and  upper  base  6  feet.    How  many  solid  feet  does  it  coutaiu  ? 

Ans.  338§  cu.  ft. 

OPERATION. 

7-  =  40  =  area  of  tlie  greater  base. 
G^'  =  30  =     "     "     "    lesser         " 
7  X  6    =  4i!  =  geometrical  meau  proportional  between  the  two  bases. 


3  )  127 


42J  =  average  area  of  the  frustum  of  the  pyramid. 

8    =  height. 


338§  cu.  feet,  Ans. 

Explanation, — In  all  problems  of  tliis  kind,  to  find  the  solidity  of  the  frustum  of  a  pyramid, 
•we  add  together  the  areas  of  the  two  b.ises,  and  a  geometrical  mean  jiroportional  between  theui, 
then  divide  the  sum  by  3  and  multiply  the  result  by  the  altitude. 

Note. — To  find  the  solidity  or  volume  of  the  frustum  of  a  pyramid  that  is  not  square  or 
rectangular,  we  multiply  the  Square  of  one  side  of  each  base,  by  the  tabular  area  set  opposite  the 
polygon  of  the  s,ame  number  of  sides  as  sliown  in  the  Table  on  page  350,  which  will  give  the  area 
of  the  two  bases.  Then,  wo  multiply  the  two  bases  together  and  extract  the  square  root  of  the 
product;  then  add  the  quotient  and  the  areas  of  the  two  bases  together  and  divide  the  sum  by  3; 
the  result  is  the  average  area  of  the  frustum  of  a  pyramid  wlilch.  multiplied  by  the  height,  gives 
the  solidity  of  the  frustum. 


TO  FIND  THE  SURFACE  OF  THE  FRUSTUM  OF  A  SQUARE  OR  REC- 
TANGULAR PYRAMID. 

PROBLEM. 

762,    How  many  square  feet  in  the  entire  surface  of  a  rectangular  pyramid 
whose  slant  height  is  8  ft.,  lower  base  7  ft.,  and  upper  base  G  feet? 

Aus.  293  sq,  ft.  surface. 

OPERATION. 

7  X  4  =  28  ft.,  perimeter  of  lower  base.    6  x  4  =  24  ft.,   perimeter  of 

upper  base. 
28  +  24  =  52  ft.,  sum  of  perimeters  of  bases.    52  x  8  =  416  4-  2  =  208 

ft.,  convex  surface. 
7-  =  49  sq.  ft.  area  of  lower  base.    6-  =  3G  sq.  ft.  area  of  upper  base. 
208  +  49  +  30  ^  293  sq.  ft.  entire  surface. 

Explanation. — In  all  problems  of  this  Icind,  we  first  find  the  convex  surface  of  the  frustum  by 
multiplying  the  sum  of  the  perimeters  of  the  two  bases  by  the  slant  height  of  the  frustum  and 
dividing  by  2. 

Then,  to  find  the  entire  surface  of  the  frustum,  add  to  the  convex  surface  the  areas  of  the 
two  bases. 

Note. — When  the  frustum  of  the  pyramid  is  not  four  sided,  the  area  of  the  two  bases  is 
found  by  multiplying  the  square  of  each  side  by  the  tabular  area  set  opposite  the  polygon  of  the 
same  number  of  sides  in  the  Table  on  page  350. 


374  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  ♦ 

TO  FIND  THE  SOLIDITY  OF  PRISMS  OR  SOLIDS  OF  DIFFERENT  SIDES. 


OPKRATION. 


PROBLEMS. 

763.     1.    Hovr  many  cubic  inches  in  an  octagonal  prism  or  solid,  one  side  of 
whose  base  is  20  inches  and  whose  length  is  60  inches  'I     Ans.  115882.2504  cu.  in. 

Explanation. — In  all  ]irol)lems  of  this 
Iciiid,  first  Btiuare  oiio  of  tbo  sides  of 
the  jiolygon.  2.  Miiltiiily  the  prod- 
Jict  by  the  tabular  uiunber  set  ojipo- 
site  the  polygon  of  the  same  uumber 
of  sides  in  the  Table,  page  350.  3. 
Multiply  the  last  product  by  the 
length. 

2.    The  side  of  a  pentagonal  prism  is  2  feet,  its  length  is  30  feet.     How  many 
cubic  feet  does  it  contain  ?  Ans.  20G.457288  cu.  ft. 


20=  =  400;  400  x  4.8284271  = 
1931.37084  square  inches  area  of  base. 
1931.37084  X  60  =  115882.2504  cu.  in. 


OPERATION    INDICATED. 


X  1.7204774  x  30. 


TO  FIND  THE  SOLID  CONTENTS  OF  A  CYLINDER. 

PROBLEMS. 

764.     1.    A  cylinder  is  8  feet  high  and  4  feet  6  inches  in  diameter.    How 
many  cubic  feet  does  it  contain  ?  Ans.  127.2348  cu.  ft. 


9 

12 

54 

9 

or, 

12 

54 

8 

8 

.7854 

cu. 

ft., 

— 

.7854 

127.2348 

127.2348  cu.  ft.. 

An 

s. 

Ans. 

Explanation. — In  all 
problems  of  this  kind, 
vre  multiply  the  square 
of  the  diameter,  by  the 
height,  which  gives  the 
solidity  of  a  rectangular 
solid  whose  height  and 
width  of  sides  are  equal 
to  the  height  and  diame- 
ter of  the  cylinder.  Then  we  multiply  by  .7854,  the  ratio  between  the  area  of  a  square  and  that 
of  a  circle,  whose  diameter  is  equal  to  one  side  of  the  square. 

Note. — The  solidity  of  a  cylinder  is  eqnal  to  the  solidity  of  a  square  prism,  whose  sides  and 
altitude  are  the  same  as  the  diameter  and  altitude  of  the  cyliuder,  multijilied  by  the  decimal  .7854. 

2.    "What  are  the  solid  contents  of  a  cylinder,  whose  altitude  is  6  feet,  and 
diameter  4  feet?  Ans.  75.3984  cu  ft. 


765.    1. 


TO  FIND  THE  SURFACE  OF  A  CYLINDER. 

PROBLEMS. 

In  the  cylinder  in  Probem  No.  1  above,  how  many  square  feet  are 


contained  on  its  entire  surface  ? 

OPERATION. 

4i  X  3.1416  =    14.1372    ft.  circumference. 

14.1372  X  8  =  113.0976    sq.ft.  area  in  convex  surface. 

4p  X  .7854  =    15.9043J  sq.  ft.  in  lower  base. 

4J=  X  .7854  =    15.9043i  sq.  ft.  in  upper  base. 


144.9063    sq.  ft.  in  surface  of  cylinder. 


Ans.  144.9003  sq.  ft. 

Explanation. — In  all  prob- 
lems of  this  kind,  we  multi- 
ply the  circumference  by  the 
height,  and  to  the  product, 
add  the  area  of  the  two 
ends  or  bases. 


MENSURATION    OF    SOLIDS. 


6/ 


7.5 


\ 


OPERATION. 

8 

8 

2 

9 

12 

54 

o 

9 

.7854 

or, 

12 

54 
.7854 

3 

3 

42.411G 

cu.  ft., 
Ans. 

42.4110  cu.  ft, 

Aus. 

2.     "UTiat  is  the  surface  of  a  cyliiuler  6  feet  long  aud  4  feet  in  diameter  T 

Alls.  100.5312  sq.  feet. 

TO  FIND  THE  SOLID  CONTENTS  OF  A  CONE. 

PROBLEMS. 

766.     1.     A  cone  is  8  feet  liigli,  and  its  base  is  4  feet  G  inches  in  diameter. 
How  many  solid  feet  does  it  contain  ?  Aus.  42.411G  cu.  ft. 


Explanation. — In  all 
problems  of  tbis  kiud, 
we  make  the  same  solu- 
tion as  in  a  cylinder  and 
then  ilivide  by  3,  becanse 
the  solidity  of  a  cone  is  J 
of  a  ci/Jiiider  whose  height 
and  diameter  are  equal  to 
that  of  the  cone. 


2.     What  is  the  solidity  of  a  cone,  the  diameter  of  whose  base  is  5  feet,  and 
altitude  12  feet  1  Ans.  78.54  cubic  feet. 


TO  FIND  THE  SURFACE  OF  A  CONE. 

PROBLEMS. 

767.     1.    How  many  square  feet  on  the  entire  surface  of  a  cone  whose  slant 
height  is  8  feet,  and  whose  base  is  4  ft.  6  inches  in  diameter? 

Ans.  71.4531  sq.  ft.  surface. 

OPERATION. 

4^  X  3.141G  =  14.1372  circumference  of  the  base. 

14.1372  X  8  =  113.0976  -^  2  =  5G.5488  sq.  ft.  convex  surface. 

4p  X  .7854  =  14.9043  sq.  ft.  area  of  base. 

5G.5488  +  14.9043  =  71.4531  entire  surface. 

Explanation. — In  all  problems  of  tbis  kind,  we  miiltipl;/  the  circumference  hy  the  elavt  height, 
Mnd  divide  the  product  by  2;  then,  to  tlie  quotient  thus  obtained,  add  the  area  of  the  base,  aud  the  result 
■will  be  the  eutire  surface. 

2.    "What  is  the  surface  of  a  cone,  the  diameter  of  whose  base  is  5  feet,  and 
slant  height  12  feet?  Aus.  113.883  sq.  ft.  surface. 


TO  FIND  THE  SOLID  CONTENTS  OF  A  FKUSTUM  OP  A  CONE. 

PROBLEM. 

768.     A  frustum  of  a  cone  is  8  feet  high,  lower  base  7  feet  in  diameter,  and 
upper  base  6  feet  in  diameter.    How  many  solid  feet  does  it  contain  ? 

Ans.  265.9888  cu.  ft. 


3/6  soule's  philosophic  practical  mathematics. 

first  operation. 

7^  =  49  =  square  of  greater  diameter. 
6=  =  30  =      "         "  lesser  "  c 

7  X  6  =  42  =  geometrical  mean  between  tlie  two 

diameters. 

3)127 


127 
.7854 

8 

205.9888 
cu.  ft., 


42J  =  area  of  the  base  of  a  rectangular 

solid,  the  sides  of  whose  bases  are  Ans 

equal  to  the  average  diameter  of 
the  frustum  of  the  cone. 

Explanation. — In  all  proWems  of  this  kind,  we  first  find,  as  is  shown  in  the  operation,  th& 
area  of  a  rectangular  solid,  the  sides  of  whoso  bases  are  e(|ual  to  the  average  diameter  of  the 
friistnni  of  tlie  coue ;  then  we  multiply  this  by  .7854,  to  reduce  the  area  of  a  square  to  the  area  of 
a  circle,  aud  then  multiply  by  the  height. 

Note. — In  the  above  snlntion,  we  found  the  geometrical  mean  between  the  diameters  of  the 
two  bases  hy  simply  multiplying  the  two  diameters  together.  This  is  by  far  the  shortest  and 
easiest  method  of  obtaining  the  geometrical  mean  between  two  numbers,  and  it  is  correct  because, 
by  demonstration,  it  is  proved  that  the  product  of  any  two  numbers  is  equal  to  the  square  root  of  the 
product  of  the  squares  of  those  two  numbers. 

The  usual  method  of  finding  the  geometrical  mean  between  the  diameters  is 
as  follows:  Square  each  diameter,  multiply  their  products  together,  extract  the 
square  root  of  the  product,  the  result  is  the  geometrical  mean  between  the  diameters 
of  the  two  bases. 

operation. 
Thus:  7^  =  49;    G^  =  30;    49  x  30  =  1704. 


■v/17t)4  =  42  the    geometrical  mean,  the  same  as   obtained  in  the  solution 

above,  by  simply  multiplying  the  two  diameters  together. 

Note. — The  last  method  would  be  used  when  the  areas  of  the  bases  and  not  the  diameters 
are  given. 

SECOND   OPERATION  BY   THE  PRISMOIDAL  FORMULA. 

7x7  =    49    =  area  of  the  lower  base. 

6x6  =    36    =  area  of  the  iqjijer  base. 


2)  13       13 


6J  X  6^  =  42J  X  4r       =  109    =  4  times  the  area  of  the  middle  section 

between  the  two  bases. 

6)254 


4:2^  r=  average  area  of  the  frustum  of  a  square 
l>yramid  of  the  same  dimensions  as  those 
of  this  frustum  of  a  cone. 
42^  X  .7854  =  33.2486  =  average  area  of  the  frustum  of  the  cone. 
33.2486  X  8  =  265.9888  =  solid  contents  or  cubic  feet. 

Note. — The  above  solution  is  in  accordance  with  the  prismoidal  formula,  by  which  the  solid- 
ity of  cubes,  rectangular  solids,  cones,  cylinders,  pyramids,  frustums  of  cones  or  pyramids  and 
several  other  forms  of  solids  may  be  determined. 

The  prismoidal  formula  is  as  follows:  Add  together  the  areas  of  the  two  ends  or  bases  and 
four  times  the  middle  section  parallel  to  them.  Then  divide  this  sum  by  6,  and  multiply  the  quo- 
tient by  the  height,  or  depth. 


MENSURATION    OF    SOLIDS.  377 

TO  FIND  THE  ENTIRE  SUEFACB  OF  THE  FEUSTUM  OF  A  CONE. 


PROBLEM. 

769.     In  the  above  problem,  wliat  is  the  entire  surface  of  the  frustum  ? 

Ans.  230.12l.':2  sq.  ft. 

OPERATION 

7  X  3.1416  ==  21.9013  =  circumference  of  greater  base.         i:xpianation.—in  all  prob- 
6  X  3.1416  =  18.8496  =  circumference  of  lesser  base.  lems  of  tk-j  kind,  nuiltiply 

the  bases  by  3.141G  and  the 
sum  of  their  products  by  the 
height ;  then  divide  tliis  ])rod- 
uct  by  2  -which  gives  sq.  ft. 
in  convex  surface ;  to  this 
result,  add  the  areas  of  the 
two  bases. 


40.84(18  X  8  =  326.7264 
326.7264  4-  2  =  163.3633  sq.  ft.  convex  surface. 
7=     X     .7854=    38.4846  sq.  ft.  of  larg:er  base. 
6^     X    .7854=  =    28.3744  S(i.  ft.  of  lesser  base. 


230.1333  sq.  ft.  in  entire  surface. 


TO  FIND  THE  SOLIDITY  OF  A   SFHERE. 

PROBLEMS. 

770.     1.    A   sphere  is  4  feet  in  diameter.     How  many  cubic  feet  does  it 
contain  ?  Ans.  33.5104  cu.  ft. 

OPERATION. 

4',  or  4  X  4  X  4  =  64  cu.  ft.  in  a  cube  which  is  4  feet  on  each  side. 


64  X  .5236  =  33.5104  cu.  ft.,  Ans. 

Hxplanation.— In  all  problems  of  this  kind,  we  cube  the  diameter  by  multiplying  it  by  itself 
3  times,  as  shown  in  tlie  oi>eration,  and  then  multiply  this  result  by  .5236,  which  is  the  ratio 
between  the  solidity  of  a  cube  aud  that  of  a  sphere,  whoso  diameter  is  equal  to  one  side  of  the 
cube. 

Note. — To  find  the  side  of  .a  cube  which  may  be  cut  from  a  given  sphere,  square  the  diarn* 
eter,  divide  by  3,  and  extract  the  square  root  of  the  quotient. 

2.    What  are  the  solid  contents  of  a  sphere  whose  diameter  is  12  feet  1 

Ans.  904.7808  cubic  feet. 


TO  FIND  THE  SURFACE  OF  A  SPHERE. 


PROBLEMS. 

771.     1.     In   Problem  No.  1  above,  what  is  the  surface  of  the  sphere? 

Ans.  50.2656  sq.  ft. 

OPERATION. 

4  X  3.1416  =  12. 5(!64  circumference.  SxpJana«07!.— In  all  problems  of  this  kind,  we 

12.5664  X  4  =  60.2656  sq.  ft.  surface.  m"ltiply   the   circumference   by    the    diameter, 

'■  and  the  result  13  the  surface. 

2.    What  is  the  surface  of  a  sphere  whose  diameter  is  12  feet  ? 

Ans.  453.3904  sq.  feet. 


378 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

TO  FIND  THE  SOLIDITY  OF  A  HEMISPHERE. 


PROBLEM. 


772.     How  many  cubic  inches  in  a  liemispkere  wliose  diameter  is  GO  iuclies 
and  radius  30  inches  T  Aus.  56548.8  cu.  in. 


OPERATION    INDICATED. 


CO 

CO 
30 
.5236 


Explanation. — In  all  problems  of  this  kind, 
multiply  the  square  of  the  diameter  hy  the 
radius,  and  multiply  the  product  by  .5236. 


TO  FIND  THE  SOLIDITY  OF  A  SEGMENT  OF  A  SPHERE, 

PROBLEMS. 

773.     3.    How  many  cubic  inches  in  a  spherical  segment  which  has  a  diam- 
eter of  60  inches  and  a  deptli  of  20  inches  *  Ans.  32463.2  cu.  in. 

OPERATION. 

CO  4-  2  =  30  in.  radius. 

30=  X  3  =  2700. 

2700  +  20^  =  3100  =  three  times  the  square  of  the  radius 

and  the  square  of  the  depth. 
3100  X  20  X  .5236  =  32463.2  the   cu.  in.  iu    the  spherical 

segment. 

ExpUnialion. — In  all  problems  of  this  kind,  to  three  times  the  square  of  the  radius,  -ire  add 
the  square  of  the  depth  or  height ;  then  multiply  this  sum  by  the  depth  and  the  product  by  .5236. 


ment  an 


Note. — When  the  segment  is  fjreater  than  a  hemisphere,  find  the  solidity  of  the  lesser 
and  subtract  the  same  from  the  solidity  of  the  entire  sphere. 


seg- 


2.    What  is  the  solidity  of  a  spherical  segment,  the  height  of  which  is  6  feet 
and  the  diameter  of  the  base  20  feet?  Ans.  1055.5776  cu.  ft. 

Note  1. — To  find  the  surface  of  a  segment  of  a  sphere,  multiply  the  height  by  the  circumfer- 
ence of  the  segment,  and  add  the  j)roduct  to  the  area  of  the  base. 

Note  2. — To  find  the  solidity  of  a  spherical  zone,  or  the  frvntvm  of  a  sphere,  add  to  the  sum  of 
the  squares  of  the  radii  of  the  two  ends  or  bases,  i^  of  the  sijuare  of  the  height  of  the  zone  or  frus- 
tum ;  then  multiply  this  sum  by  the  height  and  this  product  by  1.5708. 


MENSURATION    OF    SOLIDS. 
TO  FIND  THE  SOLIDITY  OF  A  PROLATE  SPHEEOID. 


379 


PROBLEM. 

774.     A  prolate  spheroid  has  a  transverse,  or  longer  diameter  of  8  feet,  and 
a  conjugate,  or  shorter  diameter  of  5  feet.    How  many  cubic  feet  does  it  contain  1 

Ans.  104.72  cu.  ft, 

OPERATION. 

5  =  diameter.  •  8 

5  =  diameter.  or,  5 

.7854  =  ratio  of  circle,  etc.  5 

8  =  height  or  transverse  diameter.  .5236 

2  =  ratio  between  0.  &  P.  S.  —    

104.72  cu.  ft.,   Ana. 


104.72  cu.  ft.,  Ans. 

Explanation. — In  the  first  Btatement,  we  indicate  the  solution  for  a  cylinder  of  equal  beigbt 
and  diameter  as  the  iirolate  spheroid,  and  then  multiply  by  J,  because  a  prolate  spheroid  is  equal 
to  J  (if  a  cylinder  of  equal  height  and  diameter. 

In  the  second  statement,  we  indicate  the  solution  for  a  rectangular  solid,  by  multiplying 
together  the  three  dimensions,  and  then  multiply  l)y  .5236,  which  is  the  ratio  between  the  solidity 
of  a  cube  and  that  of  a  sphere,  the  diameter  of  which  is  equal  to  one  side  of  the  cube. 


TO  FIND  THE  SURFACE  OF  A  PROLATE  SPHEROID. 

PROBLEM. 

775.     In  the  above  problem,  what  is  the  surface?         Ans.  104.7849  sq.  ft. 

OPERATION. 

8^  =  64.  5=  =25.  64  +  25  =  89  -^  2  =  44i  ;  ^"44^  =  6.6708  X  3.1416 
X  5  =  104.7849264  sq.  ft. 

Explanation. — In  all  problems  of  this  kind,  square  the  diameters,  and  multiply  the  square 
root  of  half  their  sum  by  3.1416,  and  this  product  by  the  conjugate  or  shorter  diameter.  The  result 
will  be  the  surface. 


y 


TO  FIND  THE  SOLIDITY  OF  AN  OBLATE  SPHEROID. 

PROBLEM. 

776.     An  oblate  spheroid  has  a  height  or  shorter  diameter  of  5  feet,  and  a 
width  or  longer  diameter  of  8  feet.    How  many  solid  feet  does  it  contain  ? 

Ans.  167.552  cu.  ft. 

OPERATION. 

8  =  diameter.  8 

8  =  diameter.  or,  8 

.7854  =  ratio  of  circle,  etc.  .5236 

5  =  height  or  conjugate  diameter.  5 

2  =  ratio  between  C.  &  O.  S.  

167.652  cu.  ft,  Ans. 

167.552  en.  ft.,  Ans, 
Explanation. — In  the  first  statement,  we  Indicate  the  solntion  for  a  cylinder  of  eqnal  heigbt 


38o 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


and  diameter  as  the  oblate  spheroid,  and  then  multiply  by  J,  since  an  oblate  spheroid  is  equal  to  } 
of  a  cylinder  of  equal  height  and  diameter. 

For  au  explauation  of  tho  second  statement,  see  the  explanation  in  the   problem  under 
Article  774. 


TO  FIKD   THE  SUEFACE  OF  AN.  OBLATE  SPHEROID. 

PROBLEM. 

777.    In  the  above  problem,  Low  many  square  feet  on  the  surface  ? 

An.s.  1C7.C558+  sq.  ft. 

OPERATION. 


52  =  25.     8=  =04.     25  +  04  =  89  -V-  2  =  44.J.      ■/  44J  =  6.G70S  x  3.1416 
X  8  =  1G7.G5588224  sq.  ft. 

Explanation. — In  all  problems  of  this  kind,  square  the  diameters  and  multiply  the  square 
root  of  half  the  .sum  by  3.141G,  and  this  product  by  the  transverse  or  longer  diameter.  The  result 
will  be  the  surface. 


TO  FIND  THE  SOLIDITY  OF  A  SEMI-PROLATE  SPHEROID. 


PROBLEM. 


778.  How  many  cubic  inches  in  one-half  of  a  prolate  spheroid,  the  conjugate 
diameter  A  B,  of  -which  is  00  inches,  and  the  semi-transverse  diameter  C  D,  being 
40  inches  ?  Ans.  75398.4  cu.  in. 

OPERATION    INDICATED. 


00 
60 
40 
.5236 


or  thus:  60=  x  40  x  .5236. 

Note.— See  Article  774,  page  379. 


TO  FIND  THE  SOLIDITY  OF  A  SEMI  OBLATE  SPHEROID. 

PROBLEM. 

779.    How  many  cubic  inches  in  one-half  of  an  oblate  spheroid,  -whose  longer 
diameter  A  B  is  80  inches,  and  whose  semi-conjugate  diameter  0  D  is  20  inches? 

Ans.  67020.8  cu.  in. 


MENSURATION    OF    SOLIDS. 


?8i 


80 

SO 
•20 
.5236 


OPEKATION    INDICATED. 

or  thus:  SO-  x  20  x  .5230. 

Note. — See  Article  770,  page  379. 


TO  FIND  THE  SOLIDITY  OF  A  SEG:\IENT  OF  A  PROLATE    SPHEROID. 


rROBLEM. 

780.  Ill  a  prolate  splieroid,  Laving  a  transverse  diameter  A  B  of  90  iucLes, 
and  a  conjugate  diameter  C  D  of  60  inches,  tlie  height  of  the  segment  E  B  F  is  20 
inches.     Wliat  is  the  solidity  of  the  segment?  Ans.  202S9.5  cu.  iu. 

9G" 
^  OPERATION. 


DOO" 


9G  X  3  =  288 ;  20  X  2  =  40  ;  288  - 

X  .5236  =  51941.12  inches 
Then  96=  :  GO-  :  :  51941.12  :  20289.5. 


40 


24S;  248  X  20' 


Explanation. —  la  all  problems  of  this  kind,  multiply  tbe  transverse 
diameter  by  3;  then  from  this  product,  subtract  two  times  the  height 
of  the  segment;  then  multi]dy  the  remainder  by  the  Bijuare  of  the 
height  of  the  segment  and  this  product  by  ..")236;  then  multiply  this 
result  by  the  sijuare  of  the  conjugate  djameter  and  divide  the  product 
by  the  square  of  the  transverse  diameter. 


TO  FIND  THE  SOLIDITY  OF  A  SEGMENT  OF  AN  OBLATE   SPHEROID. 


PROBLEM, 


7S1.  In  an  oblate  splieroid  the  transverse  diameter  A  B  is  96  inches,  the 
conjugate  diameter  C  D  is  CO  inches,  and  the  height  of  the  segment  E  D  F  is  10 
inches.     What  is  the  solidity  or  capacity  of  the  segment?  Ans.  8770.3  cu.  in. 

OPERATION. 

96  X  3  =  2SS;  10  x  2  =  20;  288  —  20  =268;  268 

X  10=  X  .5236  =  14032.48. 
Then  96  :  60  :  :  14032.48  :  S770.3, 

Explanation.— In  all  problems  of  this  kind,  multiply  the 
transverse  diameter  by  3,  and  from  the  product  subtract  two 
tunes  the  height  of  the  segment;  then  multiply  the  remain- 
der by  the  scpiare  of  the  height  of  the  segment  and  this 
product  by  .5236;  then  uniltiply  this  result  by  the  conjugate 
diameter  and  divide  the  product  by  the  transverse  diameter. 


Boo" 


382 


SOULES    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


EEMAEKS  ON  THE  MEASUREMENT  OF  SEGMENTS. 


The  measurement  of  segments  is  attended  with  much  difficulty  iu  the  prac- 
tical affairs  of  life.  As  shown  in  Article  773,  jjage  378,  the  measurement  of  a  seg- 
ment of  a  circle  is  a  very  simple  operation.  But,  as  shown  in  Article  780,  page  381, 
and  Article  781,  page  381,  the  measurements  of  the  segments  of  prolate  and  oblate 
spheroids  are  very  difficult  operations,  and  require  as  elements,  in  their  solutions, 
the  transverse  and  the  conjugate  diameters  of  the  spheroids.  Hence  the  difficulty 
of  the  operation  in  practice,  where  the  diameter  and  the  depth  of  the  segment  only 
are  known.  The  operations  to  measure  the  segments  of  spheres,  of  prolate,  and  of 
oblate  spheroids  arc,  as  shown  in  the  above  problems,  very  different,  and  unless  we 
know  what  kind  of  a  segment  we  have  to  measure,  we  cannot  perform  the  operation 
correctly.  And  this  knowledge  cannot,  in  all  cases,  be  determined  from  the  diam- 
eter and  the  depth  of  the  segment.  To  illustrate  this  point,  we  will  suppose  that  it 
is  required  to  find  the  capacity  of  a  segment  in  the  form  of  a  sugar  kettle,  or  a 
boiler,  whose  diameter  is  70  inches  and  depth  is  25  inches. 

Before  we  can  proceed  with  accuracy,  we  must 
'»"  know  whether  the  segment  is  that  of  a  sphere,  or  of 
a  prolate  or  of  an  oblate  spheroid. 

If  we  proceed  to  measure  it  as  a  segment  of  a 
sphere,  the  result  of  our  operation  will  be  greater 
than  if  we  had  measured  it  as  a  segment  of  a  prolate  siAeroid,  and  less  than  if 
we  had  measured  it  as  a  segment  of  an  oblate  spheroid. 

To  have  measured  it  as  a  segment  of  a  jirolate  or  of  an  oblate  spheroid,  we 
must  have  known  the  transverse  and  conjugate  diameters  of  the  figure  of  which 
the  segment  is  a  part.  And  this  knowledge  we  could  not  have  obtained  from  the 
given  diameter  and  depth  of  the  segment. 

From  the  above,  we  see  the  difficulty  in  practice  of  measuring  segments  of 
spherical  solids.  The  best  that  can  be  done  in  cases  M'here  the  figure  is  not  known, 
and  where  the  dimensions  given  are  limited  to  the  diameter  and  the  depth  of  the 
segment,  is  to  measure  it  as  a  segment  of  a  sphere,  and  then,  if  it  is  possible,  to 
judije  from  the  form  and  dimensions  to  what  class  of  segments  it  belongs;  to  allow 
a  small  deduction  in  case  it  is  a  segment  of  a  prolate  spheroid,  and  to  allow  a  small 
increase  in  case  it  is  a  segment  of  an  oblate  spheroid. 

J^OXE. — In  many  cases  where  the  capacity  of  segments  of  an  tinknown  class  is  required,  the 
most  expiMlitioiis  and  accurate  method  of  determining  the  matter,  would  be  to  discard  mathema- 
tics, and  to  measure  the  vessel  by  a  gallon  measure. 


25" 


MENSURATION    OF    SOLIDS. 


3S3 


TO  FIND  THE  SOLIDITY  OF  THE  MIDDLE  FKUSTUM  OF  A  PIIOLATE 

SPHEROID. 

PROBLEM. 

782.  Tlie  middle  frustum  of  a  prolate  spheroid  i  0,  is  70  inches  in  length,  CO 
inches  v  d,  conjugate  diameter,  aud  50  inches  at  its  ends,  ef  and  g  h.  What  is  its 
solidity  or  capacity?  Ans.  177762.2  cu.  in. 


OPERATION. 

60=  X  2  =  7200  +  502  ^  9700. 
Then   9700  x  70  x  .2618  = 
177762.2  cubic  inches. 


Explanation. — In  all  problems  of  this 
kind,  first  R<|uare  the  conjiiu;ate  rtiaiiie- 
ter  and  multiply  the  product  by  2; 
then,  to  this  product,  add  the  square  of 
the  diameter  of  one  end;  then  nnilti- 
ply  this  sum  by  the  length  of  the  frna- 
tum,  and  the  product  by  .2618. 


TO  FIND  THE  SOLIDITY  OF  THE  MIDDLE  FRUSTUM  OR  ZONE  OF  AN 

ELLIPTIC  SPINDLE. 

PROBLEM. 

783.  In  the  frustum  of  an  elliptic  spindle,  the  length  e  f  is  70  inches,  the 
greatest  diameter  a  i  is  60  inches,  the  least  diameter  50  inches,  and  the  middle 
diameter  g  h,  (equally  distant  from  the  middle  aud  end)  is  56  inches.  What  is  the 
solidity  or  capacity  ?  Ans.  170834.972  cu.  in. 


g_° 


J>   b 


OPERATION. 

00=  +  50=  =  6100  +  {5G  X  2) 
=  1S644. 

Then  18G14  x  70  x  .1309  = 
170834.972. 


Explaiialion. — In  all  problems  of  this 
kind,  a<ld  together  the  squares  of  the 
greatest  and  least  diameters,  aud  the 
S(iuaro  of  double  the  medium  diameter; 
then  multiply  the  sum  by  the  length 
and  the  product  by  .1309. 


TO  GAUGE  OR  FIND  THE  SOLIDITY  OF  CASKS  OR  BARRELS. 

PROBLEM. 

784.     In  a  cask  70  inches  long,  60  inches  conjugate  or  bung  diameter,  and  50 
inches  head  diameter,  how  many  cubic  inches  ? 

OPERATION. 


Ans.  172411.008  cu.  in. 


00  —  50  =  10,  difference  be- 
tween bung  and  head  diame- 
ters. 

10  X  .6  =  6.  4-  50  =  56  av- 
erage diameter  of  cask. 

Then  50=  x  .7854  x  70  = 
172411.008  cubic  inches. 

NoTK. — In  the  three  preceding  problems,  the  cast;  the  middle  friiKliim  of  a  prolate  spheroid, 
and  the  )Hi(((?/e//-».s'(«m  of  an  e///j)/(c  Sj>iHrf/e  are  very  similar  prisma,  and  to  sliow  tlie  dilTereut 
results  by  the  ditl'ereut  methods  of  work  wo  gave  the  same  dimeusious  to  eacli  prism,  with  the 
additional  diameter  to  the  frustum  of  the  elliptical  spindle.     The  results  are  as  follows: 

1.  By  the  method  of  measuring  middle  frustums  of  prolate  spheroids  1T77G2.2. 

2.  By  the  method  of  measuring  middle  frustum  of  eili|)tical  sjiindle  170834.972. 

3.  By  the  method  of  measuring  casks  aud  barrels  172411.008  cubic  inches. 


Explanation. — In  all  problems  of  this 
kind,  add  to  the  lesser  diameter  .5  .6 
or  .7  of  the  difference  between  the 
diameters,  according  as  the  curve  in 
the  staves  may  be  slight,  medium,  or 
above  medium. 


384  SOULE's    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 

TO  FIND  THE  SOLIDITY  OF  A  CYLINDEICAL   EING. 


rROBLEJI. 


785.     What  is  the  solidity  of  a  cylindrical  ring  that  is  3  inches  thick  and 
has  an  inner  diameter  of  15  inches?  Aiis.  399.7188  cubic  inches. 


3  +  13  =  18  X  3"  =  162  X 


OPERATION.  Explanation.— In   all    problems    of   this 

kind,  first  add  to  the  thickness  of  the  ring 

the  inner  diameter,   and    then    multiply 

2.4674  —  399.7188  cubic  inches,      this  sum  hy  the  square  of  the  thickness, 

and  the  product  by  2.4674. 

Note.— To  find  the  convex  surface  of  a  ci/Hndrical  ring  :     First  add  to  the  thickness  of  the  ring 
the  inner  diameter,  then  multiply  this  sum  by  the  thickness,  and  the  product  by  y.86U6. 


TO  FIND  THE  CUBICAL  CONTENTS  OF  A  WEDGE. 

PROBLEMS. 

786.     1.    A  wedge  is  20  inches  long,  6  inches  wide  and  4  inches  thick.    How- 
many  cubic  inches  does  it  contain  ?  Aus.  240  cubic  inches. 

FIRST    OPERATION. 
6  +  (6  X  2)  =  IS;  IS  X  20  X  4  =  1440;  1440  -4-  6  =  240  cu.  in. 

Explnnntinn. — In  all  problems  of  this  kind,  add  to  the  Icnfjth  of  the  edge  or  point,  twice  the 
length  of  tbo  back  or  head;  tben  nmltijily  tliis  amount  by  tbo  length  and  the  thickness  of  the 
wedge  and  divide  the  jiroduct  by  6,  the  quotient  will  bo  the  cubical  contents. 

Note.— When  a  wedge  is  a  true  prism  as  the  above,  the  solidity  ia  equal  to  the  product  of  its 
three  dimensions,  divided  by  2. 

2.    In  the  above  problem,  what  would  have  been  the  number  of  cubic  inches, 
if  the  edge  of  the  wedge  had  been  10  inches  long  ?  Ans.  293"5  cubic  inches. 

OPERATION. 

10  +  (6  X  2)  =  22  X  20  X  4  =  1700  ^  6  =  293§  cubic  inches. 


MEASUREMENT  OF  LOGS,  TIMBER,  LUMBER,  AND  BOARDS. 

787.  There  are  various  kinds  of  scales,  and  rules,  and  tables,  that  lumber 
dealers  have  and  often  use  in  the  measurement  of  lumber,  by  means  of  which  very 
little  calculation  is  required.  But  as  these  scales  and  rules  are  readily  understood 
on  inspection,  and  as  the  results  obtained  by  their  use  are  in  many  cases  only 
approximative,  we  therefore  omit  a  description  of  them,  and  present  accurate  meth- 
ods by  means  of  figures. 

In  the  measurement  of  round  timber,  different  customs  and  methods  prevail 
in  different  localities,  which  we  will  point  out  in  the  problems  presented  a  few  pages 
further  on. 

A  Standard  Saw  Log  is  12  feet  long  and  24  inches  in  diameter. 


ROUND    TIMBER. 


3S5 


ROUND  TIMBER. 


TO  FIND  THE  SOLIDITY  OR  CUBIC  FEET  OF  HOUND  TIMBEE  OR  LOGS. 

PROBLEMS. 

788.     1.     The  lengtli  of  a  log  is  30  feet,  and  tlie  average  girt  or  circumfer- 
ence is  7  feet;  how  many  cubic  feet  does  it  coutaiu?  Aus.  117.G  cubic  feet. 

FIRST  OPERATION, 


I 

7 
30 

8 


117f  cubic  feet,  Ans. 


Explanation. — In  all  prnljlems  of  this  character,  -we 
multiply  the  square  of  the  average  girt  or  circumfer- 
ence by  the  length,  aud  the  product  thus  obtained  by 
8,  aud  then  divide  by  100  or  point  off  two  figures  in 
100     8  this  last  product,   aud  we  Lave   the   solidity    very 

nearly.  This  method  of  work  is  shorter  and  simpler 
than  "the  usual  methods,  aud  -while  the  result  is  a 
little  in  excess  of  tlie  m.athematical  result  based  on 
the  presumption  that  the  log,  or  tree,  is  a  regular 
cylinder,  it  is  nearer  the  exact  measure  of  the  frustum  of  a  cone  than  most  of  the  usual  methods 
of  work. 

The  following  statement  gives  the  correct  mathematical  result   considering  the  log  as  a 
■cylinder : 

(7  X  7  X  .7854  x  30)  ^  (3.141C  x  3.U1G)  =  116.97SG  +  cu.  ft. 


2.  What  are  the  contents  by  the  two  above  methods,  of  a  log  62  feet  5 
inches  long,  that  girts  10  feet  9  inches  at  the  larger  end,  and  3  feet  7  inches  at  the 
emaller  end  ?  Ans,  256.403  cu.  ft.  by  the  first  method, 

255.108  cu.  ft.  by  the  second  method. 


Vt. 

10 

3 


2)  14 


OPERATION  BY  THE  FIRST  METHOD. 


In. 
9  girt  of  larger  end. 
7  girt  of  smaller  end. 


12 

12 

12 

100 


2  ai>proximate  average  girt  or 
circumference. 


86 
86 
749 


256.463 -t-  cu.  ft.,   Ans, 


OPERATION  BY  THE  SECOND  METHOD. 


12 

86 

3.1416 

12 

86 

3.1416 

,7854 

12 

749 

255.108-1-  cubic  feet,  Ans 

3,    A  log  is  50  feet  long,  40  inches  diameter  at  the  larger  end  aud  20  inr'ies 
diameter  at  the  smaller  end.    How  inauy  cubic  feet  does  it  contain,  measured  by 


?S6 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


the  first  above  method,  and  how  many  when  measured  accurately  as  the  frustum  of 
a  cone  ?  Ans.  24G.741  cu.  ft.  by  first  method. 

'2'A.5li7  cu.  ft.  accurate  ujeasurement. 

FIRST   METHOD. 

40  X  3.141G  =  125.GG40  circumference  of  larj^er  end. 
20  X  3.141G  r=    Gli.83li0  circutnterence  of  smaller  end. 
125.GG4  +  G2.S33  =  188.490  -^  13  =  94.248  approximate  average   cir- 
cumference or  girt. 
Then  (94.248=  x  50  x  8)  -^  (144  x  100)  =  240.741+  cubic  feet. 

OPEEATION  FOB  ACCURATE  MEASUREMENT  AS  A  FRUSTUM  OF  A  CONE. 

402  =  1600  ) 

202  ^    400  C  2800  -^-  3  =  933.,^  accurate  area  of  a   square    one   side  of 

40  X  20  =    800  )  which  is  equal  to  the  average  diameter  of  the  log. 

Then  (933^  x  50  x  .7854)  h-  144  =  254.527+  cubic  feet. 

Note. — This  is  tbe  only  method  that  is  mathematically  correct;  but  it  is  not  adopted  by 
lumber  dealers  when  buying  or  selling  logs.  They  make  allowance  for  crooks,  sidits  and  other 
defects,  and  hence  adopt  approximate  methods.  Logs  are  more  generally  bought  by  board  meas- 
ure, and  the  different  methods  adopted  for  board  measure  will  be  presented  under  the  heading  of 
Lumber  and  Board  ileaiure,  two  jjages  following. 


TO  KEDUCB  EOUND  TIMBER  TO  SQUARE  TIMBER, 

PROBLEM. 

789.     How  many  solid  feet  of  square  timber  are  there  in  a  stick  of  round 
timber  28  feet  long,  34  inches  iu  diameter  at  the  larger  end,  and  21  at  the  smaller 


end? 


OPERATION. 


34    in.,  diameter  of  larger  end. 

21     "  "         "  smaller  " 


Ans.  05+  cubic  feet. 
55 


3 

12 

3 

12 


28 


65JJ-|cu.ft.  Ans, 


2  )  65 

3  )  27^  in.,  approximate  average  or  mean  diameter. 

9j  =  i  deducted. 

18^  in.,  average  of  each  side  of  the  square  stick. 

Explanation. — In  working  problems  of  this  character,  different  methods  are  used  by  different- 
lumber  dealers.  In  this  solution  we  allowed  i  of  the  average  diameter  for  the  Bquaring  of  the 
stick,  and  thus  obtained  18^  inches  as  the  average  measure  of  one  side  of  the  square  stick.  Then 
having  a  square  stick  (18J  inches  by  18^  inches),  we  multiply  the  three  dimensions  together  in  the 
unit  of  feet  and  obtain  cubic  feet. 

This  method  of  work  is  generally  used,  but  where  the  timber  is  large,  and  especially  where 
there  is  but  little  taper  to  the  stick,  the  deduction  of  i  is  found  to  be  too  great,  hence  some 
dealers  deduct  but  i  of  the  average  diameter  instead  of  J. 

To  work  the  foregoing  problem  by  deducting  J,  we  produce  the  following 

figures  and  result: 

4  )  27jJ  in.,  average  diameter  as  above. 
6|  in.,  =  ^  deduction. 

20§  in.,  average  of  each  side  of  the  square  stick, 

82ifjcu,  ft.,   Ans. 


8 

165 

8 

105 

12 

12 

28 

ROUND    TIMBER. 


587 


TO    FIND    THE    CUBICAL    OK    SOLID    CONTENTS    OF    SQUAEED     OR 

FOUli-SIDED  TIMBER. 


PROBLEM. 

790.    A  piece  of  timber  is  34  feet  long,  16  iuches  ■wide,  and  15  incbes 
thick ;  how  many  solid  feet  does  it  coutaiii  ?  Ans.  56§  cubic  feet. 

OPERATION. 

34  Explanation. — In  all  problems  of  this  character,  we 

16  multiply  the  length,  width  anil  thickness  together  in 

the  same  unit  of  measure,  and  the  result  is  the  cubical 

contents. 


12 
12 


15 


56§  cubic  feet,    Ans. 


TO  FIND  THE   CUBICAL  CONTENTS  OF   SQUARED    OR    FOUR-SIDED 
TIMBER  THAT  TAPERS  REGULARLY. 


PROBLEM. 


?91.     The  larger  end  of  a  rectangular  stick  of  timber  is  25  inches  wide  by 
20  inches  thick  ;  the  smaller  16  inches  wide  by  12  inches  thick,  and  the  length  is  30 


feet;  how  many  cubic  feet  does  it  coutaiuf 


Ans.  697^2  cubic  feet. 


Larger  end 
Smaller    " 


25  X  20  = 
16  X  12  = 


41  X 


FIRST    OPERATION. 

500  sq.  in.  in  larger  end. 
192  sq.  in.  in  smaller  end. 

1312  =  4  times  the  middle  sec- 
tion  between  the  ends. 


12 
12 


334 


30 


6  )  2004 


69;^,  cubic  feet,   Ans. 


334  sq.  in.,  average  area  of  the  stick. 
Explanation. — In  all  problems  of  this  character,  -nhcro  the  taper  of  the  adjacent  sides  is  not 
in  the  same  ratio,  we  tirst  multiply  together  the  width  and  thickness  of  the  two  ends,  and  then 
add  together  tlio  widths  of  the  two  ends  and  also  the  thickness,  and  multiply  together  their  sums. 
We  then  add  the  three  products  together  and  divide  the  sum  by  6,  which  gives  the  average  area 
of  the  stick  in  square  incbes,  which  we  reduce  to  the  unit  of  feet,  by  dividing  by  12  x  12,  and 
multiply  by  the  leugth,  and  obtain  cubic  feet. 

Note. — AVhen  the  timber  does  not  taper  regularly,  measure  separately  the  parts   that  do, 
and  take  the  sum  for  the  entire  solidity. 
SECOND  OPERATION  BY  THE  PEISMOIDAL  FORMULA,  ELUCIDATED  ON  PAGE  376. 
25    X  20  =  500  sq.  in.  in  larger  end. 

16    X  12  =  192  sq.  in.  in  smaller  end. 


2  )  41     X  32 


20J  X  10  =  328  X  4  = 


1312  =  4  times  the  middle  section  between  the  2 
areas. 


6  )  2004 


Then  334  x  30 


334  =  sq.  in.  average  area  of  the  stick. 
144  =  69-iKj-  cubic  feet,  Ans. 


3S8  SOULe's    rHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

TO    FIND    THE    LENGTH   THAT    MUST    BE    CUT    OFF    OF    A    SQUAEE 

riECE  OF  TIMBER  WHICH  DOES  NOT  TAPER,   TO  OBTAIN 

A  GIVEN  SOLIDITY. 

PROBLEMS. 

792.     1.     A  piece  of  timber  is  8  iuclies  wide  and  G  inches  thick,  hovr  mucli 
must  be  cut  off  to  make  1  cubic  foot  ?  Aiis.  3G  inches. 

OPERATION. 

1  cubic  foot  =  1728  cubic  in.  8     1728 

8  X  C  =  48  square  inches.  or,  G 

1728  -^  48  =  3G  in.,   Aus. 

36    in.,  Aus. 

2.     A  piece  of  timber  is  24  inches  wide  and  16  inches  thick;  what  length  is 
required  to  make  4  cubic  feet?  Ans.  18  inches. 

OPERATION. 


24 
10 


4 
172S 


IS      inches.,   Ans. 

3.     A  stick  of  timber  is  4  iuclies  thick  and  16  inches  wide;  how  much  must 
be  cut  off  to  make  6  cubic  feet?  Ans.  13J  feet. 


LUMBER  AND  BOARD  MEASURE, 


793.  Lumber,  as  the  term  is  here  used,  inchides  boards,  planks,  joists, 
scantlings,  and  sawed  timber  of  all  kinds. 

794.  A  Standard  Board  is  one  that  is  12  feet  long,  12  inches  wide,  and  1 
inch  thick.  Hence  1  Board  Foot  is  12  inches  long,  12  inches  wide,  and  1  inch 
thick,  or  1  foot  long,  1  foot  wide,  and  1  inch  thick,  and  contains  144  square  or  board 
inches. 

Note. —  Since  1  board  fodt  contains  Imt  144  hoard  inches,  there  are  12  times  as  many  hoard 
feet  as  cubic  feet  in  lumber,  timber,  and  logs.  Hence  to  change  board  feet  to  cubic  feet,  divide  by 
12 ;  and  to  change  cubic  feet  to  board  feet,  multiply  by  12. 

795.  A  Standard  Saw  Log  is  12  feet  long  and  24  inches  in  diameter. 


BOUND  TIMBER  OR  LOGS  REDUCED  TO  BOARD  MEASURE. 
796.    There  are  various  methods  of  ascertaining  approximately  the  number 


LUMBER  AND  BOARD  MEASURE. 


589 


of  feet  of  square  edged  inch  boards  that  can  be  cut  from  a  given  log;  dififereut 
methods  are  adopted  by  diflerent  lumber  dealers,  which  give  approximate  results 
sufficiently  accurate  to  answer  practical  inirposes. 

The  actual  number  of  feet  varies  according  as  the  logs  vary  iu  not  being  per- 
fectly straight  cylinders,  and  according  to  the  manner  of  sawing  them.  If  the  log 
is  first  sawed  into  inch  boards,  and  each  board  is  then  separately  edged,  it  will  give 
several  feet  more  than  if  the  log  is  first  slabbed,  and  then  sawed  into  inch  boards. 

PROBLEM. 

797.  now  many  feet  can  be  sawed  from  a  log  20  feet  long,  and  18  inches  in 
diameter  at  the  smaller  end  ?  Aus.  245  feet. 

OPERATION  BY  THE  DOYLE  METHOD,  OR  RULE. 


16 


18 

in 

4 

in 

14 

in 

14 

20 

diameter. 

allowed  to  square  the  log. 

each  side  of  the  square  stick. 


245  sq.  feet,  Ans. 


ExpJanation. — By  this  method,  4  inches  are 
deducted  from  the  diameter  in  order  to 
reduce  the  log  to  a  square.  Then  the  square 
of  the  remainder  will  be  the  board  feet  of  a 
log  16  feet  long.  Then  since  tlie  yield  of 
logs  is  in  the  ratio  of  their  lengths,  we 
therefore  divide  by  16  to  obtain  the  board 
feet  in  a  log  1  foot  long,  and  uniUiply  by  20 
to  obtain  the  board  feet  iu  a  log  20  feet  long. 


GENEEAL  DIEECTIONS  FOR  DOYLE'S  EULE. 

798.  From  the  diameter  of  the  smaller  end  deduct  4  inches ;  then  square 
the  remainder  and  multiply  the  product  by  the  length  of  the  log  given,  and  divide 
this  product  by  10. 

The  above  method  is  the  basis  of  the  table  given  in  Scribner's  Lumber  and 
Log  Book,  which  is  xirobably  used  more  generally  than  all  other  books  of  this  kind. 
Though  this  method,  from  a  mathematical  jioint  of  view,  is  only  approximate,  it  is 
practical  iu  its  simplicity  and  is  regarded  M'itli  favor  by  lumber  dealers. 

This  method  favors  tlie  buyer  in  the  case  of  small  logs,  and  the  seller  in  the 
case  of  large  logs.  For  logs  14  inches  in  diameter,  tliis  method  gives  the  same 
result  as  squaring  the  log  and  allowing  ^  for  saw  kerf.  For  logs  35  inches  iu 
diameter,  it  gives  %  of  the  solid  contents;  and  for  logs  50  inches  in  diameter  this 
method  is  equivalent  to  allowing  nothing  for  slab  and  less  than  ^  for  saw  kerf. 

In  the  combination  of  figures,  to  produce  this  method,  it  is  estimated  that 
the  saw  cuts  :}  of  an  inch  each  time  it  goes  through  the  stick,  and  as  tlie  standard 
board  is  1  inch  thick,  it  is  clear  that  ^  of  the  stick  is  cut  away  by  the  saw,  and 
accordingly  an  allowance  is  made  therefor. 

The  Doyle  method  is  adopted  by  the  New. Orleans  lumber  dealers,  with  the 
following  regulations  regarding 


CLASSIFICATION",  INSPECTION  AND  DEDUCTION. 
799.    Logs  are  classified  as  cypress,  jiine  and  oak. 


390 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


DEDUCTIONS. 


For  tlie  liollowTmtts  of  cypress,  deduft  the  leiigth  of  the  hollow.     For  the  red 
heart  of  pine,  and  for  all  i)ecky  logs,  deduct  10  feet  from  the  length  of  all  such  logs. 
For  crooks,  knots  and  wind  shakes,  deductions  are  made  as  per  agreement. 


OTHER  METHODS. 

800.  Some  lumber  dealers  adopt  the  following  method :  Subtract  from 
the  diameter  of  the  smaller  end  expressed  in  inches,  ^  or  ^  of  itself  according  as  the 
logs  are  small  or  large,  and  multiply  the  square  of  the  remainder  by  the  length  in  feet, 
and  divide  the  product  by  12. 

PROBLEM. 

How  many  board  feet  in  a  log  22  feet  long,  and  28  inches  in  diameter  at  the 
smaller  end,  working  iirst  by  the  Doyle  method,  second  by  deducting  J,  and  third 
by  deducting  \  of  the  diameter? 

Ans.  By  the  Doyle  method,  792  board  feet ; 
by  deducting  J,  G3S|f  board  feet; 
by  deducting  \,  808J  board  feet. 

OPERATION. 

By  the  Doyle  luetUod. 

28 
4 


10 


24 
24 


792  ft.,  Ans. 


OPERATION. 

OPERATION. 

When  \  of  the   diameter  is 

When  i  of  the  diameter  is 

deducted. 

deducted. 

28 

28 

9i 

7 

18§ 

21 

21 

3 

56 

12 

22 

3 

56 

12 

O'^ 

S08J  ft.,  Ans. 

G38tf  ft.,  Ans. 

TO    FIND    THE    BOAED    OR    SQUARE    FEET    IX    PLANKS,    GIRDERS, 
SCANTLINGS,  JOISTS,  AND  SQUARE  TIMBER. 

PROBLEMS. 

801.     1.    How  many  square  feet  of  lumber  in  a  board  16  feet  4  inches  long 
and  15  inches  wide  ?  Ans.  20i\  sq.  ft. 

OPERATION. 


49 
5 


or. 


12 
12 


20  ,\,  Ans. 


196 
15 


Explanation. — In  all  problems  of 
this  kind,  we  multiply  the  length 
and  width  together  in  the  unit  of 
feet. 


20 1%,  Ans. 

Practical  Solution. — Since  the  board  is  15  inches  wide,  it  contains  IJ  square 
feet  for  each  foot  of  length.     Hence  IJ  times  IGJ  =  20-j%  square  feet.  Ans. 


LUMBER  AND  BOARD  MEASURE. 


391 


2.     A  board  is  20  feet  C  inches  long,  21  inclies  wide  at  one  end  and  15  inches 
at  the  other  end.     How  many  square  feet  does  it  contain  1  Ans.  30J  sq.  ft. 

OPERATION. 


21  in.,  wider  end. 
15  in.,  uarrower  end. 


2 
12 


2  )  3G 


41 
IS 

30|  sq.  ft.,  Ans. 


18  in.,  average  or  mean  width. 


Practical  Solution. — Since  the  average  widtli  of  the  board  is  IS  in.  wide,  it 
contains  1  i  sq.  ft.  for  each  foot  of  length.    Hence  1|  times  20J  =  30J  sq.  ft.  Ans. 

3.     How  many  square  feet  in  a  plank  24  feet  long,  22i  inches  wide,  and  3 
inches  thick  ? 

OPERATION. 

24 


2 
12 


45 
3 

135  sq.  ft.,  Ans. 


Ans.  135  sq.  ft. 

Explanation. — As  the  board  foot  is  1  inch  in  thickness, 
it  is  clear  tliat  when  the  tliickiiess  exceeds  1  inch  the 
measurement  must  be  increased  accordingly  ;  hence  when- 
ever the  thickness  exceeds  1  inch,  we  multiply  by  the 
thickness.  When  the  thickness  is  less  than  1  inch,  by 
custom,  no  deduction  is  made,  the  measurement  bekig  in 
that  case  the  same  as  if  the  lumber  was  1  inch  thick. 

Practical  Solution. — Since  the  board  is  22J  in.  wide,  and  3  in.  thick,  it  is 
practically  C7J  sq.  in.  or  5if  sq.  ft.  for  each  foot  of  length.  Hence  5jf  times  24  = 
135  sq.  ft.  Ans. 

4.  "What  is  the  number  of  board  feet  in  16  pieces  of  scantling  each  20  feet 
long,  4  inches  wide,  and  3  inches  thick?  Ans.  320  board  feet. 

Practical  Solution. — 4  in.  wide  and  3  in.  thick,  is  equivalent  to  12  in.  wide  or 
1  sq.  foot  for  each  foot  of  scantling;  and  as  there  are  IG  scantlings,  hence  20 
times  16  is  320  sq.  ft.  Ans. 

5.  How  many  board  feet  in  200  girders  each  30  feet  long,  15  inches  wide, 
and  2  inches  thick.     And  what  will  they  cost  at  $18  per  M.  ? 

Ans.  15000  board  feet ;  $270  cost. 

Practical  Solution. — 15  in.  wide,  by  2  in.  thick  =  2J  sq.  ft.  to  the  foot.  Hence 
2J  times  30  =  75  x  200  =  15000  sq.  ft. ;  and  $18  x  15m"  =  $270.  Ans. 

6.  What  will  be  the  cost  of  4  black  walnut  boards,  each  10  feet  9  inches 
long,  2SJ  inches  wide,  and  13  inches  thick,  at  $:Jo  per  M.?  Ans.  $9.S29f5. 

7.  A  piece  of  timber  is  34  feet  long,  10  inches  wide,  and  15  inches  thick. 
How  many  board  feet  does  it  contain  ?  Ans.  G80  sq.  ft. 

Practical  Solution. —  IG  in.  =  IJ  ft.  wide  by  15  in.  thick  =  20  sq.  ft.  for  each 
foot  of  length.     Hence  20  times  34  =  GSO  sq.  ft.  Ans. 

8.  A  rectangular  telegraph  pole  is  CO  feet  long,  IG  inches  square  at  the 
larger  end  and  G  inches  square  at  the  smaller  end.  How  many  cubic  and  how 
many  board  feet  does  it  contain  ?  Ans.  53f  cubic  feet;  GlGg  board  feet. 

Note.— See  Article  761,  pag«  373. 


392 


SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


9.  A  circular  telegraph  pole  is  60  feet  long,  16  inches  in  diameter  at  the 
larger  end,  and  6  inches  in  diameter  at  the  smaller  end.  Sow  many  cubic  and  how 
many  board  feet  does  it  contain,  and  what  is  the  cost  at  ISC'  per  cubic  foot? 

Ans.  42.3243i  cubic  feet;  507.892  board  feet ;  $6.34865  cost. 

Note. — See  Article  708,  page  375. 

10.  How  many  superficial  or  board  feet  in  40  scantlings,  each  15J  feet  long,  4 
inches  wide,  and  2  inches  thick  ?  Ans.  413 J  feet. 

SOLUTION   STATEMENT. 

Practical  Solution. — 4  x  2  =  8  in.  =  §  sq.  ft.     15J  X 
40  =  620  X  §  =  413i  sq.  ft. 

or  thus : 
15J  X  40  =  620  —  J  of  itself  =  (206§)  413^  sq.  ft.,  Ans. 
413^  sq.  ft.,  Ans. 

11.  How  many  square  feet  in  62  joists,  each  20  feet  long,  14  inches  wide,  and 
3  inches  thick;  and  what  will  be  the  cost  at  $30  per  thousand? 

Ans.  4340  feet.     $130.20,  cost. 

12.  How  many  square  feet  (board  measure)  in  a  square  piece  of  timber,  20 
feet  8  inches  long,  14  inches  wide,  and  9  inches  thick  1  Ans.  217  feet. 

13    "What  will  be  the  cost  of  840  feet  of  boards  at  $22.50  per  M.,  and  of  240 


2 

31 

2 

4 

2 

40 

feet  of  flooring  at  $32.75  per  M.  ? 


Ans.  $26.89i. 


14.  The  width  of  a  board  is  16  inches,  what  must  be  its  length  that  it  may 
contain  24  square  feet  ?  Ans.  18  feet. 

OPERATION. 

24  sq.  ft.,  contents  of  the  board. 
144  sq.  in.,  in  each  square  foot.         3456  sq.  in.  -^  16  in.,  the  width,  =  216  in.  -=-  12 

. =  18  feet,  the  length  of  the  board. 

3456  sq.  in.,  contents  of  the  board. 

15.  A  plank  is  2  inches  thick  and  14  feet  long,  how  wide  must  it  be  to  contain 
42  feet  ?  .  Ans.  18  inches. 

OPERATION. 

42  sq.  feet  in  the  board. 


144  sq.  inches  in  each  sq.  foot. 


6048  sq.  inches. 

14  feet,  length  of  plank. 
12  in.  in  1  foot. 


0048  4-  168  =  36  inches,  width  of  plank  1  inch 
thick. 

30  4-  2  =  18  inches,  width  of  jjlank  2  inches 
thick. 


108  inches,  length  of  plank. 

16.     A  board  is  8  inches  wide;  what  length  will  make  2  square  feet? 

Ans.  3  feet. 

OPERATION. 

■^  sq.  feet. 

288  sq.  in.  -i-  8  in.,  the  width  of  the  board, 


144  sq.  inches  in  each  sq.  foot. 
288  sq.  in.  in  2  sq.  feet. 


36  ill.  or  3  feet.    Ans. 


PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS. 


393 


Practical  Problems  in  the  Mensuration  of  Solids, 


TO   FIND    THE   NUMBER   OR   CCTBIG   FEET   AND    INCHES    IN   BOXES 
AND  SOLIDS   OF  DIFFERENT  FORMS. 


deep  I 


PROBLEMS. 

802.     1.     How  many  cubic  feet  in  a  box  6  feet  long,  3  feet  wide,  and  2  feet 
'  Ans.  3G  cubic  feet. 


OPEKATION. 

6  X  3  X  2  =  3G  cu.  ft.,    Ans. 


Explanation. — In  all  problems  of  this  character, 
we  have  but  to  multiply  the  three  dimensions 
together  iu  the  unit  of  feet. 


2.     How  many  cubic  feet  in  a  box  4  feet  3  iucbes  long,  3  feet  wide,  and  16 
inctes  deep  ?  Ans.  17  cubic  feet. 


12 
12 


51 

3 

16 


OPEKATION 
i 


or, 


17  cu.  ft.,  Ans. 


17 

3 

4 

17  cu.  ft.,  Ans. 


Explannlion. — 'U'henerer  any  of 
the  (liniensiiiMs  are  given  in  inches, 
Tve  either  divide  them  by  12,  iu 
order  to  reiluce  them  to  feet,  or 
else  we  use  the  inches  as  a  fraction 
of  a  foot,  as  shown  by  the  second 
line  statement. 


OPERATION. 


3.  How  many  solid  feet  in  8 
boxes,  each  28  inclies  long,  10  inches 
■wide,  and  6  inches  deep  ? 

Ans.  7J  solid  feet. 


12 
12 
12 


28 
10 
6 

8 

75  solid   feet,    Ans. 


4.  How  many  cubic  feet  iu  a  cylinder  6  feet  long,  and  3  feet  4  inches   in 
diameter?  Ans.  52.36  cubic  feet. 

5.  How  many  cubic   feet  in  a  frustum  of  a  cone,  whose  height  is  6  feet 
diameter  of  the  greater  end  is  4  feet  and  of  the  smaller  end  3  feet  ? 

Ans.  58.1196  cubic  feet. 
Note. — See  Problem  on  page  376. 

6.  How  many  solid  feet  in  a  sphere  that  is  30  inches  in  diLxmetor  ? 

Ans.  S.1S125  solid  feet. 

7.  How  many  cubic  inches  in  a  cannon  ball  which  is  13  inches  in  diameter? 

Ans.  1150.3492  cu.  in. 


8.     Suppose  an  orange  to  be  exactly  spherical  and  4  inches  in  diameter,  how 


394 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


many  oranges,  2  inclics  in  diameter  and  splierical  in  form,  will  it  take  to  equal  the 
larger  orange  ?  Ans.  8. 

FIRST   OPERATION. 

4^  =  Gi  eulie  of  larger  orange. 
2^  =    8     "      "  smaller      " 


SECOND 

OPERATION. 

4 

o 

•1 

'> 

i 

') 

.5236 

r5236 

33.5104 

4.1SSS 

4.1SSS  =  8  oranges,   Ans. 


04  -^  8  =  8  oranges,  Ans. 

Explannlinn. — The  first  soliit  inn  of  this  proWom  is  hased  upon  the  geometrical  fact  that  all 
solids  are  to  each  other  as  the  cubes  of  their  lil^e  dimeusious. 

In  the  .second  solution,  Tve  first  find  the  solid  contents  of  each,  and  then  we  find  how  many 
times  the  contents  of  the  larger  orange  is  enual  to  the  contents  of  the  smaller. 

9.  If  an  apple,  spherical  in  form,  and  12  inches  in  circumference,  sells  for 
10  cents,  what  is  the  proportional  value  of  3  apples,  also  sjiherical  in  form,  that  are 
5  inches  in  circumference?  Ans.  2/(2,- <?. 

10.  How  many  cubic  inches  in  a  senii-oblate  spheroid  which  is  70  inches  in 
diameter,  and  30  inches  deep  ?  Ans.  7C969. 2  cu.  inches. 

Note.—  See  Article  779. 

11.  An  orange  peddler  sells  two  oranges  which  are  each  3  inches  in  diameter, 
or  3  oranges  which  are  each  2  inches  in  diameter,  for  5/.  Allowing  the  oranges  to 
be  i^erfect  spheres,  which  is  the  better  purchase,  and  how  many  cubic  inches  of 
orange  would  be  gained  ?  Ans.  The  better  purchase  would  be  the 

2  oranges  each  3  in.  in  diameter. 
The  gain  M'ould  be  15.708  cu.  in. 

TO  FIXD  THE  COST  OF  FEEIGHT  PER  OUBIO  FOOT  OX  BOXES, 

PROBLEMS. 

803.     1.     What  will  be  the  freight  on    a  box  9  feet  3  inches  long,  4  feet  6 


inches  wide,  2  feet  10  inches  deep,  at  30  cents  a  cubic  foot  ? 


Ans.  $35.3SJ. 


2.     What  is  the  freight  on  8  boxes,  which  are  each  3  feet  3  inches  long,  2  feet 
8  inches  wide,  and  21  inches  deep,  at  20/  jjcr  cu.  foot  ?  Ans.  $24.2Gg. 


TO  FIND  THE  NUMBER  OF  BUSHELS  IN  VESSELS,  BOXES,  BINS,  ETC. 


PROBLEMS. 


804.     1.     How  many  bushels  will  a  bin  hold,  that  is  10  feet  long,  8  feet  6 


inches  wide,  and  5  feet  2  inches  deep  ? 


First  Operation. 

10 

12 

102 

62 

100 


215042 


Second  and  Contracted 
Operation. 
10 

17 
31 
.80355 


352.894-bu.  Ans. 


Atis,  352.89+  bushels. 

Explanation.  Inthe  first  operation 
we  multiply  the  three  dimensions 
together,  inthe  unit  of  inches,  which 
gives  cubic  inches,  and  then  divide 
by  2150.42  the  number  of  cubic  in. 
in  a  bushel.  By  reason  of  the  deci- 
mals in  the  2150.42  we  also  multiply 
by  100  to  cancel  the  decimals. 

In  the  second  operation  we  mul- 
tiply the  three  dimensions  together 


352.89+ bu.  Ans, 

m  the  unit  of  feet,'  and  then  multiply  the  product  by  .SO.%55,  which  is  the  decimal  quotient  of 
1728,  the  number  of  cubic  inches  in  a  cubic  loot,  divided  by  2150.42,  the  number  of  cubic 
inches  in  a  bushel.  ,       „  ,         ^       ..•  t 

This  is  a  very  desirable  contracted  method.     See  another  form  of  contraction  on  next 
page,  Problems  2  and  4. 


PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OK    SOLIDS. 


595 


2.  A  -n-agon  box  is  1 2  feet  long,  4J  feet  wide,  and  15  inches  high.    How  many 

bushels  will  it  hold  when  even  full?  Ans.  54:.24r+  bushels. 

Note. — A  very  short  way  to  find  tlie  number  of  bushels  any  box  or  vessel  will  hold  is  to 
ninltiply  the  cubic  feet  it  contains  by  45  and  divide  the  product  by  56.  Tliis  is  correct  because 
the  ratio  between  2150.4  and  1728.0  is  as  5G  to  45. 

3.  A  planter  wishes  to  build  a  bin  that  will  hold  100  bushels;  it  is  to  be  10 
feet  long  and  4J  feet  high.     What  must  be  the  width  of  the  bin  ? 


OPERATION. 


2150.42  =  cu.  in.  in  1  bu. 
100  =  bushels. 


Ans.  33.1S5+  inches. 

120  in.  =  length. 
54  in.  =  height. 


215042.00  =  cu.  in.  in  100  bu.  C4S0  sq.  inches. 

215042  -^  C4S0  =  33.185+  inches,  Ans. 

4.     How  many  bushels  in  a  cylinder  shaped  box,  whose  height  is  10  feet,  and 
diameter  10  feet?  Ans,  G31.125  bu. 


1st   OPEKATION. 

120 
120 
120 
.7854 


2150.4 


2d    OPERATION. 

10 
10 
10 

.7854 
56    45 


Explanation. — In  the  second  operation,  we  first 
malio  tlio  statement  to  find  cubic  feet  in  the 
lyliiuler;  then  instead  of  luultiplylnj;  by  1728 
to  reduce  the  cubic  feet  to  inches,  and  dividinjj 
by  2150.4  to  reduce  the  cubic  inches  to  bushels, 
wo   multiply    by   45   and   divide    by   5(i.     These 

numbers  are  obtained  by   reducini; — ^"^to   its 

'    2150.4 

lowest  terms. 

Note. — In  this  problem,  we  use  2150.4  cu.   in.  as  a  bushel  which  is,  for  ordinary  purposes, 

sufiiciently  correct. 

5.  A  sngar  kettle  in  the  form  of  a  half  of  a  prolate  spheroid  is  42  inches 
deep  and  50  inches  in  diameter.  How  many  bushels,  and  how  many  gallons  will  it 
liold  1  Ans.  25.560+  bushels.     238  gallons. 


liushela. 


2150.42 


42 
50 
50 
.5230 


or, 


25.566  bu. 
Note. — See  Article  778. 


OPERATIONS 

INB 

ICATE 

D. 

Bushel.^. 

Gallons. 

42 

50 

50 

42 

.7854 

50             c 

3 

2 

50 

2150.42 

231 

.5236 

25.566 

bu. 

23«  gals. 

Gallous. 
42 


or. 


231 


50 

50 

.7854 

2 


238  gals. 


TO  FIND  THE  NtHVIBER  OF    CUBIC  YAKDS  IX  LEVEES   OR   EXCAVA- 
TIONS. 


80.5. 


PROBLEMS. 

1.     How  many  cubic  yards  in  a  levee  80  rods  long,  CO  feet  wide  at  the 


base,  12^  feet  at  the  top,  and  5  feet  4  inches  average  depth  ? 

Ans.  9451|2-  cubic  yards. 


396 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


60    width  at  the  base. 
12^       "      "     "     top. 

2)72J 

3C;J  average  width. 


OPEKATION. 

o 

4 
V2 

07 


SO 
33 
145 
04 


Explanation. — Here 
we  first  multiply  the 
three  dimensions  to- 
gether in  the  unit  of 
feet,  and  then  divide 
by  27,  the  number  of 
cubic  feet  in  a  cubic 
yard. 


9451^1  cu.  yds.,  Ans. 

2.  A  railroad  contractor  excavated  a  cut  through  a  hill  840  feet  long,  70 
feet  -wide  on  top,  and  30  feet  wide  on  the  bottom,  average  depth  50  feet.  What 
was  the  cost  at  21/  per  cubic  yard  1  Aus.  $16333.33^. 

3.  A  tunnel  was  excavated  1350  feet  long  with  a  cross  section  or  area  of 
660  sq.  feet.    How  many  cubic  yards,  and  what  was  the  cost  at  30/  per  cu.  yard  ? 

Ans.  33000  cu.  yards.     $9900,  cost. 


TO  FIND  THE  NUMBER  OF 

806.     1.     How  many  gallons 
■wide,  and  3  feet  deep  ? 

OPERATION. 

GO 


231 


28 
36 


261.81+  gal.,  Ans. 
2.    How  many  gallons  in  a 
and  8  feet  4  inches  in  diameter  ? 


231 


GALLONS   IN   VESSELS,  CISTERNS,  ETC. 

PROBLEMS. 

will  a  box  hold,  that  is  5  feet  long,  2  feet  4  inches 
Ans.  261.81+  gallons. 

Explannation, — In  this  problem,  we  multiply  the  three 
dimensions  together  in  the  unit  of  inches,  and  divide  by 
231,  the  number  of  cubic  inches  in  a  gallon. 

cylindrical  cistern  or  tank  12  feet  4  inches  high, 

Ans.     5032  gals. 


uU  Operation. 

148 
100 
100 
.7854 

2d 
C 

Contracted 
)peratiou. 

148 

100 

100 
34 

3dC 

Ol 

3 

3 

,3 

ontracted 
leration. 

37 

25 

25 

5.8752 

4th 

0 

3 

3 

3 

Contracted 
peration. 

37 

25 

25 

5.875 

5t 

3 
3 
3 

8 

h  Contracted 
Operation. 
37  feet. 
25     " 
25     " 
47     " 

5032.0000  gals. 

5032  g.ils. 

5032  gals. 

5031.8287 

5031.8287 

Explanal'wn  to  1st  Operation. — In  all  problems  of  this  character,  Tve  multiply  the  height  iind 
the  square  of  the  diameter  together  in  the  unit  of  inches,  then  the  product  thus  produceil  by  the 
decimal  .7854,  and  then  divide  by  231,  the  number  of  cubic  inches  in  a  gallon. 

Explanation  of  f!d  Operation. — Since  .7854  is  eqnal  to  231,  .0034  times,  we  may  shorten  the 
■work  by  Biinply  multiplying  by  .0034,  omitting  the  division  by  231.  In  the  operation  we  set  34 
only,  remembering  there  will  bo  four  decimal  places  in  the  answer. 

Explanation  of  the  Sd  Operation. — In  this  operation  we  use  the  three  factoral  dimensions  in 
the  nnit  of  feet,  and  multiply  their  product  by  5.8752.  This  5.8752  is  produced  thus:  In  a  cubic 
foot  there  are  1728  cu.  inches ;  this  we  multiply  by  .7854,  which  is  tlie  area  of  a  circle  whose 
diameter  is  equal  to  one  side  of  .a  square,  and  then  divide  the  product  by  231  the  luimber  of  cubic 
inches  in  a  gallon.     Thus  the  5.8752  is  the  number  of  gallons  in  a  cubic  foot  reduced  to  a  cylinder. 

Explanation  of  the  4th  and  5th  Operations. — In  the  4th  operation  we  shorten  the  work  by  mul- 
tiplying by  5.875  instead  of  5.8752,  and  produce  an  answer  nearly  or  practically  correct.  In  the 
5th  operation  we  consider  that  as  .5.875  gallons  equals  5J  gallons,  and  as  5|  equals  V  gallons,  we 
hence,  multiply  by  47  and  divide  by  8. 

NoTK. — The  4th  and  5th  methods  are  so  nearly  correct  that  they  are  used  by  many  builders 
and  engineers.  We  prefer  the  2d  Contracted  Operation,  as  it  is  exact  and  in  the  majority  of  cases 
it  is  shorter  than  the  otlier  methods. 

3.    How  many  pints  in  a  cylindrical  vessel,  whose  height  is  14  inches  and 


diameter  12J  inclies  ? 


Ans,  59.5  pints. 


PRACTICAL    PROBLEMS    IN   THE   MENSURATION   OF   SOLIDS. 


397 


OPERATION. 
14 

2     25 

c-  Explanation. — Here  we  first  fiud  the  mimber  of 
34  gallons,  the  same  as  in  the  preceding  problem,  and 
^j  then  \ye  multiply  by  4,  which  gives  us  quarts,  and. 
then  by  '2,  which  gives  us  pints. 

59.5  pint!?,  Ans. 
4.     How  mail}-  gallons  of  water  will  a  steamboat  boiler  hold,  tliat  is  40  feet 


long,  5  feet  4  inches  iu  diameter,  allowing  for  3  Hues,  each  15  inches  in  diameter  ? 

Ans.  5583.072  gals 

OPERATION. 

TO   FIND  THE    CONTENTS    IN     GALLONS    OF    THE 

WHOLE   BOILKIl. 


40 
12 
64 
G4 

34 


0GS4.C72 
1101.600 


OPERATION. 

TO     FIND     THE     DEDUCTIONS     IN    CONSEQUENCK 

OF   THE   3  FLUES. 


40 
12 
15 
15 
34 
3 


1101.000 


5583.072  gals.,  Ans. 
5.     now  many  gallons  in  a  cistern  which  is  in  the  form    of  a  frustram  of  a 
cone,  whose  height  is  9  feet  6  inches,  lower  base  7  feet  2  inches,  and  upper  base  G 


feet  8  inches  ? 

OPERATION. 

86=  = 

7396       3 

80=  = 

64U0 

86  X  80  = 

6880 

20676 

20676 
114      . 

34 

2671.3392  gals.,  Ans. 


Ans.  2671.3392  gals. 


Explanation. — In  all  prob" 
blems  of  this  character,  we 
first  find  the  contents  or 
capacity  of  the  cistern  in 
cubic  inches,  according  to  the 
principles  elucidated  on  pp. 
376  and  396. 


Note. — 231  cancels  .7854,  giving  a  quotient  of  .0034. 
6.     How  many  quarts  will  a  bucket  hold,  that  is  10  inches  deep,  12J  inches 
in  diameter  at  the  top,  and  9J  inches  at  the  bottom  1  Ans.  16.2038  quarts. 

OPERATION. 


V2i  X 


12J=  =  156  J 
91=  =    85-a 
115  t 


H 


357 


16 

5719 

3 

10 

34 

4 

inches  deep  1 

OPERATION. 
4 

4 
6 
34 

4 


16.2038  quarts,  Ans.     . 
How  many  pints  in  a  conical  cup,  4  inches  in  diameter  at  the  top,  and  6 

Ans.  .8704  pint. 

Explanation. — Th«  contents  of  a  cone  being  i  of  the 
contents  of  a  cylinder  of  equal  height  and  diameter,  we 
proceed  the  same  as  in  the  measurement  of  a  cylinder,  and 
then  divide  by  3.  We  multiply  by  4  and  2  to  reduce  gallons 
to  pints. 


.8704  pt.,  Ans. 


398 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


8.    How  many  gallons  will  a  tub  hold  which  is  34  inches  npper  diameter,  29 
inches  lower  diameter,  and  21  inclies  deep?  Ans.  70.995i  gallons. 

9.  How  many  gallons  in  a  boiler  or  kettle, 
the  lower  part  of  which  is  in  the  form  of  a  hemis- 
phere, with  a  diameter  of  60  inches  and  a  depth 
of  30  inches,  and  the  upper  part  being  in  the  form 
of  a  frustum  of  a  cone,  whose  lower  base  is  60 
inches,  upi)er  base  70  inches  and  the  depth  or  height 
being  12  inches'!  Ans.  417.52  gallons. 

OPERATION 

TO  FIND  THK   CONTENTS  OF  THE 
FRUSTUM   PORTION. 


OrKKATION 

TO   FIND   TUB   CONTENTS   OP  THE 
HEMISI'UERE  PORTION. 


231 


60 

60 

.5236 

30 


or  thus : 


244.8  gals. 


60 
60 

.0022J 
30 

244.8  gals. 


60'  = 
70'  = 
70    X  60: 


3600 
4900 
4200 

12700 


231 


12700 

12 

.7854   or  thus ; 


172.72  gals. 


12700 

12 

.0034 

172.72   gals. 


Note.— The  decimal  .00225  is  the  quotient  of  .5236  —  231;  ami  the  tlecimal  .0034  is  the 
quotient  of  .7854  —  231. 

Explanation.— See  Artirlo  772,  page  378,  and  Article  768,  page  375,  for  a  full  explanation  for 
finding  the  contents  of  hemispheres  and  frustums  of  cones.  The  roasuu  for  dividing  by  231  is  that 
231  cubic  inches  constitute  a  gallon. 


TO  FIND  THE  NUMBER  OP  GALLONS   IN  A   HORIZONTAL  CYLINDER 

WHEN  THE  DIAMETER  AND  THE  LENGTH  OF  THE  CYLINDER 

AND  THE  DEPTH  OF  THE  LIQUID  ARE  GlYEN. 

PROBLEMS. 

807.    1.     How  many  gaUons  in  a  cylinder  IS  feet  4  inches  long,  6  feet  diam- 
eter, the  depth  of  the  liquid  being  12  inches  ?  Ans.  424.7917-f  gals. 

18'  4" 

'/' 
/ 


OPERATION. 


PRACTICAL    TROBLEMS    IN    THE    MENSURATION    OF    SOLIDS. 


399 


Fz'rsi  Siep.—Deiith  of  liquid:  12    in.  ~  the  diameter,  72   in.  =  .IGCCs  = 
tabular  number. 

Seco7id  Step. — See  Table,  page  353,  and  find  tbe  quotient  .16G  in  the  column  of 
versed  sines.  Then  take  the  area  of  segment  noted  in  the  next  column  and  multi- 
ply it  by  tbe  square  of  the  diameter,  the  result  will  be  the  area  of  the  segment  of 
the  cylinder.    Thus : 

The  tabular  area  for  .166  =  .08554 

The  tabular  area  for  .107  =  .0SG29 


Difiference  is  .00075 

.00075  X  .eg  or  .CG+  =  .00050. 
Then  .166      =  .08554 
And  ,00OG§  =  .00050 


.08604  =  the  sum  by  which  the  square  of  the  diameter  is  to 
be  multiplied. 

Third  Step.  —.08604  x  72=  =  446.0313  +  sq.  in.  in  area  of  segment. 


Fourth  Step. — To  find  gallons  : 

446.0313        X  220  in.  length  of  cylinder. 
231  cu.  in.  in. a  gallon. 


424.7917+  gallons. 


Note  1. — To  find  the  area  of  a  segment  greater  than  a  semi-circle,  find  tbe  area  of  the  lesset 
segment,  and  subtract  the  same  from  the  area  of  the  whole  circle. 

Note  2. — To  find  the  area  of  a  zone  of  a  circle,  find  the  areas  of  tbe  two  segments  and  sub- 
tract the  sum  from  tbe  area  of  the  whole  circle. 


2.  How  many  gallons  in  a  tank  of  the  following  description  and  dimen- 
sions: The  upper  portion  is  iu  the  form  a  vertical  rectangular  prism,  13  feet  5  inches 
deep,  11  feet  9  inches  long,  and  9  feet  3  inches  wide.  The  bottom  is  a  segment  of  a 
horizontal  cylinder,  the  chord  being  9  feet  3  inches,  and  the  height  or  versed  sine 
4  feet?  Ans.  13372.404  gallons. 


OPERATION 


First  Step. — To  find  the  gallons  in  the 

cubical  part  of  the  tank : 

IGl  in.  deep. 

141  in.  long. 

Ill  in.  wide. 

161  X  141  X   111         .nonooTO       ii 

TTTT^ ■■ r-  =  10908.272  gallons. 

231  cu.  in.  a  gal.  * 


400  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

Second  Stci). — To  find  tlie  diameter  of  the  circle  of  which  the  given  segment 
is  a  part : 


A  B  =  111  in.  lengtli  of  cliord. 
A  P  =  55J  in.  i  of  chord. 
P  D  =  48  in. 


(55J)  =  -|-48  =  64.17+ in.  =  the  difference  be- 
tween the  diameter  and  the  versed  sine. 

04. 17+  48  =  112.17  in.  diameter  of  circle. 

112.17  H-  2  =.  56.085  in.,  the  radius  of  the 
circle  A  0. 


Third  Stej). — To  find  the  quotient  of  the  versed  sine  divided  by  the  diameter: 
4S  4-  112.17  =  .4279  quotient. 

See  Table  on  page  354,  and  find  the  quotient  in  the  column  ot  versed  sines. 
Then  take  the  area  number  shown  in  the  next  column  on  the  right  and  multiply  it 
by  the  square  of  the  diameter.    The  result  will  be  the  area  of  the  given  segment. 

The  tabular  area  number  for  .427  =  .31996 
The  tabular  area  number  for  .428  =  .32095 


The  difference  is  .00099 

Then,  .00099  x  .9,  which  is  the  difference  between  .4279  and  .427,  =  .000891. 
Hence  .31996  +  .000891  =  .320851  the  sum  by  which  the  square  of  the  diame- 
ter is  to  be  multiplied. 

Thus,  .320851  x  112.17=  =  4036.9828+  sq.  in.  area  of  the  segment  A  D  B. 

Fourth  Step. — To  find  the  gallons  in  the  segment : 

4036.9828  x  141    in.,  length  of  segment  in  tank  _,„,  ^„„       ,, 

^ — ~ n-^ =  2464.132  gallons. 

231  cu.  in.  in  a  gallon 

Add  the  gallons  in  the  cubical  part  of  tank,  as  above  10908.272 


Total  gallons  in  the  tank  13372.404  gals. 

Note. — See  Haswell's  Eugineers'  aud  Mechauics'  Pocket  Book,  for  extended  work  on  Men- 
suration and  on  other  Bubjects. 


TO  FIND  THE  NUMBER  OP  BARRELS  IN  A  CISTERN. 

PROBLEM. 

808.    How  many  barrels  will  a  quadrilateral  cistern  hold,  whose  height  is  12 
feet,  and  width  of  each  side  5  feet  8  inches  ?  Ans.  91 1^^,-  barrels. 

OPERATION. 

144  Explanation. — In  all  problems  of  this  character,  we 

231      08  mnltiply  the  three  dimensions  together  in  the  unit  of 

08  inches,  then  divide  by  231,  which  gives  gallons;  ■vre 

C3     2  then  divide  by  31^,  the  number  of  gallons  in  a  stand- 

ard  or  conventional  barrel,  aud  thus  obtain  the  con- 

91-i*j21   bbls..  Ads.  tents  of  the  cistern  in  the  unit  of  barrels. 


*  PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS.  40 1 

TO  FIKD  THE  NUMBER   OF  CAET  LOADS   OF   EAETH  TO  FILL  LOTS. 


PROBLEMS. 

809.  1.  A  piece  or  lot  of  ground  measures 
63  feet  2J  inches  front  by  ll'O  feet  deep.  How 
many  square  yards  in  the  piece,  and  how 
many  cart  loads  of  earth  will  it  take  to  fill  the 
above  stated  lot  G  inches  deep,  the  cart  box 
cartuos.  being  5  feet  4  inches  long  on  the  top,  4  feet 

8  inches  long  on  tlie  bottom,  3  feet  6  inches  wide,  and  2  feet  deep?     How  many 
cubic  yards  of  earth  are  used  in  the  filling  1 

Ans.  842  J  sq.  yds. ;  140  |f  cu.  yds. ;  lOS-i^-  loads. 


12 
9 


OPERATION 

TO  FIND   SQUARE  YARDS. 

75S.5  =  ft.  front  of  lot. 
120     =  ft.  depth  of  lot. 


842|  sq.  yds.,  Ans. 


OPERATION 
TO   FIND  CUBIC   YARDS   TO   BE   FILLED. 


12 

12 

27 


758.5    =  ft.  front. 
120       =  ft.  depth  or  length. 
6  =  ft.  thickness  to  be  filled. 

=  en.  ft.  in  a  cu.  yd. 

140|-|  cu.  yds. 


2 

27 


OPERATION 
TO  FIND   CAPACITY  OP  THE   CART  BOX. 

7  =  ft.  wide. 

5  =  ft.  average  length  (5^  i"  +  i'  8") 

2  =  ft.  deep.  2 

1/7-  cu.  yds. 


OPERATION 
TO   FIND   NUMBER   OF   LOADS. 

140-|f  -^  1  i-  =  108  A:  loads, 
or  thus : 


54 
35 


7585 

27 

108 1\  loads,  Ans. 


2.  A  yard  is  144  feet  long  and  32  feet  4  inches  and  7  lines  wide,  American 
measure.  It  is  desired  to  fill  the  yard  27  inches  deep  with  earth.  Allowing  the 
yard  to  be  a  perfect  plane,  how  many  cubic  yards  of  earth  will  be  required  to  fill 
it ;  and  at  70/  per  cubic  yard,  what  will  be  the  cost  ? 

Ans.  388^  cu.  yds. ;  $272.00|,  cost. 


12 
12 
12 

27 


PARTIAL    OPERATION. 
144 

4GG3            (144  X  12)  x  388t^  x  27 
or, 

27  1728  X  27 


cu.  yds. 


402 


SOULE  S    PHILOSOI'IIIC    PRACTICAL    MATHEMATICS. 


TO    FIND    THE    NUMBER    OF     SQUARES    OF   EARTH   IN   GRADINGS, 
LEVEES,  AND    EXCAVATIONS. 


PROBLEM. 


810.     Ho%y  many  squares  of  earth  are  contained  in  a  piece  of  grading  or  levee, 
420  feet  long,  average  width  GO  feet  aiid  average  depth  21  ft.  4  inches '! 

Ans.  2737  J  squares. 


OPERATION. 

420 


3 

210 


G6 
G4 


2737J  squares,   Ans. 


Explanation. — As  shown  Ijy  the  Table  of  Solid  Meas- 
ure, page  247,  a  square  of  earth  is  a  cube  6x6x6  = 
216  cubic  feet.  Hence  we  multiply  the  three  dimen- 
sions of  the  grading  together  in  the  unit  of  feet  and 
divide  by  216. 


TO  FIND  THE  NUMBER  OF  CORDS  OF  WOOD  IN  RANKS. 

PROBLEMS. 

811.     1.     How  many   cords  of  wood  in  two   ranks,  each   44   feet  long  and  6 
feet  3  inches  high?  Ans.  17-1^8  cords. 


OPERATION. 

44 


12 

128 


4 


cords,    Ans. 


Explanation. — In  all  problems  of  this  character,  it^ 
is  customary,  in  the  absence  of  a  special  agreement 
to  the  contrary,  to  estimate  the  length  of  the  stick  as  4 
feet,  notwithstanding  the  fact  that  in  practice  it  i» 
ofteu  much  less  than  4  feet. 


Note. — In  solving  all  problems  of  this  character,  we  multiply  together  the  three  dimensions- 
in  the  unit  of  feet,  an<l  divide  by  128,  the  numl>er  of  cubic  feet  in  a  cord.  In  this  problem  we  also 
multiply  by  2,  for  the  reason  that  we  had  2  ranks. 

2.  A  rank  of  wood  is  4G  feet  4  inches  long,  G  feet  6  inches  high,  and  3  feet 
8  inches  deep  or  length  of  stick.    How  many  cords  does  it  contain  ? 

Ans.  9 1^5,^2  cords. 

Note.— In  commerce  the  length  of  stick,  when  less  than  4  feet,  is  estimated  as  if  it  were  4 
feet  long,  and  the  price  is  graded  accordingly. 

3.  What  will  be  the  cost  of  two  ranks  of  wood  each  rank  44  feet  long  and 
6  feet  3  inches  high,  at  $5  per  cord  1  Ans.  $85.93f . 

4.  How  many  cords  of  wood  can  be  placed  under  a  shed  that  is  40  feet  long, 
16  feet  wide  and  IS  feet  high  ?  Ans.  90  cords. 


*  PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OK    SOLIDS.  403 

TO  FIXD  THE  XUMBER  OF  CARS  REQUIRED  TO  CARRY  A  SPECIFIED 

XUMBER  OF  BUSHELS. 

PROBLEMS. 

812.     1.     How  many  cars  will  be  required  to  carry  loOOO  bushels  of  wlieat, 
each  car  carryiug  20000  iioundsT  Ans,  45  cars. 

OPERATION  INDICATED. 

15000  X  00  =  900000 ;  900000  ~  20000  =  45  cars. 

2.  If  a  freiglit  car  is  28  feet  long  and  S  feet  wide,  liow  deep  must  bulk  corn 
be,  in  order  tliat  tlie  car  may  contain  22400  ^jounds,  which  is  a  large  weight  for  a 
single  car?  Aus.  26.G6G+  inches  deep. 

OPERATION    INDICATED. 

22400  -^  56  =  400  bushels.     2150.42  x  400  =  8601C8  cubic  inches. 

28  X  12  X  8  X  12  =  32256  sq.  inches.     860168  h-  32256  =  26.666+  inches  deep. 

3.  How  many  cars  will  be  required  to  carry  120000  feet  of  lumber,  allowing 
6000  feet  per  car?  Ans.  20  cars. 

4.  How  many  trains  of  12  cars  each,  will  be  required  to  transport  COOOOO 
cubic  yards  of  gravel,  estimating  that  each  cubic  yard  weighs  3200  pounds,  and 
that  each  car  carries  18000  jiounds  ?  Ans.  8888|  trains. 

OPERATION    INDICATED. 


ISOOO 
12 


600000 
3200 

or  thus:  (6(10000  x  3200)  -h  (18000  x  12). 


TO  FIND  THE  NUMBER  OF  BRICKS   IX   WALLS   AXD   FOUNDATIONS 

PROBLEMS. 

813.     1.     How  many  bricks  in  a  wall  230  feet  long,  12  feet  9  inches  high,  and 
21  inches  thick,  allowing  20  bricks  to  the  solid  foot?  Aus.  102637^  bricks. 

OPERATION. 
230 

''I  Explanation. — Here  we  first  find  the  number  of  cubic 

21 

,,„  feet,  and  then  multiply  by  20,  the  number  of  bricks  in 

a  cubic  foot. 

102637J  bricks,  Ans. 


4 
12 


404 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  What  will  it  cost  to  build  the  foundation  walls  of  a  house  of  brick,  the 
house  being  60  feet  long  aiul  28  feet  wide,  and  the  walls  G  feet  high  and  2  feet  6 
'uches  thick,  at  $15  per  thousand,  allowing  18  bricks  per  solid  foot  ? 

Ans.  $672.30. 

OPEKATION, 

60  X  2  =  120  ft.  iu  the  two  sides. 
28  X  2  =    50  ft.  in  the  two  ends. 


170  ft.  round  the  outside  of  all  the  walls. 
2^  X  i  =    10  ft.  in  4  times  the  thickness  of  the  walls,  deducted  for  the 

corners  measured  twice. 

IGG  ft.  actual  length  of  walls. 

166  =  length  of  walls. 


1000 


6 
5 

18 
15 


=  height  " 

=  thickness     " 

=  number  of  bricks  iu  a  cubic  foot. 

=  price  per  thousand. 


$672.30 


3.  How  Diany  Milwaukee  bricks  which  are  8J  x  4J  x  2|  inches,  will  it  take 
to  build  a  wall  00  feet  long,  40  feet  high,  and  3  bricks  thick,  allowing  ^  inch  between 
bricks  for  mortar !  Ans.  45139ff  bricks. 


i  =  83  in 


FIRST   OPERATION. 

length  of  brick  and  mortar, 
thickness  of  brick  and  mortar, 
sq.  in.  surface  of  one  brick  and  mortar, 
bricks  in  one  square  foot  of  wall,  1 


83  X  2g  =  2i^M 
144  -  22§| 

thick, 
eija  X  3  =  ISf  II  bricks  iu  1  square  foot  3  bricks  thick. 
60  X  40  =  2400  X  18^1  =  45139  =  ^,  uumber  of  bricks  in  the 

wall  3  bricks  thick. 


SECOND   OPEKATION. 

GO 

40 

12 

rick 

12 

8 

99  =  3>  4J 

35 

4 

the 

21 

8 

33 

8 

45139Jf  bricks. 

TO  FIND  THE  inJMBER  OF  PERCHES  liS"  A  STOISTE  WALL. 


PROBLEMS. 

814.     1.    How  many  perches  in  a  stone  wall  60  feet  long,  8  feet  6  inches  high, 
and  2  feet  8  inches  thick  f  Ans.  54f|  perches. 

OPERATION. 
60 
102  Explanation. — ^Here  we  multiply  the  three  dimen- 

32 

.  sions  together  in  the  unit  of  feet,  and  then  divide  by 

— —  24},  the  number  of  cubic  feet  iu  a  perch. 

54||  P.,  Ans. 

2.    What  will  it  cost  to  build  a  breakwater  of  stone,  220  feet  long,  18  feet 
thick,  and  12  feet  high  at  $1.90  per  perch  ?  Ans.  $3648. 


12 
1-2 
99 


PRACTICAL    PROBLEMS    IX    THE    MENSURATION    OF    SOLIDS. 


40  5 


3.     IIow  iiuiiiy  perches  of  stone  are  contained  in  a  wall  90  feet  long,  12  feet 
higL,  and  22  iiiclies  thick?  Ans.  71^  perches. 


OPEKATION. 

90 

12 

00 


12 
99 


80   perches  in  wall. 
Sly  =  ^  of  perclies  in 
—  wall. 

71^  =  i>erches  of  stone. 


Explanation. — A  perch  is,  as  sliown  in  the  Table  of 
Gallic  Measure  011  page  247,  24i  cubic  feet.  But  in 
building  stone  walls,  2|  cubic  feet  are  allowed  per 
perch  for  mortar  and  filling.  Hence,  -when  built  in  a 
wall,  22  cubic  feet  of  stone  make  a  perch.  Then,  as 
as  2|  =  i  of  245,  we  therefore  subtract  from  the  num- 
ber of  ))erches  in  the  whole  wall,  |  of  the  same,  and 
thus  produce  the  perches  of  stone.  If  preferred,  the 
perches  of  stone  may  be  obtained  by  multiplying  the 
whole  number  of  perches  by  22,  and  dividing  the  prod- 
uct by  24i. 


Note. — It  is  estimated  that  three  peeks  of  lime  and  4  bushels  of  sand  are  used  in  a  perch. 

"WIDTHS  OF  WALLS. 
A  wall  1  brick  or  layer  thick  is  called,  by  brick  masons,  a    4  J  inch  wall. 

9   a    ((    ('(     ((    a         i(     li         it               II              tt     n"       II  ti 

Q  ti          It          tt             tl          ti          ti             tl          ti                it                it   yj          ic  tt 

A        ti          it          tl             tt          li          it               I          it                tt                it   10    it  tt 

^  it       it       it         a       II       ii         tl       it           it            It  22   (<  *< 

R      tt        tt         tt           tt        tt        tt           ti         ti             tt             tt  2Q        tt  t 


TO    FIND    THE    NUMBER   OP    CUBIC    YARDS,    CUBIC     FEET,     CUBIC 

INCHES,  BUSHELS  OR  GALLONS,  IN  DIFFERENT  KINDS 

OF  VESSELS  AND   SOLIDS. 


PROBLEMS. 


815.    1.    A  rectangular  bos  is  G  ft.  3  in.  long,  3  feet  wide,  and  4  feet  6  inches 
high  or  deep.     How  many  cubic  yards,  how  many  cubic  feet,  and  how  many  cubic 
inches  does  it  contain  ?    Also  how  many  bushels,  and  how  many  gallons  will  it  holdt 
Ans.  3i  cu.  yds.;  S4|  cu.  ft;  145800  cu.  in.;  67.8-|-bu.;  G31.17— gals. 


: 

Tofii 
4 

2 
27 

First. 

d  cu.  yds. 

25 

3 

9 

3i  cu.  yds. 

s 

To  fit 
4 
2 

OPE 

econd. 

id  cu.  ft.               To 

25 

3 

9 

84|  en.  ft. 

RATIONS   INDICATED 

Third.                                  Four 
(indcu.  in.                       Tolind 

75 
30 
54                         2150.42 

th. 

bu. 
75 
36 
54 

67.8+ 

bu. 

Tof 

231 

Fifth, 
nd  gills. 

75 

36 

54 

145800  cu.  in. 

631.17— 

gals 

Explanation. — To  find  cubic  yards,  we  multiply  the  three  dimensions  together  in  the  unit  of 
feet,  and  then  divide  by  27,  because  27  cubic  feet  make  1  cubic  yard. 

To  find  cubic  feet,  we  multiply  together  the  three  dimensions  in  the  unit  of  feet. 

To  find  cubic  inches,  we  multiply  together  the  three  dimensions  in  the  unit  of  inches. 

To  find  bushels,  we  find  the  number  of  cubic  inches  and  then  divide  by  2150.42,  because 
2150.42  cubic  inches  make  a  bushel. 

To  find  gallons,  we  find  the  number  of  cubic  inches  and  then  divide  by  231,  because  231 
cubic  inches  make  a  gallon. 

2.  A  box  is  5  feet  6  in.  long,  3  feet  4  in.  wide,  and  2  ft.  8  in.  deep.    How 
many  cubic  feet  does  it  contain  and  how  many  gallons  will  it  hold  ? 

Ans.  4Sf  cu.  feet;  365f  gallons. 

3.  A  cellar,  in  the  form  of  an  inverted  frustum  of  a  rectangular  pyramid,  is 
40  ft.  4  inches  long  at  the  top  and  30  ft.  long  at  the  bottom ;  it  is  24  ft.  wide  at  the  top 


4o6 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


aud  18  feet  8  inches  wide  at  the  bottom ;  and  it  is  7  feet  C  inches  deep.  How  many 
of  each,  cnbic  yards,  cnbi('  feet,  and  cnhic  inches  does  it  contain?  Also,  how  many 
bnshels,  and  liow  many  gallons  will  it  hold  ?         Ans.  209if  3.  eu.  yds. ;  5001^-  cu.  ft.; 


40. 
oO 


X  -'4       = 

X    18-;%  = 


9782400  cu.  in. ;  4549.06+  bu. ;  42348i,Vf  gals. 

OPERATION   BY  THE  PRISMOIDAL  FORMULA. 

908    =  area  of  the  top  of  the  cellar  in  sq.  ft. 
500    =     "     "     "     bottom  of  the  cellar  iu  sq.  ft. 


2)  70-A-       42-1^ 

33-,%  X  21-^2  =  750=-  X  4  =  3000^ 


G  )  452S3 


4  times  the   middle   section   between    the 
two  areas. 


754ff  =  average  area  of  the  cellar. 


754tf  X  7|  (ft.  deep)  =  50011  cubic  feet. 
5001i  cu.  ft.  4-  27  =  209Jrf  f  en.  yards. 
5001^  cu.  ft.  X  1728  (l-u.  in.)  =  97S2400  cubic  inches. 
97S2400    cu.  in.  -4-  2150.42  =  4549.00+  bu. 


9782400    cu. in.  -^  231 


42348-1/1  gals. 


Note  1. — If  it  is  desired,  the  different  dimensions  in  the  ahove  problem  may  all  be  reduced 
to  inches  and  the  worlc  ]ierl'ornicd  in  the  same  manner,  tlins  avoiding  ninch  fractional  work. 

NoTK  2. — The  abo\o  solution  is  in  accordance  with  the  ]irisnioi(lal  lorninla,  liy-\vhirh  the 
Solidity  of  cubes,  rectangular  solids,  cones,  cylinders,  pyramids,  frustums  of  coues  or  pyramids 
and  several  other  forms  of  solids  may  he  determined. 

Note  3. — For  an  explauatiou  of  the  Prismoidal  Formula,  see  Article  768,  pages  375  and  376. 

4.  How  many  cubic  yards  were  excavated  from  a  cellar  92  feet  long  and  50 
feet  wide  at  the  top,  80  feet  long  and  44  feet  wide  at  the  bottom,  and  8  feet  4  inches 
deep?  Ans.  1291  ja  cu.  yds. 

5.  A  cylinder  is  7  feet  4  inches  high  and  3  feet  5  inches  in  duimeter.  How 
inany  of  each,  cubic  yards,  cubic  feet,  and  cubic  inches  does  it  contain?  Also,  how 
many  bushels  and  how  many  gallons  will  it  hold,  and  allowing  that  a  cubic  foot  of 
water  weighs  62J  pounds,  what  would  be  the  weight  of  water  the  cylinder  could 
contain?  Ans.  2.4902+  cu.  yds.;  07.2353+  cu.  ft.;  11G1S2.C512  cu.  in.; 

54.028  —  bu. ;  502.9552  gals. ;  4202.2081  pounds. 


Cn.  T(l3. 

3 


12 
12 


41 
41 
.7854 


2.4902+  cu.  yds. 


(;m.  111. 
88 


41 
41 

.7854 


116182.0512  cu.  in. 


OPERATIONS    INDICATED. 

Cu.  Ft.                                     Weight  of 

3 

Water. 
22 

3 

22                                12 

41 

12 

41                                12 

41 

12 

41 

.7854 

.7854                             2 

125 

07.2353+  cu.  ft. 

4202.2081  pounds. 

B 

a.                                                        Ga 

88 

s. 

88 

41 

41 

41 

41 

.7854                          231 

2150.42 

.78.54 

54.028—  bu. 

502.9552  gals. 

Note. — By  inspection,  we  find  that  231  always  cancels  .7854  and  gives  a  quotient  of  .0034. 


PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS. 


407 


6.  How  many  gallons  in  a  cylinder  that  is  9  feet  3  inches  high  and  8  feet 
4 inches  in  diameter?  Ans.  3774  gals. 

7.  A  cone  is  134  inches  high  and  45  inches  in  diameter  at  the  base.  How 
many  of  each,  cubic  yards,  cubic  feet,  and  cubic  inches  does  it  contain  !  Also,  if 
it  is  a  vessel,  and  tliese  dimensions  are  inside  measurements,  how  many  bushels  and 
how  many  gallons  will  it  hold  ?  Ans.  1.5226+  cu.  yds. ;  41.1108 —  cu.  ft. ; 

71039.43  cu.  in. ;  33.035+  bu. ; 


307.53  gals. 


OPERATIONS    INDICATED. 


Cu. 

rds. 

Cu. 

Ft. 

Cu. 

In. 

liu. 

Gala. 

12 

134 

134 

12 

45 

12 

134 

134 

45 

134 

12 

45 

12 

45 

45 

45 

45 

.7854 

12 

45 

45 

.7854 

45 

3 

.7854 

.7854 

3 

3 

.7854 

27 

3 

3 

2150.42 

231 

1.5226+ 

41.1108— 

71039.43 

33.035+ 

307.53 

cu.  yds. 

cu.  ft. 

cu.  in. 

bu. 

gals 

8.  How  many  cubic  inches  in  a  cone  that  is  10  feet  2  inches  high  and  8  feet 
3  inches  in  diameter  at  the  base  ?  Ans.  313040.019G  cu.  in. 

9.  A  cistern  in  the  form  of  a  frustum  of  a  cone  is  10  feet  2  inches  high,  9 
feet  6  inches  diameter  of  lower  base  and  8  feet  G  inches  diameter  of  upper  base. 
How  many  bushels,  and  how  many  gallons  will  it  hold,  and  allowing  that  a  gallon 
of  water  weighs  8J  jiounds,  (8.3389  is  the  accurate  weight  of  a  gallon)  what  would 
be  the  weight  of  the  water  the  cistern  could  hold  ? 

Ans.  520.20+  bu. ;  4843.2048  gals. 
403G0.04  pounds. 


114=  =  12996 

102-  =  10404 

114  X  102    =  11628 


3I*ERATI 

ONS    INDICATED. 

Bushels. 

Gallons. 

11C76 

11676 

122 

122 

.7854 

.7854 

2150.42 

bu. 

231 

520.2G+ 

4843.204S 

3  )  35028 
11676 

4843.2048  X  Si  =  40360.04  pounds. 

10.     A  rectangular  pyramid  is  6  feet  high,  and  has  a  base  4  feet  3  inches  by  3 
feet  9  inches.     How  many  of  each,  cubic  yds.,  cubic  feet,  and  cubic  inches  does  it 
contain  ?     And  if  hollow,  how  many  bushels  and  how  many  gallons  will  it  hold  ? 
Aus.  lif  cu.  yds. ;  31^  cu.  ft. ;  55080  cu.  in.;  25.61+  bu. ;  238fA  gals. 

OPERATIONS  INDICATED. 


Cq. 

Yds. 

Cu. 

Ft. 

Cu. 

In. 

Bn. 

Ga 

3. 

6 

72 

72 

4 

17 

6 

72 

51 

51 

4 

15 

4 

17 

51 

45 

45 

3 

4 

15 

45 

3 

3 

'21 

3 

3 

2150.42 

231 

l^icu 

yds. 

3Ii  cu.  ft. 

55080  cu.  in 

25.61+  bu 

238?  f 

4oS 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TO  FIND  THE  HEICxnT  OR  THE  DIAMETER  OF  CYLINDRICAL  VESSELS 
TO  CONTAIN  A  SPECIFIED  NUMBER  OF  GALLONS. 

PROBLEMS. 

816.     1.    The  diameter  of  a  cylindrical  cistern  is  6  feet  8  iiiclies.    What  must 
be  its  height  to  hold  3000  gallons  ?  Ans.  137.867G  in.  height. 

OPERATION    INDICATED. 

3000  gallons. 
231  cu.  in.  in  1  gallon. 


693000  cu.  in.  in  a  cistern  to  hold  3000  gallons. 


80  in.  diameter  of  cistern. 

80  "         " 


GIOO  =  square  of  diameter. 
.7854  =  ratio  of  the  area  of  a  square  to  the  area  of  a  circle. 


502G.5600  =  area  of  a  section  of  the  cistern. 
693000  4-  5026.50  =  137.8076  inches  height. 


or  thus : 


80 

80 

.7851 


3000 
231 


137.8676  in.  =  11  feet,  5.8676  inches. 

2.    The  height  of  a  cylindrical  cistern  is  7  feet  6  inches.    What  must  be  the 
diameter  so  that  it  will  hold  2000  gallons  ?  Ans.  80.84  inches  diameter. 

OPERATION   INDICATED. 


90 

.7854 


V  6535.9477 


2000 
231 


6535.9477 

=  80.84  +  inches.,  Ans. 


PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS.  409 

GAUGING  COAL. 

817.  A  barge  of  coal  measures  as  follows:  Whole  length  from  a  to  h  173' 
6",  width  of  boat  25'  10",  depth  8'  9-,%".  The  length  of  bulkhead  or  rake  Xo.  1  is 
79"  and  the  depth  is  02".  The  length  of  bulkhead  No.  2  is  72"  and  its  depth  is 
102".     How  many  of  each,  bushels  and  barrels,  does  it  contain  1 

Ans.  24096.1008  bushels.     9267.731  barrels. 


(Liued  allowing  sitle  uf  boat). 

OPERATION 

TO  FIND   TIIK   CUBIC   IN'CHKS   IN'   THE   m'LKHEADS. 


No.  1,  79"  leiiijth  x     92"  depth  =    72GS  sq.  in. 
No.  2,  72"  length  x  102"  depth  =     7314  sq.  in. 


Total  sq.  in.  in  the  area  of  length  and  depth  of  the  two  bulkheads   14812 
14612  X  310"  width  =  4529720  cu.  in.  in  the  bulkheads. 


GEN'ERAI.  OPERATION-. 
173'     6"  =  lenj;th  of  ljo;it. 
1'  =  allowance  for  rail. 


172'     6"  =  length  including  bulkheads. 

12'     7  '  =  length  of  bulkheads,  (79"  +  72"  =  12'  7"). 

159'  11"  =  length  less  bulkheads. 

12  =  in.  in  1  foot. 

1919"  =:  length  in  inches  of  main  boat. 

310"  =  width  of  boat. 


594890      =  sq.  in.  are.a  of  main  boat. 
105.8    =  depth  of  main  boat. 

62939362  =  cu.  in.  in  main  boat. 
4529720  =  cu.  in.  in  bulkheads. 

67469082  =  en.  in.  in  main  boat  and  bulkheads. 
2698763  =  4°J   dednctious,  for  gunnels  cross-ties,   pnmps,  the   adjustment   of    the    bulk- 

heads,  etc. 

64770319  =  cu.  in.  net  in  the  whole  boat. 

64770319  en.  in.  —  2688  cu.  in.,  the  number  of  cubic  inches  allowed  for  a  bushel  of  coal, 
=  24096.10l'8  bushels,  24096.1008  -H  2.6  bushels,  the  number  of  bushels  allowed  for  a  barrel  of 
coal,  =  9267.7310  barrels. 

Note. — The  depth  of  the  bulkheads  or  rakes  varies  according  to  the  constraction  of  different  boats. 
REMARKS  ON  THE  ABOVE  SOLUTION. 

The  above  solution  is  made  in  accordance  with  the  custom  of  the  official  gaugers  of  coal  for 
Louisiana.  The  allowance  of  2688  cu.  in.  for  a  bushel,  and  2.6  bushels  for  <a  barrel,  is  practically 
in  accordance  witli  the  law  of  Louisiana,  which  says,  that  SJ-  bushels  of  the  United  States  standard 
shall  be  a  barrel  of  coal.  Accordingly,  2150.42  x  3J  =  6988.865  cu.  in.  constitute  a  barrel  of  coal. 
See  pages  248  and  249  of  this  book. 

{Continued.) 


4IO 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


The  al)ovo  Gt770f!I0  on.  in.  ilivided  l>y  (5988.863  cu.  in.,  gives  9267.R448  l>arrels. 

The  ro.sult  (ibtaiiied  l)y  dividing  the  bnshels  by  2.6  was  11267. 7310  barrels. 

The  difference  is  .0862 

The  excess  by  the  2.6  method  in   a  quantity  of  9267  barrels,  is  so  insigniiicant  that  the 
method  may  be  regarded  as  jiractically  correct. 

ANOTHER  METHOD. 

The  following  method  of  me.asnring  or  ganging  coal  is  nsed  in  Ohio  and  some  other  states: 

1.  Find  the  enbic  inches  in  the  barge  exclusive  of  the  rakes*,  and  including  the  gunnels, 
timbers,  etc. 

2.  To  this  add  the  enbic  inches  in  the  r.nkes. 

3.  Then  find  the  actual  cubic  inches  in  the  timbers,  pumps,  etc.,  and  deduct  the  same  from 
the  total  cnliic  inches  in  the  barge. 

4.  Divide  tln^  remainder  by  2688,  the  numher  of  cubic  inches  in  a  hushel  of  coal,  iu  Ohio, 
Pennsylvania  and  Kentucky  and  several  other  states — but  not  iu  Louisiana. 

t  The  following  operation  elucidates  this  method. 


Long. 


DIMENSIONS   OF   BARGE. 

Wide.        Deep. 


Body     1350 
Itakes     185 


298 
298 


90  6-7  =  36551828  Cubic  inches. 
76i       =    4231227      " 


Deductions  for  timbers,  etc., 


407830.55 
1129646 


2688  )  39653409  ( 14752  bushels 
in  the  barge. 


*Rake3  are  the  ends  of  the  bar;:e  or  boat, 
not  perpendicular  to  the  keel  or  bottom.  Tlie 
word  Imlkhead  is  often  used  in  tlio  sense  of 
rake.     It  \^a3  .so  used  in  the  preceding  problem. 

IThis  operation  statement  is  an  actual  meas- 
urement of  a  barge  of  coal  by  Mr.  E.  A.  BuI^N- 
SUJE,  Coal  Measurer  of  Cincinnati,  through 
whose  courtesy  wo  present  the  method. 


APPROXIMATE  METHOD.S  OF  FINDING  THE  TONS  OF  HAY  IN  STACKS,  MOWS, 

LOADS  AND   WINDROWS. 

818.  The  only  accurate  metlio<l  of  measuring  hay  is  to  weigh  it.  But  as  this  cannot 
always  be  done,  a]>i)roxiinate  methods  are  sometimes  used. 

The  observation  and  exiierience  of  farmers  and  dealers  in  haj*  have  shown  the  following 
approximate  facts : 

1.  That  hay,  well  settled  iu  stacks  or  mows,  will  weigh  a  ton  for  every  15  cubic  yards. 

2.  That  hay,  slightly  settled  iu  stacks,  mows  or  loads,  will  weigh  a  ton  for  every  20  cubic 
yards. 

3  That  good,  average,  meadow  hay  unsettled  in  windrows,  will  weigh  a  ton  for  every  25 
cubic  yards. 

4.     That  hay,  when  baled,  will  weigh  a  ton  for  every  10  cubic  yards. 

Note. — There  is  a  difference  in  the  weiirbt  of  equal  bullcs  of  different  kinds  of  Iiay.  new  or  old,  clover,  timothy  or 
other  kinds,  and  hence  some  allowance  should  bo  made  therefor. 

TO  FIND  HOW  MANY  TONS  OF  HAY  IN  CIRCULAR  STACKS. 

810;  /)irfcrio)?s.— Multiply  the  square  of  the  circumference  in  yards  by  4  times  the  height 
in  yards,  and  divide  by  1500;  or  point  otf  two  places  and  divide  by  15. 

PROBLEM. 

How  many  tons  iu  a  circular  stack  whose  circumference  is  30  yards,  and  whose  height  is 
10  yards  t  ^"'*-  ^■t  tons. 

OPKRATION. 

30"  =  900  X  (4  X  10)  =  36000;  36000  ^  1500  =  24  tons,  Ans. 


*  PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OK    SOLIDS.  4II 

TO  FIND  now  MAXY  TOXS  OF  HAY  IX  RECTA^^GULAE  AND  SQUARE 

STACKS. 

820.     Directions. — Multiply  tlie  length  in  yards,  by  the  width  in  yards;  then 
multiply  this  product  by  half  the  height  in  yards,  and  divide  the  last  product  by  15, 

PROBLEM. 

IIow  many  tons  in  a  square  stack  40  yards  long  and  10  yards  wide  and  9 
yards  high  1  Ans.  li:0  tons. 

OPERATION. 

40  X  10  =  400;  400  X  4i  =  ISOO  -^  15  =  120  tons,  Ans. 


TO  FIND  HOW  MANY  TONS  OF  HAY  IN  A  MOW. 

821.     Directions. — Multii)ly  the  length,  width  and  heiglit  in  yards,  aud  divide 
the  product  by  15. 

PROBLEM. 

How  many  tons  in  a  well  settled  mow  whose  length  is  60  feet,  width  40  feet 
and  height  or  depth  30  feet  ?  Ans.  177J  tons. 

FIRST  OPERATION.  SECOND  OPERATION. 

60  ft.  =  20  yds.;  40  ft.  =  13^  yds.;  30  ft.  =  10  yds. 
20  X  13i  X  10  =  2CG6|  cu.  yds.  -=-  15  =  177^ 
tons. 

Note. — If  the  mow  in  very  liigb,  as  it  often  is  where  horse 

or  steam  hav  fork.s  are  used,  the  hav  will  weigh  more  -i  ^■-t_  4.„,.„     a  _„ 

per  cubic  yard  and  hence  the  divisor   should   be   13  ^"»  ^""*'  ^"*' 

or  14. 


TO  FIND  THE  TONS  OF  HAY  IN  LOADS. 

822.     i)irec/((»!S.— Multiply  the  length,   width   and  height,   in   yards,    and 
divide  by  20. 

PROBLEM. 

How  many  tons  of  hay  in  a  load  IS  feet  long,  10  feet  wide  and  6  feet  high  ? 

Ans.  2  tons. 

FIRST    OPERATION.  SECOND   OPERATION. 

18  ft.  -  3  =  6  yds. ;  10  ft.  -^  3  =  3J  yds.;  6  ft.-r 

3  =  2  yds. 
6  X  3J  X  2  =  40  cu.  yds.  .^  20  =  2  tons. 
or  thus : 

IS  X  10  X  6 


3 

60 

3 

40 

3 

30 

o 

3 

IS 

3 

10 

3 

6 

20 

2  tons,  Ans 

3  X  3  X  3  X  20 


=  2  tons. 


412  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  ♦ 

.     TO  FIND  TnE  TONS  OF  IIAY  IN  A  WINDROW. 

823.     IHrecUons. — Multiply  tlie  lenjjth,  width  and  lieigiit  together,  in  yards 
and  divide  by  25. 

PROBLEM. 

How  many  tons  of  hay  in  a  windrow  COO  ft.  long,  6  ft.  wide  and  5  ft,  high  ? 

Ans.  2C§  tons. 

FIRST   OPERATION.  SECOND   OPERATION. 


600  ft.  4-  3  =  200   yds. ;   6   ft.  -^  3  =  2   yds. ;    5 


3     COO 
3     6 

3     5  ft.  -^  3  =  12  yds. 

200  X  2  X  1§  =  6GC§  cu.  yds.  -4-  25  =  2G§  tons. 

2C§  tons,  Ans. 

TO   FIND  TUE  APPEOXIMATE  NUMBER  OF  BUSHELS  OF    CORN    IN 

CRIBS. 

PROBLEMS. 

824.  1.  How  many  bushels  of  shelled  corn  in  a  crib  whose  inside  measure 
is  10  feet  0  inches  high,  12  feet  long,  S  feet  wide  at  the  top,  and  6  feet  wide  at  the 
bottom,  allowing  2  cubic  feet  of  corn  iu  the  ear  to  make  a  bushel  1 

Ans.  441  bushels. 

OPERATION. 

12  ft,  length. 

7  ft.,  average  width  (8  +  6  -^  2).  or  thus :  12  x  7  x  lOJ 

2    21  ft.,  lieiglit  (lOi  ft).  =  441  bushels. 

2  cu.  ft.  allowed  for  a  bushel.  2 

441  bushels,  Ans, 

Note. — It  is  found  by  experiment  that  from  2  to  2}  cubic  feet  of  corn  on  the  cob  is  required 
to  give  one  bushel  of  shelled  corn. 

2.  How  many  bushels  of  shelled  corn  in  a  crib  of  corn  in  the  shuck,  the  crib 
being  10  by  8  feet  at  the  bottom,  15  by  12  feet  at  the  top  and  11  feet  high,  allowing 
3i  cu.  ft.  for  a  bushel  1!  Ans.  418  bushels. 

OPERATION. 

Bot.  measure,  10  x     8  =    80    sq.  ft.  on  the  bottom. 
Top  "       15  X  12  =  180    sq.  ft.  on  the  top. 

Suni  of  top or  thus :     3 J 

and  bottom,    25  x  20  =  500    sq.  ft.  in  4  times  the  middle  — 

section  between  top 

G  )  7C0  and  bottom. 


126g 
11 


418  bushels. 


12G§  sq.  ft.  average  area  of  crib. 
126§  X  11  ft.  height  4-  SJ  =  418  bushels. 

This  crib  is  iu  the  form  of  a  frustum  of  a  pyramid. 

Note  1. —  See  Article  791,  page  387,  for  an  explanation  of  the  solution  of  Frustums  of 
Pyramidal  Solids.     Also  see  page  373. 

NoTB  2.— It  is  found  by  experiment  that  from  34  to  3|  cubic  feet  of  corn  iu  the  shuck  is 
required  to  give  one  bushel  of  shelled  corn. 


^:0k- 


*  PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS.  413 

3.  A  corn  crib,  filled  with  corn  on  tlie  ear,  is  20  x  8  feet  at  the  top,  and  16 
X  G  feet  at  the  bottom  and  11  feet  high.  Allowing  two  even  bushels  of  corn  on  the 
ear  to  make  one  bushel  of  shelled  corn,  how  many  bushels  of  shelled  corn  are  there 
in  the  crib  ?  Ans.  6962  bushels. 

KoTE  1. — The  form  of  this  crib  is  a  frustum  of  a  pyramid. 

Note  2. — For  accurate  methods  of  finding  the  number  of  bushels  or  gallons  in  boxes,  bins, 
tanks,  etc.,  see  page  394. 

Note  3. — Most  corn  cribs  have  parallel  and  vertical  sides  and  ends.  In  all  such  cases,  mul- 
tiply the  lenj;th,  width,  and  height  together  in  the  unit  of  feet,  thus  producing  cubic  feet.  Then 
divide  by  2  or  2i  as  above  explained. 


MEASUREMENT  OF  GRAIK  WHEN  m  EEGULAE  HEAPS. 

PROBLEJI. 

825.     A  farmer  has  a  hcai>  of  grain  in  a  conical  form,  the  diameter  of  whicli 
is  14:  feet  4  inches  and  the  depth  5  feet  3  inches ;  how  many  bushels  does  it  contain  ? 

Ans.  22G.906  bu. 

OPERATION. 

172  Explanation. — The  first  part  of  this  operation   is  the 

272  same  as  for  a  cylinder,  and  then  wo  divide  by  3  for  the 

fr,  reason  that  the  contents  of  a  cone  are  J  of  a  cylinder  of 

^^  equal  height  and  diameter.     Had  tlie  grain  been  heai>ed 

.78o4  against  the  side  of  a  wall  or  building,  we  would  divide 

the  result  of  this  operation  by  2,  because  in  such  a  case 

the  heap  would  be  but  i  a  cone.     Had  it  been  placed  in  a 

oop  nnp   V,  A     a        righij  angled  corner,  we  would  divide  the  result  of  this 

...-O.JUO   DU.,  jins.      operation  by  4,  for  the  reason  that  the  pile  would  then  be 

i  of  a  cone. 


2150.4 


GAUGnfG-  BAEEELS,  CASKS,  ETC. 

826.  Gauging  is  a  method  of  finding  the  contents,  or  cubical  capacities  of 
casks,  barrels,  etc.  In  practice,  the  measurement  or  gauging  of  casks  and  barrels 
is  generally  jjerformed  by  means  of  iustrimieuts  known  as  diagonal  and  ullage  or 
''wanting"  rods. 


THE  USE  OF  THE  DIAGOITAL  AND  ULLAGE  EODS. 

827.  To  gauge  barrels  or  casks  with  these  rods,  place  the  diagonal  rod  into 
the  bung  and  extend  the  pointed  end  to  the  lower  side  of  each  end  of  the  barrel  or 
cask,  and  thus  find  the  average  or  mean  diagonal  distance ;  then  on  the  side  of  the 
rod  at  the  point  of  the  mean  diagonal  distance,  will  be  found  the  number  repre- 
senting the  full  capacity  of  the  vessel  in  gallons.  If  the  barrel  or  cask  is  not 
full,  then  the  ullage  or  "wanting"  rod  is  inserted  vertically  into  the  bung,  and 
at  the  highest  iioint  where  the  liquid  touches  this  rod  will  be  found  a  number 


414 


souLE  s  riiiLosopiirc  practical  mathematics. 


represciitiiiji  the  imniLcr  of  pillons  "wanting"  to  make  tlie  barrel  full.  Tlie 
dift'ercuce  between  the  full  capacity  of  the  barrel,  as  shown  by  the  first  rod,  and 
the  gallons  "wanting"  will  be  the  contents  of  the  barrel  or  cask. 

The  diagonal  rod  may  be  prepared  by  finding  the  bung  diagonal  of  a  one 
gallon  barrel  of  medium  curve,  which  is  7.154-  inches,  and  then  to  find  the  diagonal 
number  for  each  successive  number  of  gallons,  multiply  the  7.154-  inches  by  th© 
cube  root  of  the  successive  numbers,  2,  S,  4,  5,  etc.,  of  gallons.  Thus,  to  find  th© 
diagonal  number  of  inches  for  S  gallons,  we  proceed  as  follows:  i^'^Z^2-  then 7.154- 
X  2  =  14.04-  inches. 

There  is  no  way  of  finding  the  exact  capacity  of  casks,  barrels,  etc.,  for  th© 
reason  that  they  are  not  constructed  in  exact  conformity  with  any  regular  mathe- 
matical figure.  The  curvature  of  the  staves,  for  nearly  all  kinds  of  casks  and 
barrels,  is  slightly  different,  and  hence  all  measuremeHts  of  their  contents  are  only 
approximately  correct. 

The  capacity  of  a  cask  is  equivalent  to  that  of  a  cylinder  having  the  sam© 
length  and  a  diameter  equal  to  the  mean  diameter  of  the  cask. 

To  find  the  contents  of  a  cask  or  barrel  without  the  aid  of  the  diagonal  rods, 
we  first  measure  the  length  and  the  head  and  bung,  or  center  diameters, 
inside  measure.  Then,  having  the  two  diameters,  we  approximate  the  mean 
diameter  according  to  the  quantity  of  curve,  as  follows :  When  the  curve  is  but 
slight,  we  add  J  or  .5  of  the  difference  between  the  two  diameters  to  the  lesser  dia- 
meter; when  the  staves  have  a  medium  curve,  we  add  .0,  and  when  they  are  above 
a  medium  cui've,  we  add  ,65  or  .7. 

Having  thus  produced  the  average  or  mean  diameter,  we  proceed  the  same 
as  in  the  measurement  of  a  cylinder. 


A  barrel  with  slUihUij 
curved  staves. 


A  barrel  with  medium 
curved  staves. 


A  barrel  with  ahore 
medium  curved  staves. 


TROBLEMS. 

1.     A  barrel  is  20  inches  long,  17  inches  in  diameter  at  the  head,  and   20 
inches  in  diameter  at  the  bung  or  center.    The  staves  have  a  medium  curve.    How 


many  gallons  will  it  hold  ? 

20  in.,  center  diameter. 
17  in.,  head  diameter. 

3  in.,  difference. 

.0  adjustment  for  curve  in  staves, 

l.S  =  .6  of  the  difference. 
17     head  diameter  added. 


Ans.  31.2444-  gallons. 


OPERATION. 

1S.8 
1S.8 
.7854 
26 


231 


or  thus : 


31.244096  gals.,  Ans. 


IS.S 
18.8 
.0034 
26 


31.244090  gals. 


18.8  average  or  mean  diameter. 

Note. — By  the  use  of  the  gauging  rods,  a  barrel  of  the  above  diiueusions  gauged  31-]-  gallons. 


PRACTICAL    PROBLEMS    IN    THE    MENSURATION    OF    SOLIDS. 


4x5 


2.  The  inside  measurements  of  a  certain  barrel  are  as  follows :  LengtL,  from 
cliinie  to  f liinie,  30  inches ;  bung  diameter,  22.8  inclies ;  head  diameter,  19.5  inches. 
How  many  gallons  will  it  contain  by  the  mathematical  method  of  ganging? 

Aus.  47.0018+  gallons. 

OPERATION. 

21.4S 


22.8"  =  bung  diameter. 
19.5"  =  head  diameter. 

3.3"  =  difference. 
.6 


1.98"  =  .0  of  the  difference. 
19.5"    =  head  diameter. 


21.48"  =  average  or  mean  diameter. 


231 


21.48 
.7854 
30 


or  thus : 


47.0618+   gals. 


21.48 
21.48 
.0034 
30 

47.0618+ 

gallons. 


COMPAEISOX  OF  MEASUEEMENTS. 

828.  A  barrel  of  the  above  dimensions  was  gauged  by  the  gauging  rods 
and  gave  a  capacity  of  47+  gallons. 

The  same  barrel,  when  measured  by  a  gallon  measure,  held  47  gallons  and  1 
gill. 

When  the  same  barrel  was  measured  with  a  head  diameter  of  19.4  inches, 
the  other  dimensions  being  the  same  as  above,  it  showed  a  capacity  of  46.8867+ 
gallons. 

1.  A  cask  is  40  inches  long,  31i  inches  head  diametei',  and  38|  inches  bung 
diameter.    What  is  its  caiiacity  in  wine  and  beer  gallons  1 

Ans.  181.93+  gallons  wine.     149.02+  gallons  beer. 

OPERATION. 


385    in.,  bung  diameter. 
3li    in.,  head  diameter. 

7:^    in.,  difference. 


38.75 
31.50 


or. 


5.02 


.7  of  the  difference. 


31.5      head  diameter  added. 

30.575  average  or  mean  diameter. 
OPERATION 

TO   FI.VD  WINK   GALLONS. 

30.575 


7.25 
.7 

5.075 
31.5 

231 


30.575 

.7854 

40 

181.9313+  gals.,  Ans. 


36.575 

OPERATION 

TO  FIND    r.EEIl   GALLONS. 

.30.575 


282 


30.575 

.7854 

40 

149.0288+  gals.,  Ans. 


41 6  soule's  philosophic  practical  mathematics.  * 

TONNAGE  OF  VESSELS. 

829.  The  Tonnage  of  a  vessel  is  the  number  of  tons  weight  it  will  carry 
with  safety. 

The  carrying  capacity  of  vessels,  that  is,  the  amount  of  space  available  for 
stowing  away  the  cargo  is  reclioned  in  tons.  The  adopted  measurement  is  one 
hundred  cubic  feet  of  space  to  a  ton.  This  measurement  does  not  include  the  deck 
space,  where  part  of  the  cargo  is  sometimes  carried.  The  registered  tonnage  is 
always  less  than  the  actual  tonnage. 

To  determine  accurately  the  tonnage  of  a  vessel  or  steamboat,  is  a  very 
difficidt  problem.  The  methods  in  practical  use  do  not  give  accurate  results.  The 
best  method  probably  would  be  to  divide  the  vessel  into  several  sections,  and  then, 
by  numerous  measurements,  find  the  average  length,  width  and  depth  of  each  sec- 
tion, and  having  by  this  means  found  the  contents  in  cubic  feet,  the  actual  tonnage 
is  fouud  by  dividing  the  cubic  feet  by  40. 

The  following  is  the  United  States  method  of  finding  the  tonnage  of  vessels : 

Measure  in  feet,  above  the  upper  deck,  the  length  of  the  vessel,  from  the 
forepart  of  the  main  stem  to  the  afterpart  of  the  sternpost.  Then  measure  the 
breadth  taken  at  the  widest  part  above  the  main  wale  on  the  outside,  and  the  depth 
from  the  under-side  of  the  deck-plank  to  the  ceiling  in  the  hold.  Then,  from  the 
length  take  three-fifths  of  the  breadth,  and  multiply  the  remainder  by  the  breadth 
and  depth,  and  the  product  divided  by  95  will  give  the  tonnage  of  a  single-decker. 

If  the  vessel  is  double-decked,  substitute  half  of  the  breadth  for  the  depth, 
and  proceed  as  directed  for  single-decked  vessels. 

The  following  is  the  carpenters'  method  of  measuring  tlie  tonnage  of  vessels: 

Measure  the  length,  breadtli,  and  depth,  all  in  feet.  Divide  the  continued 
product  of  the  dimensions  by  95,  and  the  quotient  will  be  the  tonnage. 

If  the  vessel  is  double-decked,  half  the  breadth  is  taken  as  the  depth. 

Note. — For  more  extended  work  on  the  Tonnage  of  Vessels  and  Steamboats,  see  Haswell's 
Book  for  EnKiiii'era  and  Mechanics,  or  the  Merchants'  and  ShiiJmasters'  Manual,  or  the  Cyclopedia 
Britannica  or  the  Century  Dictionary. 


DRILL    EXERCISE    IN    MENSURATION    OF    SOLIDS. 


417 


DRILL  EXERCISE  IN  MENSURATION  OF  SOLIDS. 


830.  Statements  showing  the  capacity  in  gallons  of  each  of  the  following  fig- 
ures. The  dimensions  of  each  are  given  above,  and  the  statement  to  obtain  gallons 
is  given  below  each  figure : 


Height  6  feet . 
Xacli  side  40  in. 


Height  6  feet. 
Diameter  40  in. 


Height  6  feet. 

Diameter  of  base 

40  inches. 


Height  C  feet. 
Tpper  Diain.  3U  in. 
Lower  Diam.  40  in. 


3 

302  _  900  231 
40-  =  1(500 
40  X  :iO  ^  1200 

3700 

3 


Gals. 
3700 
.7854 


Height  C  feet. 

Each  side  of  base 

40  inches. 


Heiglit  0  feet. 
Side  (»!'  uitper  base 

30  iiiclies. 

Side  of  lower  base 

40  inches. 


CiIs. 


3 

231 


3700 


30^  =    900 

40-  =  1600 

40  X  30  =  1200 


3700 
3 


Diameter 
40  inches. 


231 


Gals. 
40 
40 
40 
.5236 


Transverse  Diam- 
eter 0  ft. 
Conjugate  do40in 


Transverse  Diani  C  ft. 
Conjugate        ■"  40  in. 


Diameter  CO  in.        Depth  30  in. 


Diameter  Go  in.         Depth  40  in. 
SO" 


Gals. 


231 


60 
60 

.5236 
40 


Diameter  60  in.        Depth  20  in. 


60  —  2  =  30 
30-  X  3  =  2700 
2700  +  20^  =  3100 


231 


Gala. 
3100 
20 
.5236 


4i8 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


See  Problem  and  Solution, 
page  398. 


See  Article  780. 


Sou  Ax-ticlo  781. 


Leneth  61)  in. 
I5iui;i  l)i;ini.  40  in. 
Head  Diam.  32  iu. 


40  —  32  =     8 
8  X  .7=rj.6 
32+;i.6=37.6 
37.6-  X  60  X  .0034  = 
...    KalloiiK. 


go 


/^' 


A    b 
See  Article  783. 


See  Article  782. 


See  Article  785. 


Note. — To  find  the  capacity  in  bushels  of  the  above  figures,  divide  by  2150.4  or  2150.42, 
in8tea<l  of  231. 

To  Cud  the  number  of  cubic  feet  instead  of  gallons,  divide  by  12,  12,  and  12  or  1728,  instead 
of  231. 


<^^ 


MECHANICAL    POWERS. 


419 


MECHANICAL  POWERS, 


831.  The  Mechanical  Powers  consist  of  six  simple  machines,  or  the 
elements  of  machinery,  by  the  scientific  use  of  which  we  are  enabled  with  a  given 
force  and  velocity,  to  accumnlate  a  greater  force  with  less  velocity,  or  a  less  force 
with  a  greater  velocity;  or  in  other  words,  to  chanr/e  either  the  direction  or  the  inten- 
sity of  a  force,  or  both  direction  ((ltd  intensiti/.  Tliese  Powers  are  called  the  Lever, 
the  Wheel  and  Axle,  the  Pulley,  the  Inclined  Plane,  the  Wed(je,  and  the  Screw. 

To  fully  understand  the  nature  of  a  machine,  four  things  must  be  considered, 
viz.:  1st,  The /orce  or  ^)rvre>- whifh  acts  on,  or  is  api)licd  to  the  ma(;liine;  2d,  The 
resistance  or  iceii/ht  which  is  to  be  overcome  or  moved  by  the  power;  3d,  The  center 
of  motion,  called  the  Fulcrum,  which  is  the  point  over  which  the  power  and  resist- 
ance act;  and  4th,  The  respective  velocities  of  the  i>ower  aiid  the  resistance. 

A  machine  is  in  eqnilibriuni  when  the  power  and  the  resistance  are  equal ; 
in  which  case  it  is  either  at  rest  or  in  a  state  of  uniform  motion. 


THE  LEVER. 

832.     The  Lever  is  a  rod  or  bar  of  metal,  wood,  or  other  nnitcrial,  which  is 
capable  of  moving  around  the  f((lcrum. 

Levers  may  be   either   straight   or  bent;    simple   or   compound.     They   are 
usually  divided  into  three  classes,  according  to  the  j)osition  of  the  fulcrum  in  rela- 
tion to  the  weight  and  power. 
1 


H 


\V F 


A 


Figure  1,  is  a  lever  of  the  first  class,  where  the 
fulcrum  is  between  the  weight  and  the  power. 


Figure  2,  is  a  lever  of  the  second  class,   where 
the  weight  is  between  the  fulcrum  and  the  power. 


;  V- 


t 


Figure  3,  is  a  lever  of  the  third  class,  where  the 
power  is  between  the  fulcrum  and  the  weight. 


I 


420 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


The  Arms  of  a  lever  are  tlie  lines  ou  each  side  of  the  fulcrmn,  at  rijjht 
angles  to  the  direction  of  the  iiower  and  the  weight. 

In  the  three  figures  given  above,  the  levers  being  horizontal  :ind  the  forces 

vertical,  the  arms  of  the  lever  are  evidently,  in  eacih  case,  the  portions  into  which 

it  is  divided.     If  however,  the  h^ver  is  bent  or  is  inclined  to  the  direction  of  either, 

*  or  botli,  of  the  forces,  then  the  arms  are  the  perpen- 

,.-'\  diculars  between  the  fulcrum  and  the  directions  of  the 

,,--''         \  forces.     Thus,  in  Figuke  4,  the  power  acting  in  the 

p  ,--''  \  direction  B  P,  the  momentum  or  force  of  tlie  power  is 

A  \         not  expressed  by  P  x  A  F,  but  by  P  x  B  F.     The 

,  \      distance  from  the  fidcrum  is  called  the  hrenuie. 

When  the  forces  act  perpendicularly  to  the  arms   of  a   straight  lever,    an 
equilibrium  is  produced  when  the  weight  multiplied  by  its  distance  from  the  ful- 
crum is  equal  to  the  power  multiplied  by  its  distance  from  the  fulcrum ;  or,  in  other 
Avords,  the  weight  is  to  tlie  power,  as  the  distance  from  the  power  to  the  fulcrum  is   I 
to  the  distance  from  the  weight  to  the  fulcrum. 

Application  of  the  Lever.— Machines  and  utensils  in  daily  use  furnish  us 
familiar  examples  of  the  three  classes  of  levers: 

Of  the  first  class,  we  name  the  crowbar  and  ^ 

I)oker,  when  used  to  raise  the  load  or  weight  on 
their  points;  scissors,  snuffers  and  pincers  are 
pairs  of  levers  of  this  class.  The  point  C,  Fig- 
ure 5,  which  connects  them  being  the  fulcrum. 

The  power  is  applied  at  the  handles,  and  the  resistance  is  the  object  between 
the  blades. 

Another  example  of  the  bent  lever 
is  seen  in  the  ordinary  truck,  Figure 
G,  used  for  moving  heavy  goods  a 
short  distance.  In  this  machine,  the 
axis  of  the  wheels,  F,  is  the  fulcrum, 
against  which  the  foot  is  placed,  while 
the  weight,  at  R,  is  raised  off  the 
ground  by  the  hand,  applied  at  P. 

The  Scale  Beam,  or  balance,  is  one  of  the  most  useful  applications  of  the 
first  class  of  levers.  The  Ijcain  is  a  lever  poised  at  its  centre  on  a  knife-edge  of 
steel,  rt.  Fig.  7.  From  its  ends,  A  B,  are  suspended  the  scale  pans,  C  and  E.  The 
center  of  gravity,  »«,  is  placed  below  the  fulcrum,  a,  to  secure  a  horizontal  position 
of  the  beam  M'lien  in  equilibrium.  If  it  coincided  with  the  fulcrum,  the  balance 
would  rest  equally  well  in  all  positions,  and  if  it  were  above  the  fulcrum  the  beam 
would  be  upset  by  a  slight  disturbance. 


MECHANICAL    POWERS. 


421 


The  Steelyard  is  a  lever  of  tlie  first  cluss, 
with  uiRMiual  arms.  The  mass,  Q,  to  be  weighed,  is 
attached  to  tlie  short  arm,  A,  Figure  8,  and  it  is 
couiiteriioised  by  a  constant  weight,  G,  shifted  upon 
the  longer  arm,  marked  with  not(-lies  to  indicate 
pounds  and  ounces,  until  equilibrium  is  obtained. 
It  is  evident  that  a  pound  weight  at  G  will  balance 
as  many  pounds  at  (^  as  the  distance  G  0  is  greater 
tlian  A  G. 

Levers  of  the  second  class  occur  less  frequent- 
ly. A  pair  of  nut-crackers  with  the  fulcrum  at 
the  joint  G,  Figttre  !),  is  a  double  lever  of  this 
class.  An  oar  is  another  example;  the  water  is 
the  fulcrum,  the  boat  is  the  weight,  and  the  hand 

the  power.     A  door  moving  on  its  hinges,  and  a  wheelbarrow,  are  other  examples 
of  levers  of  the  second  class. 

In  levers  of  the  third  class,  the  power,  being  nearer  the  fulcrum,  is  always 
greater  than  the  weight.  On  account  of  this  mechanical  disadvantage,  it  is  used 
only  when  considerable  velocity  is  required,  or  the  resistance  is  small. 

The  common  fire-tongs,  sugar-tongs,  and  i" 

sheep-shears  are  double  levers  of  this  class. 
The  most  striking  illustrations  of  this  class  of 
levers  are  seen  in  the  animal  kingdom.  The 
compact  form  and  beautiful  symmetry  of  ani- 
mals depend  on  tlie  fact  that  their  limbs  are 
such  levers.  The  socket  of  the  bone,  a,  FiG- 
XJKE  10,  is  the  fulcrum  ;  a  strong  muscle,  b  c, 
attached  near  the  socket,  is  the  power,  and  the 

Aveight  of  the  limb  and  whatever  resistance,  w,  may  oppose  to  motion,  is  the  weight. 
The  fore-aim  and  hand  are  raised  through  a  space  of  one  foot,  by  tlie  contraction 
of  a  muscle  a]>plied  lu'ar  the  elbow,  moving  through  less  than  1-11!  that  space.  The 
muscle,  therefore,  exerts  12  times  the  force  with  which  the  hand  moves.  The 
muscular  system  is  the  exact  inversion  of  the  system  of  rigging  a  shiji.  The  yards 
are  moved  through  snuill  spaces  witli  great  force,  by  haiiling  in  a  great  length  of 
rope  witli  small  force;  but  the  limbs  are  moved  through  great  S])aces  with  compar- 
atively little  force,  by  the  contraction  of  muscles  through  small  spaces  with  very 
great  force. 

EXAMPLES. 

1.  In  a  lever  of  tlie  first  kind,  the  fulcrum  is  placed  10  inches  from  the 
weight,  the  long  arm  is  8  feet;  if  a  ]iower  of  150  pounds  be  applied  at  the  end  of 
the  long  arm,  what  weiglit  will  it  balance  at  the  end  of  the  short  arm  1 

Aus.  1110  pounds. 


OPERATION. 

150  lbs.,  iiower. 
90  in.,  long  arm. 
or,     10 


10)  14400 

1440    lbs.. 


Ans. 


150 
OG 

1440  lbs 


Ans. 


Explannlion. — Since  the  product  of  the 
•weight  by  the  distance  it  is  from  the  fulcrum, 
is  equal  to  the  product  of  the  power  by  the 
distance  it  is  from  the  fulcrum,  -we  therefore 
multiply  the  power  and  its  distance  from  the 
fulcrum  together,  and  divide  the  product  liy 
the  length  of  the  short  arm,  and  obtain  the 
required  weight. 


422 


souLE  s  niiLosopinc  practical  mathematics. 


2.     Ill  a  level'  of  the  second  kind,  if  a  ■vpeif;:'lit  of  1200  pounds  is  placed  2  feet 
from  the  end  resting  upon  the  fulcrum,  what  power  will  it  recxxure  to  sustain  this 

Ans.  lS9-i2,-lbs. 

OPERATION. 


weight,  the  length  of  the  lever  being  12  feet  8  inches  ? 


1200  lbs.,  weight, 
24  in.,  distance  of  weight  from  fulcrum. 


or, 


153 


2SS00  product,  wliich 


1200 
2-1 


152  (inches  length  of  lever)  gives 

189,^7,  lbs.,  the  required  power. 
3.     In  a  lever  of  the  tliird  kind,  what  power  will  it  require   to   sustain   a 
weight  of  80  pounds,  the  distance  from  the  fulcrum  to  the  power  being  4  feet  4 
inches,  and  from  the  fulcrum  to  the  weiglit  G  feet?  A-ns.  110]!}  lbs. 

OPERATION. 

so 

72 


80  lbs.,  weight. 

7'>  ' 


'<li  in.,  length  of  lever. 


or, 


5760  product,  which  -^  52  gives  llOj!}  lbs.,  the  required  power.  HOla  Ans. 

4.  If  the  orifice  of  the  safety-valve  of  a  steam  boiler  is  2  inches  in  diameter 
and  5  inches  from  the  fulcrum  of  the  lever,  how  far  from  the  orifice  must  a  weight 
of  CO  pounds  be  placed  so  as  to  just  equal  the  force  of  the  steam  when  at  a  press- 
ure of  80  pounds  to  the  square  inch,  making  no  allowance  for  the  concave  surface 


of  the  boiler  or  weight  of  lever  ? 


Aus.  15.944  in. 


OPERATION. 


7854 


3.1416    sq.  in.  on  surface  of  safety-valve. 
80  lbs.  pressure  to  the  square  inch. 


251.3280  pressure  on  safety-valve. 

5  in.  distance  from  fulcrum  to  the  orifice. 


60  )  1256.6400  product  -H  the  weiglit  gives 

20.9440  in.  distance  of  weight  from  the  fulcrum. 
5  in.  distance  from  fulcrum  to  orifice  deducted. 


15.944    in.  distance  of  the  weight  from  the  orifice. 


THE  WHEEL  AND  AXLE. 

833.  The  common  lever  is  chiefly  employed  to  raise  weights  through  small  ^ 
spaces,  by  a  succession  of  short  iutermitting 
efforts.  After  the  weight  has  beeu  raised,  it 
must  be  supported  in  its  new  iwsitiou  until  the 
lever  can  be  again  adjusted,  to  repeat  the  action. 
The  wheel  and  axle  is  a  modification  of  the  lever, 
which  corrects  this  defect ;  and,  since  it  converts 
the  intermitting  action  of  the  lever  into  a  con- 
tinuous motion,  it  is  sometimes  called  the  perpet- 
ual lever. 


MECHANICAL    POWERS. 


423 


This  niacliine  consists  of  a  cylinder  called  the  axle,  turning  ou  a  centre,  and 
connected  with  a  wheel  of  much  greater  diameter.  The  power 
is  api>lied  to  the  circumference  of  the  wheel,  and  the  weight  is 
attached  to  a  rope  wound  around  the  axle  in  a  contrary  direc- 
tion. Instead  of  the  whole  wheel,  the  power  may  be  applied  to 
a  handle  named  a  winch,  or  to  one  or  more  spokes  inserted  in 
the  axle.  When  the  axle  is  horizontal,  the  machine  is  called 
a  windlass.  Figure  11,  when  it  is  vertical  it  forms  the  capstan 
PiGURE  12,  used  ou  shipboard,  chiefly  to  raise   the   anchor. 

The  head  of  the  capstan  is  pierced  with  holes,  in  each  of  which  a  lever  can  be 
placed,  so  that  many  men  can  work  at  the  same  time,  exerting  a  great  force,  as  is 
often  necessary  in  raising  an  anchor,  while  the  recoil  of  the  weight  is  arrested  by 
a  catch  at  the  bottom. 


The  law  of  equilibrium  is  the  same  as  in  the  lever. 
Draw  from  the  center,  or  fulcrum  c,  FiG^'RE  13,  the 
straight  lines  c  b  and  c  a,  or  c  a',  to  the  points  on  which 
the  weight  and  power  act ;  a  c  b,  or  a'  c  b  is  evidently  a 
lever  of  the  first  class,  in  which  the  short  arm  c  6  is  the 
radius  of  tlie  axle,  and  c  a,  or 
radius  of  the  wheel.    Heuce, 


c  a 


,  the  long  arm,  is  the 
r  X  a  c  =  W  X  c  b. 


or, 


P:  W 


a  c. 


That  is  to  say,  the  icheel  and  axle  are  in  eqtdlibrium,  when  the  power  is  to  the 
weight  as  the  radius  of  the  axle  is  to  the  radius  of  the  icheel. 

In  one  revolution  of  the  machine,  the  power  moves  through  a  space  equal  to 
the  circumference  of  the  wheel,  and  the  weight  moves  through  a  space  equal  to  the 
circumference  of  the  axle ;  hence  the  power  and  weight  are  inversely  as  their 
velocities,  or  the  sjiaces  they  describe. 


PROBLEMS. 

1.  What  must  be  the  radius  of  a  wheel,  or  the  length  of  a  crank,  by  which 
SOO  pounds  suspended  from  an  axle  of  4  inches  radius  may  be  kept  in  balance  by 
a  power  of  10  pounds  1  Ans.  4  feet  2  inches. 


OPERATION. 


4      inches  radius. 
500  pounds  weight. 


40 


or. 


I      4p  )  2000  product  4-  40  the  i^ower. 


4 
500 


50      in.  =  4  ft.  2  in.,  Ans. 


I  50    in.  =  4  ft.  2  in.,  radius  of  wheel  or  length  of  crank. 

I  Explanation. — The  operation  of  this  work  is  based  upon  the  principle  above  elucidated,  that 

the  power  is  to  the  weight  as  the  radius  of  the  axle  is  to  the  7-adius  of  the  wheel  or  the  length  of  the  crank. 


424 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  What  must  be  tlic  <liiiineter  of 
ail  axle  in  order  that  a  power  of  150 
pounds,  iii)])li('d  at  the  end  of  a  crank  28 
inches  lon^;',  nieasurinj^  from  the  center 
of  the  axle,  will  balance  800  pounds? 
Ans.     lOi  inches. 


OPERATION. 

28    in.,  length  of  crank. 
150  pounds  power. 


( 


800  )  4200  product  4-  the  weight. 
5^    in.,  radius  of  axle. 


lOi    in.,  diameter  of  axle. 

3.  If  the  anchor  of  a  vessel  weighs  3000  pounds,  what  force  will  be  required 
from  each  of  5  sailors  to  raise  the  same,  working  at  the  capstan  with  bars  5i  feet 
long,  the  diameter  of  the  capstan  being  16  inches,  allowing  that  each  bar  enters 
the  capstan  4  inches?  Ans.  08^^  pounds. 

OPERATION. 

6  ft.    G  in.  length  of  bar. 

8  in.  radius  of  capstan. 


C  ft.    2  in. 

4  in.  distance  the  bar  enters  the  capstan. 


5  ft.  10  in.  =  70  in.,  length  of  bar  from  the  center  of  capstan. 

IC  4-  2  =  8        in.,  radius  of  capstan. 
3000  lbs.,  weight  of  anchor. 


70  )  24000  product  ~  the  length  of  bar. 


6  )     342f  lbs.,  power  required  -4-  5,  the  number  of  sailors. 


6Sf  lbs.,  force  required  of  each  sailor. 


THE  PULLEY. 


831.  Fixed  Pulley. — The  usual  form  of  this  machine  is  a 
small  wheel,  turning  on  its  axis,  and  having  a  groove  on  its 
edge,  to  admit  a  flexible  rope  or  chain.  In  the  simple  fixed 
pulley.  Figure  14,  there  is  no  mechanical  advantage,  except 
that  which  may  arise  from  changing  the  direction  of  the 
power.  Whatever  force  is  exerted  at  P,  is  transmitted,  with- 
out increase  or  diminution  (except  from  friction  and  the 
rigidity  of  the  rope),  to  the  resistance  at  the  other  end  of  the 
cord.  From  the  axis  C,  draw  C  a  and  C  b,  radii  of  the  wheel, 
at  right  angles  to  the  direction  of  the  forces;  a  C  b  represents 
a  lever  of  the  first  class,  with  equal  arms ;  hence,  in  equilibri- 
um, ihe  poicer  and  tveigJd  must  be  equal,  and  they  describe  equal 
spaces. 


r 


w 


MECHANICAL    POWERS. 


Movable  Pulley. — "When  the  block  or  frame  i.s  not  fixed, 
the  i)ulley  is  said  to  be  iiiovable.  Tlie  weight  is  suspended 
from  the  axis  of  the  movable  pulley,  and  the  cord  is  fastened 
at  one  end,  and  passing;  over  a  fixed  jmlley,  is  acted  on  by  the 
l)ower  at  the  other.  In  this  arrangement,  Figure  15,  it  is 
plain  that  the  weight  is  supported  equally  by  the  power  and 
the  beam  at  D.  For  the  pulley  acts  as  a  lever  of  the  second 
class,  whose  arms  are  to  each  other  as  ]  :  2;  the  fulcrum  is  at 
b,  b  cis  the  leverage  of  the  weight,  and  b  a  the  leverage  of 
the  power.  The  diameter  b  a  is  twice  the  radius  b  c,  therefore 
equilibrium  will  obtain  tchen  the  power  is  equal  to  one-half  of 
the  tceight:  i.  e., 

P  :  W  =bc:ba  =  1:2, 
therefore, 


To  raise  the  weight  one  foot,  each  side  of  the  cord  must  be  shortened  one 
foot,  and  the  power,  consecjuently,  passes  over  two  feet.  The  sijace  traversed  by 
the  power  is  twice  the  space  described  by  the  weight. 

Conipouud  Pulleys. — Sometimes  compound  pulleys  are  nsed, 
each  consisting  of  a  block  which  contains  two  or  more  single  pulleys, 
generally  placed  side  by  side,  in  separate  mortises  of  the  block.  Such 
an  arrangement  is  shown  iu  Figure  16.  The  weight  is  attached  to 
the  movable  block,  and  the  fixed  one  only  serves  to  give  the  power 
the  required  direction.  It  is  easily  seen  that  the  jjower  required  at 
P  is  just  the  same  as  would  be  required  at  any  point  between  A  and 
B.  The  weight  is  divided  equally  among  the  pulleys  of  the  movable 
block,  and,  of  course,  among  the  cords  passing  around  them  ;  and  as 
the  power  recpiired  to  sustain  a  given  weight  is  diminished  one-half 
by  a  single  movable  pulley,  it  follows  that  such  a  system  will  be  in 
equilibrijim  when  the innrcr  is  equal  to  the  weight  divided  by  the  number 
of  cords,  or  by  twice  the  number  of  movable  2)ullcys. 


P:  W=l 


F  =  K 


In  this  system,  as  in  the  single  movable  pulley,  the  space  through  which  the 
weight  is  raised,  is  as  much  less  than  the  space  through  which  the  power  descends, 
as  the  weight  is  greater  than  the  jjower. 

F  :  W  =  velocity  of  weight :  velocity  of  power. 


If  the  power  is  moved  through  G  feet,  fig.  16,  each  division  of  the  cord  in 
which  the  movable  block  hangs  will  be  shortened  one  foot,  and  the  weight  raised 
one  foot. 


426 


SOULES    PHILOSOPHIC     I'RACTICAI,    MATHEMATICS. 


Another  system  of  pulleys  is  represented  in  FiGUKE  17. 
Ill  this  arraiijjeiiieiit  each  pulley  hangs  by  a  separate  cord, 
one  end  of  which  is  attached  to  a  lixed  support,  and  the  other 
to  the  adjacent  pulley.  The  etfect  of  t  lie  i)ower  is  rapidly 
augmented,  being  doubled  by  eiiv.h  movable  luilley  added  to 
the  system.  The  uumbers  placed  near  the  cords  show  what 
part  of  the  weiglit  is  sustained  by  each,  and  by  eacli  i)nll('y. 
Such  a  system,  however,  is  of  little  practical  use,  on  account 
of  its  limited  range.  In  the  common  block  system,  fig.  16  (iu 
practice  tlie  pulleys  or  sheaves  of  each  block  are  placed  side 
by  side,  to  save  room,  here  tliey  are  separated  for  sake  of 
clearness),  the  motion  may  continue  until  the  movable  block 
touches  the  fixed  one;  but  in  tliis  only  till  D  and  E  come 
together,  at  which  time  A  will  have  been  raised  only  J  of 
that  distance. 


P :  W 


or, 


PROBLEMS. 


1.  What   power   will   supjiort   a   weiglit   of  1200   pounds   iu   two   movable 
pulleys  ?  Ans.  300  pounds. 

OPERATION. 

1200  weight  4-  4,  twice  the  uumber  of  movable  pulleys,  =  300  lbs.,  Ans. 

2.  In  a  compound  pulley  (see  fig.  16),  containing  three  single  ])ulleys  in  the 
movable  and  immovable  blocks,  what  power  will  support  3000  pounds  ? 

Ans.  500  pounds. 

OPERATION. 

3000  ^  6  =  500  pounds,  Ans. 

3.  In  a  block  and  tackle  with  4  sheaves  in  the  fixed  block  and  3   in   the 
movable  block,  what  power  will  be  required  to  support  5000  pounds  1 

Ans.  Tlif  pounds. 

OPERATION. 


5000  -^  7  =  714f  pounds,  Ans. 
Note. — The  imlleys  in  a  block  and  tackle  are  called  sheaves. 


MECHANICAL     POWERS. 
THE  IJSrCLmED  PLA]^E, 


427 


835.  This  mechanical  power  is  commonly  used  wlienever  heavy  loads, 
especially  such  as  may  be  rolled,  are  to  be  r;)ised  a  moderate  height.  In  this  way 
casks  are  moved  in  and  out  of  cellars,  and  loaded  upon  carts.  The  common  dray  is 
itself  an  incline  plane  (as  is 
clearly  seen  by  inspecting  FiG- 
TJRElS).  Suppose  a  cask  weigh- 
ing 500  lbs.  is  to  be  raised  4 
feet  by  means  of  a  plank  12  feet 
long;  it  is  plain,  that  while  the 
cask  ascends  only  4  feet,  the 
power  must  exert  itself  through 
12  feet,  and  hence,  12  :  4  =  500  : 


IG65,  the  force  necessary  to  roll  the  cask. 


In  mechanics,  the  inclined  plane  is  a  hard,  smooth  surface,  inclined  obliquely 
to  the  resistance.  The  length  of  the  plane  is  K  S,  figure  19,  S  T  its  height,  and  B 
T  its  base.    The  power  may  be  applied, 

a — In  a  direction  parallel  to  the  length  ; 
h — Or  parallel  to  the  base. 
c — Or  in  any  other  direction. 

In  each  case  the  conditions  of  equilibrium  may  be  derived  from  those  of  the 
lever. 

Application  of  the  power  parallel  to  the  length  of  the  Inclined  plane. — 

When  a  body  is  placed  upon  an  inclined  plane.  Figure  19,  its  weight,  which  is  the 
resistance  to  be  overcome,  acts  in  the  direc- 
tion of  the  force  of  gravity,  namely  in  the 
perpendicular  b  a.  Let  the  power,  P,  act,  by 
means  of  the  cord,  in  the  direction  a  c  par- 
allel to  the' inclined  jjlane  R  S,  then  from  the 
point  rt,  draw  a  d  at  right  angles  with  the  in- 
clined plane,  and  complete  the  parallelogram 
a  ch  d.  The  force  of  gravity  will  be  resolved 
into  two  other  forces ;  one  represented  by  h 
c  causing  pressure  on  the  inclined  plane;  the 
other,  represented  by  c  a,  tending  to  cause 
motion  down  the  inclined  plane.  This  latter  force  is  to  be  balanced  by  the  power 
applied  to  move  the  body.    The  body  will  therefore  be  sustained  Mhen 


P  :  W  =  ac-.ah; 
and  since  the  triangles  ale  and  li  S  T  are  similar, 


or. 


P:  TF=ST:11S; 
ST 


P  =  ir  X 


RS. 


figure. 


This  may  be  illustrated  by  an  apparatus  constructed  like  that  shown  in  the 


428 


SOULES    PHILOSOPHIC    PKACTICAL    MATHEMATICS. 


If  the  direction  of  the  poicer  is  parallel  to  the  inclined  plane,  equilibrium  irill 
obtain  u-hen  the  poicer  is  to  the  weight  as  the  height  of  the  plane  is  to  its  length. 
"NVliilo  the  weight  is  raised  through  a  space  equal  to  the  vertical  height  of  the 
l)lauc,  the  jiower  imist  move  through  a  space  equal  to  its  Icuglh.  If  the  lengtli  of 
SI  plane  is  10  feet,  and  its  height  2  feet,  P  must  move  10  feet  while  W  is  raised  3 
feet;  lieiice  the  jrower  aud  weight  are  inversely  as  their  velocities. 


Applicatiou  of  Iho  power  parallel  to  the  base  of  the  inclined  plane. — In 


the  second  case,  let  the  power  act  in  the  direction  of  a 
r.  Figure  20,  parallel  to  13  C,  the  base  of  tlie  plane; 
and  draw  the  lines  b  a  and  b  c  jierpendicular  to  tlie  di- 
rection of  the  power  aud  weight;  then  a  b  c  is  a.  bent 
lever,  having  its  fulcrum  at  b,  aud  equilibrium  will  take 
place  when 


r  :  W  =  b  c:  ab; 
the  triangles  a  b  c  and  ABC  being  similar, 


or, 


P:  ir=  AB:E  C; 


If  the  direction  of  the  power  is  parallel  to  the  base  of  the  plane,  equilibrium 
will  obtain  ivhen  the  power  is  to  the  weight  as  the  height  of  the  plane  is  to  the  length 
of  its  base. 

in  this  case,  the  space  described  by  the  i)ower  is  to  the  space  described  by 
the  weight  as  the  base  of  the  plane  is  to  its  height. 

PROBLEMS. 

1.  The  lengtli  of  an  inclined  plane  is  10  feet  and  its  height  3  feet,  what 
power  applied  parallel  to  the  plane  will  be  required  to  balance  a  weight  of  500 
pounds?  Aus.  150  pounds. 


OPERATION. 


500  ])ounds,  weight. 
3  feet,  height. 


or. 


10 


500 


1500  product  4-  the  length  of  tlie  iilane,  10  ft.,  gives  150 
lbs.,  the  power  required. 

2.     The  height  of  an  inclined  plane  is  12  feet  and  the  length  15  feet,  what 
•weight  applied  parallel  to  the  plane  will  support  a  power  of  150  pounds? 

Ans.  1S7J  pounds. 


OPERATION. 


150  pounds,  power. 
15  ieet,  length  of  plane. 


or, 


12 


2250  product  -^  the  height,  12  feet,  gives  187^   lbs.,   the 
required  weight. 


150 
15 

1S7J 


MECHANICAL    POWERS. 


429 


3.  The  base  of  an  inclined  plane  is  12  feet  and  the  lieight  4  feet,  what  power 
applied  parallel  to  the  base  of  the  inclined  plane  will  support  a  Aveight  of  50(? 
pounds?  Aus.  IGCg  pounds. 


OPEKATION. 


500  ])onnds,  weight. 
4  feet,  height. 


or, 


12 


2000  product  4-  the  length  of  the  base,  12   feet, 
gives  16Cg  lbs.,  the  i)ower  required. 


500 
4 

1665 


or,      500  X 


THE   WEDGE. 


B  d 


836.  Instead  of  lifting  a  load  by  moving  it  along  an  inclined  plane,  the 
same  result  may  be  obtained  by  moving  the  plane  under  the  load.  When  used  in 
this  manner,  the  inclined  plane  is  called  a  Medge.  It  is  custom- 
ary, however,  to  join  two  planes  base  to  base.  In  Figure  21,  A 
B  is  called  the  back  of  the  wedge,  A  C  and  B  C  its  sides,  and  d 
O  its  length.  The  poMcr  is  applied  to  the  back  of  the  wedge,  so 
as  to  drive  it  between  two  bodies,  and  overcome  their  resistance. 

The  resistance  may  act  at  right  angles  to  the  length  or  to  the 
sides  of  the  wedge.     In  the  first  case,  it  resembles  an  inclined 
plane,  when  the  power  is  parallel  to  the  base ;  and  hence  the  forces 
will  he  -in  equilibrium  when  the  power  is  to  the  resistance  as  the  bach  of  the  wedt/e  is 
to  its  leuf/th.    In  the  second  case,  it  is  similar  to  a  plane  when  the  power  is  iiarallel 
to  the  length  ;  and  therefore  in  equilibrium  tchcn  the  poicer  is  to  the  resistance  as  the 
back  of  the  wcdeje  is  to  its  side. 

If  the  wedge  is  isosceles,  the  power  is  to  the  weight  as  half  the  back  of  the 
wedge  is  to  one  of  its  sides. 

The  power  is  supposed  to  move  through  a  space  equal  to  the  length  of  the 
wedge,  while  the  resistance  yields  to  the  extent  of  its  breadth. 

Application  of  the  Medge. — As  a  mechanical  power,  the  wedge  is  used  only 
where  great  force  is  to  be  exerted  in  a  limited  space.  In  oil-mills,  the  seeds  from 
which  vegetable  oils  are  obtained  are  sometimes  compressed  with  enormous  force 
by  means  of  a  wedge.  It  is  everywhere  employed  to  split  masses  of  stone  and 
timber.  The  edges  of  all  cutting  tools,  as  saws,  knives,  chisels,  razors,  shears, 
etc.,  and  the  points  of  piercing  instruments,  as  awls,  nails,  pins,  needles,  etc.,  are 
modified  wedges.  Chisels  intended  to  cut  wood,  have  their  edge  at  an  angle  of 
about  30  deg.;  for  cutting  brass,  from  50  deg.  to  60  deg. ;  and  for  iron,  about  80 
deg.  to  90  deg.  The  softer  or  more  yielding  the  substance  to  be  divided  the  more 
acute  the  wedge  may  be  constructed.  In  general,  tools  which  are  urged  by  press- 
Tire,  admit  of  being  sharper  than  those  which  are  driven  by  a  blow. 


430  SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS.  * 

The  theory  of  the  ■wedge  gives  but  very  little  aid  in  estimating  its  effects,  as 
it  talies  no  account  of  friction,  whicli  so  largely  modities  the  results,  and  the  pro- 
portiou  betweeu  pressure  and  a  blow  canuot  be  detined. 

PROBLEMS. 

1.  The  liead  of  a  wedge  is  2  inches  thick  and  the  length  of  each  inclined 
side  is  8  inches ;  what  will  be  the  measure  of  its  efiect  if  struck  by  a  force  equal  to 
200  pounds?  An s.  1  COO  pounds. 

OPERATION. 

200  pounds  force. 

8  inches,  length  of  inclined  side. 


1600  product  4-  1,  which  is  J  the  thickness  of  the  head  of  the   wedge,   gives 
1600  pounds,  the  measure  of  the  effect  required. 

2.  The  head  of  a  wedge  is  4  inches  thick  and  the  length  of  each  inclined 
side  is  15  inches;  Mhat  pressure,  applied  to  the  head  of  the  wedge,  is  required  to 
drive  it  between  two  surfaces,  tlie  resisting  force  of  which  is  equal  to  300  pounds? 

Ans.  40  i)ounds. 

OPERATION. 

300  pounds,  resisting  force. 

2  inches,  .J  of  the  thickness  of  the  wedge. 

600  product  -^  la,  the  length  of  the  inclined  side,  gives  40  pounds,  the  pressure 
required. 


TOE    SCREW. 

837.  Tlie  screw  is  a  spiral  tliread,  or  groove,  winding  round  a  cylinder,  so  as 
to  cut  all  the  lines  drawn  on  its  surface,  i)arallel  to  its  axis  at  tlie  same  angle. 

The  screw  has  the  same  relation  to  the  ordinary  inclined  plane,  that  a  spiral 
staircase  has  to  a  straight  one. 

The  thread  of  a  screw  projects  from  the  surface  of  the  cylinder,  and  is  de- 
signed to  tit  into  a  siiiral  groove,  cut  in  the  interior  of  a  block  called  the  nut;  a 
lever  is  also  fixed  in  the  head  of  the  cylinder  to  which  the  power  is  applied.  The 
combination  of  these  parts  forms  the  mechanical  power  technically  called  the  .screw. 

In  working  the  screw,  the  resistance  acts  on  the  inclined  face  of  the  thread, 
and  the  power  parallel  to  the  base  of  the  screw.  This  corresponds  to  the  case  iu 
■which  the  direction  of  the  power  is  parallel  to  the  base  of  the  inclined  plane. 


MECHANICAL    POWERS. 


431 


Equililirhnn  will,  therefore,  take  place  when  the  power  is  to  the  resistance  as  the  dis- 
tance between  the  threads  of  the  screw  is  to  the  circumference  described  by  the  poicer. 

P:  ^Y  =h:2  Rn.    and     P  =  W  x  .-4-; 

h  being  the  distance  between  the  threads,  E  the  radius  of  the  screw,  or  of  the 
length  of  the  lever  attached  to  it,  and  t  the  ratio  of  the  circumference  of  a  circle 
to  its  diameter. 

During  each  revolution  the  power  describes  a  circle,  whose  circumference 
depeods  on  the  length  of  the  lever,  but  the  end  of  the  scre'w  advances  only  the 
distance  between  two  threads;  tlius  in  this,  as  in  all  cases  of  the  nse  of  machines, 
what  is  gained  in  power  is  lost  in  velocity. 

Mechanical  efficiency  and  applications  of  the  screw. — The  mechanical 
efficiency  of  the  screw  is  augmented,  either  by  increasing  the  length  of  the  lever,  or 
by  lessening  the  distance  between  the  threads.  If  the 
tlireads  of  a  screw  are  i  of  an  inch  apart,  and  the  jwwer 
describes  a  circle  of  5  feet  (120  half-inches)  circumfer- 
ence, a  power  of  1  pound  will  balance  a  resistance  of  120 
pounds;  if  the  threads  are  ^  inch  apart,  1  pound  "will 
balance  240  pounds,  the  efficiency  being  doubled. 

Fine  screws  are  therefore  more  powerful  than  coarse 
ones.  The  applications  of  this  most  useful  mechanical 
l>ower  are  too  numerous  to  mention,  but  no  more  fre- 
quent or  important  example  of  its  use  can  be  named 
than  is  seen  in  its  use  in  presses  of  nearly  all  kinds.  A 
good  illusti-ation  of  wliich  is  seen  in  the  copying  press, 
FlGITEE  22. 

The  endless  screw  is  a  contrivance  by  which  a  slow  motion  is  inqiarted  to  a 
wheel,  as  shown  in  Figure  23.  The  threads  of  the 
scrcAV  act  upon  tlie  cogs  of  the  wheel,  and  serve  to  move 
the  weight  Q,  attaclied  to  the  axis  M  L.  If  we  call  the 
radius  of  the  circle  described  by  the  winch  D  B  =  r, 
and  let  h  =  the  distance  between  the  threads  of  the 
screw,  we  shall  have  the  power  of  the  screw 


23 


ifHSPi'-I^H^^S' 


^' 


ioso 


-? 


Let  R  =  UF,  and  R'  =  M  L, 

and  the  power  of  the  wheel  and  axle  will 
Then  W:  P  =2.r  X  B:h  X  R'; 


% 


R 

w 


-H 


w 


P  X 

hR' 


Therefore  the  weight  is  to  the  power  as  the  circumference  of  the  circle 
described  by  the  winch  D,  multiplied  by  the  radius  of  the  wheel,  is  to  the  distance 
between  the  threads  of  the  screw  multiplied  by  the  radius  of  the  axis. 


432 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


PROBLEMS. 

1.  The  distance  between  the  threads  of  a  screw  is  ^  of  an  inch,  and  the 
cii-cuniference  described  by  tlie  handle,  or  lever,  is  4  feet;  what  weight  can  be 
raised  by  a  i)Ower  of  2o0  i)oiinds,  applied  at  the  end  of  the  lever? 

Aus.  48000  i)0uud8. 

OPEllATION. 

4    feet  circumference  of  handle. 
12     inches  in  1  foot. 


48         " 
4     i  " 


4     " 
1  inch. 


192     ^  "        "  48  inches. 
2J0     pounds  power. 


4S000      "      Ans. 
2.     The  distance  between  the  pitch  threads  of  a  screw  is  |  of  an  inch,  and 
the  length  of  the  crank  or  lever  is  IG  inches,  what  weight  can  be  raised  by  a  power 
of  400  pounds  applied  at  the  extremity  of  the  lever?  Aus.  321099.84  pounds. 

OPERATION. 

16    inches  length  of  lever 
2 


32    inches  diameter  of  the  circle  described  by  the  lever. 
3.141G 


100.5312    inches  circumference  of  the  circle  described  by  the  lever. 
8     J  inches  in  1  inch. 


804.2490     ^  inches  in  100.5312  inches. 
400     pounds  power. 


321099.8400    pounds,  Ans. 

3.  The  distance  between  the  threads  of  a  screw  is  |  of  an  inch,  the  handle 
or  lever  is  10  feet  long,  what  power  will  be  rerpiired  to  sustain  a  weight  of  50000 
pounds?  *"°    ''«^''-'7j_  ,.«„,wi^ 

OPERATION. 


Ans.  26.5257+  pounds. 


10 


feet  length  of  handle. 


20 
3.141G 

62.8320 
12 

753.9840 
5 

2  )  3769.9200 


1884.96 


or. 


=  i  in. 


=  i' 


in. 


=  «  in. 


10 

2 

3.1416 

12 
5 


50000 


50000  -r-  1884.96  =  26.5257+  pounds,  Ans. 


MECHANICAL    POWERS. 


433 


In  the  foregoing  work  on  Meclianical  Powers,  much  of  tlie  work,  especially 
tlie  introduction  to  the  subjects  treated,  was  taken  from  Silliman's  Principles  of 
Physics,  a  book  of  great  worth,  and  to  which  we  refer  for  a  more  comprehensive 
discussion  of  the  subject  here  briefly  treated. 


'^'^^■'^tini^^^ir^^' 


^i-ary  (^rm-TL^lJ  ^ 


838.  In  the  above  diagram  of  wheels,  A  is  a  pinion  with  14  teeth  and  makes 
50  revolutions  per  minute.  B  is  a  spur  wheel  with  42  teeth  and  is  connected  with 
pinion  A.  C  is  a  pinion  with  28  teeth  and  is  carried  by  the  shaft  of  wheel  B.  D 
is  a  spur  wheel  with  104  teeth  and  is  engaged  with  pinion  C.  E  is  a  pinion  with 
24  teeth  and  is  carried  by  engaging  wheel  B.  F  is  a  spur  wheel  with  112  teeth 
engaged  with  pinion  E.  A  is  the  engine  shaft  pinion  at  50  revolutions  per  minute 
and  is  therefore  the  pinion  motor.  "What  is  the  cycle  of  the  whole  double  train  in 
time,  and  how  many  revolutions  will  each  wheel  make  during  the  cycle? 

Ans.  lOEJ  min.  cycle  of  the  whole  train. 

A  makes  546  revolutions. 

B,  C  and  E  each  make  182  revolutions. 

D  makes  49  revolutions. 

r  makes  39  revolutions. 

OPERATION. 

A  makes  50  revolutions  per  minute. 
B  fj  X  50  =  -3^  revolutions  per  minute. 
Note. — C  ami  E  have  the  same  time  of  revolution  as  B. 

D  i-o\  X  -/  =  ^-9-  revolutions  per  minute. 

'1  1  2     X        3       7 

Since  A  makes  50  revolutions  in  one  minute,  it  will  make  1  revolution  in  -^ij 
minute;  in  like  manner  B,  D  and  F  will  make  one  revolution  in  5%,  -i^,\  and  -/g 
minutes  respectively. 

The  L.  C.  M.  of  1-,  ^„  -ft\  and  A  =  W  or  10f| 

Since  A  makes  1  revolution  in  ^V  min.,  in  1054  min. 
revolutions  as  lOf  f  is  equal  to  5V  which  is  546  times ;  in  like  manner  B  (also  C  and  E), 
X)  and  F  will  make  respectively  182,  49  and  39  revolutions  during  the  cycle. 
Note. — See  Problem  5,  page  215. 


Ans.  to  1st.  question. 
,  it  will  make  as  many 


839.  Per  Cent  is  derived  from  the  Latin  plirase  per  centum,  and  means 
literally  on  or  of  the  htindrcd.     Or  more  briefly,  the  hundredth. 

840.  The  Sign  of  Per  Cent  is  ffo.    Thus  5%  is  read  5  per  cent. 

8-1:1.  Percentage  is  a  term  ai^plied  to  all  calculations  in  which  100  is  used 
as  the  unit  of  measure  or  the  basis  of  comparison.  It  may  be  applied  to  all  tran- 
sactions of  business,  to  all  departments  of  science,  and  to  every  thing  in  nature. 
It  is  in  general  use  in  almost  every  department  of  jiractical  life. 

Thus  we  say  20%  gain  or  loss;  10%  of  the  number  or  quantity;  25%  better 
or  lower  quality;  20  is  80%  of  25;  30%  increase  or  decrease  of  the  products  of 
field,  mine,  or  factory ;  the  general  won  75%  of  the  battles  fought  and  lost  45%  of 
his  army;  the  illiteracy  of  the  State  is  10%  of  the  population;  the  scholarship  of 
the  student  is  98  % ,  etc. 

842.  In  the  application  or  the  practical  computation  of  Percentage,  the 
following  elements  are  considered,  viz:  the  Base,  the  Rate,  the  Percentage,  and  th© 
Amount  or  Difference. 

843.  The  Base  is  the  number  on  which  the  jier  cent  is  calculated. 
Thus,  in  the  expression,  "  What  is  5%  of  70  ?"  the  base  is  70. 

844.  The  Rate  is  the  number  of  hundredths  to  be  taken. 
Thus,  in  the  question,  "What  is  5%  of  70  ?"  the  rate  is  5. 

845.  The  Percentage  is  the  result  obtained  by  taking  the  given  per  cent  of 
the  base. 

Thus,  in  the  statement,  5%  of  70  is  3.50,  the  percentage  is  3.50. 

846.  The  Amount  is  the  srim  of  the  base  and  the  pc>-ce?i<«r/e. 

Thus,  if  the  base  is  70,  and  the  percentage  is  3.5,  the  atnount  is  70  +  3.5  =^ 
73.5, 

847.  The  Difference  is  the  remainder  obtained  by  subtracting  the  percen- 
tage from  the  base. 

Thus,  if  tlie  base  is  70  and  the  percentage  is  3.5,  the  difference  is  70  —  3.5 
=  66.5. 

Note. — Tlio  relationship  between  the  above  elements  or  terms  is  such,  that  when  any  two  of 
"iiliem  are  given,  the  others  <*an  be  found. 

In  all  problems  of  per  cent,  two  of  these  elements  must  be  stated. 

848.  Since  per  cent  means  on  the  hundred,  it  is  clear  that  1%  of  any  number 
or  thing  is  the  hundredth  part  of  it;  and  if  1%  is  the  hundredth  part,  then  2%,  3%, 
10%,  25%,  etc.,  are  2,  ?>,  10,  25,  etc.,  tunes  as  many  or  as  much. 

Thus,  1%  of  $400  is  $4,  and  2%  would  be  twice  as  much,  which  is  $8;  and 

(434) 


*  PERCENTAGE.  435 

39J,  of  200  i^ounds  would  be  2  jiounds,  aud  10^^  would  be  10  times  as  much,  wbicli 
is  20  ijouiids. 

Per  cent  may  be  expressed  as  a  common  or  a  decimal  fraction. 

Thus,  2%  =  T-J-j,  or  .02;  3%  =  ^fo,  or  .03  ;  25%  =  -,",%  or  .25;  62  J  %  =  U^, 
or  .C2i;  100%  =  ifa,  or  1 ;  125%  =  iff,  or  1.25;  250%  =  ^a  or  2.50. 

849.  PER  THOUSAND,  PER  MILLE  OR  PER  M. 

In  some  of  tbe  professions  and  in  some  lines  of  business,  tbe  unit  of  measure 
or  the  basis  of  comparison  is  1000. 

Thus,  the  statistics  and  the  tables  of  mortality  are  given  per  M  ;  the  rate  of 
exchange  is  often  quoted  per  M;  the  price  of  lumber  is  quoted  per  M  feet;  and  the 
price  of  many  kinds  of  articles  of  commerce  is  quoted  per  M. 

Note. — M,  is  the  Latin  numeral  meaning  1000. 

PROBLEMS   IN   PEKCENTAGE. 

850.  1.     What  is  6%  of  150  bales?  Ans.  9  bales. 

OPERATION. 

1.50  Explanation. —In  all^jer  cent  problems  of  tliis  kind,  we  first  find  1%  by 

6  pointing  off  2  places,  and  then  multiply  by  the  given  rate  per  cent.     We 

reason  as  follows  :  1%  of  any  thing  is  the  hundredth  part  of  it;  hence  1% 

9.00  Ans.  of  150  bales  is  1.50  bales,  and  6°^  is  6  times  as  much,  -which  is  9  bales. 

2.  What  is  4%  of  $510 1  Ans.  $20.40. 

OPERATION. 
,$5.10  Explanation  .—First  find  l"o   by   dividing  by   100,    which  is   done  by 

I  pointing  off  2  places,   and  then  niultijdy  by  4.     AVe  reason  thus,   1%  of 

$510  is  $5.10,  and  4"o  is  -t  times  as  much,  which  is  $20.40. 
[  In  all  percentage  operations,  we  always  find  1",^  first,   and  then  the  % 

$20.40  Ans.  required;  Ijy  this  system,  we  will  always  have  in  the  answer  as  many 

decimal  figures  as  we  had  in  the  multixjlicand.     We  should  never  decimal 
the  rate  per  cent. 

3.  What  is  40%  of  375  oranges  ?  Ans.  150  oranges. 

OPERATION. 

3.75 
40 


150.00  Ans. 

4.  A  stock  raiser  had  860  head  of  cattle  and  they  increased  80%.     What 
was  the  increase,  and  how  many  has  he  now  1 

Ans.  688  increase.     1548  present  stock. 

5.  A  man  has  $485420,  and   gives  to  various  charitable  and  educational 
,  institutions  25%.     How  much  has  he  left  ?  Ans.  $364065. 

6.  A  factor  sold  48524  pounds  of  cotton  at  12i/  per  pound,    and  charged 
;  2J%  commission.     What  was  his  commi.ssion  ?  Ans.  $151.0375. 

7.  The  population  of  a  city,  ten  years  ago,  was  200000.     During  the  past 
ten  years  it  has  increased  12i%.     Wliat  is  the  present  population  ?     Ans.  225000. 

8.  A  planter  owned  550  sheep,  and  in  1  year  they  increased  50%  ;  how  many 
did  he  then  have  ?  Ans.  825. 


43^  soule's  philosophic  practical  mathematics.  * 

9.  What  is  3J%  of  $1720.15?  Ans.  $55.90+ 

NoTK. — In  7iraetice,  whenever  the  decimal  or  fraction  of  a  cent,  in  the  answer,  exceeds  i,  it 
is  counted  as  a  whole  cent,  and  when  it  is  less  than  i,  it  is  not  counted. 

10.  What  is  12J%  of  a  cargo  that  weighs  1280  tons  10  cwt.  3  qrs.  and  20 
pounds?  Ans.  ICO  T.  2  cwt.  0  qr.  11  lbs.  14  oz. 

851.  When  the  rate  per  cent,  is  an  aliquot  part  of  100,  the  ^)er  ce^it  may  be 
obtained  l)y  a  simple  division  operation.  Tims,  to  obtain  25%  of  940,  we  divide  by 
4,  for  the  reason  that  25  on  the  hundred,  or  1"^%,  is  ^,  and  ^  of  940  is  235,  the 
answei'. 

To  aid  the  calculator  in  this  contracted  work,  we  present  the  following 


852. 


T^I3LE. 


ii%  = 
H%-- 
H%-- 

H%-- 

61%  = 

10  %  = 

12i%: 
13^%  = 
1G§%  = 

183%  = 
20  %  = 
25  %  = 
26§%  = 

33^%  = 
35  %  = 

37i%: 

40  %  = 
411%  = 
43f%  = 
45  %  = 
46§%  = 


X\_  — 

-1- 

fill 

1  00  — 

80 

dll 

J-S  — 

.J_ 

100  — 

75 

100  — 

ih 

I?  ^. 

.J_ 

1  0  "(7  — 

4  0 

.a'ru  — 

1 

100  — 

SO 

.a'i  _ 

1 

1  1)  0  — 

1  6 

-«=J_ — 

.!_ 

10  0  — 

1  5 

6i-,L 

100  — 

iV 

.±0    _ 

J. 

10  0  — 

1  0 

1-ih  — 

i 

100  — 

ias  — 

100  — 

Ts 

i«5i  — 

J. 

100  — 

1; 

X>>\  — 

:i 

100  — 

1  (i 

2(1    

1 

10  0  — 

¥ 

2  5    

i 

100  — 

2BS  — 

4- 

1  00  — 

1  5 

iilk  — 

5 

1  oa  — 

'1  6 

ajis  — 
100  — 

i 

-35    

7 

10  0  — 

U  0 

3  7  1^ 

1  00  — 

t 

.i-0__ 

^ 

100  — 

5 

A1?J  — 

5 

10  0  — 

1 2 

AS'A  — 

7 

1  0  0  — 

1 1; 

4  5    

0 

1  0  0  — 

20 

4  Rfa 

7 

100  — 

rs 

and  conversely  sV=  li% 

■7^6=  n7o 

h=  3^% 
-A=  6i% 

-.\=  8J% 

-l\=ll'  % 

*=12J% 

*=16|% 

-A=l8f% 

J  =20  % 
i  =25  % 

iV=2<3i% 

4=33r/' 

2-% =35  % 
t=37i% 
I-  =40  % 
A=41i% 
i=43a% 

2^0=45  % 
A=46§% 


50    %: 

53i%: 
t>5  %: 
5Gi%: 
5Si%: 
02^%  = 

05  %  = 
GGi%  = 
G8!|%  = 
73^%  = 
75  %  = 
81i%: 

83i%: 
85    %: 

87i%  = 
9l|%  = 

033%  = 

95  %  = 

100%: 

150%: 
200%: 


.  5.0.  . 
-100- 

-10  0" 

-  JL-l 
-10  0- 
_  Jj^G^^ 
-100" 
-5Bii. 
-100- 

H  2  k 

-  1  Oo- 
_  ^5  _ 
-100- 

-iiiJii- 

-  1  00- 

-  &&h  - 

-  1  o'o- 
-X-aS- 
-  1  0  0  - 

-  7.5  _ 
-100- 

-100- 
-83il- 
-100- 

-  &•>-  - 

-10  0- 
-100- 

-a.j.s>- 

-100- 

oas. 

-  1  0  o  - 


•10  0- 

- _i  no  _ 

■10  0- 

.15  0. 
•1  0  0- 


J  ;  and  conversely  J  =50  % 

A=53i% 

M=55  % 
i-^6=56i% 
i^2=58i% 
|=62i% 
M=65  % 
§  =«C§% 

S  =75  % 

ii=8U% 
t=83^% 
ii=85  % 
^=87i% 
ii=91i% 
it=93J% 
it=93a% 

it=95  % 
the  whole  number,  or  the 

whole  of  a  thing, 
one  and  one-half  times  the 

whole  number, 
two  times  the  whole  num- 
ber. 


■15) 
.11. 
■20  ) 
.  il    . 

■  1G> 
-   X    . 

■  1  2  J 

=  1; 

-IS  . 

-■.II J 


-  I  5  ) 

-i; 

-IS  . 
-10) 

-  5.   . 

-  «  J 
-17  . 

-  20  ) 

--i; 

-XI  • 

-1  ^ ) 
-11  • 

-15) 
-15  . 
-16) 
-li  • 

-  20  ) 


u 
u 
ii. 
a 
u 
u 

a 
u 
it 
u 
u 
II 
li 
it 
li 


u 
il 
il 
II 
u 
u 
II 
il 
ii 
il 
il 
II 
II 
il 
il 
a 
li 


ALIQUOT  PAETS  OF  1000. 


62  J  %  =  -h 

1S7J%  =  ,a. 
250  %  =   i 

41G§%  = 

437i%  = 

83i%  =-,\ 
125   %  =   i 
133i%  =  ft 
166§%  =   i 

2GG§%  =  -i 
312^%  =  -,\. 
333i%  =   i 
375  %  =   1 

4GG|%  = 
533i%  = 
5G2^%  = 
583i%  = 

G25  %  = 
eGG§% 
G87i% 
733J% 
750  % 
812  J  % 


3 

1  L 


la 

1 6 


833^%  = 
8GG§%  = 
875  %  = 
91G§%  = 
933J%  =  if 
937J%  =  H 


n 


PERCENTAGE. 


437 


PROBLEMS. 

1.     What  is  20%  of  $4204.20  ? 

1st   OPERATION.  2(1    OPERATION. 


$42.64.20 
20 

$852.8400  Alls. 


$4204.20 

$  852.84  Aus. 


Wliatis2i%  of  $500? 
What  is  10  7p  of  450  pounds  1 
What  is  12i'/c  of  1000  chickens? 
What  is  20%  of  444  apples  ? 


Aus.  $852.84. 


Explanation. — Since  20°o  is  i 
of  the  amouut,  we  therefore, 
ill  the  second  operation,  divide 
by  5  and  thus  obtain  the  cor- 
rect result. 

Alls.  $12.50. 

Ans.-  45  pounds. 

Ans.  200  chickens. 

Ans.  88 i  apples. 


CLASSIFIED   PROBLEMS. 


853.  To  find  the  Percentage  and  the  Amount  or  Difference,  irhen  the  Base  and 
the  Rate  Per  Cent  are  given. 

Or,  in  Commercial  ProhlemSf  to  fliid  the  selliiig  price,  irhen  the  cost  and  the 
gain  or  the  loss  per  cent  are  given. 


1.     What  is  20%  of  550? 

OPERATION. 


5.50 


JO 


or  5  I  550 


110.00  Ans. 


110,  Ans. 


Ans.  110. 


Explanation. — In  all  problems  of  this  kind, 
for  reasons  above  given,  ve  first  divide  by 
100,  ■which  is  done  Ijy  pointing  off  two  places 
and  then  multiply  ))y  tlie  given  rate.  Or, 
■when  the  rate  is  an  aliquot  part  of  100,  ■we 
take  such  a  part  of  the  number  as  the  rate 
jier  cent  is  part  of  100. 


2.     What  is  25%  increase  of  500  pounds,  and  what  is  the  amount? 

Ans.  125  percentage  increase,  and  625  amount, 

OPERATION. 

5.00 


or 


125.00 

125  +  500 


500 

125  =  jiercentage. 
625  pounds,  amount. 


Explanation. — In  this  problem,  we  find  the 
percentage  as  above,  and  then  add  the  same 
to  the  base. 


854.     From  the  foregoing-  elucidations  we  derive  the  follo'wiiig  general  direc- 
tions for  finding  the  Percentage  and  the  Amount  or  Difierence : 

1.  To  find  the  Percentage,  divide  the  base  hi/  100,  and  then  multiply  hy   the 
rate.     Or,  tahc  sitch  a  part  of  the  base  as  the  rate  is  part  of  100. 

2.  To  find  the  Amount,  add  the  percentage  to  the  base. 

3.  To  find  the  Difference,  subtract  the  percentage  from  the  base. 

Note. — When  the  base  is  a  compound  number,  reduce  the  ■whole  number  to  the  lowest  named 
denomination ;  or,  reduce  the  lower  denominations  to  a  decimal  of  the  highest. 


438  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

3.  What  is  1G§%  of  1272?  Ans.  212. 

4.  What  is  11%  of  38r)0  hats?  Ans.  G54.50  hats. 

5.  What  is  40%  of  $1011.15?  Ans.  $G44.4G. 

G.     Bought  goods  at  24/,  and  sold  them  at  a  profit  of  25%.     What  was  the 
selling  price?  Ans.  30/. 

7.  Goods  cost  $2.40  a  yard,  and  sokl  at  20%  loss.  What  was  the  selling 
price?  Ans.  $1.92. 

8.  What  is  52%  of  $514.63?  Ans.  $207.60+. 

9.  What  is  62  J  %  of  3240  peaches?  Ans.  2025  peaches. 

10.  What  is  162%  of  GOO  books?  Ans.  100  books. 

11.  A  steamer  running  at  a  speed  of  12  miles  an  hour  increased  her  speed 
12J%.     What  was  her  speed  after  the  increase?  Ans.  13.i  miles  per  hour. 

12.  Bought  2462  bushels  of  corn  and  sold  33  J  %  of  it.     How  many  bushels 
were  left?  Ans.  1641 J  bushels. 

13.  A  capitalist  had  $85000  and  invested  25%  of  it.     How  many  dollars  did 
he  invest?  Ans.  $21250. 

14.  The  product  of  a  factory  in  1894,  was  94400000  pounds,  and  in  1895,  it 
was  12  J  %  less.     How  many  jiounds  was  the  deficiency?       Ans.  11800000  pounds. 

15.  Which  is  greater,  and  what  is  the  difference  between  8%  on  1200,  and 
7%  on  1300?  Ans.  1st,  8%  on  1200;  2nd,  5. 

IG.    A  bankrupt  merchant  who  owed  $43000,  paid  21  % .    How  much  did  he  pay  ? 

Ans.  $9030. 

17.  A  man  had  $1400  and  spent  10%  of  it,  and  then  20%  of  the  "I'emainder. 
How  much  did  he  spend  ?  Ans.  $392. 

18.  Paid  $8.50  per  barrel  for  flour  and  sold  it  at  a  gain  of  10%.     What  was 
the  selling  price  ?  Ans.  $9.35. 

19.  Paid  $15  for  a  coat  and  sold  it  at  20%  gain.     What  was  the  profit? 

Ans.  $3. 

20.  Goods  cost  $22.50  per  dozen,  and  sold  at  33J%  gain.     What  was  the  gain 
per  dozen  ?  Ans.  $7.50. 

21.  A  grocer  bought  160  dozen  eggs  and  sold  8J%  of  them.     How  many  did 
he  sell,  and  how  many  dozen  are  left?  Ans.  13J  doz.  sold;  14G|  doz.  left. 

22.  A  resides  IJ  miles  from  school;  he  walked  37J%    of  the   distance   and 
rode  the  remainder.     How  many  feet  did  he  ride  ?  Ans.  4950  feet. 

855.  To  find  the  Rate  Per  Cent,  when  the  Base  and  the  Amount  or  Difference 
are  given,  or  when  the  Base  and  the  Perceiitaf/e  are  given. 

Or  in  Commercial  problems,  to  find  the  Rate  Per  Cent  when  the  cost  and  the 
selling  price,  or  the  Cost  and  the  Gain  or  Loss  are  given. 


PERCENTAGE. 


439 


The  base  is  40  aud  the  amouut  is  50. 


What  was  the  rate  per  cent  ? 

Ans.  25%. 


40 


OPEKATION. 

40  =  base. 

50  =  amount,  or  the  base 
—  +  the  PerceTitagc. 

10  =  Perceutage,  gain,  increase. 


a 


Explanation  — In  all  problems  of  this  kind,  we 
first  tind  tbe  percentage,  increase,  or  decrease 
the  gain  or  loss,  as  shown  in  the  operation. 
Then  by  inspection  and  reason,  we  see  that  it 
was  the  base  40  which  produced  tlje  10  percent- 
age or  gain.  Wo  then  reason  as  follows:  Since 
,,.|^  40  gained  10,  1  will  gain  the  40th  part,  and  100 

will  gain  100  times  as  much,  which  is  25. 

It  may  and  should  he  here  asked,  where  we 
get  the  100  and  how  do  we  know  that  the  result 
represents  per  cent?  AVe  answer:  The  100 
comes  from  the  problem  itself.  The  c]nestion  of  the  ]iroblem,  "  TThnt  was  the  rate  per  cent,"  if 
expressed  as  it  is  understood,  would  read:  What  was  tlie  gain  or  increase  on  100?  By  this  we  see 
where  the  100  comes  from,  and  as  the  reusoniug  showed  that  the  result  was  a  gain  on  the  100,  it  is 
therefore  clear  that  it  is  gain  per  cent. 

2.     The  popuhition  of  a  city  in  1885,  -^vas  68000  ;  in  18SG,  it  -n-as  72700.     What 


gain  or  increase  on  100. 


•was  the  rate  per  cent  increase  ? 

OPERATION. 

GSOOO  =  base. 

727G0  =  amount,   base  +  P.    or 


increase. 


68000 


4:700  =  Percentage   =    increase 

or  gain, 
100 


Ans.  7%. 


Explanation, — The  reasoning  for  this  problem 
is  the  same  as  in  the  preceding  one,  and  hence 
is  omitted. 


7  =  increase  or  gam  on  eveiy 
100,  and  hence  is  7%. 

3.    A  hat  cost  $3.50  and  sold  for  $• 


I 


OPERATION. 

$3.50  —  cost  =  base. 
4.20  —  selling  price  =  amount. 


350 


70/  =  gain  =  percentage. 
100 


20  =  gain  on  every  100/ and  hence 
is  rate  %  gain. 

4.     Goods  cost  $3.50  and  sold  for  $ 

OPERATION. 

^3.50  =  cost  =  base. 
2. SO  =  selling  price  =  difference. 


.20.     What  was  the  rate  per  cent  gain  ? 

Ans.  20%. 

Explanation. — As  in  the  second  problem  above, 
we  first  find  the  gain  or  percentage.  We  then 
reason  thus :  Since  $3.50  gain  70c.,  Ic.  will  gain 
the  350th  part  and  100c.  will  gain  100  times  as 
much,  which  is  20. 

12.80.    What  was  the  rate  per  cent  loss  ? 

Ans.  20%. 


350 


70/  =  loss  =  percentage. 
100 


Explanation. — In  this  problem,  we  first  find  the 
loss  or  percentage,  and  then  reason  as  in  the 
above  problem. 


20  =  loss  on  every  100/  and  hence 
is  rate  %  loss. 


440  soule's  philosophic  practical  mathematics.  • 

5.     Goods  cost  $22  aud  sold  at  a  profit  of  $5.50.    What  was  tlie  per  cent 


gain 


OPERATION. 


22.00 


5.50  =  gain  =  percentage. 
100 


25  =  gain  =  gain  %. 


Ans.  25%. 

Explanation. — In  this  problem,  we  have  the 
gain  or  percentage  given,  and  hence  have  bub 
to  make  the  proportional  statement,  the  reason- 
ing for  which  is  the  same  as  above. 


6.    Yesterday  the  thermometer  registered  76  degrees;  to-day  it  stands  11 
degrees  lower.    What  is  the  per  cent  decrease  in  temperature?  Ans.  li^^f/o. 

opebation. 

11    =  percentage  — -  decrease.  Explanation.— The  reasoning  for  the  propor- 


70 


100 


14 1%  %  decrease  in  temperature. 


tional  statement  of  this  problem  is  the  same  in 
the  main  as  in  problem  3,  and  is  therefore 
omitted. 


856.     To  find  ichatper  cent  one  number  is  of  another,  or  the  Base  and  Percent- 
age  given  to  find  the  %  or  Mate. 


1.     What  per  cent  of  15  is  12  ?     Or,  12  is  what  per  cent  of  15 1     Ans.  80%. 
operation. 

Explanation. — In  this  problem,  15  is  the  base  or  unit  of 
comparison ;  and  to  produce  the  required  proportional  num- 
bers to  solve  the  question,  we  find  lOO^'o  of  it,  and  reason 
thus:  Since  lOOj!^  gives  15,  conversely  15  required  100%; 
hence  1  will  require  the  15th  part  and  12  will  require  12  times  as 
much,  which  is  80%. 


100%  of  15  is  15. 
100% 
12 
80%  Ans. 


15 


OPERATION. 

100%  of  12  is  12. 


2.     What  per  cent  of  12  is  15  ? 
or  15  is  what  per  cent  of  12  ? 

Ans.  125%. 


12 


100 

15 

125%  Ans. 


857.     To  find  what  %  one  niimher  is  greater  than  another,  or  the  Amount  and 
Base  given  to  find  the  <^/o  or  Rate. 

1.     What  per  cent  is  15  more  than  12  ?     Or,  3  is  what  %  of  12  ?  Ans.  25%. 

OPERATION. 
100%  of  12  is  12.  Explanation. — Here  12  is  the  base  or  unit  of  comparison; 


12 


'0 

00 


25%  Ans. 


and  to  produce  the  required  proportional  numbers  to  solve 
the  problem,  we  find  100%  of  12  and  reason  thus :  Since  100% 
gives  12,  conversely  12  requires  100%;  and  since  12  requires 
100%,  1  will  require  the  yV  part  and  3  will  require  3  times  as 
much. 


*  PERCENTAGE.  44I 

858.     To  find  what  fo  one  number  is  less  than  another,  or  the  Difference  and 
Base  given  to  find  the  %  or  Bate. 

OPERATION. 

100%  of  15  is  15. 


1.     What  per  cent  is  12  less  than  15  * 
or  3  is  what  %  of  15  ?  Aus.  20%.  15 


100 
3 

20%  Ans. 


859.  From  the  foregoing  elucidations,  we  derive  the  following  directions 
for  finding  the  rate  per  cent : 

1.  To  find  the  Rate  %  when  the  Base  and  Amount,  or  the  Base  and  the  Differ- 
ence, or  the  Cost  and  the  Selling  Price  are  given,  first  find  the  percentage — the  increase 
or  decrease,  the  gain  or  loss — and  then  male  the  proportional  statement  shown  in  the 
operations  ;  or  thus  :  The  base  or  cost :  the  increase  or  decrease,  the  gain  or  the  loss  :  : 
100:  the  required  rate  per  cent. 

2.  When  the  base  and  the  percentage  of  increase  or  decrease,  or  the  base  and  the 
gain  or  loss,  are  given,  then  malce  the  above  proportional  statement  at  once. 

Or,  if  it  is  desired  to  ignore  all  processes  of  reasoning,  thus  :  Multiply  the 
increase  or  decrease,  the  gain  or  loss,  by  100,  and  divide  by  the  base  or  cost, 

PROBLEMS. 

1.  The  base  is  1750 ;  the  percentage  is  43.75.    What  is  the  rate  %  ? 

Ans.  2J%. 

2.  A  yard  of  cloth  cost  16/  and  sold  for  18/.     What  was  the  gain  %  ? 

Ans.  124%. 

3.  Paid  $120  for  a  horse  and  sold  it  at  a  profit  of  $22.  What  was  the 
gain  %  ?  Aus.  18J%. 

4.  In  a  population  of  240000,  there  were  6214  deaths  in  12  nioiitlis.  What 
was  the  rate  per  cent?  Ans.  2.58|J%. 

5.  What  per  cent  of  54  is  6 ?  'Ans.  lli%. 

6.  What  per  cent  of  $540  is  $67.50  ?  Ans.  12i%. 

7.  $2.50  is  what  %  of  $60.40?  Ans.  4M%- 

8.  9  is  what  %  of  216  ?  Ans.  4^%. 

9.  An  invoice  of  goods  cost  in  New  York  $3840 ;  the  freight  was  $57.60. 
What  was  the  rate  %  ?  Ans.  1.4%. 

10.  On  a  bill  of  $421.80,  a  discount  of  $21.09  was  allowed.     What  was  the  %, 
discount?  Ans.  5%. 

11.  What  %  of  a  number  is  |  of  it?  Ans.  024%. 

OPERATION. 

Let  100  =  the  number.     Then  f  of  100  is  624,  and  624  on  the  hundred  is  624%. 

12.  What  %  of  a  number  is  5%  of  15%  of  it?  Ans.  5%,  or  .75%. 


442  soule's  philosophic  practical  mathematics.  * 

13.  What  %  of  a  number  is  10%  of  20%  of  25%  of  it!     Ans.  J%,  or  .5%. 

14.  15%  of  CO  is  what  %  of  75  ?  Ans.  12%. 

15.  40  is  what  per  cent  of  8  ?  Ans.  500%. 

16.  G  is  what  ^o  of  24?  Ans.  25%. 

17.  A  jjassenger  train  runs  25  miles  an  hour ;  an  express  train  runs  35  miles 
an  hour.  What  %  slower  does  the  passenger  train  run,  and  what  %  faster  does 
the  express  train  run,  when  compared  with  each  other? 

Ans.  Pass,  train  runs  2S;J-%  slower.     Ex.  train  runs  40%  faster. 

18.  Single  tickets  cost  25/ ;  in  packages  of  10  they  cost  20/  each.  What  is 
the  %  gaiu  by  buying  by  the  package,  and  what  is  the  %  loss  by  buying  by  the 
single  ticket?  Ans.  25%  gain  by  buying  by  the  package. 

20%  loss  by        "  "      single  ticket. 

19.  What  %  of  a  number  is  ^  of  it?  Ans.  CS%. 

20.  What  is  2%  of  12i%  of  CG§%  of  25%  of  a  number?  Ans.  ^i  %. 

OPEEATION. 

100  assumed  as  the  number.    Then  ^§0  X  i^g^  x  Vi^  X  i%%;  or  -^to  X  -Mo  X  f^  X 
I'o'o )  or  50   X   s  X  g  X   4  =  Tfoo^- 

Then  o^^oo  of  100  is  /^  and  „\  on  the  hundred  is  a^^- 

21.  In  the  above  problem,  what  would  be  the  percentage  on  $4800  ? 

Ans.  $2. 

OPEEATION. 

ij%  of  $4800  is  $2  or  thus:  25%  of  $4800  is  $1200  :  GG|%  of  $1200  is  $800  :  12J% 
of  $800  is  $100;  2%  of  $100  is  $2,  answer. 

22.  What  is  48%  of  1C§%  of  37i%  of  G2^%  of  a  number?  Ans.  1^%. 

23.  The  cotton  croj)  of  the  Southern  States  ending  the  fiscal  year,  Sept.  1, 
1884,  was  5713200  bales ;  and  for  the  fiscal  year  ending  Sept.  1,  1885,  was  5655900 
bales.  1.  What  %  more  Avas  produced  in  1SS4  than  in  1885  ?  2.  What  %  less  iu 
1885  than  in  1884  ?  3.  What  %  is  5713200  bales  of  5G55900  bales  ?  4.  What  %  is 
5655900  bales  of  5713200  ?  Ans.  1.0131+  %  more  was  produced  in  1884. 

1.00290+  %  less     "  "  1885. 

101.0131+  %,  5713200  is  of  5655900. 
98.99705+  %,  5655900  is  of  5713200. 

24.  A  man  has  due  him  $45,  and  compounds  on  receipt  of  $36.  What  %  did 
he  lose?  Ans.  20%. 

25.  A  broker  bought  bonds  at  90/  on  the  dollar  and  sold  them  at  95/  on  the 
dollar.     What  %  did  he  gain?  Ans.  5f%.  ' 

26.  In  a  j^ear  of  365  days,  67  days  are  rainy.  What  %  of  the  days  are  not 
rainy?  .  Ans.  81fJ%. 

27.  According  to  the  Carlisle  mortality  tables,  43  persons  of  every  5879  of  25 
years  of  age  die  annually.     What  is  the  rate  %  of  deaths?  Ans.  .731+%. 

860.  To  find  the  Base  tchen  the  Amount  or  Difference  and  the  Bate  per  cent 
Increase  or  I)ecrease  are  given. 

Or,  in  Commercial  Froblcms,  to  find  the  cost  when  the  Selling  price  and  the 
Oain  or  Loss  per  cent  are  given. 


PERCENTAGE. 


443 


1.     The  manufactured  value  of  goods  is  $2100,  which  is  20%  more  thau  the 
value  of  the  raw  material.     ~V\Tiat  was  the  value  of  the  raw  material? 

Alls.  $1750. 

FIRST    OPERATION. 


$100  =:  assumed   base   or   value   of  raw 
material. 
20  =  20%  increase  value. 


$120  =  mauufactured  value. 
$ 

assumed  base. 


120 


100  = 

2100 


$1750  cost  of  raw  material. 


Explcmation. — By  considering  tlio  problem,  we 
see  that  the  $2100  is  the  aniouut  of  the  valne  of 
the  raw  material  and  the  20)';^  cost  to  manufac- 
ture the  goods.     AVe  also  see  that  the  203'q   was 
calculated  on   the   value   of  the   raw   material, 
and  not  on  the  $2100,  the  value  of  the  manufac- 
tured goods.     Hence  there  are  no  figures  in  the 
problem  upon  whose  face  we  can  calculate  the 
20%   cost    to    manufacture.     We    therefore,    as 
shown  in  the  operation,  assume  100,  as  the  base 
or  value  of  the  raw  material,   and  on   this   we 
calculate    and   add    thereto    tlio    20,"^    cost    to 
manufacture.     This  gives  an  amount  of  $120,   as 
the  manufactured  value  of  goods  from  a  base  or 
raw  material  value  of  $100. 
Now  with  these  values  which  contain  the  same  ratio  of  base  and  amount,  or  of  raw  material 
and  of  manufactured  goods,  that  exists  between  the  $2100  of  manufactured  goods  and  the  required 
value  of  the  raw  material  from  which  it  was   produced,    we  make   the   proportional   statement, 
shown  by  the  operation  and  obtain  the  required  base  or  raw  material  value. 

lu  making  the  solution  statement,  we  place  the  $100  assumed  base  or  value  of  raw  material 
on  the  line  and  reason  thus:  Since  $120  amount  or  manufactured  value  at  a  gain  of  20%"  reqnire 
$100  base  or  value  of  raw  material,  $1  will  rec|uire  the  120th  part,  and  $2100  amount  or  manufac- 
tured value  will  require  2100  times  as  much. 

In  assuming  a  number  to  represent  cost,  it  is  immaterial  so  far  as  correct  results  are 
concerned,  what  number  wo  assume  ;  but  we  always  select  100  for  the  reason  that  ]kt  cent  being 
on  the  hundred,  our  operation  is  facilitated  to  a  greater  extent  by  100  than  by  any  other  number. 

The  foregoing  reasoning  and  solution  are  applicable  to  all  problems  of  like  character, 
regardless  of  the  rate  of  gain  or  loss  per  cent.  But  when  the  gain  or  loss  ]>er  cent  is  an  aliquot 
part  of  100,  the  operation  may  be  very  much  shortened  by  the  following  process  of  work  and 
reasoning : 


SECOND   OPERATION. 


6  )  $2100 


350  =  i  of  base  or  value  of  r.  m. 
5 


$1750  =  base  or  value  of  r.  m. 


Explanation. — In  this  solution,  since  the  rate 
P,,'  is  au  alicjuot  of  100,  we  reason  thus:  Since 
the  amount,  $2100,  contains  a  gain  of  20°^',  and 
since  20?o  is  equal  to  y  of  a  thing,  the  $2100  is 
therefore  the  base  or  valne  of  the  raw  material, 
plus  I  of  the  same,  which  makes  !;,  and  since 
$2100j  is  f  of  the  base,  -k  of  the  base  is  the  J 
part,  and  J  or  tho  whole  base  is  5  times  as  much. 


2.     Sold  goods  for  $40  and  gained  25%.    What  was  the  cost  of  the  goods  ? 

Aus.  $32 


FIRST    OPERATION. 


$100  =  cost  assumed. 
25  =  25%  gain  added. 


J125  =  selling  price. 
■     $ 

I  100 
125  I  40 

I  $32  cost,  Ans. 


Explanation. — Here,  omitting  the  explanatory 
remarks  given  in  the  preceding  problem,  we 
assume  $100  as  the  base  or  cost  and  reason  thus : 
To  gain  25°^  we  must  sell  what  cost  $100  for 
$125;  then  since  $125  selling  price  at  a  gain  of 
25%,  required  $100  cost,  $1  selling  price  will 
require  the  125th  p.art,  and  $40  selling  price  will 
require  40  times  as  much,  which  is  |32. 


444 


SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS. 


SECOND    OPERATION. 

5  )  $40  =  selliiif!:  price. 


^  of  cost. 


32  cost,  A  IIS. 


Explanation. — For  this  operation,  we  reason 
thus  :  f>ince  the  $40  selling  price  contains  a  gain 
of  25,",^  and  since  25," „'  is  i  of  tlie  cost,  the  $40  is 
hence  f  of  the  cost ;  and  since  $40  is  J  of  the  cost, 
i  of  the  cost  is  the  i  part,  which  is  $8 ;  and  J  or 
the  -whole  cost  is  4  times  as  much,  which  is  $32. 


3.     Sold  goods  for  $30  and  lost  25%.     What  did  the  goods  cost?  Aus.  $40. 


FIRST   OPERATION. 

=  cost  assumed. 
25  =  25%  loss  deducted. 

$75  =  selling  i>rice. 
75 


100 
30 


cost,  Ans. 


SECOND   OPERATION. 


3  )  $30  =  selling  price. 
10  =  ^  of  cost. 


$40  cost,  Ans. 


Explanation. — The  reasoning  for  the  proportion- 
al statement  is  as  follows :  Since  $75  selling 
price  at  a  loss  of  2o%  required  $100  cost,  |1 
selling  price  will  require  the  75th  part,  and  $30 
selling  price  will  require  30  times  as  much. 

Note. — In  practice  the  line  statement  is  all 
that  needs  to  be  made.  The  assumed  Ijase  or  cost 
and  the  amount  or  selling  price  resulting  there- 
from, should  be  mentally  performed. 


Explanation. — In  this  solution,  we  reason  as 
follows  :  Since  the  $30  selling  price  is  the  cost 
less  25^„',  it  is  therefore  J  of  the  cost;  and  since 
t  of  the  cost  is  $30,  i  is  the  4  part  which  is  $10, 
and  f  or  the  whole  cost  is  4  times  as  much  which 
is  $40. 


861.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  finding  the  base  or  cost  when  the  amount,  or  diflereuce,  or  selling  price, 
and  the  rate  per  cent  are  given  : 

1.  To  find  the  Base  or  Cost,  first  assitme  100  to  represent  base  or  cost;  on  it 
calculate  the  given  rate  %  to  find  the  percentafje ;  then  either  add  it  to  or  subtract  it 
from  the  base  or  cost  according  as  the  rate  %  is  a  gain  or  a  loss,  an  increase  or  a 
decrease,  and  thus  pi-oduce  an  amount  or  difference,  which  has  the  same  ratio  of  value 
to  the  100  assumed  base  or  cost  as  the  given  Amount  or  Difference  has  to  the  required 
base  or  cost.  Now,  irith  this  ratio  of  values  and  the  given  amount  or  difference,  make 
the  proportional  statement  shoicn  in  the  operation  ;  or  thus  :  The  produced  amount  or 
difference :  the  assumed  base  or  cost :  :  the  given  amount  or  difference :  the  required 
base  or  cost. 

2.  If  it  is  desired  to  ignore  all  processes  of  reasoning  in  the  solution,   the 
following  brief  and  arbitrary  directions  will  solve  this  class  of  problems  : 

Multiply  the  given  amount  or  difference  or  selling  price  by  100,  and  divide  by  100 
plus  the  gain  %  or  minus  the  loss  %. 

PROBLEMS. 

4.     Sold  goods  for  $20.62J  and  gained  25%.    What  did  the  goods  cost  1 

Ans.  $16.50. 


*  PERCENTAGE.  446 

5.  Sold  a  watcli  for  $70  and  lost  20%.     What  did  it  cost?        Ans.  $87.50. 

6.  The  rent  of  a  house  is  $67.50  per  month,  which  is  12i%  advance  on  last 
year's  rent.    What -^vas  the  rent  last  year  ?  Ans.  $00. 

7.  The  amount  is  266|,  the  rate  %  is  33J.    What  is  the  base  ?      Ans.  200. 

8.  The  difference  is  400,  the  rate  %  is  20.     What  is  the  base  ?      Ans.  500. 

9.  What  number  increased  by  10%  of  itself  is  1815?  Ans.  1050. 

10.  $18.17  is  15%  more  than  what  sum?  Ans.  $15.80. 

11.  $4.50  is  50%  less  than  what  sum  ?  Ans.  $9. 

12.  A  produce  merchant  sold  corn  at  004/  per  bushel,  which  is  10%  more 
than  he  paid  for  it.     What  did  he  i^ay  for  it  ?  Ans.  55c. 

13.  A  carpenter,  after  using  80%  of  his  lumber  had  500  feet  on  hand.  How 
many  feet  had  he  at  first  ?  Ans.  2500  feet. 

14.  A  shoe  factory  manufactui-ed  2000  pairs  of  shoes,  M'hicli  weighed  5500 
pounds,  not  considering  the  thread  and  nails.  There  was  a  waste  of  S^%  in 
manufacturing.     How  many  pounds  of  leather  were  used  in  making  the  shoes? 

Ans.  0(100  pounds. 

15.  Sold  sugar  at  S^f  per  pound,  and  gained  6 J  per  cent.     "VNTiat  did  it  cost? 

Ans.  8('. 
10.     My  merchandise   account  is   debited  with   $78500    and    credited    with 


u- 


ooo 


5.    The  gain  %  on  sales  has  been  12i.    What  was  the  cost  of  the  sales,  and 
how  much  merchandise  have  I  on  hand  1  Ans.  $04200,  cost  of  sales. 

$14300,  mdse.  on  hand. 

17.  A  planter  lost  by  a  storm  20%  of  his  grain  and  has  29120  bushels  left. 
How  many  bushels  had  he  before  the  storm,  and  how  many  bushels  did  he  lose  ? 

Ans    304(  0  bu.  before  the  storm.     7280  bu.  lost. 

18.  Sold  goods  for  $100  and  gained  40%.     What  did  thev  cost? 

Ans.  $71.42J. 

19.  Sold  goods  for  $175  and  gained  150%.     What  did  they  cost? 

Ans.  $70. 

20.  Wliat  number  increased  by  25%  of  itself  is  $535  ?  Ans.  $428. 

21.  $72  is  20%  more  than  what  sum?  Ans.  $60. 

22.  What  number  diminished  by  25%  of  itself  is  510  ?  Ans.  080. 

23.  A  firm  lost  14%  of  its  capital   by  the  failure   of  another   house.     The 
remaining  capital  was  $278864.40.     What  was  the  whole  capital  ?     Ans.  $324201. 

STATEMENT. 


86 


100  =  capital. 
278864.46. 


24.  A  firm  pays  $9600  rent  for  its  store,  which  is  20%  more  than   it   paid 
last  year.     What  was  last  year's  rent  ?  Ans.  $8000. 

25.  If  40  boxes  of  peaches  be  sold  for  $113.75,  at  a  gain  of  62J%,  what  was 
the  cost  and  selling  price  per  box  1  Ans.  $1.75  cost.     $2.84|  selling  price. 


446 


soule's  riiiLosopiiic  i'ractical  mathematics. 


862.     To  find  the  Base  icheu  the  liatc iicr  cent  and  jfcrcentaf/e  are  given. 
Or,  in  Coinmervial  Problems,  to  find  the  cost  ichen  the  Rate  per  cent  and  tlie 
Gain  or  Loss  are  given. 

1.    The  rate  jier  cent  is  8  aud  the  percentage  is  36.    What  is  the  base  ? 

Ans.  450. 

OPERATION. 


100  =  assumed  base. 
8(/o  calculated  on  assumed  base. 


8.00 


percentage  on  assumed  base, 
assumed  base. 


100 
3G 


450  base,  Ans. 


Explanation. — In  all  problems  of  this  kind,  we 
first  assume  100  base  and  iind  the  percentage 
thereon  at  the  given  rate  °q.  Uy  this  work  we 
produce  a  base  and  a,  percentage  containing  the 
same  ratio  that  exists  between  tlie  given  per- 
centage and  the  required  base.  Having  tljese 
ratio  numbers,  we  then  make  a  proportional 
statement  as  sliown  in  the  operation. 

In  this  problem  we  find  the  percentage  to  bo 
8.  Tlien  placing  the  assumed  base  on  the  state- 
ment line,  we  reason  thus :  Since  100  base  at  8"^ 
produced  8  percentage,  conversely  8  percentage 


required  100  base ;  and  since  8  percentage  requires  100  base,  1  percentage  will  require  the  8th  part, 
and  36  percentage  will  reiiuire  36  times  as  much,  which  is  450. 

Note. — In  practice,  only  that  part  of  the  operation  shown  by  the  statement  line  should  be 
made ;  the  other  work  shouhl  be  mentally  performed. 


2.     40  is  5%  of  what  number  ! 


Ans.  800. 


OPERATION  INDICATED. 

100 
40 


20. 


3.     On  a  sale  of  a  lot  of  goods,  a  loss  of  $17  was  sustained.     Tlie  <fo  loss  was 
What  was  the  cost?  Ans.  $85. 

OPERATION. 

100  =  assumed  cost. 

20% 


^20.00  =  loss  on  assumed  cost. 

$ 
100 

20    i; 


Explanation. — In  this  problem,  the  reasoning  is 
the  same  as  in  the  first  problem,  and  hence  is 
omitted. 


$85  cost,  Ans. 


863.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  lindiiig  the  base  or  cost  when  the  rate  per  cent  and  the  jjerceutage  or  gain 
or  loss  are  given  : 

1.  Ti>  find'  the  Ba.te  or  Cost,  first  assnme  100  to  represent  base  or  cost,  and  then 
find  the  percentage  or  gain  or  loss  thereon  at  the  given  Hate  %.  Then,  for  reasons 
given  in  the  explanation  of  the  first  problem,  make  the  proportional  statement  as  shown 
in  the  operation. 


*  PERCENTAGE.  447 

2.     Jf  it  is  desired  to  ignore  all  reasoning  in  the  solution,    then  multiply   the 
^percentage,  gain  or  loss,  by  100,  and  divide  by  the  given  rate  "Jo- 

PROBLEMS. 

4.  The  rate  per  ceut  is  10  and  tlie  pei'centage  is  108.     What  is  the  base  ? 

Ans.  1080, 

5.  3021  is  12i%  of  what  number?  Ans.  24168. 

C.     Goods  were  sold  at  a  profit  of  25%  and  a  gain  of  $5. GO   was  realized. 
What  was  the  cost  %  Ans.  $22.40. 

7.  Sold  a  watch  for  $26.50  more  than  it  cost  and  gained  50%.     What  did  it 
cost,  and  what  was  the  selling  price  ?  Ans.  $53  cost.     $79.50  selling  price. 

8.  City  taxes  are  2%,  and  a  man  paid  $886.42.     What  was   the   assessed 
value  of  his  iiroperty  ?  Ans.  $44321. 

9-     sHd  i'^  J%  of  what  number  ?  Ans.  \\. 

10.  8J  is  1%  of  what  number  ?  Ans.  S50. 

11.  State  taxes  are  .6%,  and  a  man  paid  $67.50.    What  is  the  assessed  Aulue 
of  his  property  ?  Ans.  $1 12.50. 

12.  5%  of  $180  is  \1Y/c  of  what  sum  ?  Ans.  $72. 

13.  A  man  lost  $35.88,  which  Avas  8%  of  what  he  had  at  first.     How  much 
did  he  have  left  %  Ans.  $412.62. 

14.  A  planter  sold  2430   barrels   ot   :iiolasses   and   had  25%    of  his  yearly 
product  left.     What  was  his  yearly  product  f  Ans.  3240  barrels. 

864.  To  find  the  Gain  or  Loss  per  cent,  ichcn  the  Cost  and  the  Selling  price 
are  given. 

1.  Goods  cost  $3  and  sold  for  $4.     What  was  the  gain  %  ?        Ans.  33J%. 
Note. — For  the  operation  and  reasoning  for  tliis  class  of  problems,  see  Article  855. 

2.  Bought  a  horse  for  $180  and  sold  it  for  .*160.     What  per  cent  was  lost? 

Ams.  11^%. 

3.  Sold  flour  which  cost  $7.50  at  $8  per  barrel.    What  was  the  gain  per 
cent?  Ans.  6§%,. 

4.  Bought  shirts  at  $16.50  per  dozen  and  sold  them  at  $1.75  a  piece.     What 
%  did  I  gain  ?  Ans.  27  A  %. 

5.  Bought  3  apples  for  lO/'  and  sold  them  at  5/  each.     What  %  did  I  gain  ? 

Ans.  50%. 

6.  Bought  tea  at  $1.25  per  lb.  and  sold  it  at  7iC  per  oz.     What  %  did  I  lose? 

Ans.  4%. 

865.  To  find  the  Cost,  when  the  Gain  per  cent  and  the  Selling  price  or  the  Gain 
or  Loss  are  given 

1.     Sold  goods  for  $15  and  gained  25%.    What  did  they  cost  ?      Ans.  $12. 

KoTE. — For  the  operation  and  reason  for  the  solution  of  this  class  of  prohlems,  see  Article 
%9. 


448 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  Sold  a  vest  for  $2  more  that  it  cost  and  gaiued  33 J%.     "What  did  it  cost? 

Aus.  $C. 

OPERATION    INDICATED. 

100 

100  cost  assumed,  33^^^^  of  100  =  33^  ;  then  33^^     2 

3.  If  3/  i^er  pound  is  lost  by  selling  coffee  at  a  loss  of  20%,  what  was  the 
cost  ?  Aus.  15/  per  lb. 

4.  Sold  goods  for  $19.80  per  dozen  and  lost  10%.     What  was  the  cost  per 
siugle  article  ?  Ans.  $1.83J. 

5.  A  loss  of  If  per  pound  was  sustained  by  selling  at  7  cs'/o  loss.    What 
•was  the  cost?  Ans.  13/. 

C.     Sold  com  at  5G J/ iier  bushel  and  gained  12 J%.     What  did  it  cost  ? 

Ans.  50/. 

7.  Sold  gloves  at  a  profit  of  25/  per  pair,  and  gained  30%.     What  was  the 
cost  per  dozen  ?  Ans.  $10. 

8.  Sold  butter  at  35/  a  pound  and  lost  C§%.    What  was  the  cost  ? 

Ans.  37|/. 

9.  I  gained  $15  by  selling  a  horse  at  16|%  gain.    What  did  the  horse  cost 
me  I  Ans.  $90. 

10.  A  clothier  sold  a  coat  for  $18  and  thereby  lost  11^%.     What  did  the  coat 
cost?  Ans.  $20.25. 

11.  A  dealer  asked  $25  for  a  suit  of  clothes,  but  took  off"  10%  to  effect  a  sale, 
and  thereby  gained  12J%.     What  did  the  suit  cost  him  ?  Ans.  $20. 

12.  14|%  was  lost  by  selling  silk  after  deducting  25%    from   $4  per  yard 
asking  price.     What  was  the  cost  per  yard  ?  Ans.  $3.50. 

ONE  SELLING  PEICE.  AND  THE  GAIN  OR  LOSS  PEE  CENT  OF  THE  SAME  BEING  GIVEN, 
TO  FIND  WHAT  IT  SHOULD  BE  SOLD  FOE  IN  OEDEE  TO  GAIN  OE  LOSE  A 

DIFFERENT  PEE  CENT. 

866.     1.     Sold  merchandise  for  $19G  and  lost  30  per  cent;   what  should  it 


have  been  sold  for  to  gain  20  per  cent? 

PIEST   OPERATION. 

$  0 

100 

70     19G 


Ans.  $336. 


$280  cost. 

56  =  20%  gain. 


336  Aus. 


SECOND   OPERATION. 


$100  cost  assumed. 

20  =  20  7o  gam  added. 


$100  cost  assumed. 
30  =  30  7o  loss. 


70 


s.  p. 

$120 

196 

$336  Ans. 


$120  selling  price.  $70  selling  price. 

Note. — In  practice,  the  assuming  of  figures  and  the  adding  to  and  deducting  from  them 
should  always  he  mental  work. 

2.     Sold  goods  for  18/  and  lost  10  per  cent ;  what  should  they  have  been  sold 
for,  to  gain  10  per  cent  1  Ans.  22/. 


PERCENTAGE. 


449 


3.  Sold  hats  for  $4.50.  and  gained  50  per  cent;   at  what  price  should  they 
have  been  sold  to  gain  25  per  cent  ?  Ans.  $3.75. 

4.  Sold  boots  at  $50  per  dozen  and  lost  1C§  per  cent ;  at  what  jmce  per  pair 
should  they  have  been  sold  to  gain  20  per  cent?  Aiis.  $G. 

5.  Sold  goods  for  21^/  and  lost  15  per  cent  ?  for  what  shoidd  they  have  been 
sold  to  lose  10  per  cent  1  Ans.  22J/. 

TWO   SELLING  PRICES,   AND   THE   GAIN  OR  LOSS   PER   CENT  OF  THE   FIRST  BEING 
GIVEN,  TO  FIND  THE  GAIN  OR  LOSS  PER  CENT  OF  THE  SECOND. 

8G7.     1.     Sold  goods  at  $12  and  lost  25  per  cent;  what  per  cent  woidd  have 
been  gained  or  lost  had  they  been  sold  for  $18  ?  Ans.  12i%  gain. 

FIRST   OPERATION.  SECOND   OPERATION. 

$0  G  $ 


75 


100 
12 

$16  cost. 
18  selling  price. 


^2  gain. 


IG 


$2 
100 


2XP 


12  J  %  gain.  Ans. 


75 


112J 
100 


12i%  Ans. 

2.  If  20  per  cent  is  gained  by  selling  goods  at  24/,  what  per  cent  would  be 
gained  if  the  goods  were  sold  for  30^-  ?  Ans.  50%. 

3.  Sold  flour  at  $4.80  per  barrel  and  lost  40  per  cent,   what  per  cent  would 
have  been  gained  or  lost  if  the  flour  had  been  sold  at  $7.80  i>ev  barrel  ? 

Ans.  2J%. 

4.  If  sugar  is  sold  at  12|/  per  pound,  and  a  gain  of  12J^  per  cent  realized, 
what  would  be  the  gain  per  cent  if  the  sugar  was  sold  at  13 J/  ? 

Ans.  20i\%  gain. 

TO  COMPUTE  PER  (JENT  ON  ENGLISH  MONEY. 


868.     1.    What  is  3%  of  £  647  18s.  5d.  ? 

FIRST  OPERATION.      SECOND  OPERATION. 


Ans.  £19.  8s.  9d. 


£   6.47.18.05  Explanation. — lu  the  first  operation, 

Q  we  divide  the  £.  s.   and  d.   each   by 

100   as   shown  hy   the   decimal  points 

and  tlms  obtain  IJ!,,''.     Then  we  mul- 

0,%.  £19.43.15.03  tiply  by  3  and  obtain  3,%'.     We  then 

20  reduce  the  hundredths  of  £.  8.  and  d. 

to   shillings   and  pence,    according   to 

the  principles  of  denominate  numbera, 
as  shown  on  page  260. 

In  the  second  operation,  we  fovmdl^j 
as  in  the  first  operation,  and  then,  when 
multiplying  by  the  required  rate  per 
cent,  we  reduced  mentally  the  .15d  to 
hundredths  of  shillings,  by  dividing 
ty  12,  and  the  .55s.  to  hundredths  of  pounds,  by  dividing  by  20.  Then  we  reduced  the  Uundredths 
of  £.  s.  and  d.  to  shillings  and  pence  as  in  the  first  operation. 


£  6.47.18.05 
3 

£19.41.54.15 
20 

8S.74 
12 

9d.03 


8s.75 
12 

9d.03 


450 


SOULES    rillLOSOPHlC    PRACTICAL    MATHEMATICS. 


TO  EEDUCE  SHILLINGS  AND  PENCE  TO  THE  DECIMAL  OF  A  POUND, 
AND  TO  EEDUCE  THE   DECIMAL  OF  A  POUND 
TO  SHILLINGS  AND  PENCE. 
869.     1.    Reduce  12s.  6d.  to  tlie  decimal  of  a  pouud. 

SECOND    OPEKATION. 


FIRST  OPERATION. 


12s, 
12 


6d. 


£0.62 


Multiply  the  shillings  by  .05  and  tlie  pence  by  .004^  and 
add  the  results  together.    Thus : 


240)  150  (.625 


12s.     6d. 

.05    .004^    or  thus, 


12s. 
5  (hund'ths) 


6d. 

4^  (thous'ths) 


60  (hund'ths)+25  (thous'ths)=£.625' 


OPERATION. 


.60+.025=£.625, 

Note. — The  multipliers  in  the  second  operation  are  proiluced  thus: 

Since  20s.  =  £1.,  Is.  =  £v^,-,  or  .05  of  a  £. 

Since  LMOd.  =  £1,  Id.  =  £jb,  or  .004^01" a  £. 
In  like  manner  the  decimal  of  a  farthing  would  be  £5^!,,  or  -OOl/f  of  a  £. 

2.     Reduce  .625  of  a  pound  to  shillings  and  pence. 

Explanation. — In  the  2d  operation  of  the  first  prob- 
lem, we  showed  how  tlie  multiplier.^,  .05  and  .004^^, 
were  obtained  and  used  to  reduce  shillings  and  pencei 
to  the  decimal  of  a  jiound.  Now  since  the  decimal 
was  produced  by  these  niultiplers  it  is  clear  that  by 
thus,  using  them  as  diyisors,  we  will  reproduce  the  shilling* 

giirj  and  pence.     We  first  divide  the  first  two  figures  of 

"  the  decimal,  .625  by  .05  which  gives  12s.  and  .02  over. 

We  then  divide  the  .025  by  .004^  and  obtain  6  ponce. 
12s.  6d.  lu  jiractice,  the  operation  is  )ierfornied  by  dividing 

the  first  two   figures  of    the  decimal   by  5  and  the 
remainder,  if  any,  and  the  third  decimal  figure  by  4J, 

PROBLEMS. 

Ans.  £.796. 

Ans.  13s.  7d. 

Ans.  £19  Ss.  9d. 


Divide  .62 

by  5 


12s. 


Divide  .025 


1  = 

^  j  =  6d. 


3. 
4. 
5. 


Reduce  15s.  lid.  to  the  decimal  of  a  pound. 
Reduce  .679  of  a  pound  to  sliillings  and  pence. 
What  is  •S'/o  of  £617  18s  5d.  1 

SOLUTION    OF    THE    FIFTH    PROBLEM. 


Fir.ST   OPKKATION. 

£647.  18s.  5d.=£    647.921 


£19.43.763 
20 

8s.  .75260 

12 

9d.   .0312 


SKCOXI>    IIPEIJATION. 

£647  18s.  5d. 


£19.41.54.15 
20     . 


8s.  74 
12 

9d.  03 


THIIiD   OPERATION. 

£647  18s.  5d. 

£19.43.15  3 
20 

8s.  75 
12 

9d.  03 


Ans.  £130  lis.  Id. 

Ans.  £1  Os.  0.13d. 

Ans.  3f. 

Ans.  £300  Is.  Od. 


6.  Whatis5%  of  £2614  Is.  2d.? 

7.  What  is  1  %  of  £100  Is.  Id.  ? 

8.  What  is  J  per  cent,  of  12s.  7d.  ? 

9.  What  is  60  per  cent,  of  £500  Is.  8d.  ? 

Note — See  pages  484,  569,  6G1,  744  to  758  for  further  work  in  English  money. 

PROFIT  AND   LOSS. 
870.     Profit  and  Loss  are  terms  -which  express  the  gain  or  loss  in  business 
transacted,  and  which  involve  the  principles  of  percentage  in  all  its  various  com- 
binations, as  applied  under  difiereiit  subjects,  such  as  Commission,  Brokerage,  etc. 
The  subject  is  of  great  ])ractical  value  and  should  receive  the  tlioughtful 
attention  of  the  business  aspirant. 


*  ''   PERCENTAGE.  45 1 

Profit  or  Gain  is  the  sum  resulting  from  ttusiness  or  from  business  trans- 
actions in  excess  of  the  Cost  and  Expense  of  the  Business. 

Loss  is  the  sum  spent  or  incurred  to  carry  on  business  or  to  effect  business 
transactions,  in  excess  of  the  receipts  or  income  of  the  business. 

.  MAEKII^G  GOODS. 
871.  To  Mark  Goods  at  a  Given  Gain  or  Loss  Per  Cent. 

1.  Bought  butter  at  25/  per  pound.  At  what  price  must  it  be  sold  to  gain 
25  per  cent  ?  Ans.  31^/. 

25 

OPEKATION. 

25%  =  ,\%  =  i 

311/,  Ans. 

2.  Bought  cheese  at  15/  per  pound.  At  what  price  must  it  be  sold  to  lose 
10  per  cent  ?  Ans.  13J/. 


01  =  25%  gain. 


OPERATION. 

inrf    __    _- 

100    —    10 


10%   =   J-A   —   X 


15 

1.5  =  10%  loss. 


13.5/,  or  13J/,  Ans. 


Note. — Wlien  the  per  cent,  gain  or  loss  is  an  aliqnot  part  of  100,  all  we  have  to  do,  ■when 
marking  goods,  is  to  add  to  or  subtract  from  the  cost,  such  a  part  of  itself  as  the  rate  jier  cent,  is 
part  of  100. 

PEACTICAL  OPERATIONS  IIST  MARKING  GOODS. 

872.  In  marking  goods,  it  is  the  custom  of  most  merchants  to  use  a  private 
mark,  either  of  arbitrary  characters  or  letters,  instead  of  figures,  to  show  the  cost 
and  selling  price,  or,  as  some  mark  goods,  to  show  the  cost,  the  wholesale  and 
the  retail  price.  If  letters  are  used,  any  word  or  phrase  that  contains  10  different 
letters  may  be  selected. 

Any  of  the  following  words  or  phrases  may  be  adopted  for  a  Marking  word 
or  Key  word : 

Cash  Profit.         Regulation.       Importance.  Black  Horse.         Hard  Moneys. 

Now  be  sharp.    Washingtou.    My  Wife's  Hat.    Dont  be  lazy.        Brick  House. 
Buy  for  Cash.     Corn  Basket.    Purchasing.  God  help  us.  X.  Lord  Save  it. 

To  illustrate  the  work,  we  will  select  the  words  "  Cash  Profit,"  and  with  the 
ten  letters  represent  the  ten  figures  as  follows : 

12     3     4        567     8     90 

CASHPKOFIT 

To  avoid  the  repetition  of  any  letter  or  character,  an  extra  letter  or  character 
called  a  "Repeater,"  is  used.  By  this  means,  the  uninitiated  are  less  likely  to 
discover  the  key  word  or  mark. 

"When  the  repeater  is  used,  it  should  follow  the  letter  or  character  which  it 
repeats. 


452 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


With  the  above  key  word,  and  l  as  a  repeater,  what  is  the  mark  for  the  cost' 
and  selling  price  of  goods  that  cost  $18.00,  on  which  20%  gain  is  required? 

Ans.  O  F  E  T  cost.     A  L  S  A  Selling  price. 

Instead  of  using  one  mark,  as  above,  for  cost  and  selling  i)rice,  it  is  the 
custom  of  many  merchants  to  use  two  diflferent  marks,  one  for  cost  and  one  for 
selling  price. 

To  illustrate  this  work,  we  will  use  the  above  mark.  Gash  Profit,  for  the  selling 
price  mark,  and  select  the  word  Begulation  for  the  cost  price  mark.  Our  marks 
then  stand  as  follows : 

Selling  Price  Mark. 
23     4        567890 

ASn  PROFIT 

With  tbese  key  words,  and  B  as  a  repeater,  what  is  the  mark  for  the  cost  and 
selling  price  of  goods  that  cost  50/  per  yard  and  that  are  held  for  sale  at  a  gain  of 
10  per  cent  ?  Ans.  L  N  cost.     P  B  selling  price. 

Note. — To  distinguish  dollars  from  cents,  some  use  the  dollar  or  cent  mark,  some  the  decimal 
point,  and  some  make  no  designation  of  either  in  the  mark,  but  consider  the  mark  as  all  cents. 

Instead  of  using  words  for  marks,  many  merchants  use  characters.  Any  ten 
characters  may  be  selected  to  represent  the  ten  figures.  The  following  double 
cross  is  frequently  itsed,  thus : 


Cost  Price  Marl: 

1234567890 

1 

REGULAT      ION 

C 

The  double  cro.ss  may  be 

with  0  as  the 
tenth  charac- 
ter and  X  as- 
a  repeater. 

12     3 

JUL 

4    5    6 

J   D    C 

7     8 

1  n 

9     0 

r  0 

used 

3 
5 

n  I 
4 

*> 

Kiny  ways,  thus: 

G         7    8    9 

7         0    12 

8 

9 

1        5 

4 

3 

With  Buy  for  Cash,  as  a  cost  mark,  using  D  as  a  repeater,  and  with  the 
characters  shown  in  the  first  double  cross  above  as  the  mark  for  the  selling  ijrice, 
mark  the  cost  and  selling  price  of  the  following  articles,  at  the  gain  or  loss  per 
cent,  given : 

F  D   H 


Cost  $  2.50  ®  20%  gain,  mark,  ^  o  x 
"  .18  ®  25%  gain,  mark, 

"      24.00  ®  12|%  gain,  mark. 


Cost  $  4.40  ®  10%  loss,  mark,    ^  r  c 
"      65.00  Qij  40%  loss,  mark, 
"  .15  ®  33  J  %  loss,  mark. 


Mark  in  letters  from  the  Key  of  Biii/  for  Cash,  with  D  for  a  repeater,  Cost 
and  Selling  Price  of  the  following  : 

u  f  h 

Ans.  Ijnra 


Cost  $  2.40  ©  25%    gain, 
"  .16  ®  12i%  gain, 

15.00  ®  33|%  gain, 
16.50  -a  30%  gain, 


u 
it 


Cost  $  3.00  ®  10%  loss. 
"  4.50  ®  40%  loss. 
"  .10  ®  20%  loss. 

"      75.00  ®    5%  loss. 


Find  the  selling  price  of  the  following: 

1.  Of  goods  which  cost  18/  and  sold  at  33J%  gain.  Ans.  24/. 

2.  Of  goods  which  cost  $240  and  sold  at  40%  gain.  Ans.  $336. 

3.  Of  goods  which  cost  $4.25  and  sold  at  16§%  gain.  Ans.  $4.95|. 


i 


PERCENTAGE. 


453 


4.  Of  goods  which  cost  $14.50  and  sold  at  30%  loss.  Ans.  $10.15. 

5.  A  merchant  paid  $108  per  dozen  for  coats.    At  what  price  per  coat  must 
he  sell  them  to  gain  oO^J,  ?  '  Ans.  $13.50. 

6.  Bought  silk  for  $1.60  per  yard.     At  what  price  must  it  be  sold  to  gain 
140%  1  Ans.  $3.84, 

A  PUZZLIKG  PEOBLBM. 

87.3.     A  nest  of  five  tubs  cost  $3.00.    What  should  be  the  selling  price  of 
each  tub,  to  gain  33J%  on  cost? 

Opekation  by  Assumed  Eaxios  op  Value,  to  obtain  One  of  Many 

Answers. 

$3.00  cost  +  33J%  =  $4.00,  selling  price  for  the  nest  of  5  tubs.     2/  +  3/  + 
5/  +  S/'  +  12/  =  30 c,  which  is  the  sum  of  the  assumed  ratio  of  values.     Then  the 
following  proportional  statements  give  the  respective  value  of  each  tub : 
First  or  smallest  Tub.     Second  Tub.  Third  Tub.  Fourth  Tub.  Fifth  Tub. 


30 


2f 
4.00 


30 


2G§/,  Ans. 


3/ 
4.00 

40/,  Ans. 


30 


5  c' 
4.00 


30 


4.00 


30 


$1.06§,  Ans. 


12/ 
4.00 

$1.60,  Ans. 


66§/,  Ans. 

Practically,  where  the  nickel  is  the  smallest  coin  used  in  trade,  the  prices 
would  be  as  follows  : 

1st  tub,  25/  ;  2d  tub,  40/;  3d  tub,  65/;  4th  tub,  $1.10;  5th  tub,  $1.60. 

Second  Operation  by  Proportion,  according  to  the  Serial  Numbers 

OP  the  Tubs. 


Ifo.  of  smallest  tub,  1 
"     "  2d     "  "      2 

"     "  3d     "  "      3 

"     "  4th   "  "      4 

"     "  largest       '•      5 

Sum  of  serial  Nos.    15 


The  following  statements  give  the  Tahie  of  the  respective  tubs  : 

Third  Tub.  Fourth  Tub.  rifth|Tub. 


First  or 
Smallest  Tub. 


I  4.00 
15  I  I 

I  26sc. 


Second  Tub. 

$ 

I  4.00 
15     2 


I  4.00 
15     3 


I  4.00 
15     4 


I  4.00 
15     5 


53  ic 


80c. 


1.065c. 


I  1.33ic 

Note. — Other  ratios  of  value  may  be  assumed  according  to  the  judgment  of  the  calculator. 
Or,  if  it  is  deemed  proper,  the  prices  may  be  made  proportional  to  the  capacity  or  volume  of  the 

respective  tubs. 

2.    What  must  be  the  selling  price  per  box  of  a  nest  of  7  boxes  which  cost 
$15,  the  gain  %  being  25  ? 

IN  THE  FOLLOWING   ARTICLES    THE   LETTERS    INDICATE  THE  COST 

PRICE. 

874.    Write  the  cost  in  figures,  and  write  the  selling  price  in  letters,  at  a 
gain  of  20%,  using  Dont  be  lazy  as  the  key,  and  any  other  letter  as  a  repeater. 


1.  Hats,  $da.yc  per  doz. 

2.  Gloves,  $o.cb  per  pair. 

3.  Cufls,  $o.by  per  doz. 


4.  Oranges,  be/  per  doz. 

5.  Books,  $11. yc  each. 

6.  Shirts,  $d.ob  each. 


7.  Flour,  $t.lb  per  bbl. 

8.  Coffee,  od/  per  lb. 

9.  Sheeting,  dnif  jjer  yd. 
Ans.  to  the  last.  13J/  cost.     164/  selling  price. 

Ans.  to  the  first,  $18.00  cost.     $od.ey  selling  i)rice. 


454  soule's  philosophic  practical  mathematics.  ■* 

A   EAPID   PEACTICAL  PEOCESS   OF   FINDING   THE    SELLING   PEICE 

PEE  SINGLE  AETICLE,  AT  A  CEETAIN  PEOFIT  WHEN  THE 

COST  PEICE  IS  PEE  DOZEN. 

875.  1.  Bought  shoes  at  $7.50  per  dozen ;  at  wliat  price  per  pair  must  I 
sell  them  to  gain  20%,  I  Ans.  75/. 

In  all  questions  of  this  kind,  when  the  goods  are  bought  by  the  dozen  and 
the  gain  per  cent  is  20,  we  simply  diviile  by  ten  or  remove  the  decimal  point  one 
place  to  the  left,  and  the  result  is  the  selling  price  per  article.  Tliis  is  a  very- 
important  contraction  for  merchants  who  buy  at  auction,  as  by  it  they  can  instantly 
determine  at  each  bid  what  the  selling  price  must  be  per  article,  in  order  to  give  a 
fair  profit. 

The  philosojihy  of  this  work  is  simple  and  will  be  easily  comprehended  by 
the  elucidation  of  the  following  problem  : 

If  shirts  cost  $22.50  per  dozen,  what  must  they  be  sold  for,  a  piece,  to  gain 
20%  »  Ans.  $2.25. 

operation. 

$  Explanation, — For  reasons  often  repeated  we    assume    $100 

ygrf  cost,  and  reason  as  follows:  In  order  to  gain  20"^  we  must  sell 

tnii      oo  nft  what  cost  $100  for  $120.     This  selling  ])rice  we   place   on   the 

lOp      w-^.OU  statement  line,    and   continue   the   reason   thus:    If  !flOO   cost 

]Lp  require  $120  selling  price,  |1  cost  -will  require  the  lOdth  jiart, 

and  22i",i"u  cost  will  re(iuire  22-,Vly  times  as  much.     Then  as  the 

result  of  this  statement  will  give  the  selling  price  of  a  dozen,  and  as  we  wish  to  obtain  the  selling 

price  for  a  single  article,  we  therefore  divide  l)y  12.     This  completes  the  reasoning  and  statement, 

and  by  the  same  we  see  why  we  remove  the  decimal  point.     First  cancel  the  0  on  the  120  and  one  0 

on  the  100;  then  cancel  the  12  on  each  side  of  the  line,  and  we  have  the  cost  price  remaining  on 

the  right  and  10  on  the  left.     It  is  obvious  that  this  will  always  be  the  case  when  the  )ier  cent 

gain  is  20,  and  hence  the  work  of  dividing  the  cost  by  10,  which  is  done  by  simply  removing  the 

decimal  point,  to  find  the  selling  price  per  article,  is  made  clear  and  comprehensible. 


876. 

T  ^  B  L  T^- 

FOB 

DIFFERENT 

RATES     PEE 

CENT    TO    BE    USED 

ACCORDING     TO 

THE 

ABOVE 

PRINCIPLES  : 

To  gain  20  %  remove  the  decimal 

point  one 

lilace 

to  the  left 

25  % 

u 

a 

and  add 

1 
2-4 

itself. 

26  % 

u 

a 

u 

1 
■20 

28  % 

li 

a 

li 

1 

TS 

30  % 

it 

a 

li 

1  3 

32   % 

(( 

ii 

11 

-h 

33i% 

u 

li 

11 

X 
9 

35   % 

li 

11 

11 

i 

•37  J  % 

a 

11 

11 

1 

7 

40  % 

a 

a 

li 

1 
0 

44  % 

a 

a 

11 

1 

5 

50  % 

a 

11 

11 

i 

CO   % 

li 

ii 

11 

h 

80  % 

a 

4( 

11 

k 

i2i% 

li 

a 

and  subtract 

-h 

ic§% 

u 

li 

'( 

■h 

18f% 

11 

a 

t; 

-«\ 

10   % 

a 

ii 

11 

A 

•Note.— This 

s  practically,  not  accurately  correct. 

According 

to  the  above 

what  would  be  the 

price  per 

article  to 

gain 

20 

%  of 

the  following  goods  bought  by 

the 

dozen : 

Hats,  $32. 

Gloves,  $18.50 

.      Shoes, 

$44.50. 

Coats, 

$150.      Pants 

,  $37.50. 

Knives,  $3.75.    Chairs,  $21.25 

*  PERCENTAGE.  455 

TO  FIND  THE  PRICE  AT  WHICH   GOODS  MUST   BE   MARKED   SO   THAT  A  GIVEN  PER 

CENT  OF  GAIN  OR  LOSS  MAY  BE  REALIZED  AFTER  ALLOWING  FOR 

STATED  DISCOUNTS,  BAD  DEBTS,  WASTE,  COMMISSION,  ETC. 

877.     1.     A  mert'baiit  marks  lii.s  goods  at  40%  gain,  but  supijlies  liis  whole- 
sale buyers  at  a  discount  of  20%  from  list  price.     What  is  his  gaiu  %  1 

Aus.  12%  gain. 

OPERATION. 

$100  =  cost. 

40  =  gaiu  %.  $112  =  selling  price. 


100  =  cost. 


$140  =  list  price. 


iS  =  207,,  discoxiut.  $  12  =  gain,  and  as  this  gaiu   is 

—  on   the    hundred    it    is 


$112  =  selling  i^rice.  therefore  gain  %. 

2.  If  a  merchant  marks  his  goods  at  25  per  cent  profit  and  eflects  sales  at 
20  i^er  cent  discount  on  retail  i)rice,  does  he  gain  or  lose  ?  Ans.  Neither. 

3.  If  a  merchant  marks  his  goods  at  50  per  cent  profit  and  effects  sales  at 
40  per  cent  discount  on  retail  price,  does  he  gain  or  lose,  and  if  so  how  nuich  ? 

Ans.  Loses  10%. 

4.  A  merchant  wishes  to  increase  the  price  of  an  article  which  he  now  sells 
for  $9.  so  that  he  can  deduct  10%  and  still  receive  the  present  price ;  what  must  be 
the  advanced  price  ?  Aus.  $10, 

OPEKATION. 


$100      advanced  price  assumed. 
10%,  deducted. 

90 

selling  price. 


100 
9 

$10  Ans. 

5.  A  merchant  wishes  to  mark  his  goods  in  such  a  manner  that  when  he 
sells  at  wholesale  he  may  discount  20%  from  his  retail  price  and  still  gain  10%  on 
cost.  Accordingly,  what  must  be  the  retail  price  of  goods  that  cost  $1.G0  per 
yard?  Ans.  $2.20. 

OPERATION. 

$1.00  =  cost.  .  i 

10  =  10%,  gain. 
80 


70  =  wholesale  price. 


100  =  assumed  retail  price. 
1.7C 


12.20  =  retail  price,  Ans. 

6.  What  must  be  the  asking  price  of  goods  that  cost  60/  per  pound,  so  that 
10%  may  be  deducted  and  a  gain  of  20%  still  be  realized  1  Ans.  80/. 

7.  What  must  be  the  asking  price  of  goods  that  cost  $15,  so  that  40%  may 
be  deducted  and  the  goods  sold  at  cost?  Ans.  $25. 

8.  Goods  cost  45/  per  pound ;  what  must  they  be  marked  so  that  by  deduc- 
ting 25%  sales  may  be  effected  at  a  loss  of  10%  ?  Ans.  54/. 


456 


SOULE  3    rillLOSOPIIIC    I'RACTICAL    MATHEMATICS. 


9.     Paid  IS/'  per  dozen  for  eggs.     Allowing  5'/^  for  breakage  and  10%  for 


uncollectable  sales,  liow  mncU  per  dozen  must  tliey  be  sold  for  to  gain  14''/^  on  first 

cost?  Aus.  24/. 

OPERATION  OPERATION  OPERATION 

to   find   si'lliiif;   price  to  allow  10"^  lor  un-  to  allow  for  5^'^  Ortlius: 

to  gain  14^^  collectible  ■  sales.  breakage. 


18('. 
U% 

.0252 
.18 

.20o2 
10.    Goods  cost 


90 


100 

,2052 

.2280/. 


100 
.2280 

$.2100  Aus. 


90 

95 


114 
100 

18 

24  (' 
Aus, 


What  must  be  the  asking  price  so  that  15%  may  be 
deducted  and  a  gain  of  25%  realized  on  first  cost?  Aus.  $14.70^". 

11.  Goods  cost  $12.  What  must  be  asked  for  tliem  so  that  20%  may  be 
deducted  from  the  asking  price,  6^%  be  allowed  for  waste,  10%  allowed  for  "bad 
debts,"  and  25%  gain  made  on  first  cost?  Aus.  $22f. 


OPERATION. 


$12 


3  =25%  gain.     933 


100 

15 

16 


90 


100 
IG 

17J 


SO 


100 

17i 


Or, 


933 

90 

80 


125 

100 

100 

12 


$  22| 

12.  Bought  a  cargo  of  coffee  at  10/ a  pound.  Paid  10%  duty.  What  must 
I  ask  for  the  coffee  per  pound,  so  that  I  may  fall  10%  on  my  asking  price,  allow 
10%  for  loss  by  shrinkage  of  coffee,  lose  10%  of  sales  iu  bad  debts,  and  still  gain 
10%  on  my  investment  (full  cost)  ?  Ans,  164|-|/. 


TO  MARK  DIFFEEBNT  LOTS  OF  GOODS  IN  AN"  INVOICE. 


878.  Bought  in  New  York  an  invoice  of  goods  amounting  to  $2400.  Paid 
freight  and  other  charges  to  deliver  them  in  New  Orleans,  $102.  In  the  invoice 
are  8  dozen  hats  which  cost  in  New  York  $27  per  dozen.  1-.  What  was  the  %  of 
the  charges  ?  2.  What  was  the  cost  of  1  hat  in  New  Orleans  ?  3.  What  is  the 
retail  price  per  hat,  from  which  20%  may  be  deducted  and  the  hats  sold  at  whole- 
sale at  a  gain  of  25%  on  New  Orleans  cost  ? 

Ans.  4J%  charges. 

$2.34+cost  of  1  hat  in  New  Orleans. 
$3.06 — retail  price  of  1  hat. 

MARKING  GOODS,  BASED  ON  FACTORY  COST  AND  ON    SELLING 

PRICE,    ETC. 

879.     Goods  cost  at  the  factory,  $1400.     Boxing,  $4.     Drayage  to  dock,  $9.50. 
Freight  to  destination,  $30.     Drayage  from  wharf  to  store,  $9.     (1).     What  is  the 


PERCENTAGE.  467 

floor  cost  of  the  goods ?  (2).  What  is  tbe percentage  of  tbe  charges?  (3).  What 
price  must  the  goods  be  marked  to  gain  'M<^/o  on  the  floor  cost?  (4).  Based  on 
floor  cost,  what  must  the  goods  be  sold  for,  so  that  the  gain  will  be  20%  of  the 
selling  price?  (5).  "When  the  goods  are  marked  to  gain  20%  on  the  sales,  wliat 
per  cent  is  such  gain  ou  the  factory  cost?        Aus.  (1).     $1432.50    (2).  0^% 

(3).     $1743.00     (4).     $1815.62.5     (5).     29.6875% 

OPERATIONS. 

(1).     To  find  floor  cost.                 (2).     To  find   percentage  of  (3).     To   find  gain  of  20%  on 

charges.  floor  cost. 
1400. 

4.                                                     I    52.50  $14152. 50=floor  cost. 

9.50                                       1400  I  100.  •«10.50=20»o  gain. 


30. 

9.  I      3.75  or  3|%                       $1743.00=selling  price  to  gain 

20%  ou  floor  cost. 

$1452.50 

(4).     To  find  gain  of  20"^  on                                  (5)     To  find  wbat  tlie  20%  gain  on 

the  sales  over  floor  cost.  sales  is  on  factory  cost. 

I  S  100.  $1815.625  =  .selling  price. 

80  I    1452.50  1400.000  — factory  cost. 


I  $1815.625=  selling  price  to  $  415.625  =  gain— $1400=29.6875%'. 

gain  20%  on  sales. 

Goods  cost  at  the  factory,  $130.  (1).  Estimating  that  the  expenses  to  place 
the  goods  in  the  store  of  the  ])urchaser  and  the  discount  allowed  to  customers 
equal  12%  of  Factory  cost,  what  must  the  goods  be  marked  to  gain  10%  on  the 
Store  cost?  (2).  Based  on  floor  cost,  what  must  the  goods  be  marked  and  sold  for 
80  that  the  gain  will  be  10%  of  the  selling  price  ?  Aus.     (1).     $1(10.16 

(2).    $161,771 

OPERATIONS. 
FIRST.  SKCOND. 

$130.  ■  I  $100. 

15.60=12%.  $90  I    145.60 


$145. 60=flonr  cost.  |  $161.77g=Selling   price   to 

14.56=10%'  ou   floor  coat.  gain  10%  on  sales. 

$160.16=price  marked. 

TO  IMPORT  AND  MARK  DIFFERENT  LOTS  OF  ENGLISH  GOODS. 
'  879J.  Imported  from  England  an  Invoice  of  goods  amounting  to  £1450.  8s.  4d.  Paid  freight 
$190.50.  Insurance  $132  15.  In  the  invoice  were  40  dozen  pocket  knives  which  cost  £160.  2s.  8d. 
The  duty  on  the  knives  was  25%  ad  valorem.  What  was  the  invoice  cost  iu  United  States  Money, 
aad  what  must  be  the  retail  selling  price  per  knife  to  allow  20%  discount  and  to  sell  at  wholesale 
at  a  gain  of  25%  on  the  fnll  importing  cost  ?  Ans.    $3.2Si. 

OPKRATION. 

£1450-8-4=£1450,416+x4.8665=$7058  45  niv.  cost  of  goods  in  United  States  Money. 

£160  2-8=£160.133+x4.8665=$779.29         "  "         knives 

$190.-,0+$182.15=*322.65,  Insurance  and  freight. 

(322.65x100)— $70.5S. 45=4. 57+%  charges  on  invoice. 

$779.29x4.57%=$35.61  cliarges  on  knives. 

$779.29x25%=$194.82  duties  on  knives. 

$779. . '9+3.1. i)l+$194.82=S1009, 72  full  cost  of  knives  in  New  Orleans. 

$1009.72— 40=$25. 24  cost  of  one  dozen  knives  in  New  Orleans. 

$25.24^12=.$2  10  cost  of  one  knife  in  New  Orleans. 

$2. 10+25";,  =$2,624  wholesale  jirico  per  knii'e  in  New  Orleans. 

($2.62ixl00)— 80=$3  28i  retail  price  of  one  knife  in  New  Orleans. 

Remark.  —To  find  iniportint:  rn.st.  each  lot  of  cooils  in  a  foreifm  invoice  ninst  he  treated  in  manner  as  above  shown, 
iae  care  being  given  tu  the  dltfereut  rates  ot  duty  ou  the  different  lines  and  the  grade  or  qualities  ot  the  gooils. 


458 


souLE  s  niii.osorHic  practical  mathematics. 


TRADE  DISCOUNTS  AND  LIST  PRICES. 
880,     Trade  Discounts  are  allowances  or  rebates  made  by  merchants,  manufacturers  and  other 
classes  of  business  men  from  their  regular  marked  or  list  prices. 

It  is  customary  for  many  classes  of  business  men  to  have  fixed  price  lists  of  their  goods 
based  on  the  highest  marljet  rate,  and  when  the  market  declines,  instead  of  changing  the  fixed 
price  list,  they  allow  a  discount  or  a  series  of  discounts  from  the  list  prices.  These  discounts  will 
be  shown  in  the  problems   following. 

Merchants  and  Manufacturers  also  frequently  allow  certain  discounts  or  rebates  from  the 
retail  price,  proportioned  to  the  amount  of  the  bill  sold. 

Some  lines  of  goods  are  sold  at  "Timk  Prices,"  conditioned  on  certain  discounts  if  paid 
prior  to  the  time  specified.  These  terms  and  conditions  are  often  printed  on  the  "Bill  Heads," 
and  read  as  follows  :  "Terms  4  months,  or  in  30  days  less  5  i)er  cent.";  or  "Terms  90  days  or  2  per 
cent,  discount  in  30  days,  or  3  jier  cent,  discount  in  10  days,"  etc. 

In  m.any  cases,  where  goods  are  sold  on  Time  prices,  the  rate  per  cent,  allowed  is  a  matter  to 
be  agreed  upon  by  the  parties,  and  is  generally  5  per  cent.,  10  per  cent.,  or  25  per  cent.,  according 
to  the  length  of  time  that  the  bill  is  paid  before  due,  the  hazard  of  the  credit,  the  profit  realized 
on  the  sale  of  the  goods,  the  necessities  of  the  parties  for  money,  and  the  rate  of  interest  at  which 
money  can  be  obtained  at  the  bank. 

The  discount  is  allowed  on  the  face  of  the  bill  or  invoice  without  regard  to  time,  notwith- 
Btanding  that  time  and  the  rate  of  interest  on  money  were  considered  in  determining  the  rate  of 
discount. 

With  a  view  to  give  a  better  understanding  of  this  system  of  discounting,  we  will  state  that 

!%■  discount  on  bills  maturing  in  30  days  is  equal  to  12  per  cent,  interest. 

2%  discount  on  bills  maturing  in  30  days  is  equal  to  24  per  cent,  interest. 

3%  discount  on  bills  maturing  in  CO  days  is  ecjual  to  18  per  cent,  interest. 

5%  discount  on  bills  maturing  in  3  months  is  equal  to  20  per  cent,  interest. 

5%"  discount  on  bills  nuituring  in  4  months  is  equal  to  15  per  cent,    interest. 

5%  discount  on  bills  maturing  in  6  months  is  equal  to  10  per  cent,  interest 

5%  discount  on  bills  maturing  in  12  months  is  equal  to  5  per  cent,  interest. 
When  no  rate  of  discount  has  been  agreed  upon,  and  the  bills  are  paid  before  maturity, 
business  men  usually  allow  and  deduct  from  the  face  of  the  bill  the  current  rate  of  interest  for 
the  time  the  bill  was  paid  before  due. 

PROBLEMS. 

1.     A  bill  of  goods  at  list  price  amounts  to  $462.50. 
what  is  the  net  amount! 

OPERATION. 

$46.2,50  =  10?^ 
23.125  =    5% 


A  discount  of  15  per  cent,  is  allowed. 
Ana.  $393.12i. 

$462.50 
69.37i 


$69,375  =  15%  discount.  $393. 12J  Answer. 

Note. — To  find  15  per  cent,  multiply  by  10  and  add  |  of  the  jiroduct. 

2    An  invoice  of  merchandise  at  regular  list  prices  amovints  to  $3480.     What  is  the  net 
amount,  the  series  of  trade  discounts  being  25  per  cent.,  10  per  cent  and  5  per  cent  ? 

Ans.  $2231.55. 

OPERATION. 


$3480  =  list  price. 
870  =  25"'  discount. 


$2610 

261  =  10,V  discount. 


$2349 

117.45  =  5%  discount. 


Explanation. — From  the  list  price,  we  first  deduct 
the  25  per  cent,  discount,  and  from  the  successive 
remainders  the  other  per  cents  are  deducted. 

The  order  in  which  the  series  of  discounts  is 
deducted  does  not  change  the  result. 


$2231.55  =  net  amount. 
In  this  manner,  the  discount  may  be  found  in  any  series. 


*  PERCENTAGEo  459 

TO  FrN"D  THE  SIXGLE  OR  EQUIVALENT  DISCOU-N^T,  FEOM  A  SEEIES 

OF    DISCOUNTS. 

881.    1.    lu  tlie  above  problem,  -what  single  discount  on  tlie  list  price  is 
equivalent  to  the  series  of  25  per  cent.,  10  per  cent,  and  5  per  cent.  ?    Ans.  3o|  % . 

FIRST   OPEKATION. 

$100  =  list  price.  Discounts.  Explanation.-^  We  first  assume 

25  =  25  per  cent,     discount  25  $100  as  the  list  price,   and  then 

deduct    therefrom    the    25    per 

rj-  cent,  discount;   then   from   the 

remainder  $75,    deduct    10    per 
7.50  =  10  per  cent,    discount         7.50  ce"t.  discount;   then   from   the 

second  remainder  §(57. 50,  deduct 

,_  _„  *  *      5    per    cent,     discmint,     ■nhicli 

'"■•^0  leaves  a  net  price  of  64J.     This 

Q  <?7^n  —  ~i  npr   pPTit     r1i«pnnnt      ^  ^T^i  subtracted    from    the    assumed 

a.6too  _  o  per  cent,  discount    c..d<i>  ^jp^,  j;^^  ^,^,5,,^^  ^i^^,^  ^3.^  ^^.^ 

count,     which     being    on     the 


$  64.1250  =  net  price.  35.875%  =35|%      hundred,  is  35|  per  cent,   dis- 

.  count.     Or,  by  adding  the  sever- 

100.  —  list  price,  as  above.  al   discounts   as   shown   in    the 

operation,   we  also   obtain  35J 


35.875  per  cent.  =  equivalent  single  discount.         P^r  <^'=°t-  discount. 

A  series  of  any  number  of  discounts,  may  be  worked  in  the  same  manner. 

SECOND   OPERATION. 

100  —  25%  =  .75  (  Then  75  x  90  x  05  =  .G41250  of  100,  or 
100  —  10%  =  .90  }  64.125  per  cent,  and  100  —  64.125  =  35.875 
100  —    5%  =  .95  (  percent,  the  single  equivalent  discount. 

A  series  of  any  number  of  discounts  may  be  worked  in  the  same  manner, 

PROBLEMS. 

2.  The  list  price  of  a  bill  is  $1781.40 ;   the  rates  of  discount  are  33J  per 
cent.,  20  jier  cent.,  10  per  cent,  and  5  per  cent.     What  is  the  net  amount  I 

Alls.  $812.3184. 

3.  A  merchant  sold  a  bill  of  goods  amounting  to  $530  on  which  he  allowed 
40  per  cent,  and  ^  discount.    What  was  the  net  value  of  the  bill?  Ans.  $265. 

4.  Which  is  the  greater  discount,  and  what  is  the  difference  on  a  sale  of 
goods  amounting  to  $1420,  40  per  cent.,  or  25  per  cent,  and  15  per  cent.  ? 

Ans.  40  pr.  ct.  is  $53.25  greater. 

5.  What  is  the  equivalent  per  cent,  of  J,  33^  per  cent.,  ^,  and  10  jyev  cent, 
discount  1  Ans.  76  per  cent. 

6.  A  manufacturer  after  allowing  J  and  20%  discount,  received  for  a  lot  of 
goods  $3360.     What  was  the  catalogue  or  list  price  ?  Ans.  $6300. 


460  soule's  philosophic  tractical  mathematics. 

[  7.  ] 
PRANK  DTIMAS, 
TEEMS— Cash;  lor^off. 


New  Orleans,  August  12,  1895. 
Bot.  of  C.  McGuigin. 


462  lbs.  Prime 


375 

(( 

Good 

180 

it 

Fair 

216 

(i 

Ordinary 

74 

It 

Commou 

Coffee,  -        -        -         -         -         ®  21  c. 

"  ...._"  22  c. 

"  _-..-"  21  c. 

"  "  20ic. 

"  "  19ie. 

Loss  10^0  discount.  ..... 


97 

02 

82 

50 

37 

80 

44 

28 

14 

43 

$248  43 


[8.] 


JONES  &.  \rEl^?,, 

TEEMS — Cash ;  10  per  cent,  and  5  per  cent  discount. 


New  Orleans,  Jan'y  21,  1895. 
Bot.  of  Stewart  &  Henderson. 


34  Ijlils.  La.  Oranges,  lar.,  .......  ®  $5.75 

27  Ijoxes  Messiua  Lemons,  ......  "           6.00 

{>3  cases  Malaga  Grapes,  ...-..-  "           I.75 

45  l)oxe3  California  Tears,  ....--  "          4. 50 

5  mats  Dates,  593  li)s.,             "               7ic. 

Less  10;^  and  5;^  oflf. 


714 


73 


[9.] 


J.  M.  BTTTCHEE, 


TEEMS — Cash ;  1-^  per  cent,  and  10  per  cent,  discount. 


New  Orleans,  Jan'y  31,  1895. 
Bot.  of  J.  B.  Cundiff, 


875  bbls.  Nea.  Potatoes, 
440      "     P.  B.  Potatoes, 
325      "    Perfection  Fknir, 

1324      "     St.  L,  XX  Flour, 
112      "     Family  Clear  Pork, 
G50      "    Prime'Pork, 
220  kegs  Pig  Feet, 
124  half  l)bls.  F.  M,  Beef, 

1S72  lbs.  Choice  Ham, 
289     "    B.  Bacon, 
106  Pig  Tongues, 


® 


DIs.  121;^  and  lO^i". 


I  4.25 
3.87^ 
8.50 
6.62J 
17,50 
13.75 
7.50 
11. 
14  c. 
9ic. 


31167 


27 


I 


PERCENTAGE. 
[10.] 


461 


GEO.  B.  BRACKETT  &  CO., 

TEEMS— Cash;  16;  percent,  off. 


New  Orleajjs,  Nov.  17,  1895. 
Bot.  of  R.  Spencer  Soule. 


1427  bu. 

856  " 

420  " 

3145  " 

1040  " 


No.  1  Winter  AVlieat,               --....        @  |;l.5,5 

No.  2  Winter  Wheat,               "  1.47 

111.  No.  1  White  do.                   ..-..-"  I.4I 

W.  Corn, "  .70 

B.  Oats. "  .55 

Less  1615'^. 


683r 


87 


[11.] 

New  Orleans,  August  12,  1895. 
E.  L.  CHAPPUIS, 

Bot.  of  W.  Reed  Carradine. 

TEEMS — 4  months  or  5  per  cent,  and  2J  per  cent,  discount  in  10  day.s. 


Aug. 


12 


42  bills.  Flour,  "Winter  Patents  " 

16  '•  "      Extra  Fancy 

27  "  "      Fancy 

60  "  "      Choice 

36  "  "       Family 

70  "  "       Minnesota  Patents 

89  "  "      Minn.  Bakers'  Grade    - 

79  "  Corn  Meal,       -        .        .        . 

12  "  Grits,        -        -        .        .        . 

10  "  Hominy,            .... 

5  "  Rye  Flour,       -        -        .        . 

9  "  Oat  Meal,         .... 


Discount  S^o'  and  2A%. 


® 


?5.25 

220 

50 

( 

5.00 

80 

00 

4.75 

128 

25 

4.50 

270 

00 

4.75 

171 

00 

5.75 

402 

50 

5.14 

457 

46 

3.20 

252 

80 

3.10 

37 

20 

2.65 

26 

50 

3.90 

19 

50 

5.75 

51 

• 

75 

2117 

r46 


Note  1. — 'I'lio  above  bill  was  paid  G  days  after  the  purchase  was  made. 

See  pafie  457,  fur  an  explanation  of  Trade  Discounts. 

When  Trad<i  Discounts  are  not  allowed  and  bills  are  paid  before  maturity,  interest  for  the 
unexpired  time,  usually  allowed  the  payor. 

Note  2. — For  other  bills  coverinj;  many  lines  of  business,  see  pages  283  to  295.  Also,  see 
pages  278  to  283,  for  Rapid  Practical  Methods  of  Bill  Computations. 

12.     A  dealer  Iwught  goods  at  25%  and  10%  discount  on  list  prices  and  sold 


at  the  list  price.     What  was  his  gaiu  j;er  ceut^ 

OPERATION. 

Let  $100  =  tne  list  price. 

25  $100    =  sales  at  list  price. 
61^  =  cost  of  list  price. 


Aus.  4S./.^ 

Then  the  following 
statemeut : 


$  32J  =  gain  ou  cost. 


o 

135 


05 

2 
100 


$  G7^  cost.  48oV  Ans. 

13.  A  merchant  bought  an  invoice  of  goods  Sept.  15,  1890,  amounting  to 
$1200,  on  the  following  terras:  '-4  months  or  5%  discount  in  30  days."  The  bill 
was  paid  Oct.  13,  1890.     How  much  was  paid  ?  Ans.  $1140. 


462  SOULE's    I'lIILOSOPIIIC    PRACTICAL    MATHEMATICS.  * 

14.     Goods  are  bought  at  33;^%   and  20%  discount  from  list  prices,  and  sold 
at  30 fo  and  i\,-  oil".     What  is  the  gain  per  cent?  Aus.  1^1%. 

OPERATION. 

=  assumed  as  list  jirice.  100    =  assumed  list  price. 

20  =  20%  discount  on  purchase.         30    =  30%  discount  ou  sales. 


80  70 

26|  =  33J%  discount.  7    =  10%  discount  on  sales. 


53J  =  cost.  $  63    =  selling  price  of  $53^  cost. 

53^  =  cost. 


$     9§  =  gain  on  $53J  cost. 
Then  if  $53J  gain  $0§,  what  will  100  gain  ?  'Ans.  18J%. 

15.  What  %  is  gained  by  selling  hats  at  $3  each,  that  cost  $42  per  dozen 
■with  1G§%  and  20%  off?  Ans.  284%. 

IG.  What  was  the  list  price  of  goods  that  sold  for  $1.20  per  gallon,  with 
50%,  10%  and  12J%  discount?  Ans.  $3.20. 

17.  A  business  man  insured  property  to  the  amount  of  $05000  at  1^%.  He 
was  allowed  a  rebate  of  20%  for  cash.     What  was  the  net  iiremiuni  paid? 

Ans.  $910. 

18.  A  wholesale  dealer  sold  a  bill  of  goods  amounting  to  $2000,  on  6  months' 
credit.  A  few  days  after  the  sale  had  been  completed,  the  purchaser  i)roi)osed  to 
pay  the  bill  if  the  wholesale  dealer  would  allow  him  5  per  cent  discount.  His  prop- 
osition was  acceiited  and  the  bill  paid.     What  amount  of  money  was  required  ? 

Ans.  $1900. 

OPERATION. 

$2000  face  of  bill. 
100  =  5%  discount. 


$1900  amount  required  to  pay  the  bill. 

19.  At  the  same  time  tliat  the  above  transaction  was  made,  the  same  wliole- 
sale  dealer  sold  to  aiiotlier  merchant,  on  C  months'  credit,  goods  amounting  to 
$3000.  The  purcliaser  of  this  bill  jn'oposed  to  pay  $1900,  provided  tlie  wholesale 
dealer  would  allow  liim  5  per  cent  discount  on  the  amouTit  of  invoice  paid,  which 
the  wholesale  dealer  agreed  to  do,  and  the  $1900  was  paid.  What  amount  does  the 
purchaser  of  the  goods  still  owe?  Ans.  $1000. 


PERCENTAGE, 


463 


OPEEATION. 


$100  invoice  assumed. 

0  =  5%  discount  deducted. 


)5  casli. 


95 


100 
1900 

L'OOO 
3000 

$1000  Ans. 


Explanation. — The  solution  of  this  problem  is 
entirely  ditterent  from  the  solution  of  the  pre- 
ceding one.  The  discount  being  allowed  on  the 
amount  of  the  invoice  that  the  money  will  pay, 
and  as  the  -nholo  of  the  invoice  is  not  paid,  we 
cannot  tlierclore  calculate  on  the  $1900  cash  or 
on  the  face  of  the  $3000  invoice.  AVe  are  hence 
obliged  to  assume  some  number  to  represent 
invoice,  and  iipon  which  we  may  calculate  and 
deduct  the  5  per  cent  discount,  and  thus  obtain 
the  necessary  equivalent  values  with  which  to 
make  a  solution.  As  shown  in  the  oiieration, 
we  assume  flOO  to  represent  the  invoice,  and 
then  by  deducting  the  5  per  cent  discount  we 
have  in  the  remainder  $95  as  tlie  cash  e(juivalent 
of  the  |I100  invoice;  or  as  the  amount  of  cash 
required   to    pay    $100    invoice.     Having    these 


equivalent  Talues,  we  place  the  $100  assumed  invoice  on  our  statement  line,  and  reason  thus:  If 
$95  cash  will  pay  §100  invoice,  $1  cash  will  pay  the  95th  part,  and  §1900  cash  will  pay  1900  times  as 
much.  The  result  of  which  is  §2000.  Then  liaving  the  amount  of  invoice  that  $1900  will  ])ay  or  is 
equal  to,  we  deduct  the  same  from  the  whole  invoice,  §3000,  and  have  in  the  remainder  $1000,  the 
balance  that  the  purchaser  still  owes. 

20.  A  country  mercliaiit  purchased  $4800  worth  of  goods,  on  4  months'  time. 
He  then  proposed  to  pay  cash  if  the  city  merchant  woidd  discount  5  per  cent  on 
the  face  of  the  bill,  which  he  agreed  to  do.     What  amount  will  pay  the  bill  ? 

Ans.  $4617. 

21.  Another  country  merchant  jiurchased  $8450  worth  of  merchandise,  on  4 
months'  time.  lie  then  proposed  to  i>ay  $5000  ca.sh  on  condition  that  he  be  allowed 
5  per  cent  discount  on  the  amount  of  bill  that  he  pays,  which  proposition  is 
accepted.     What  amount  does  he  still  owe  ?  Ans.  $3180.84. 

22.  A  retailer  bought  from  a  wine  merchant,  on  3  months'  credit,  goods 
amounting  to  $1500,  and  from  a  grocery  merchant,  on  3  months'  credit,  goods 
amounting  to  $1800.  But,  without  waiting  for  the  maturity  of  the  bills,  he  pays 
the  wine  merchant  cash  in  full  less  5  per  cent  discount  allowed,  and  to  the  grocery 
merchant  he  pays  $1000  cash,  for  which  he  is  to  be  allowed  5  per  cent  discount  on 
the  amount  of  invoice  that  it  will  pay.  How  much  did  he  pay  the  wine  merchant? 
And  how  much  does  he  still  owe  the  grocery  merchant  ? 

Ans.  $1425,  he  paid  the  wine  merchant. 

$747.37,  he  owes  the  grocery  merchant. 

23.  A  debtor  pays  his  merchant  $500  in  nickles  at  7.}%  discount.  How  nmch 
credit  does  he  receive  1  Ans.  $462.50. 

24.  A  debtor  owes  $555  and  pays  the  same  in  nickles  at  7J%  discount.  What 
amount  of  nickles  will  it  take  to  pay  the  debt  ?  Ans.  $600.00. 


464 


SOULE  s  riiiLnsopiiic   practical  mathematics. 


882. 


TABLES  OF  TRADE  DISCOUNTS. 

Table  No.  1. 


2  \>T.  ct.  auil    1  pr. 

ft. 

. 

. 

=z 

2n  pr. 

3       " 

2       ' 

' 

_ 

. 

^ 

4*?,       " 

5      "        ' 

3      ' 

' 

- 

. 

^  = 

in    " 

10      " 

5       ' 

' 

. 

. 

'  = 

Ui        " 

10      " 

5      • 

'       ami  2i  pr.  ct. 

. 

. 

= 

lOvi      " 

15       " 

5      ' 

' 

- 

. 

= 

l!»i         " 

15       " 

10      ' 

' 

. 

- 

= 

23^ 

15       " 

10      ' 

*       ami  5    pr.  ct. 

- 

. 

= 

2713       " 

20       •' 

5       ' 

* 

_ 

. 

=: 

24 

20       " 

10       ' 

' 

_ 

. 

= 

28 

20      " 

10       ' 

'       and  5    pr.  ct. 

- 

. 

= 

311 

25       " 

5       * 

' 

_ 

_ 

= 

28i         " 

25       " 

10       ' 

' 

- 

. 

= 

32^ 

25      " 

10       ' 

'       and  5    pr.  ct. 

- 

. 

^ 

355         " 

25      " 

20       ' 

' 

- 

. 

=^ 

40           " 

25      " 

20       ' 

'      and  10  jir.  ct. 

- 

- 

= 

46 

25      " 

20       ' 

'        "     10      "      and 

5  pr  ct. 

- 

=^ 

■18,'a       " 

30      " 

5       ' 

' 

- 

. 

= 

33} 

30      " 

10       ' 

< 

- 

. 

=; 

37 

30      " 

20       • 

' 

_ 

- 

= 

44 

30      " 

25       ' 

' 

- 

. 

= 

47i 

30      " 

25       ' 

'       and  20  pr.  ct. 

- 

■ 



58 

30      " 

25       ' 

••    20      "      amllOpr.  ct. 

- 

^= 

62i- 

30      " 

25       ' 

1        i<    20      "        " 

10       "      and 

5  pr. 

Ct.  = 

G4r3u     " 

35      " 

5       ' 

' 

. 

. 

= 

38i 

35      "        ' 

10       ' 

' 

- 

. 

= 

41*         " 

35      '• 

20       ' 

' 

- 

. 

= 

48           " 

35      " 

20       ' 

'       and  10  pr.  ct. 

- 

- 

:^ 

53^ 

35      " 

20       ' 

•■     10       "      and 

5  pr.  ct. 

- 

= 

5ny     " 

40      " 

5      ' 

' 

. 

= 

43          " 

40      " 

10      ' 

' 

- 

r= 

46          " 

40      " 

15       ' 

' 

- 

=r 

49          " 

40      " 

20      ' 

' 

. 

=^ 

52          " 

40      " 

30      ' 

' 

. 

= 

58          " 

40       " 

30      ' 

'      and  10  pr.  ct. 

. 

= 

62i 

40      " 

30       ' 

'         ••     20       " 

- 

= 

66  J 

40      " 

30       ' 

"    20      "      and  10  pr.  ct. 

■ 

1= 

G9iJ      " 

50      " 

5      ' 

' 

- 

. 

=z 

52J         " 

50      " 

20      ' 

' 

. 

- 

z= 

60          " 

50      " 

30      ' 

' 

. 

- 

= 

65          " 

50      " 

30      ' 

'       and  20  pr.  ct. 

- 

- 

= 

72          " 

50      " 

30       ' 

'         "     20       "      and  10  pr.  ct. 

- 

;= 

74* 

60      " 

10      ' 

' 

- 

- 

=■ 

64 

60      " 

20      ' 

' 

- 

- 

= 

68          " 

60      " 

20      ' 

'       and  10  pr.  ct. 

- 

- 

= 

71i        " 

70      " 

20      ' 

' 

- 

. 

= 

76 

70      " 

40      ' 

' 

- 

. 

= 

82 

70      " 

40      ' 

'       and  20  pr.  ct. 

. 

- 

:= 

85  J        " 

80      '•' 

30      ' 

' 

. 

. 

= 

86 

80      " 

50       ' 

' 

. 

. 

= 

90 

80      " 

50       ' 

'       and  30  pr.  ct. 

. 

- 

= 

93          " 

90      "        ' 

5       ' 

' 

- 

- 

= 

90} 

90      " 

10      ' 

' 

- 

- 

= 

91          " 

ct.  ofi'  or 


97fi'iy  pr.  ct.  ot 
95rf^- 

85.i 

83-;; 

80| 
76i 
■72i5 
76 
72 
68'^ 
7U 
67  i 
64i 
60 
54 
51  rt, 
66i 
63 
56 
52i 
43 

61t 
58i 
52 
46* 
44j5 
57 
54 
51 
48 
42 
37;* 
33^ 
30..'V 
47} 
40 
35 
28 
25i 
36 
32 
28^ 
24 
18 
141 
14 
10 
7 

9 


APPLICATION  OF  THE  ABOVE  TABLE. 
A  merchant  sold  goods  amounting:  to  $321.80,  and  allowed  20%,  10% 
5<fo  off-     What  was  the  net  amount  of  the  bill? 


and 


FIRST   OPERATION. 
$321.80 

31^%  the  equivalent  of  20%', 

10°^  and  5'!^. 

f99.7580  =  3i;f  of  »32L86. 
1.9308  =  I  of  $321.80. 


.$101.6888  =  discount. 


$220.11      =  net  amount. 


SECOND    OPERATION. 


$321.80 

68f%  =  ;„-  on. 

$218.8240  =  68,"^  of  $321.80. 
1.2872  =  I  of  1321.80. 


$220.11      =  net  amount. 


PERCENTAGE. 
Table  No.  2. 


465 


Mark    your 

Aud  your  net 

Mark  your 

And  your  net 

goods 

Aud  take  off 

gain 

goods 

And  take  off 

gain 

at  a  gain  of 

will  be 

at  a  gain  of 

■will  be 

=      0^ 

9                    0/ 

2-,=,r  % 

50,",f 

30            % 

gain  5         % 

10     " 

2i 

1\      " 

50" 

25 

Vl\      " 

12i  " 

3 

9i     " 

50" 

20 

20       " 

161  " 

5             " 

m  " 

50" 

15 

27i     " 

20     " 

10 

8       " 

50" 

10 

35       " 

20     •' 

7i 

11     ". 

50" 

5 

42i     " 

20     " 

5             " 

14       •*• 

50" 

20    &10" 

8       " 

25     " 

15 

6J     " 

60  " 

33i 

6|     " 

25     " 

10 

12i     " 

60  " 

30 

12       " 

25     " 

5 

18f     " 

60" 

25 

20        " 

25     " 

10    &5   " 

6|     '' 

60  " 

20 

28       " 

25     " 

20 

0       " 

60  " 

15 

36       " 

30     " 

20 

4       " 

60  " 

10 

44       " 

30     " 

15 

lOi     " 

60  " 

5 

52       " 

30     " 

10             " 

17       " 

60  " 

25    &10" 

8       " 

30     •' 

5 

23i     " 

70  " 

40 

2        " 

30     " 

15    &  5   " 

413   " 

70  •• 

33i 

13i     " 

33  J  " 

25 

0       " 

70" 

30 

19       " 

33i  " 

20             " 

6f     " 

70  " 

25 

27J     " 

33i  " 

15             " 

13i     " 

70  " 

20             '•■ 

36       " 

33i  " 

10 

20       " 

70" 

30    ifc5    " 

132^    " 

33J  •' 

5             " 

26f     " 

80  " 

40 

8       " 

33  J-  '^ 

10  &    5    &2i" 

11.=^   " 

80  " 

33t 

20       " 

40     " 

25             " 

5       " 

80" 

30 

26       " 

40     " 

20             " 

12       '* 

80  " 

25 

35       " 

40     " 

15             " 

19       " 

80  " 

30    &10" 

13.4    « 

40     " 

10 

26       " 

90  " 

40 

14       •' 

40     " 

5             " 

33       " 

90  " 

33  J-           " 

26J     " 

40     ■' 

15  &    5    &2J" 

lOiSr' 

90  " 

30 

33       " 

50     " 

40              " 

lose   10       " 

90  " 

20  &  10    &    5  " 

29IJ  " 

50    " 

33i 

0       " 

APPLICATION  OF  THE  TABLE. 


Goods  cost  $12.50  aud  were  marked  at  20%  gain,  from  ■which  a  discount  of 


10%  was  allowed.    What  was  the  net  gain  j)er  cent? 

FIRST   OPERATION. 


Cost 
20%  gain 


$12.50    $13.50  selling  price. 
2.50       12.50  cost  price. 


Marked  price  $15.00    $  1.00  gain. 
10%  discount      1.50 


12J 


Selling  price  $13.50 


1 

100 

8% 


Ans.  8%  net  gain. 

SECOND    OPERATION 

$100  cost  assumed. 
20%  gain. 


120  marked  price. 
12  =  10%  discount. 


$108  selling  price. 
100  cost. 


$    8  gain  on   a  hundred 
aud  hence  8%. 


466 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


Table  No.  3. 


If  vou  are  allowed 

Aud  sell  at  the  list 

If  you  are  allowed 

And  sell  at  the  list 

a  (lisconnt  I'rom 

price 

a  discount  Iroui 

price 

the  list  price  of 

Yonr  <r:un  will  lie 

tlio  list  pricft  of 

your  fjiiiu  will  he 

2J 

m  ro 

331                    ^0 

50     «i 

5 

5i     " 

40 

66f    " 

5    &    2J 

7U?  " 

40    &20&10" 

131i»  " 

8J 

9,V    " 

50 

100      " 

10 

lU     " 
16|»f  " 

60 

150      " 

10    &.    5            " 

66  J 

200      " 

m 

14?      " 

70 

233i-    " 

15 

17H    " 

75 

300      " 

161 

20        " 

80                     " 

400      " 

20                         " 

25 

20  &    10    &5  " 

46M-  " 

25                      " 

m   " 

30 

m   " 

APPLICATION  OF  THE  TABLE. 


The  list  price  of  goods  is  $500,  from  which  5%  and  2J%  discount  is  allowed. 


What  is  the  net  gain  per  cent  ? 

FIRST   OPERATION. 

$500    list  price. 

25    =5%.        365  X  100 


Ans.  mi%. 


I/O 

115  =  2J%. 


4634 


SECOND    OPERATION. 

$100    list  price  assumed. 
5    =5%.  8 

. 741 

95 

2f  =  2J%.,  _ 


$463J  cost. 
500 


925  =  cost. 


59 

8 
100 


'mfc- 


$  3C5  gain. 


'3  = 


883. 


TO  FIXD  THE  VALUE  PEE  TOX  OF  SUGAE  CANE. 


Note. — For  the  leadiuf;  facts  regarding;  this  new  suhject,   we  are  iudehted  to  Prof.  W.  C. 
Stubbs,  Director  of  the  La.  Sugar  and  State  Experimental  Station. 

The  method  to  determine  the  price  to  pay  for  Sugar  Cane,  varies  somewhat 
with  different  manufacturers  of  sugar  in  the  different  sections  of  the  sugar  district 
of  Louisiana.  The  following  work  and  statements  will  make  clear  the  various 
methods  used : 

Cane  is  usually  sold  uj^on  three  factors,  as  follows :  1.  The  price  of  refined 
sugar  in  the  New  Orleans  Market.  2.  The  i^ercentage  of  juice  extracted  from  the 
cane.  3.  The  percentage  of  sucrose  contained  in  the  cane.  The  value  pf  the  cane 
being  a  certain  price  for  each  of  these  elements  or  factors.  Tlie  element  of  glucose 
is  not  considered  as  a  material  factor  in  determining  the  price  of  cane. 

The  average  chemical  analysis  of  Louisiana  sugar  cane  is  about  90  per  cent 

juice  and  10%  fibre.    The  juice  contains  on  an  average  about  12%  of  sucrose,  2% 

of  glucose,  and  1  to  l.J%  of  solids  not  sugar. 

Note. — A'cry  f(>w  of  the  sugar  mills  in  Louisiana  extract  more  than  an  average  of  72  to  80"^ 
of  juice  from  the  cane. 


PERCENTAGE.  467 

QUANTITATIVE  CHEMICAL  ANALYSIS. 

884.  Quantitative  Chemical  Analysis  consists  in  the  determination  and 
separation  of  the  relative  proportions  of  the  component  parts  of  ■which  a  substance 
is  composed. 

The  several  parts  are  expressed  in  per  cents  of  the  whole  substance.  In  the 
analysis  of  cane  juice,  molasses,  etc.,  the  component  parts  are,  as  above  stated, 
sucrose,  glucose  and  solids,  that  are  not  sugar. 

The  Co-efficient  of  Purity  is  the  per  cent  that  the  sucrose  is  of  the  whole 
solids  extracted  from  the  cane.  Thus,  as  in  the  above  average  in  which  the  juice 
contains  12%  sucrose,  2^^  glucose  and  li%  of  foreign  solids,  not  sugar,  the 
co-efiflcient  of  purity  is  foiind  as  follows : 

12  +  2  +  li  =  15.i  ;  then,  12  x  100  -^  loi  =  77iJ%. 

The  Glucose  Ratio  is  the  per  cent  that  the  glucose  is  of  the  sucrose,  and  is 
found  as  follows,  using  the  above  figures:     (2  x  100)  -^  12  =  16^%. 

The  per  cent  of  the  foreign  element,  not  sugar,  is  found  as  follows :  (1^  x  100) 

^15J  =  9|i%. 

Fertilizers.  The  component  parts  generally  considered  in  fertilizing  sub. 
stances  are  potash,  nitrogen  and  phosphoric  acid. 

THE  SUCEOSE  MULTIPLIER  AND  HOW  TO  FIND  IT. 

885.  A  Sucrose  Multiplier  is  a  number  used  as  a  factor  in  determining 
the  price  of  sugar  cane  per  ton,  and  is  found  as  follows  : 

1.  By  estimating  an  average  extraction  of  80%  of  juice  from  the  cane.  2. 
By  allowing  the  seller  50%  of  the  juice.  3.  By  estimating  an  average  of  10  or 
10  J  %  sucrose  in  the  juice. 

First  Operation  to  find  a  Sucrose  Multiplier. 

886.  1  ton  of  cane  =  2000  lbs.  80%  of  which  is  ICOO  lbs.  of  juice.  J  of 
1600  lbs.  =  800  lbs.  juice  allowed  the  seller.     (800  or  8  is  the  sucrose  multii^lier). 

Hence,  800  x  10A%  x  4^,  the  price  of  Prime  Yellow  Clarified  Sugar  in  New 
Orleans,  gives  $3.30,  the  price  per  ton  of  cane.  These  figures  practicalized  become 
8  X  lOJ  X  4  =  $3.30,  the  price  per  ton  of  cane.  (The  two  O's  in  the  800  and  the 
%  in  the  10i%  being  omitted). 

Second  Operation  to  find  a  Multipliee. 

887.  1  ton  =  2000  lbs.,  80%  =  1600  lbs.  h-  2  =  800.  10%  of  800  =  80 
pounds  sucrose  or  sugar.     (80  is  the  sucrose  midtiplier). 


468  soule's  philosophic  practical  mathematics.  * 

Hence,  80  x  i^,  the  price  of  Prime  Yellow  Clarified  Sugar  in  New  Orleans, 
is  $3.20,  price  per  ton  of  cane. 

Or  thus,  10A%  of  800  =  84.     Hence  S4  x  4/.  =  $3.30,  the  price  per  ton  of 

cane. 

Note. — According  to  tlie  above  method  and  principles,  when  the  ^^  of  juice  extracted  is  70, 
72,  7.5,  83i,  86  or  90,  the  ninltipliers  wonld  be  respectively,  7  or  70,  7.2  or  72,  7.5  or  7.5,  8.3i  or  83J^, 
8.5  or  85  and  9  or  90  accordiuj;  to  the  method  used.  In  like  manner,  the  sucrose  multiplier  would 
be  found  for  other  jier  cents  of  juice  extracted. 

PKOBLEMS. 

888.  1.  If  the  price  of  refined  sugar  in  Kew  Orleans  is  4  cents  per  pound 
ajid  the  ])ercentage  of  sucrose  in  cane  juice  i.s  12  per  cent,  what  would  be  the  value 
of  cane  per  ton,  at  8  cents  per  unit  of  sucrose  and  of  the  price  of  refined  sugar? 

Ans.  $3.84. 
operation.  8/  X  12  X  4  =  $3.84  per  ton. 

2.  A  purchaser  of  sugar  cane  says  to  a  planter  "  I  will  give  you  8J  cents 
per  unit  for  each  1%  of  sucrose  that  your  cane  contains,  multiplied  by  the  i:)rice  per 
pound  of  refined  sugar  iu  New  Orleans."  If  the  price  of  refined  sugar  in  New 
Orleans  is  4J^  per  pound,  and  the  juice  contains  13.4  per  cent  sucrose,  what  is  the 
value  of  the  cane  per  ton  1  Ans.  $5.1255. 

opekation. 


8J  X  13.4  X  4i  =  $5,125+  or  thus. 


9 
13.4 

17 


NOTK. — The  8^-  cents  per  unit  iu  this  problem,  equals  So^!^  (8.5)  of  juice  extraction  as  shown 
by  the  above  operations  to  hud  multipliers. 

889.  Another  method  of  determining  the  price  of  sugar  cane  per  ton. 

In  some  of  the  sugar  districts  of  Louisiana,  the  price  or  value  of  cane  per 
ton  is  determined  by  allowing  85^  pounds  of  sugar  per  ton  of  cane,  and  to  estimate 
its  value  at  the  New  Orleans  market  price  for  Prime  Yellow  Clarified  Sugar. 

PROBLEMS. 

890.  1.  Allowing  83J  pounds  of  sugar  per  ton  of  cane,  what  is  cane  worth 
when  Prime  Yellow  Clarified  Sugar  is  quoted  at  3^^/  in  New  Orleans? 

Ans.  $2.C5|. 

OPERATION.  16      51 

3     250 
S^a,  X  83J  =  $2.C5f  or,  thus:  —     

$2.G5| 

Note. — The  83J  multiplier  hern  used,  is  found  as  shown  above  in  the  "second  operation  to 
find  a  multiplier,"  substituting  83^°^^  for  juice  extracted  in  place  of  the  80  there  used. 

2.     Using  the  figures  of  the  above  problem,  and  allowing  that  sugar  cane  in 

Louisiana  will  produce  152  pounds  per  ton,  what  would  be  the  gross  profit  to  a  mill 

that  purchased  ajid  manufactured  41G2  tons  of  cane.    No  allowance  to  be  made  for 

molasses  or  syrup  that  may  be  made,  or  the  expenses  to  manufacture,  or  to  market? 

Ans.  $nion.57f. 

Note. — 150  to  160  pounds  of  sugar  per  ton  of  cane  is  the  average  for  the  cane  of  Louisiana. 


PERCENTAGE. 


469 


OPERATION   TO  FIKD   COST. 


OPERATION  TO  FIND   SAIES  AND  GAIN. 


16 

o 
O 


51 

250 

4162 


16 


51 

152 
4102 


$20164.89 
11055.31i 

$11055.31^  $20164.89  $  9109.57a  gain. 

Note. — Some  purphasers  of  sugar  cane  pay  per  tou,  80  cents,  more  or  less,  for  eacli  cent  o^ 
the  New  Orleans  price  of  Prime  Yellow  Clarilied  Sugar. 

This  80  wliich  is  nscd  as  a  multiplier  of  the  Ke-w  Orleans  price  of  Prime  Yellow  Clarified 
Sugar,  i.s  found  as  above  shown  in  tlie  secouil  operation.  Article  887,  to  find  a  multiplier  allowing 
805' 0'  of  extraction  of  juice  from  the  cane. 

3.  How  many  pounds  of  sucrose  are  there  in  1840  pounds  of  caue  juice 
■wliicli  contains  13.0%  of  sucrose?  Ans.  250.24  pounds. 

4.  If  tlie  solids  in  the  juice  of  cane  arel5J%,  and  tlie  co-efficient  of  purity  is 
77^3%,  Low  many  jiounds  of  sucrose  are  there  in  660  gallons  of  juice,  allo\ring  8  J 
pounds  to  the  gallon  ?  Ans.  660  pounds  of  sucrose. 

FIRST    OPERATION.  SECOND   OPERATION. 

77H  X  15i%  =  12%  sucrose.  660  x  8^  =  5500  pounds. 

660  X  8>^  =  6500  pounds  of  juice.  5500  x  15i  =  852.50  lbs.  solids. 

5500  X  12%  =  660  pounds  of  sucrose.  852.50  x  7'7iJ  —  600  lbs.  sucrose. 

5.  In  the  above  problem,  if  the  glucose  ratio  vras  16§%,  hovr  many  pounds 
of  glucose  would  there  be  ?  Ans.  110  pounds. 

OPERATION. 

10§  X  12  =  2%  glucose. 

660  X  8J  =  5500  pounds  of  juice. 

5500  X  2%  =  110  pounds  of  glucose. 

6.  How  many  pounds  of  sucrose  are  there  iu  6|  tons  of  svtgar  cane,  of  whicli 
86J%  is  juice,  and  14%  of  the  juice  is  sucrose!  Ans.  1513.75  pounds. 

OPERATION. 

2000  X  ej  =  12500  lbs.  x  86J%  =  10812.5  lbs.  X  14%  =  1513.75  lbs. 

7.  From  8  tons  and  500  jKninds  of  cane  which  was  passed  through  a  mill, 
the  bagasse  weighed  3630  pounds.    What  was  the  per  cent  of  juice  extracted  ? 

Ans.  78%. 

OPERATIONS   INDICATED. 

% 

100 

3030 

or  thus : 


16500 


16500 
3630 


16500 


Wc/  \''^fo  extracted. 


12870 


22  f 
100% 

8.     12870  pounds  of  juice  were  extracted  from  a  lot  of  cane ;  if  the  rate  per 
cent  of  extraction  was  78%,  how  many  tons  of  cane  were  there  7       Ans.  8^  tons. 

OPERATION  INDICATED. 
100 
12870 

or  (12870  X  100)  ~  (78  x  2000). 


78 
2000 


470  soule's  philosophic  practical  mathematics.  * 

9.  In  oi  tons  of  cotton  seed  fertilizing  meal,  there  is  6i%  of  nitrogen,  3J% 
of  pliosplioric  acid  and  1.9%  of  i)otash.  How  many  pounds  of  each  I'ei-tilizing 
ingredieuf?  Ans.  715  pounds  of  nitrogen. 

3-lo.75  pounds  of  phosphoric  acid. 
209  pounds  of  i>otash. 
10.     In  3400  j)ounds  of  bone  meal,  there  are  ICOi  pounds  of  nitrogen  and  790.8 
pounds  of  phosphoric  acid  ;  what  is  the  ])er  cent  of  eacli  ? 

Ans.  4.72+%  uitrogen.     23.20+%  phosphoric  acid,   • 

MISCELLANEOUS  PRACTICAL  QUESTIONS  IN  PERCENTAGE. 

891.  1.  Bought  G  dozen  shirts  for  $129.G0  ;  at  M'hat  price  per  shirt  must  I 
sell  them  to  gain  50  per  cent!  Ans.  $2.70. 

2.  Bought  butter  at  48/  perixmnd;  at  what  price  per  ouuce  must  I  sell  it 
to  gain  66|  per  cent  ?  Aus.  5/. 

3.  Bought  floiir  at  .$7.50  i)er  barrel ;  at  what  price  must  it  be  sold  to  lose  10 
per  cent  ?  Ans.  $G  75. 

4.  Sold  goods  to  the  amount  of  $250  and  gained  25  per  cent ;  what  did  they 
cost?  Ans.  $200. 

5.  Sold  flour  at  $8.75  and  lost  12i  per  cent ;  what  did  it  cost?       Ans.  $10. 

G.  Bought  gloves  at  $4  per  dozeu  and  sold  them  at  50/  per  i>air ;  what  per 
cent  did  I  gain  1  Ans.  50%. 

7.  My  merchandise  account  is  debited  $145000  and  credited  $174000.  There 
is  no  merchandise  on  hand.     What  ])er  cent  did  I  g:iin  on  sales?  Ans.  20%. 

8.  My  merchandise  account  is  debited  $280000  and  credited  $314212.50.  I 
have  on  hand  $44797.50.     What  is  my  gain,  and  what  my  gain  i)er  cent? 

Ans.  $79010  gain.     28,,'Vo%  gain. 

9.  A  manufacturer  received  a  certain  quantity  of  cotton  to  be  manufactured 
into  cloth.  He  delivered  182400  yards,  averaging  4J  ounces  to  the  yard.  Allowing 
that  there  was  a  waste  of  10  per  cent  in  manufacturing  tlie  cloth,  how  niany  ptmnds 
of  cotton  did  he  receive?  Ans.  57000  pounds. 

10.  I  sell  a  certaiu  quality  and  style  of  goods  at  $1.10  per  yard.  My  rival  in 
trade  sells  the  same  quality  and  style  at  $1.25  per  yard.  What  jier  cent  more  does 
he  charge  for  his  goods  than  I  ?  Ans.  13fy%. 

11.  I  sell  flour  at  $8.50  per  barrel.  My  rival  sells  the  same  kind  at  $9. 
What  per  cent  less  do  I  sell  than  he?  Ans.  5|%. 

12.  The  assessment  of  real  and  personal  property  of  a  city  amounts  to 
$150,000,000.  The  various  ex])enses  of  tlie  municipal  government  are  a])proximated 
and  estimated  at  $2,800,000.  What  must  be  the  per  cent  tax  assessment  to  raise 
the  amount?  Ans.  ].86§%. 

13.  Sold  coffee  at  ISC  per  pound  and  lost  10  \>er  cent;  what  should  I  have 
sold  it  for,  to  have  gained  12^  per  cent  ?  Ans.  22^/. 

14.  Sold  butter  at  35/  per  pound  and  gained  25  jier  cent;  what  per  cent 
would  I  have  gained,  had  I  sold  it  at  30/  ?  Ans.  7|%. 


PERCENTAGE. 


471 


15.  Sold  potatoes  at  $3.15  per  barrel  aud  lost  10  per  cent;  what  per  ceut 
would  I  have  gained  or  lost  bad  I  sold  them  at  $i.  ?  Ans.  14f  %  gain. 

10.  Sold  rice  at  SJ/  and  lost  8J  per  cent;  wliat  per  cent  would  I  have  gained 
or  lost,  had  I  sold  it  at  9  c  !  Ans.  1%  loss. 

17.  AVhiskey  which  cost  90/  per  gallon  is  compounded  with  water  in  the 
proportion  of  1  gallon  of  water  to  2  gallons  of  whiskey,  and  the  mixture  is  sold  for 
Sop  per  gallon.     What  %  is  gained?  Ans.  ■ilg'/^. 

IS.  A  faithful  accountant,  who  was  receiving  a  compensation  of  $720  per 
year,  had  his  salary  increased  33J%.     How  nuich  does  he  now  receive  per  mouth  ? 

Ans.  $80. 

19.  A  firm  i)aid  $300  per  month  rent  and  was  raised  25%.  What  is  their 
current  rent?  Ans.  $375. 

20.  A  clerk  who  received  $125  jier  month  had  his  salary  reduced  20%.  What 
did  he  then  receive  ?  Ans.  $100. 

21.  A  man  who  was  receiving  $2.50  per  day  demanded  an  increase  of  40%. 
What  would  then  be  his  daily  pay  ?  Ans.  $3.50. 

22.  A  merchant  compounded,  in  equal  qviantities,  sugar  that  cost  6/  per 
pound  and  sugar  that  cost  9/  per  pound.  He  theu  sold  the  mixture  at  9;'  per 
pound.    What  %  did  he  gain?  Ans,  20%. 

OPERATION. 


1  pound  at  6/  =  6/. 

1         "       "  9/  =  9/. 


2  )  15/. 


9/    sales. 
7J/  cost. 


15 


3  gain. 
100 


-'0%  gain,  Ans. 


Uf  gain. 

1  lb.  of  the  com- 
pound cost       7J/. 

23.  A  merchant  bought  5  bbls.  flour  for  $40  and  sold  the  same  at  a  gain  of 
25%.  He  then  bought  5  bbls.  more  for  $40,  and  sold  the  same  at  a  loss  of  25%. 
Did  he  gam  or  lose,  and  if  so,  how  much  ?  Ans.  He  neither  gained  nor  lost. 

OPERATION. 

$40  cost  ®  25%  gain  gives  $50  selling  price  =  $10  gain. 
$40  sold  ®  25%  loss  gives  $30  selling  price  =  $10  loss. 

24.  A  merchant  sold  5  bbls.  of  inferior  flour  for  $40  and  gained  25%.  He 
then  sold  5  bbls.  of  superior  flour  for  $40  aud  lost  25%.  Did  he  gain  or  lose  in  the 
two  transactions,  and  if  so  how  much?  Ans.  He  lost 

OPERATION. 

$40  selling  price  ®  25%  gain  gives  $32    cost  =  $  8    gain. 
$40  selling  price  ®  25%  loss  gives   $53J  cost  =  $13J  loss. 


Net  loss  $  5^ 
25.     A  grocer  bought  10  barrels  of  molasses  each  containing  40  gallons,  at 
50/  per  gallon.     He  then  put  the  molasses  in  kegs  holding  8  gallons  each,  and  sold 
the  same  as  10  gallon  kegs,  at  55/  per  gallon.     What  was  his  profit,  and  what  % 
did  he  gam?  Ans.  $75  profit;  37i%  gain. 


472 


SOUI.F.  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


26.    A  fruit  dealer  sold  4  peaches  for  5/  and  gained  56J%. 
he  have  gained  by  selling  5  for  G  cents  ? 


What  %  would 
Ans.  50%. 


5n-4  =  li/ 


13Gi 


100  cost. 
1/  cost. 


li/- 


OPERATION. 

-f/  =  f/gain. 


2 
5 
100  ( 

50%  gain. 


156|  sales. 

4 

6 


150  sales, 
50^0  gain. 

27.  A.  and  B.  are  two  merchants ;  they  desire  to  barter  rice  and  sugar.  A. 
has  rice,  market  value  6/'  per  pound;  and  B.  has  sugar,  market  value  8/  per 
pound.  At  the  time  of  the  exchange  or  barter,  A.  suggests  to  B.  that  in  order  to 
influence  the  market  reports,  lie  will  place  his  rice  at  7/  per  pound  and  that  B. 
shall  advance  his  sugar  accordingly.  B.  accepts  the  proposition.  What  should  be 
the  exchange  price  of  B's  sugar  1  Ans.  9J/  iier  pound. 


7/  =  exchange  value  of 

A's  rice. 
6/  =  market    value    of 
—  same. 

1/  =  the    increase    on 
game. 


6/  +  1^  =  7/  =  exchange 
value  of  A's  rice. 


FIKST   OPEEATION. 

/ 

1  =  gain. 
6     100 


1G§%  gain. 


SECOND   OPERATION, 


G 


1  =  gain  on  6/. 

8 


8/  =  market    value     of 

B's  sugar. 
n/  =  1G§%  gain. 

9J/  =  exchange  value. 


8^  +  1  J/  =  9J/  =  exchange 
value  of  B's  sugar. 


1^/  gain  on  8/. 


28.  Suppose  in  the  above  problem,  that  B.  had  proposed  to  reduce  his  sugar 
1/  per  pound,  and  that  A.  should  reduce  his  rice  accordingly.  What  would  be  the 
exchange  price  of  A's  rice  1  Ans.  5  J/. 

29.  Suppose  in  the  above  jiroblem  that  A.  and  B.  had  each  raised  1/  on  the 
market  value  of  their  rice  and  sugar,  how  much  %  would  B.  have  lost,  and  what  % 
■would  A.  have  gained  ?  Ans.  B.  would  have  lost  3f  %. 

A.       "        "     gained  3fa%. 

OPERATION    TO   FIND   B'S    %    LOSS. 


9J/  =  correct  exchange  value,  as  above. 

9  /  =  incorrect  exchange  value,  as  supposed. 

^^  —  loss  by  incorrect  exchange  value. 


3 

28 


1  =  loss. 

3 

100 


34%  loss.    Ans. 


* 


PERCENTAGE. 
OPERATION   TO   FIND   A'S    %    GAIN. 


473 


9/  =  B's  selling  price. 
6 


/ 


6f/  =  price  A.  sliould 
have  sold  for  when 
B.  sold  for  9/. 


7/    =  A's  incorrect  selling 

price. 
G^c'  r=  A's    correct    selling 

price. 

^c  =  A's  gain  by   selling 
at  7/. 


1  =  gain. 

4 

100 


3-2-f  %  gain,  Ans, 


EXPLANATION    FOR    SECOND    OPERATION. 


8ff,  B's  market  value  +  1/  =  9/,  B's  selling  price ;  then  since  Sc  sell  for  9/, 
1/  -will  sell  for  the  ^  part,  and  G/,  A's  market  value,  will  sell  for  C  times  as  many, 
which  is  G^/ ;  then  since  A.  really  sold  for  7/',  when  he  should  have  sold  for  CJc,  he 
gained  7/  —  63/,  =  ^^ ■  audit  6^/  gain  :J/,  1/  will  gain  the  GJ  or  -^^  part,  and 
100/  will  gain  100  times  as  much,  which  is  3^4/  or  %. 

30.  Bought  coflee  at  15/  i>er  pound  and  sold  it  at  IS/;  what  would  I  gain  by 
buying  $75  worth  of  coffee  T  Ans.  $15. 

31.  Bought  coffee  at  15/ per  pound  and  sold  it  at  18/;  what  did  I  gain  by 
selling$75  worth  of  coffee?  Ans.  $12.50. 

32.  A  real  estate  siieculator  sold  a  house  and  lot  for  $7200  and  gained  20 
per  cent.  He  then  invested  the  $7200  in  merchandise  which  was  sold  at  a  loss  of 
20  per  cent.  Was  the  result  of  the  two  trausactious  a  gain  or  a  loss,  and  if  so  how 
much?  Ans.  $240  loss. 

33.  A  grocer  sold  a  barrel  of  flour  for  $8.75  and  gained  25  per  cent.  Soon 
after  he  sold  to  the  amount  of  $G37  and  gained  40  i)er  cent.  What  was  the  last  sold 
at  j)er  barrel,  and  how  many  barrels  were  there?  Ans.  65  bbls.  at  $9.80. 

34.  A  general  fought  32  battles  and  lost  4 ;  what  per  cent  did  he  lose  ? 

Ans.  12J%. 

35.  A  merchant  has  due  him  $32  and  compounds  the  claim  on  receii^t  of  $4; 
what  per  cent  did  lie  lose  ?  Ans.  87i%. 

3G.  The  net  assets  of  a  bankrupt  are  $72000,  which  is  15  per  cent  of  his 
liabilities.    What  are  his  liabilities  ?  Ans.  $480000. 

37.  r-.A  person  pays  $30  per  month  for  board,  which  is  20  jjer  cent  of  his 
monthly  salary.     What  is  his  salary  per  mouth  ?  Ans.  $150. 

38.  Paid  $6  for  sawing  4  cords  of  wood,  which  was  1G§  per  cent  of  its  cost; 


what  did  it  cost  per  cord  ? 


Ans. 


STATEMENT. 


50 

4 


$  cost. 
100 
3 
G 


39.  A  mother  having  a  basket  of  oranges,  took  out  50  per  cent  for  her 
children ;  of  these  she  gave  40  jier  cent  to  her  son  who  gave  ^  of  his  to  his  sister, 
who  thus  received  3  oranges.    How  many  were  there  in  the  basket  at  first  ? 

Ans.  30. 

OPERATION. 

Let  100%  =  the  whole  number  of  oranges.    50%  =  the  number  taken  out. 


474 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


40%  of  50%  =  207^  =  given  to  son.     A  of  20%  =  10%  given  to  sister.     10%  =  3 
oranges.     Then  10%  =  3,  100%  =  30  oranges,  answer. 

Or  thus : 
100%  =  the  whole  number  of  oranges.  •  50%  =  J  taken  out.     40%  =  f  of  J 
=  i  given  to  son.     J  x  i  =  n,  =  given  to  sister,     ^-g  =  3,  then  IS  =  30,  the  whole 
number  in  the  basket. 

40.  A  merchant  marks  his  goods  50  per  cent  above  cost,  but  supplies  his 
wholesale  jiatrons  at  25  i)er  cent  discount  on  retail  iirice ;  what  per  cent  gain  does 
he  make?  Ans    l'^%. 

41.  A  merchant  asked  for  a  lot  of  goods  20  per  cent  more  than  they  cost,  but 
they  being  a  little  soiled  he  closed  the  sale  at  10  per  cent  less  than  his  asking  price, 
and  realized  a  ])rofit  of  $112.     What  did  his  goods  cost?     Wliat  Avas  his   asking 


price  ?     And  for  how  much  did  he  sell  them  ? 


20 


OPERATION. 


$120 
12 

$108 
100 


Ans.  $1400  cost. 

$1080  asking  price. 

$1513  was  the  selling  price. 


100 
112 


$1400  cost. 

280  =  20%  gain. 


$S  gain. 


$1680  asking  price. 
168  =  10%  discount. 


$1512  selling  price. 

42.  If  the  cost  of  purchasing  and  shipping  goods  from  New  York  to  Kew 
Orleans  is  8  per  cent  ou.Kew  York  cost,  what  per  cent  gain  on  the  first  cost  must 
the  NeAT  Orleans  merchant  mark  his  goods  in  order  to  clear  a  profit  of  33J  per  cent 
on  full  cost  ?  What  amount  of  goods  must  he  purchase  to  make  a  net  profit  of 
$35200  on  New  York  cost  ?     What  would  be  his  New  Orleans  profit  1 

Ans.  He  must  mark  his  goods  at  44%  gain. 
He  must  purchase  $80000  worth. 
His  N.  O.  profit  would  be  $2«S00. 
OPERATION. 


$100      cost. 
8%  freight. 

$108     total  cost. 
30      =  33i%. 

X;  ^/ 

$     c 
I'Od  25 
3.^^00  3200 

$80000      Aus. 

$80000 

8% 

$  0400 
3"»'^00 

$144      N.  O.  s.  P. 
100 

$28800    Aus. 

gam. 

43.  A  merchant  bought  flour  at  $7.50  per  barrel;  what  must  he  ask  for  it  in 
order  that  he  may  fall  10  per  cent  on  his  asking  price  and  still  gain  20  per  cent  on 
the  cost  ?  Ans. 


PERCENTAGE. 


475 


44.  Sokl  a  barrel  of  oranges  for  $G  and  gained  20^^.  I  tlien  invested  the  $G 
in  merchandise  which  1  sold  at  a  loss  of  '20(/c.  What  ^^•as  the  gain  or  loss  by  the 
two  transactions?  Aus.  20/ loss. 

45.  How  many  apples  must  1  buy  so  that  after  allowing  25%  of  them  to  be 
eaten  and  20%  of  the  remainder  to  be  given  away,  I  may  sell  just  1  dozen  1 

Ans.  20  apples. 

OPEKATION. 


ta 


100 
12 

IG 


80 


100 
16 

2Q      Ans. 


or. 


80 


100 
100 
12 


46.  A  fruit  dealer  has  nears  worth  5/  a  piece,  but  will  sell  6  for  25/*.  What 
per  cent  would  be  gained  b},  buying  6  for  25/,  and  what  per  cent  would  be  lost  by 
buying  them  separately  at  5/  each?  Ans.  1°,  20%  gam;  2°,  16§%  loss. 


6  )_25 
5 


41=25 


OPEEATION. 

6      5 
6 

100 


o 
100 


1G§%  loss. 


1/  20  %gain. 

47.  The  New  Orleans  Cotton  Factory  receives  from  a  merchant  200  bales  or 
90000  lbs.  of  Middling  Cotton  to  manufacture  into  sheetings.  The  charge  for  man- 
ufacturing is  fixed  at  3/  per  yard,  and  payment  is  to  be  made  in  Middling  Cotton 
at  12/  per  pound.  After  the  factory  had  delivered  to  the  merchant  128S00  yards, 
which  weighs  8  ounces  per  yard,  a  settlement  was  made  on  the  following  conditions: 
It  was  agreed  to  allow  12i  per  cent  waste  in  the  manufacture ;  the  factory  to  retain 
tbe  cotton  remaining  on  Land  at  12c'.,  and  for  the  balance  due  to  receive  Middling 
Fair  instead  of  Middling  at  13/.  How  many  pounds  of  cotton  does  the  merchant 
owe  the  factory  ?  .  Aus.  14584 1%  lbs. 

.     OPERATION. 


87i 


100 
64400 

73600     lbs.  cotton. 
90000         "         '• 


16 


12SS00  yds. 

8 


G4400    lbs. 


16400     lbs.  on  hand. 
12/ 


128800      yards. 
3(f 


$1968.00 


$3864.00 
1968.00 


$1896.00  -i-  13/  =  14584iS-  lbs. 

48.  Botight  coffee  at  19/  per  i>ound;  allowing  that  it  will  fall  short  5  per 
cent  in  weighing  and  that  8  per  cent  of  the  sales  will  be  uncollectible,  for  how 
much  per  pound  must  it  be  sold  to  clear  15  per  cent  jiroiit  on  the  cost  ? 

Ans.  25/. 


70                          SOULES    PHILOSOPHIC 

PRACTICAL    MATHEMATICS. 

, 

OPEKATION. 

19/      cost. 
285  =  157o  gain.       95 

100 

.2185       92 

100 

23             or,       95 

n.) 

115 

100       or, 
19 

100 
95 
92 

115 

100 

100 

19 

.2185/. 

23;'. 

25/  Ans. 

49.  A  ffrocer  bought  at  auction  200  baskets  of  champagne  for  $2800 ;  what 
must  he  ask  for  it  per  basket  so  that  he  may  discount  20  per  cent  on  his  asking 
price,  allow  10  per  cent  of  sales  to  be  bad  pay,  2J  per  cent  commission  for  selling, 
and  still  make  a  profit  of  100  per  cent  on  cost  ?  Ans.  $40. 


FIRST  OPERATION. 


SECOND  OPERATION. 


100 

80 

200 

87i 

100 

100 

2800 

200 

$40    Ans. 

80 


100 
200 

250 


?n 


;p0  8 

250 


2000 


2851 


100 

m 


2000 


$40    Ans. 


50.  A  merchant  sold  a  pair  of  shoes  for  $3.20  and  lost  20  per  cent;  what 
should  he  have  asked  for  liis  shoes  in  order  that  he  might  have  fallen  20  per  cent 
on  his  asking  price  and  still  have  gained  20  per  cent  on  cost?  Ans.  $0. 

51.  If  it  costs  a  merchant  5%  over  factory  price,  to  put  his  goods  in  store, 
and  he  sells  at  a  profit  of  U>'/o  on  full  or  store  cost,  and  pays  a  tax  of  -^u'/c  on 
sales,  how  much  goods  must  he  sell  per  month,  to  equal  a  salary  of  $300  jier  month  ? 
And  what  %  did  he  actually  make  on  the  goods,  including  tax  as  part  of  cost,  in 
addition  to  the  5%  ;  and  what  amount  must  be  invested  to  produce  $300  per  month  ? 

Ans.  10.44225%  net  gain  on  store  cost. 
$3310.59+  monthly  sales. 
9.938+  %  gain  iuQluding  tax  as  part  of  cost. 
$2872.94+  amt.  to  be  invested  to  gain  $300  per  mo, 

OPERATION. 

$100  =  assumed  factory  cost.  Cost  at  factory,  $100. 

5  =  5%  to  i)lace  goods  in  store.  Charges  to  put  in  store,  5. 

Tax  on  sales,  .05775 

$105  =  store  cost.  

10.50  =  10%  gain  on  store  cost.  Total  cost,  $105.05775 


$115.50        =  selling  price. 

.05775  —  ■:}g  %  tax  on  sales. 


105.05775 


$115  44225  =  net  receipts  after  paying  tax. 
105.  =  store  cost. 


10.44225%  net  gain  on  store  cost. 


10.44225 
100 

9.939+%  gain  including 
tax  as  part 
of  cost. 


10.44225 


115.44225 
300 

$3316.59125  monthly  sales. 


115.44225 


100  cost. 
3316.59125 

$2872.94+  amount  to  be  in- 
vested to  gain 
$300  per  mo. 


PERCENTAGE. 


477 


52.  An  importer  sold  J  of  a  cargo  at  22J%-  gain;  J  of  remainder  at  lOg^ 
gain;  J  of  remainder  at  10%  loss,  and  the  balance  at  40%  loss.  There  was  a  profit 
of  $335.40  on  all.    What  was  the  cost  of  the  cargo  "I 


OPERATION. 

Let  $100  =  the  cost  of  cargo. 

Then  ^  =  $33^  at  22|%  gain  =  $  74  gain. 

Then  4  of  remainder  §  =  J    =    33 J  at  10|%  gain  =      5%  gain. 

Then  |  of  remainder  |  =  ji,  =      s|  at  10  %  loss    = 

Then  the  remainder  ^  =    25    at  40  %  loss    = 


$100 


10 


I  loss, 
loss. 


$13 iV  gain  $10|  loss, 


Then  $13  Jg  - 
Then  $2| :  $100 : 


I  =  $2|  net  gain. 

$335.40  :  $15093  cost  of  cargo. 


or  thus, 


20 


100 

'J 

335.40 

$15093.00 


53.  A  live  stock  dealer  sold  two  horses,  each  at  the  same  price.  On  one 
horse  he  gained  25  per  cent,  and  on  the  other  he  lost  25  jier  cent.  His  net  loss  on 
the  two  sales  was  $12.     What  was  the  cost  of  each  horse  1 

Ans.  $72  cost  of  1st  horse.     $120  cost  of  2nd  horse. 

OPERATION — FIRST   PART. 

1.  Assume  that  one  horse  was  sold  for  $100  at  a  gain  of  25  per  cent. 

2.  "  "  "         "  "      $100  at  a  loss  of  25  per  cent. 

Then  find  the  cost  of  each  horse,  and  the  loss  on  the  sale  of  both  horses  j 
then  proceed  as  shown  in  the  operation. 


Statement  to  find  cost  of  first 


125 


100 
100 

$80,  cost  1st  horse. 


OPERATION — SECOND   PART. 

Statement  to  find  cost  of  second 
Lorse  soUl  'iz^  25"o  loss. 

100 
,;j      100 

$133J,  cost  2nd  horse. 


$  80    cost  first  horse. 
133J^  cost  second  horse 


$213J  cost  of  both  horses 
200    assumed  selling])rice 

■        of  both  horses. 

$13 J  loss  on  sale  of  both 
horses. 


statement  to  find  cost  of  the 
two  horses  actually  sold. 


13i 


213J 

12 


OPERATION— THIRD   PART. 

Statement  to  find  cost  of  first 
horse  actually  sold. 


213J 


80 
192 


Statement  to  find  cost  of 
second  horse  actually  sold 


213J 


133J 
192 

$120 


478 


SOULE  S    rillLOSOPHIC    rRACTICAL,    MATHEMATICS. 


SECOND   OPEKATION. 


Let  $1  =  sale. 


^  or  25%  gain  on  cost  =  |  or  20%  on  sales. 
i  or  25%  loss  on  cost  =  J  or  33J%  on  sales. 


OPERATION  FOR  FIRST   HORSE. 

$1     =  selling  price, 
less      +  =  25%  gain  on  cost, 

$i  =  cost  of  1st  horse. 


OPERATION  FOB   SECOND   HORSE. 

$1     =  selling  jiriee. 
I)lns      J  =  25%  loss  on  cost. 


I  ^  =  cost  of  2ud  horse. 


i  loss  —  1  gain  =  -fy-  = 


12 
15 


then,      $90  =  sale  of  1st  horse. 
—  i  =     IS  =  gain  on  1st  horse. 


$90  =  sale  of  each  horse. 

and, 


+  i 


72  =  cost  of  1st  horse. 


=  sale  of  2nd  horse. 
30  =  loss  on  2ud  horse. 

$120  =  cost  of  2nd  horse. 

54.  A.  and  B.  are  both  merchants  and  conducting  the  same  character  of 
business.  A.  desires  to  exchange  or  barter  $9384  ■worth  of  sugar  for  $9384  worth  of 
rice.  The  market  price  of  the  sugar  is  11/,  and  the  market  price  of  the  rice  is  8/. 
B.  agrees  to  exchange  but  suggests  to  A.  that  each  increase  |/  per  pound  with  a 
view  to  increase  the  market  price  for  future  sales ;  to  this  proposition  A.  cheerfully 
assents,  and  tbey  exchange  at  the  increased  x>rice,  A's  sugar  at  11^/  and  B's  rice  at 
8J/,  to  the  amount  of  $9384  worth. 

By  reason  of  the  i<^  increase  of  price,  what  was  their  respective  gain  and  loss 
in  dollars;  their  gain  and  loss  in  pounds;  their  relative  gain  and  loss  per  cent; 
their  respective  real  gain  and  loss  i)er  cent,  and  their  respective  real  investment; 
and  at  what  price  should  A.  have  valued  his  sugar  in  order  not  to  have  lost  by  the 
J/  increase  on  B's  rice  ? 

Ans.  $144,  A's  loss  and  B's  gain. 

1309 /j  lbs.  sugar,  or  1800  lbs.  rice,  A.  loses  and  B.  gains. 
Its'!  %)  -^^^  relative  loss  per  cent  and  B's  relative  gain  per  cent. 
liif  %)  -^'s  real  loss  per  cent. 
lf|%,  B's  real  gain  per  cent. 
$8970,  A's  real  investment. 
$8832,  B's  real  investment. 

llje/j  -^'  should  have  valued  his  sugar,  in  order  not  to  have  lost 
by  reason  of  the  J/  increase  on  B's  rice. 


PERCENTAGE.  479 

CONTEACTED  SOLUTION. 

.00  —  lie.    =  8i>309n  lbs.  siignr,  should  exchange,  A.  sells  and  B.  receives. 
$9384.00  -:-  llic.  =  81G0O      lbs.  sugar,  did  exchange,  A.  sells  and  B.  receives. 

3709iJ^i-  lbs.  sugar,  A.  gains  and  B.  loses. 

$9384.00  ~  8c.    =  117300  lbs.  rice,  should  exchange,  A.  receives  and  B.  sells. 
$9384.00  -^  8ie.  =  110400  lbs.  rice,  did  exchange,  A.  receives  and  B.  sells. 


6900  lbs.  rice,  A.  loses  and  B.  gains. 
Then, 

6900     lbs.  rice  at  8c.       =:  $552  =  A's  loss  and  B's  gain. 
And, 

3709,V  lbs.  sugar  at  lie.  =  $408  =  A's  gain  and  B's  loss. 

$144  =  A's  net  loss  and  B's  net  gain. 
Also, 

ic  —  ll|c.  =  ji'jC.  =  A's  gain  and  B's  loss  on  1  cent  of  sugar. 
And, 

^c.  -^    8|c.  =  -jVc  =:  A's  loss  and  B's  gain  on  1  cent  of  rice. 
Then, 
■)Vc.  —  >/:iC.  =  ;j9tC.  =  A's  net  loss  and  B's  net  gain  on  Ic.  of  barter. 

AVhich  equals  IjSi  °o  =  A's  relative  loss  per  cent  and  B's.  relative  gain  per  cent. 

The  real  gain  and  loss  jier  cent  is  found  by  increasing  the  relative  per  cent  in  tlie  same  ratio 
as  the  original  jirices,  thus : 

^ai''-  X  yj"  ^  T»7<^-  ^  1t8?°o  =  ■'^'^  Teal  loss  per  cent. 

uItC  X  ^g-  =  ihc-  =  ltb°o  =  E's  real  gain  per  cent. 
Also, 

$9384  X  Aji  T=  $8976  =  A's  real  investment. 
And, 

$9384  X  j^  =  $8832  =  B's  real  investment. 
Then, 

1144.00  -^  lie.  =  1309 ,'f  lbs.  sugar,  A's  loss  and  B's  gain. 
And, 

$144.00  —    8c.  =  1800  lbs.  rice,  B's  gain  and  A's  loss. 
Finally, 

B's  ic.  increase  on  8c.  =  6}5u  ;  and  6}  per  cent  increase  on  A's  lie.  =  \lc.  which  added  to  the 
lie.  gives  lll^c.  as  the  price  that  A.  should  have  valued  his  sugar  in  making  the  exchange. 


<^^^^^ 


commission  and  Brokerage.^^^ 


892.  Commission  is  the  sum  charged  by  an  agent,  factor,  correspondent, 
broker  or  commission  merchant,  for  buying  or  .selling  goods  or  other  ^jroperty,  col- 
lecting debts,  making  advances,  or  transacting  other  business  of  a  similar  nature. 

The  sum  thus  charged  is  generally  calculated  at  a  certain  rate  per  cent  on 
the  amount  of  purchase,  sale,  collection,  advance  or  other  business  transacted. 

The  party  who  transacts  business  for  others  is  called  the  Agent,  Broker, 
Collector,  Solicitor,  Factor  or  Commission  Merchant,  according  to  the  nature 
of  the  business. 

The  party  for  whom  the  business  is  transacted  is  called  the  Principal. 

Note  1. — Commission  merchants  or  fiiotors  nre  usually  placed  in  possession  of  the  goods 
bought  or  sold,  and  transact  the  business  in  their  own  names. 

NoTK  2. — Brokers  do  not  have  possession  of  the  goods  or  securities  bought  or  sold  and 
transact  the  business  iu  the  name  of  those  who  employ  them 

893.  A  Consignment  is  a  quantity  of  goods  shipped  or  sent  by  one  person 
or  firm  to  another,  to  be  sold  on  commission  for  account  of  the  shipper. 

894-.     A  Consignor  is  a  iierson  who  ships  goods  to  be  sold  ou  commission  for 

his  account.  tr 

■=  'if  X 

895.  A  Consignee  is  a  person  to  whom  goods  a^e  shipped  to  be  sold  for 

account  of  another.  =  a   ^ 

896.  An  Account  Sales,  is  an  itemized  stateme^i,t  rendered  by  the  Agent 
or  Consignee  to  the  Shipper  or  Consignor,  showing  in  (tetiail  the  sales,  the  charges 
of  all  kinds,  and  the  net  proceeds. 

897.  The  Net  Proceeds  of  a  consignment  is  the  amount  of  money  due  the 
consignor,  after  deducting  from  the  total  sales  the  commission  and  all  other  charges. 

898.  An  Invoice  or  Account  of  Purchase  is  an  itemized  statement  rendered 
by  an  agent  to  his  i^rincipal  showing  in  detail  the  cost  of  goods  bought,  jier  his 
order,  and  all  expenses  or  charges  attending  tlie  purchase  and  shipment. 

899.  Guarantee  is  a  promise  or  obligation  by  which  the  Agent  becomes 
surety  for  the  jiaymeiit  of  goods  sold  on  credit,  or  for  the  performance  of  some  duty 
devolving  upon  others,  or  for  the  grade  or  quality  of  goods  bcmght ;  and  for  this 
responsibility  a  iiercentum  charge  is  made,  which  is  called  Guarantee. 

THE  KATE  OF  COMMISSION  OR  BROKERAGE  TO  CHAllGE. 

900.  The  rate  per  cent  commission  or  brokerage  charged  for  transacting 
business  for  others  is  usually  fixed  by  the  Chamber  of  Commerce  or  Board  of  Trade 

^■.  (480) 


COMMISSION     AND    BROKERAGE. 


481 


of  tlie  city  where  the  person  transacting  tlie  business  resides,  and  it  varies  according 
to  the  nature  and  extent  of  the  business.  A  less  per  cent  is  generally  charged  by  a 
brolier  tlian  by  a  commission  merchant,  because  the  work  he  has  to  perform  requires 
less  labor  and  time,  and  his  expenses  for  office  rent,  services,  employees,  etc.,  are 
Jess. 

Tlie  charges  made  by  brokers,  and  the  customs  regulating  the  business  of 
brokers,  are  generally  fixed  by  the  Association  of  Brokers  and  will  be  fully 
l^resented  under  the  subject  of  Stocks  and  Bonds.  We  will  here  remark  that  it  is 
the  custom  of  the  New  Orleans  brokers  to  charge  on  the  par  value  of  Gold,  Stocks 
or  Bonds,  and  not  on  the  purchase  or  selling  price,  excejit  as  follows:  on  stocks 
selling  at  and  over  $50  jier  share,  50/  jier  share  is  charged ;  on  stocks  selling  at 
$25  and  under  $50  per  share,  25/  per  share  is  charged ;  and  for  stocks  selling  below 
$25  per  share,  a  charge  is  made  as  per  agreement. 

901.  The  principles  of  percentage  apply  to  Commission  and  Brokerage  and 
the  questions  are  classed  as  follows : 

TO  FIND  THE  COMMISSION  WHEN  THE   COST  AND   SELLING  PEICE 
AND  THE  PEE  CENT  OF  C0M:\[ISSI0N  OR  BEOKERAGE 

ARE  GIVEN. 

PROBLEMS. 

1.  An  agent  sold  $2845  worth  of  goods  and  charged  2  per  cent  commission. 
What  was  his  commissiou  °l  Ans.  $56.90. 


OPERATION. 

$28.45 
0 


$5G.90  Ans. 


Explanation. — As  was  explained  in  the  first  problem  of 
percentage,  page  435,  we  lirst  get  1  per  cent  by  (lividing  by 
100,  which  is  done  by  placing  the  decimal  point  between  the 
tens  and  hundreds  figures,  then  find  2  i^er  cent  by  multi- 
plying by  2. 


2.     Sold  5  hhds.,  5418  lbs.  sugar  at  12i/,  and  charged  2i  per  cent  commission. 
What  is  my  commissiou  and  what  is  tlie  net  proceeds  ? 

Ans.  $10.93  commission.     $060.32  net  proceeds. 


OPERATION. 

i 

54180° 

I 

$677.25°  value  of  sugar. 

$  16.93    =  2J%  commissiou. 

$660.32    net  proceeds. 

Explanation. — In  this  problem,  we  first  find  the  value  of 
the  sugar,  and  then  the  2J  per  cent  of  the  same.  The 
imiltiplicatiou  of  the  5418  lbs.  by  12^  is  performed,  for 
reasons  given  on  jiages  94  and  436,  by  multiplying  by  100,  as 
indicated  by  the  two  small  naughts,  and  then  dividing  by 
8;  or  to  express  it  ditferently,  divide  by  8  and  carry  the 
division  two  places.  The  multiplication  of  $677  25  by  2J  is 
performed  by  multiplying  by  10,  as  indicated  by  the  small 
naught,  and  then  dividing  by  4,  as  explained  on  pages 
94  and  436. 

3.    A  real  estate  agent  sold  a  piece  of  real  estate  for  $28465,  and  charged  1 


per  cent  commission.    What  was  his  commission  ? 


Ans.  $284.6.5. 


482  soule's  philosophic  tractical  mathematics.  * 

4.  Collected  $2842.80,  and  charged  IJ  per  cent  brokerage.  What  is  luy 
brokerage?  Aus.  $35.53J. 

5.  Bought  20  shares  of  $100  each  of  Germania  Bank  stock,  at  15  per  cent 
preminni,  and  paid  brokerage  J  per  cent.  What  was  the  brokerage  and  the  cost  of 
the  stock?  Ans.  $10  brokerage.     $2.U0  cost  of  stock. 

Note. — Brokerage  is  charged  on  ]iar  value. 

6.  Exchanged  through  a  broker  $1280  fractional  currency  for  bills  of  a  larger 
denomination.    The  broker  charged  ^  per  cent;  what  was  the  brokerage  ? 

Ans.  $3.20. 

7.  Sold  208  shares  of  $100  each  of  National  Bank  stock  at  $92i.  Paid 
brokerage  J  per  cent.    What  are  my  net  proceeds,  and  what  is  the  brokerage? 

Ans.  $19084  net  proceeds.     $104  brokerage. 

8.  A  commission  merchant  has  10  hhds.,  12000  lbs.  of  sugar  on  commission 
and  sells  it  through  a  broker  at  8/  per  pound.  His  commission  is  2i  per  cent,  and 
the  brokerage  is  1^  per  cent.  All  other  charges  amount  to  $27.  What  are  the  net 
proceeds,  the  commission  and  the  brokerage  T 

Ans.  $36  commission  and  brokerage.     $897  net  proceeds. 


TO  FIND   THE   AMOUNT  TO  BE  INVESTED  AND    THE    COMMISSION, 
WHEN  BOTH  ARE  INCLUDED  IN  A  GIVEN  SUM. 

PROBLEMS. 

902.  1.  Received  $10000  from  a  correspondent  with  instructions  to  invests 
the  proceeds,  after  taking  my  commission  out,  in  merchandise  for  his  account.  I 
charge  2  per  cent  commission.  What  is  my  commission  and  what  amount  do  I 
invest  ?  Ans.  $196.08  commission.     $9803.92  invested. 

OPEEATION. 

$100  assumed  investment.  Explanation. — We  first  observe  that  as   coin- 

O  Ocf    finTTiTnisairvn  mission  is  allowed  OH  the  amount  invested  only, 

;;  ~      /"  commission.  ^^   ^^^^^^^^^   ralculate   on    the    $10000,    tor    that 

^  .       ,         .  .  would   be   getting   eonimission   on   the    amount 

$102  cash  required  to  invest  $100.  invested  plus  the  eommissiou  due  ns.    According- 

ly, for  reasons  ]ireviou,sIy  given  in   percentage, 
$    Invested  pages  443  and  4fi3,   we  assume  $100  to  represent 

the  amount   invested,   and   on   this   amount   we 
i/^r>      -tr^^nr,        A/^o^o  r.  Calculate  and  add  to  tlie  same   the   2   per   cent 

102      10000    ($9803.92  commission,  which  gives  |K«  as  the  cash  equiva- 

lent   of  J;100   investment  and    the    commissicm, 

for  making  the  same.  With  these  figures,  it  is 
clear  that  we  can  invest  as  many  hundred 
dollars  as  $10000  is  equal  to  $102 ;  or  philosophic- 


ally, placing  the  $100  assumed  investment  on  the  statement  line,  we  reason  thus:  If  $102  cash  will 
invest  $100,  $1  will  invest  the  102d  part,  and  $10000  will  invest  10000  times  as  much,  which  is 
$9803.92.  On  this  amount  we  are  allowed  commission  which  at  2  per  cent  is,  to  the  nearest  unit  of 
a  cent,  |196.08,  The  commission  may  also  be  found  by  deducting  the  investment  from  the  Avhole 
amount. 


*  '  COMMISSION    AND    BROKERAGE.  483 

2.  A  country  merchant  purchased  $8450  worth  of  merchandise,  ou  4  months' 
time.  He  then  proposes  to  pay  §5000  cash  on  condition  that  he  he  allowed  5  per 
cent  discount  ou  the  amount  of  bill  that  he  pays,  which  proposition  is  accepted. 
What  amount  does  he  still  owe  1  Ans.  $318G.S4. 

3.  The  net  proceeds  of  a  consignment  due  a  consignor  is  $2511.25,  which  I 
am  instructed  to  invest,  after  deducting  commission,  in  sugar  at  10  cents  per 
pound.  I  charge  2i  per  cent  commission.  How  many  pounds  of  sugar  can  be 
purchased  ?  Ans.  24500  lbs. 

4.  A  wine  dealer  remitted  to  a  merchandise  broker  in  New  York  $1965.60  to 

be  invested  in  claret.    The  broker  charges  2^  jjer  cent  commission  for  investing 

and  2J  per  cent  for  guaranteeing  the  quality,  both  of  which  are  to  be  deducted  from 

the  $1965.60.    How  much  money  was  invested  in  wine,  and  how  many  boxes  were 

there,  the  price  per  box  being  $6.50 !  Ans.  $1872  investment. 

288  boxes. 


TO   FIND   THE   AMOUNT  INVESTED,   OR  TOTAL  SALES,    WHEN    THE 
COMMISSION  AND  THE  RATE  PER  CENT   COMMISSION 

ARE   GIVEN. 

PKOBLEMS. 

903.     1.    An  agent's  commission  for  selling  goods  was  $56.90,  the  rate  per 
cent  was  2.     What  was  the  amount  or  gross  sales  1  Aus.  $2845. 

OPERATION. 
^100  assumed  sales.  '  Explanatio,,. ~In   all   problems   of  this    kind, 

'^  '"  assume  $100  and  find  the  commission  or  hroker- 

$2.00  commission  on  $100.  '»g<^  th  reon.     Then  with  the  relationshij.  num- 

bers thus  obtained,  make  a  proportional  state- 
ment as  shown  in  the  operation,  reasoning  thus : 
Since  $2  commission  requires  $100  sales,  $1 
commission  will  require  the  \  part,  and  $56.90 


100 
56.90 


$2845.00  Ans.  will  require  56.90  times  as  much. 

2.  A  commission  merchant  received  $27.88  commission  for  selling  merchan- 
dise, and  guaranteeing  the  payment  thereof.  The  rate  of  commission  was  5  per 
cent  and  the  rate  of  the  guarantee  was  3  per  cent.     What  was  the  amount  sold  ? 

Ans.  $348.50. 

3.  A  broker  received  $71.25  for  selling  gold  at  \  per  cent  brokerage.  What 
amount  did  he  sell  ?  Ans.  $28500, 


484 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TO  FIKD  THE  RATE  PER  CENT  COMMISSION  OR  BROKERAGE  WHEN 

THE  COMMISSION  AND  TUB  AMOUNT   OP  SALES,  COLLECTIONS 

OR  INVESTMENTS  ARE  GIVEN. 

PROBLEMS. 

904.     1.     An  agent  collected  $3100  and  his  commission  was  $85 ;  wliat  was 

Alls,  -SY/o. 


the  rate  of  commission  ? 

FIRST   OPERATION. 

100%  rate  of  coiniiiission  assnmed. 
100%  of  $3i00  is  $3400. 


% 


3400 


100 

85 


IhJ/c  Ans. 

SECOND   OPERATION. 


3400 


85 
100 


Explanation. — In  this  solution,  we  assume  100 
per  cent  uuil  find  IDO  i)er  cent  of  the  amount. 
Then  ■with  the  relationship  numbers  thus  obtain- 
ed, we  make  a  proportional  statement  as  shown 
in  the  operation,  reasoning  thus:  Since  $3400 
require  100  per  cent,  $1  will  require  the  3400th 
part,  and  $85  will  require  85  times  as  much. 

Ex]>Ianation. — In  this  second  solution,  we 
reason  as  follows:  Since  the  commission  on 
$3400  is  |85,  the  commission  ou  $1  would  be  the 

3400th  part,  and  ou  illOO,  it   would  be  100  times 

J2.J  =  2J%.  .as  much. 

2.  A  broker  received  $180.25  for  collecting  $7210.  What  per  cent  broker- 
age did  he  charge  ?  Ans.  2.J%. 

3.  A  commission  merchant  bought  per  order,  150  barrels  flonr  at  $8  a 
barrel.  He  paid  drayage  and  freight  .'i<42.  Tlie  wliole  invoice  cost  $1279.20.  What 
was  his  commission  and  rate  per  cent  of  comiiiission  '! 

Ans.  $37.20  commission.     3%  rate  of  commission. 

4.  An  attorney  collected  $2210  and  sent  me  $2099.50.  What  rate  per  cent 
did  he  charge?  Ans.  5%. 

5.  Paid  $002.16  taxes  on  property  assessed  at  $23100.  What  was  the  rate 
percent?  Ans.  2.0%. 

TO  FIND  THE  COMMISSION  OR  BROKERAGE  AT  ANY  RATE  PER 
CENT  ON  ENGLISH  CURRENCY. 


905.     1.    What  is  4  per  cent  of  £5800  IGs.  7d.  ? 


FIRST    OPERATION. 

£58.00  10s.  7d. 

i'/o 


£234.40  .64  .28 
20 


8  .C4s. 
12 


7  .96d. 


SECOND    OPERATION. 

£58.60  IGs.  7d. 

4% 

£234.43    6    4 
20 


Ans.  £234  8s.  8d. 

THIRD   OPERATION. 
£5800  16s.        7d. 
5  4i 

.80+  .0291 

See  page  450  for  reducing 
shillings  and  pence  to  the  deci- 
mal of  a  pound. 

£5860.829 

4% 

£234.43316 
20 

8.66320s 
12 

7.9584d 
Explanation. — In  the /rs(  operation,  we  reasoned  as  follows:  1  per  cent  of  £5860  168.  7d.  is 


8  .06s. 
12 


7  .96d. 


*  COMMISSION    AND    BROKERAGE.  485 

£58,60  .16s.  .07d..  and  4  per  cent  is  4  times  as  much,  -which  is  in  the  first  operation  £234.40  .64s. 
and  .28d.  These  hundredths  of  pounds,  shillings  and  pence  ive  then  reduce  to  shillings  and  pence 
according  to  the  principles  of  denominate  numbers,  as  shown  in  the  operation. 

In  the  Sfconrf  operation,  -ne  first  found  1  per  cent  in  the  same  manner  as  in  the  first  oper.a- 
tinn,  hilt  when  we  ninltii)lie(l  by  4,  we  reduced  the  .2Sd.  to  hundredths  of  shillings  by  dividing  by 
11',  which  gave  ns  .02s.  and  .04d.  We  then  miiltiplicd  the  .16s.  by  4,  and  ju-oduced  .64s.,  to  whici 
we  added  the  .02s.  and  obtained  66s  ,  which  we  reduced  to  hundredths  of  pounds  by  dividing  by 
20,  which  gave  ns  .03  pounds  and  ()6s.  We  then  reduced  the  .43  pounds  .06  shillings  and  .04  pence 
to  shillings  and  jience,  as  explained  in  the  first  operation. 

In  the  IJiird  operation,  we  first  reduce  the  shillings  and  pence  to  the  decimal  of  a  pound  and 
computed  4  jier  cent  on  the  whole  amount.  We  then  reduce  the  decimal  of  the  pound  to  shillings 
and  pence. 

2.  What  is  34  per  cent  of  £78  4.s.  lOd.  !  Ans.  £2  14s.  Od. 

3.  What  is  1  per  cent  of  £2(i0  l!».s.  3d.  ?  Ans.  10s.  Od.  2f. 

4.  What  is  2J  per  cent  on  £42G.  lis.  7d.  ?  Ans,  £1013s.  3d. 

Note  1.— To  compute  2^  per  cent,  first  compute  it  at  5  per  cent  and  then  divide  by  2.  Or 
compute  it  at  100  per  cent  and  divide  by  40,  as  2+  per  cent  is  /u.  Or  find  10  per  cent  and  divide  by 
4,  as  2i  IS  i  of  10. 

Note  2. — See  Interest  on  English  Money. 

MISCELLANEOUS  PROBLEMS. 

906.  1.  What  is  the  commission  for  selling  $1620.80  -worth  of  goods  at  2i 
percent?  Ans.  $40.52. 

2.  A  commission  merchant  invested  $5050  for  a  planter,  and  charged  1  pei 
cent  commission.     What  was  his  commission  ?  Ans.  $50.50. 

3.  A  mer<!hant  for  -whom  I  do  business  remits  me  $5050  ^vith  instructions  to 
invest  the  proceeds,  after  deducting  1  per  cent  commission,  in  certain  goods.  What 
is  my  commission  ?  Ans.  $50. 

4.  r.onght  $5000  gold  at  $112^,  and  sold  the  same  at  15^  per  cent  premium. 
Paid  ^  per  cent  brokerage  for  purchasing  and  J  per  cent  for  selling.  How  much 
did  I  gain?  Ans.  $112.50  currency. 

5.  Bought  $5000  of  bank  stock  at  $112^,  and  sold  the  same  at  10:^  per  cent 
premium.  Paid  J  per  cent  brokerage  for  purchasing  and  ^  per  cent  for  selling. 
How  mucli  did  I  lose?  Ans.  $137.50  currency. 

C.  Invested  through  a  broker  $10825  in  U.  S.  Bonds  at  108  per  cent.  Paid 
^  per  cent  brokerage.    What  was  the  brokerage,  and  what  the  face  of  the  bonds  ? 

Ans.  $25  brokerage.     $10000  face  of  bonds. 

OPERATION. 

$100    par  value  of  bonds  assumed.  $  bonds. 


8    =8%  iiremium. 
^  =  ifo  brokerage.  433 


100 

4 

10825 


$108J  =  cost  of  $100  bonds. 

$10000  bonds. 

25  brokerage. 

7.  A  broker  collected  $14812.75  and  re-invested  the  proceeds  in  securities  at 
par.  He  charged  according  to  custom  J  per  cent  for  collecting  on  the  amount 
collected,  and  ^  per  cent  for  investing  on  the  amount  invested.  What  was  his 
brokerage?  Ans.  $147.39+, 


486 


SOULE  S    PHILOSOPHIC    rRACTICAl,    MATHEMATICS. 


8.  The  commission  for  selling  books  is  25  per  cent  and  an  agent  received 
$'235.     What  was  the  amount  of  sales?  Ans.  $940. 

9.  An  auctioneer  paid  me  $2954.80  net  proceeds  for  goods  sold.  The  charges 
for  selling  were  $28.1i0  aud  his  commission  5  per  cent.  What  was  the  amount  of 
sales?  Ans.  $3140. 

10.  A  salesman  sold  $6837  of  clothing  upon  which  there  is  a  jiroflt  of  33i^ 
per  cent,  of  which  he  received  10  per  cent  commission.     What  was  his  commission? 

A  us.  $227.90. 

11.  A  salesman  sold  $4800  worth  of  goods  and  received  $240,  commission. 
What  was  the  rate  of  commission  I  Ans.  5%. 

12.  A  commission  merchant  received  $2000  to  be  invested  in  supplies.  He 
charges  2i  per  cent  commission  for  investing,  1  per  cent  is  allowed  for  charges  on 
the  sui)i">lies,  and  2  per  cent  for  insurance  on  cost  i)lus  10  per  cent  of  cost.  What 
sum  was  invested  in  supplies,  and  how  nuich  was  Ins  commission  ? 

Ans.  $1891.30  invested  in  supplies.     $47.70  —  commission. 

OPERATION. 


=  Ist  cost  assumed. 
1  =  1])q  charges. 


$101    =  cost  of  supplies. 
2^  =  ^A  commissiou. 


$101  =  cost  of  supplies. 
10  =  ^d'  of  cost  of  supplies. 


$10.10  =  10;^  of  cost. 

101        =  cost  of  supplies  added. 


Illl.lU  =  cost  -t-  10%. 
2  =  9jf  insurance. 


Aggregate  cost  of  $100  supplies. 


$100,000  : 
1.000  : 
2.525  : 
1  qoo  . 


:  cost  assumed. 
:  charges. 
:  couiniission. 
: insurance. 


105.747 


$2.2220  =  2%  insurance. 
$ 

100  =  cost  assumed. 

2000.000 


$105,747  =  cost  to  l)uy   and 
ship  $100  supplies. 

RECAPITULATION. 

$1891.30  =  1st  cost  of  supplies. 
18.91  =  l^V  charges. 
47.76  =  2i%  commission. 
42.03  =  2%  insurance.    ■ 


$1891.30  . 

18.91  : 


:  invested  in  supplies. 
!%■  charges. 


$1910.21  =  cost  and  charges  of  supplies. 
2i  =  %  commission. 


$2000.00  =  sum  received. 


$47.755J  commission. 

OPERATION  TO   FIND   INSURANCE. 

$1910.21  =  cost  and  charges  of  supplies. 
191.02  =  10%. 


'$2101.23  =  sum  insured. 
2  :=  %  insurance. 


credit,   goods 
credit,   goods 


$42.0246  =  insurance. 

13.  A  retailer  bought  from  a  wine  merchant,  on  3  mouths' 
amounting  to  $1500,  and  from  a  grocery  mer(;hant,  on  3  months' 
amounting  to  $1800.  l>ut,  witliout  waiting  for  the  maturity  of  the  bills,  he  pays 
the  wine  merchant  cash  in  full  less  5  per  cent  disecnint  allowed,  and  to  the  grocery 
merchant  he  i)ays  $1000  cash,  for  wliich  he  is  to  be  allowed  5  ])er  cent  discimnt  on 
the  amount  of  invoice  that  it  will  pay.  Uow  much  did  he  pay  the  wine  merchant? 
And  how  much  does  he  still  owe  the  grocery  merchant? 

Ans,  $1425,  he  paid  the  wine  merchant. 

$747.37,  he  owes  the  grocery  merchant. 

14.  A  merchant  received  180  bales  of  cotton  which  average  450  pounds  to 
the  bale.  The  cotton  was  sold  at  16§/^  per  pound,  and  2J  per  cent  commission  and 
J  per  cent  brokerage  were  charged  for  selling.  Allowing  ^  jier  cent  loss  on  the 
cotton  received  for  sampling,  classing  and  loose,  and  approximating  the  other 
charges  at  $750.20,  what  amount  of  goods  could  be  purchased  with  the  net  pro- 
ceeds, after  deducting  2A  per  cent  commission  on  the  face  of  the  money  invested  ? 

Ans.  $12000.83. 


lankiruptcy. 


907.  Baukruptcy  is  a  failure  iu  business  and  tlie  condition  of  being  insol- 
vent or  unable  to  i>ay  indebtedness. 

Note. — The  word  liankmpt  is  of  Italian  origin  and  is  derived  from  the  custom  in  many  towns, 
in  the  middle  ages,  of  breaking  the  counter  or  bench  (baucus),  in  the  public  exchange  or  market 
place,  that  had  been  occupied  by  merchants  who  had  subsequently  failed. 

90S.  A  Bankrupt  or  Insolvent  is  a  person  or  business  man  wlio  breaks  or 
fails,  or  becomes  unable  to  pay  his  debts  in  the  ordinary  or  regular  course  of  trade. 

909.  An  Assignee  is  a  person  that  is  elected  by  the  creditors  of  the  bauk- 
rupt,  or  ai)pointe(l  by  the  court,  to  receive  the  assets  of  tlie  bankrupt,  pay  the 
expenses  of  the  assignment,  and  use  the  proceeds  of  the  property  in  paying  the 
creditors  of  the  bankruiit,  as  far  as  the  money  will  allow,  according  to  law  and 
equity. 

910.  An  Assignment  is  the  transfer  of  the  property  of  a  bankrupt  to  the 
assignee. 

911.  A  Schedule  is  a  list  of  all  the  assets  and  liabilities  of  the  bankrupt 
containing  the  names  and  the  place  of  business  or  residence  of  all  his  debtors 
and  CREDITORS,  and  the  amount  due  from  or  to  eacli. 

912.  A  Debtor  is  a  person  or  firm  who  owes  the  bankrupt. 

913.  A  Creditor  is  a  persoii  or  firm  whom  the  bankrupt  owes. 

914.  The  Assets  or  Resources  of  a  bankrupt  are  the  entire  property  except 
such  as  may  be  exempt  by  law. 

915.  The  Liabilities  are  what  the  bankrupt  owes  at  the  time  of  his  failure. 

916.  Assets  that  can  be  converted  into  cash  at  their  stated  valuation  are 
called  AVAILABLE  ASSETS ;  and  those  that  cannot  be  immediately  converted  into 
cash  without  a  discount  or  rebate  are  called  nominal  assets. 

917.  Preferred  Creditors  are  persons  whom  the  law  allows  to  collect  their 
claims  in  full,  such  as  operatives,  and  employes  to  whom  limited  sums  are  due  for 
services  rendered  within  a  few  months  of  the  assignment. 

918.  A  Dividend  is  the  sum  or  per  cent  paid  to  the  creditors  by  the  assignee 
out  of  the  assets  of  the  bankrupt.  Dividends  of  a  part  of  the  funds  of  a  bankrupt 
are  often  made  before  the  final  settlement  of  the  estate. 

919.  A  Discliarge  is  an  order  or  decree  by  a  court  of  bankruptcy,  evidenced 
by  a  certificate  therefrom,  that  the  bankrupt  has  complied  with  all  the  requirements 
of  law  governing  cases  of  bankruptcy,  and  is  therefore  released  or  discharged  from 
all  his  debts. 

(487) 


488 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


PROBLEMS. 

1.  The  assets  of  a  bankrupt  amount  to  $203413.39.  The  liabilities  aniouut 
to  $S7l'5-t(J.  Estimating  that  tlje  exjjenses  for  settling  the  estate  will  amount  to 
$738r),  and  allowing  the  assignee  2  ^ler  cent  on  the  full  amount  of  assets,  what  per 
ceut  will  the  bankrupt  estate  pay?  Aus.  22%. 

OPERATION. 

$203413.39  assets. 

1  14.j3.27  expenses  and  assignee's  com. 


$191900.12  net  proceeds. 


872-540.00 


191900.12 
100 


22%    Ans. 


Explanaiion. — From  tlie  total  assets,  we  deduct 
tlie  expense's  and  assignee's  charges,  and  thus 
obtain  the  net  proceeds.  We  then  observe  that 
$872546,  the  liabilities,  .ire  to  receive  $1!)1960.12, 
and  to  fin<l  tlie  late  per  cent  that  $191960.12  is 
of  $872546  we  place  the  net  proceeds  on  the 
statement  line,  and  reason  thus:  I f  |872,')46  00 
are  to  receive  .$191960.12,  $1  -will  receive  the 
872546th  part,  an<l  $100  will  receive  100  times 
as  much,  which  is  22  per  cent.  Or,  we  may 
first  a.ssnme  and  i>lace  ou  our  st.itenicnt  line  100 
per  ceut,  and  reason  as  iu  Articles  856  and  857. 


2.  A.  and  B.  failed.  Their  assets  were  $0720,  liabilities  $30750,  expenses 
$250.  The  assignee  charged  5  j)cr  cent  on  the  assets.  What  was  his  commission 
and  what  per  cent  did  the  creditors  receive?  Ans.  .S336  commission. 

19.947+%. 

3.  In  the  above  problem,  what  would  have  been  the  assignee's  commission, 
and  the  per  cent  the  creditors  would  have  rec('i\ed  if  the  assignee  had  charged  5 
per  cent  on  the  amount  he  paid  to  creditors  and  for  expenses? 

Ans.  $320  commission.     20%  received  by  creditors. 

OPERATION. 


105 


100 
0720 


$0400 

250  expenses. 


30750 


$0400  paid  out. 

5% 


$0150  ijaid  to  creditors. 


0150 
100 

20%  paid  to 

creditors. 


$320.00  commission. 


4.  X.  and  Y.  failed.  The  net  proceeds,  after  deducting  $440  expenses  and 
2J  per  cent  cf)mmission  of  the  assignee,  were  $7,355.  Tlie  liabilities  were  $47970. 
What  ])«■  cent  will  the  creditors  receive  and  what  was  the  assignee's  commission 
on  the  total  amount  of  resources?  Ans.  15.74%.     $205  commission. 


7555 
440 

$7995 


97^ 


100 
7995 


OPERATION. 

8200 

2J%  commission. 


47970 


$8200  total  assets.        $205.00  commission. 


100 


15.74% 


5.  A  firm  becomes  bankrupt;  its  liabilities  amount  to  $340180,  and  its 
assets  amount  to  $73210.58.  Allowing  tlie  assignee  5  per  cent  commission  on  the 
first  $1000,  2J  per  cent  on  the  next  $5000,  tind  1  per  cent  on  the  remainder  of  the 
assets,  and  estimating  the  other  expciises  at  $2810,  what  rate  per  cent  will  the 
estate  p;iy,  and  how  nuich  will  a  creditor  receive  to  whom  141319.05  are  due? 

Ans.  20.4478  +  %.     $8448.95  +,  the  creditor  receives. 

0.  A  baiikrupt  estate  paid  a  dividend  of  IS.J  per  cent.  The  amount  received 
by  a  creditor,  to  whom  ^  of  the  liabilities  were  due,  was  $7400.  What  were  the 
liabilities  of  the  estate  ?  Ans.  $320000. 


laxes  and  Licenses. 


920.  Taxes  are  sums  of  money  assessed  according  to  law  on  persons, 
property,  incomes,  or  products,  for  the  use  of  the  nation,  State,  county  or  parish, 
incorporate  city,  or  society. 

921.  A  Poll  Tax  or  Capitation  Tax  is  a  certain  sum  assessed  on  persons 
without  regard  to  property.  It  is  generally  assessed  by  the  State  on  the  legal 
voters. 

922.  A  Property  Tax  or  Direct  Tax  is  a  certain  sum  assessed  on  the 
estimated  value  of  real  and  personal  properiy.  Tliis  tax  is  usually  a  certain  per 
cent  or  a  specified  number  of  cents  or  mills  on  $100.  In  some  cities  and  States  the 
rate  per  cent  tax  is  limited. 

923.  An  Indirect  or  Excise  Tax  is  a  tax  on  articles  of  consumption  la 
their  ti-ansfer  from  one  person  to  another.  It  is  also  levied  on  licenses  to  pursue 
certain  trades  and  to  deal  in  certain  commodities. 

92-t.  Real  Property  or  Real  Estate  consists  of  houses,  lands  and  all 
immovable  property. 

925.  Personal  Property  consists  of  money,  merchandise,  stocks,  mortgages, 
tiotes  receivable,  furniture,  live  stock,  etc.,  etc. 

926.  An  Assessor  is  a  person  appointed  to  appraise  the  value  of  property 
subject  to  taxation. 

927.  A  Collector  or  Receiver  of  Taxes  is  a  public  or  government  officer, 
Who  has  been  elected  or  apj)ointed,  to  collect  or  receive  taxes. 

928.  An  Assessment  Roll  is  a  list  or  schedule  containing  the  names  of  all 
persons  liable  to  taxation  in  the  State,  parish,  or  city  for  which  tlie  tax  is  assessed, 
and  the  valuation  of  each  person's  taxable  property.  The  assessment  roll  must  be 
prepared  before  making  an  assessment  of  taxes. 

929.  A  License  is  a  special  tax  or  charge  imposed  by  the  national.  State,  or 
municipal  authorities,  upon  merchants,  bankers,  manufactiirers,  and  nearly  all 
classes  of  business  men,  for  the  privilege  of  iirosecuting  their  various  occnpations 
and  businesses  within  the  limits  of  the  authority  imposing  the  tax  and  granting  the 
license.  A  proper  and  equitable  assessment  of  this  tax  requires  much  wisdom  and 
experience,  a  faithful  regard  to  duty,  and  unalloyed  honesty. 

930.  U.  S.  Income  Tax  is  at  present,  1895,  an  assessment  made  by  the 
general  government  on  all  incomes  exceeding  $1000,  with  special  exceptions. 

NoTK. — The  Income  Tax  lias  been  declared  unconstitutional  by  the  U.  S.  Sn]irenie  Conrf. 

931.  The  Budget  is  a  financial  statement  of  items  showing  the  amount  of 
taxes  to  be  raised.  It  is  also  a  financial  statement  with  business  men,  showing  the 
items  of  expenditure  of  the  business  for  the  current  or  succeeding  fiscal  year. 

(489) 


490 


SOULE  S    PHILOSOPHIC    PRACTICAL.    MATHEMATICS. 


933.  Tlie  method  of  assessing  taxes,  tbougli  in  some  respects  slightly 
different  in  some  States  and  cities,  is  nevertheless  virtually  the  same. 

In  making  an  assessment  of  taxes  the  first  thing  in  order,  after  the  assess- 
ment roll  has  been  prepared,  is  to  ascertain  the  amount  of  tax  required  to  be  raised, 
to  ■which  amount  must  be  added  the  expenses  to  collect  the  same,  and  the  sum  or 
per  centum  alhnved  as  being  uncollectible.  From  the  sum  total  of  these  three- 
amounts  are  to  be  deducted  any  special  tax,  the  amount  of  poll  tax,  or  the  amount 
of  license  tax,  or  all  according  as  the  assessment  is  one  for  a  Stiite,  jtarish.  or  city. 

Instead  of  increasing  the  amount  of  tax  to  be  raised  by  the  per  centum 
allowed  for  non-collectible  tax,  we  may  deduct  the  same  per  cent  IVom  the  taxable 
property  or  assessment  roll,  and  use  the  remainder  as  the  amount  to  calculate  the 
rate  per  cent  ni)on. 

Having  these  facts  and  figures,  we  then  find  the  per  cent  of  tax. 

PROBLEMS. 

933.  1.  The  budget  for  the  estimated  expenditures  and  liabilities  of  a 
State,  for  the  coming  year,  amounts  to  $45(10000.  The  inventory  or  assessment  roll 
amounts  to  $325000000.  Allowing  $600000  to  be  received  for  licenses  and  $50000 
from  polls,  and  estimating  that  10  per  cent  of  the  assessment  will  be  uncollectible, 
and  that  5  per  cent  will  be  required  to  pay  the  expenses  for  collection,  what  will  be 
the  rate  per  cent  of  taxes  on  the  assessment  roll  ?  Ans.  1.3855  -f  %. 

OPERATION. 


$4500000  amount  of  ImdRet. 

GuOOOO  licenses  and  polls  deducted. 

4i385OO00  amount,  of  not  tax  to  be  raised. 

Statement  to  find  tlio  amount  in  be  assessed  to  allow 
10  Der  cent  tor  nou-eoilections. 


90 


95 


100 
3850000 

$4277777.77  total  amount  to  be  assessed. 

Statement  to  tiud  tbe  aniomit  to  be  assessed  and 
allow  5  ijer  cent  t'uinmission  tor  collecting. 

$ 

100 

4277777.77 


Explanation. — From  the  amount  of  the  budget, 
we  first  deduct  tbe  amount  of  the  license  and 
poll  taxes,  which  leaves  .t  balance  of  $3850000. 
We  theu  find  the  amount  to  be  assessed  in  order 
to  allow  10  per  cent  for  non-collectible  tax. 
To  do  this  wo  first  assume  $100  to  represent  the 
amount  to  bo  assessed,  from  which  we  deduct  the 
10  per  cent  allowed  for  non-collections,  and 
have  in  the  remainder  $90,  the  net  collections  as 
the  equivalent  of  $100  tax  to  be  assessed.  We 
then  reason  thus:  If  $90  cash  collections  require 
$100  tax  to  be  assessed,  $1  will  require  the  90th 
part,  and  $.3850000  will  require  38,50000  times  as 
much.  This  gives  us  $1277777.77  as  the  amount 
of  tax  to  be  assessed  iu  order  to  allow  10  per  cent 
for  iion-colli'ctions  and  still  realize  $3850000 
cash.  Wo  next  find  the  amount  to  be  assessed 
in  order  to  allow  5  per  cent  commission  for 
collection  charges.  To  do  this  "^^o  again  assume 
$100  to  represent  the  amount  to  be  assessed,  and 
I'riim  this  we  deduct  the  5  per  cent  commission, 
and  obtain  $95  cash  as  the  equivalent  of  $100 
assessment ;  then  placing  the  $100  assumed 
assessment  on  the  statement  line,  we  reason 
thus:  If  $95  ca.sh  in  order  to  pay  5  per  cent 
commission  require  $100  assessment,  $1  cash 
will  reciuiro  the  95th  part,  .ind  $4277777.77  will 
require  4277777.77  times  as  much.  This  gives 
ns  ihe  amount  of  tax  to  be  assessed,  which  is  $4502923.97.  Having  now  the  total  amount  of  tax  to 
1)3  assessed,  and  the  total  amount  of  ])ioperty  as  shown  by  the  assessment  roll  on  which  the  assess- 
ment is  to  be  made,  we  can  find  the  rate  per  cent  of  tax.  To  do  this  we  first  observe  that 
$325000000  are  to  pay  or  produce  $4502923.97,  and  to  ascertain  the  rate  per  cent  we  place  the 
$4502923.97  on  thestatement  line  and  reason  thus :  If  $325000000  are  to  pay  or  produce  $4502923.97, 
|l  will  pay  or  produce  the  325000000th  part,  and  $100,  100  times  as  much.  "  The  result  of  this  stato- 
icent  is  tbe  correct  answer  of  the  problem,  carried,  to  four  decimals. 
In  practice  If  per  cent  or  l^^^  per  cent  would  probably  be  used. 


$4502923.97  total  amount  to  be  assessed. 
Statement  to  find  tbe  rate  per  cent  tax. 


325000000 


4502923.97 

100 

1.3855  +  ^o"  Ans. 


*  TAXES    AND    LICENSES.  491 

2.  The  taxable  property  of  a  village  amounts  to  $924000.  Tbe  tax  to  be 
raised  including  collection  cbarges  amounts  to  $23492.  Allowing  that  $5000  will  be 
received  for  licenses  and  si^ecial  taxes,  and  making  no  allowance  for  non-collections, 
wliat  is  the  rate  per  cent,  and  how  much  will  be  the  tax  of  Mr.  A.,  whose  real  and 
personal  property  amounts  to  $15380?  Ans.  2%,  rate  of  tax. 

$307.00  Mr.  A's  tax. 

3.  The  budget  for  the  estimated  expenditures  and  liabilities  of  a  city  are  as 
follows : 

EXPENDITURES. 

Salaries  of  city  officers,  clerks,  collectors,  etc.,              .....  $454,580 

"Wages  of  street,  wharfage,  drainage,  etc.,  departments,              -        -  424,000 

Office  expenses,                -         -                 29,900 

Printing,          ....                 50.000 

General  service,               -         -                 1,282,700 

Contingent,              50,000 

Deficit  of  previous  year, 475,000 

Total  estimate  of  expenditures, $2,772,180 

LIABILITIES. 

Bonds  maturing  and  interest  on  same,  -        -         -         $1,808,508 

7,^  %  certificates  to  redeem, 800, G(X) 

Special  bonds,  to  be  paid, 70,000 

Total  amount  of  liabilities, $2,739,108 

lotal  estimate  of  exi^endituies  and  liabilities,  -        -        $5,511,288 

The  assessment  roll  for  real  and  personal  property  of  the  city,  on  which 
the  tax  is  to  be  assessed,  amounts  to  $139,848,204.  Allowing  10  per  cent  for  non- 
collections,  and  estimating  that  the  receipts  from  new  issues  of  bonds,  from  the 
markets,  wharves,  licenses,  and  all  other  revenues  of  the  city  will  equal  the  amount 
of  liabilities,  what  will  be  the  rate  per  cent  tax  ?  Ans.  2.2025  +  %. 

4.  I  am  assessed  as  follows :  Eeal  estate  $18600.  Personal  property  $4000 
Money  at  interest  $2500.  Poll  tax  $2.  The  rate  per  cent  for  city  tax  is  2  per  cent 
and  for  the  State,  6  jier  mille;  there  is  also  a  special  city  tax  of  45  mills  on  every 
hundred  dollars.  How  much  is  my  tax,  for  each,  the  city.  State  and  special  taxes, 
and  what  was  the  total  amount  ?  Ans.  $502.00    city. 

11.29J  special. 
150.60    State. 
2.00    poll. 


$665.89J  total  amouiit. 


49^ 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


5.  A  citizen  paid  $502.00  for  City  taxes,  including  $2  poll  tax  and  the  rate 
per  cent  was  2.     What  was  Lis  property  assessed  for  1  Ans.  $25000. 

G.  If  a  property  liolder  pays  $003  for  city,  State  and  special  taxes,  and  the 
rate  for  City  is  2A  per  cent,  for  State,  ^  i>er  cent,  and  for  special  05  mills  on  the 
hundred  dollars,  what  is  the  assessed  value  of  his  property?  Ans.  $20000. 

7.  A  town  desires  to  raise  a  tax  of  2J  mills  on  the  dollar.  There  are  700 
polls  at  $1,  tlie  personal  and  real  property  is  assessed  at  $1,400,000.  What  is  the 
rate  of  tax  and  the  amount  of  tax  of  a  citizen  whose  property,  personal  and  real, 
is  assessed  at  $11550 1  Ans.  .2%  rate  of  tax. 

$21.10  tax  of  citizen,  including  i>oll  tax. 

934.  To  facilitate  the  operation  of  apportioning  taxes  among  a  large  number 
of  persons,  the  tax  on  $1,  $2,  $3,  etc.,  at  the  ascertained  rate  per  cent  may  be  found 
first  and  arranged  as  in  the  following  table : 


TABLE   AT   THREE   ANB   ONE-HALF  MILLS   TO   THE  DOLLAR. 


VALUE 

AMOUNT 

VALUE 

AMOUNT 

VALUE 

AMOUNT 

VALUE 

AMOUNT 

OK, 

OK 

OF 

OF 

PROPERTY. 

OF  TAX. 

PKOPERTY. 

OK  TAX. 

PKOPEUTY. 

OF  TAX. 

PROPERTY. 

OF  TAX. 

$  1 

.0035 

$  10 

.035 

$  100 

.35 

$  1000 

$  3.50 

<■> 

.007 

2(1 

.07 

200 

.70 

2000 

7.00 

3 

.0105 

30 

.105 

300 

$1.05 

3000 

10.50 

4 

.014 

40 

.14 

400 

1.40 

4000 

14.00 

5 

.0175 

50 

.175 

500 

1.75 

5000 

17.50 

6 

.021 

()0 

.21 

COO 

2.10 

GOOO 

21.00 

7 

.0245 

70 

.245 

700 

2.45 

7000 

24.50 

8 

.028 

SO 

.28 

800 

2.80 

8000 

28.00 

!l 

.0315 

90 

.315 

900 

3.15 

9000 

31.50 

10 

.035 

100 

.35 

1000 

3.50 

10000 

35.00 

old,  Currency,  Silver  and  Uncurrent  Money. 


^ 


935.  Premium  is  the  per  cent  tliat  an  amount  is  increased. 

936.  Discount  is  tlie  iier  cent  that  an  amount  is  decreased. 

937.  Currency  is  some  circulating  medium  representing  money,  used  as  a 
substitute  for  gold  or  the  fixed  measure  of  value. 

938.  Uncurrent  Money  is  some  circulating  medium  that  is  valued  less 
thau  gold,  or  less  than  the  fixed  measures  of  value.  The  coin  of  some  foreign 
countries  is  also  classed  as  uncurrent  money. 

When  one  currency  is  at  a  certain  per  cent  premium  over  another,  the  per 
cent  discount  on  the  currency  of  lesser  A'alue  is  not  the  same  as  the  i)er  cent 
premium  on  that  of  greater  value;  thus,  when  gold  is  25  per  cent  premium,  the 
corresponding  discount  on  currency  is  but  20  jier  cent;  and  Trhen  gold  is  200  per 
cent  premium,  $100  in  currency  is  worth  $33J  iu  gold,  or  GCg  i)er  cent  discount. 


TO  FIND   THE    COEEESPONDING   DISCOUNT   ON   PAPER   CUEEENCY 
OE  ON  SILVEE  WHEN  WE  HAVE  THE  PEEMIUM  ON  GOLD. 


939.     1.     If  gold  is  25  per  cent  i)reniium,  what  is  the  corresponding  discount 
on  currency!  Ans.  20%. 

FIRST    SOLUTION. 


OPERATION. 

$100  gold  assumed. 
25  =  25%  premium. 


$125  currency  value  of  $100  gold. 

$  Gold. 
100 


125 


100 

$80  gold  value  of  $100  currency. 
100 


$20  =  20%  discount,  Ans. 


Explanation. — Here  we  are  constrained 
to  assume  some  niimher  to  work  from. 
Accordingly,  for  re:is()ns  often  given  in 
similar  questions,  we  assume  $100  gold 
and  to  the  same  a<Iil  tlie  25  per  cent 
premium,  and  liy  tliis  work  we  obtain 
§125  as  tlio  equivalent,  currency  value  of 
$100  gol<l,  and  having  these  equivalent 
values  we  place  the  $100  gold  on  the 
statement  line,  and  reason  thus:  If  $125 
currency  is  worth  $"00  gold,  $1  currency 
is  wortli  the  ll'51h  part,  and $li:0 currency 
instead  of  $1  is  worth  lOOtinjes  as  much. 
The  result  of  this  reasoniug  and  state- 
ment,gives  us  $80,  as  tlie  gold  value  of 
$100  currency,  which  deducted  from  $100 
currency  gives  20  per  cent  discount  on 
currency. 


(493) 


494 


SOULE  S    rillLOSOPHIC    PRACTICAL,    MATHEMATICS. 


SECOND    SOLUTION. 


OPERATION. 

$100  gold  assumed. 
2o  =  2.j%  preiniiun. 


$ll.'5  currency  value  of  $100  gold. 


125 


silver  ? 


Explanation, — The  first  part  of  this 
solution  is  the  same  as  that  given  in  the 
first  solution,  so  far  as  ascertaining  the 
.  equivalent  value  in  currency  of  $100  gold 
at  25  per  cent  premium.  Then  instead  of 
finding  the  value  of  $100  currency,  we 
find  the  per  cent  at  once  hy  placing  the 
25  per  cent  on  the  statement  line,  and 
reasoning  thus:  If  $125  currency  give  25 
])er  cent  disco\int,  $1  currency  will  give 
the  125th  part,  and  $100  will  give  100 
times  as  much.  It  will  be  observed  that 
25  per  cent  premium  on  $100  gold  is  the 
corresponding  discount  on  its  eqtiivalent 
$125  currency,  which  the  result  shows  to 
be  20  per  cent. 

Gold  is  12J  per  cent  premium;  wliat  is  the  corresponding  discount  on 

Ans.  11^%. 


25 
100 


20%  discount,  Ans. 


3.  Gold  is  150  per  cent  i)remium;  what  is  the  corresponding  discount  on 
pajier  currency  ?  Ans.  G0%. 

4.  Gold  is  500  per  cent  premium;  what  is  the  corresponding  discount  on 
paper  currency  ?  Ans.  83J%. 


TO    FIND    THE     COREESPONDING    PEEMIUM    ON    GOLD    WHEN    WE 
HAVE  THE  DISCOUNT  ON  CURRENCY  OR  ON  SILVER. 


940.     1.     If  currency  is  20  per  cent  discount,    what   is   the   corresponding 
premium  ou  gold?  Ans.  25^, 

FIRST    SOLUTION. 
OPERATION. 


$100  currency  assumed. 

20  =  209^  discount  deducted. 


80 


$80  value  of  $100  currency  in  gold. 

$  Cur. 
100 

100 


$125      value  of  $100  gold  in  currency. 
100      deducted  from  same  gives. 


$25  =  25%  premium,  Ans. 


Explanation. — In  the  solution  of  this 
problem,  we  first  assume  $100  currency 
and  then  find  the  equivalent  value  of  the 
same  at  20  per  cent  discount  in  gold. 
Having  these  two  values,  we  reason 
thus;  If  $80  gold  is  equal  to  $100  ciir- 
rency,  $1  gold  is  equal  to  the  80th  part, 
and  $100  gold  is  equal  to  100  times  a3 
much.  The  result  of  this  reasoning  gives 
$125  currency  as  the  value  of  $100  gold. 
Then,  by  taking  the  difference  between 
this  value  and  $100,  we  have  the  25  per 
cent  premium  on  gold,  which  corresponds 
to  or  is  the  equivalent  of  20  per  cent  dis- 
count on  currency. 


GOLD,     CURRENCY,    SILVER    AND     UNCURRENT    MONEY. 


495 


SECOND  SOLUTION, 
OPERATION. 

currency  assvinied. 
20  =  20%  discount  deducted. 


SO  value  of  §100  currency  in  gold. 


SO 


20 
100 
25%  premium,  Ans. 


Explanation. — In  this  solution,  ■n-e  first 
find  tlio  equivalent  value  in  gold  of  $100 
curreniy  at  20  jier  cent  discount,  in  the 
same  manner  as  in  the  tirst  solution.  We 
then  jilace  the  20  per  cent  discmuit  on  the 
$100  currency  on  our  statement  line,  and 
reason  thus :  If  880  gold  give,  20  ])er  cent 
premium,  $1  gold  will  give  the  80th  part, 
and  $100  gold  will  give  100  times  as 
much.  It  will  be  observed  that  20  per 
cent  discount  on  $100  currency  is  the 
corresponding  premium  on  (its  equiva- 
lent) $80  g(dd,  which  the  result  shows  to 
he  25  per  cent. 


2.  The   discount   on   silver   is   40   per    cent;    what    is    the    corre.sponding 
premium  on  gold  ?  Ans.  UOg^  . 

3.  The  discount  ou  currency  is   90  per  cent;    what   is   the   corresponding 
premium  on  gold  ?  Ans.  900%. 

4.  If  the  discount  on  currency  is  99i^j  per  cent,  what  is  the  corresponding 
premium  on  gold ?  Ans.  10000%. 

"1.     If  the  discount  on  ciu-rency  is  100  per  cent;  what  is  the  corresponding 
premium  on  gold  1  Aus.  Infinite. 


TO    FIND    THE    COREESPONDING    PREMIUM    OiT    CURRE^^CY    OK    OX 
SILVER  WHEiST  WE  HAVE  THE  DISCOUNT  OX  UXCURRENT  MONEY. 


941.  In  comparing  gold  and  currency,  or  currency  and  uncurrent  money, 
currency  is  by  the  custom  of  bankers  and  money  dealers  used  as  the  standard  unit 
of  value. 

1.  When  the  old  paper  money  of  the  city  of  New  Orleans  was  at  10  per 
cent  discount,  what  was  the  corresx)ondiug  premium  on  currency  'I 

Ans.  lli%  premium. 

OPERATION. 

$100  currency  or  silver  assumed. 
10  =  10%  discount. 


$90  value  of  $100  uncurrent  money. 


90 


10 
100 


Explanation. — The  reasoning  for  this 
problem  is  the  same  as  the  second  solution 
of  the  preceding  question.  Hence  it  is 
omitted. 


11^%  premium,  Ans. 

2.  When  uncurrent  money  is  50  per  cent  discount,  what  is  the  corresponding 
premium  ou  currency  or  on  silver?  Ans.  100%  premium. 

3.  The  discount  on  uncurrent  paper  money  is  5  per  cent;  what  is  the  cor- 
responding premium  ou  current  money  ?  Ans.  5i%%. 


49^  SOULE's    rillLOSOFHIC    rRACTICAI.    MATHEMATICS.  * 

TO  FIND  THE  VALUE  OF  GOLD  IN  CUREENCY  OR  IN  SILVER  OR  TO 
FIND  WHAT  A:M0UNT  OF  CUIIRENCY  OR  OF  SILVER  IT 
WILL  REQUIRE  TO  PAY  A  CERTAIN  DEBT  IN 
GOLD,  WHEN  THE  GOLD  IS  AT  A  CER- 
TAIN PER  CENT  PREMIUM. 

942.     1.     I  owe  a  gold  debt  of  $2000.     Gold  is  15  \wr  cent  premium.     What 
amount  of  currency  Avill  it  require  to  pay  the  debt  ?  Ans.  $2300. 

OPERATION. 

$2000  Explanation. — It  is  clear  that  as  the  premium  is  on 

1  jrj  the  gold,  $100  gold   is   equal   to   $115   currency,    aud 

Leuce  it  will  reijuire  as  many  times  $115  as  $2000  is 

equal  to  $100,   which   is  20;    or   as   is  shown    in    the 

$2300.00  Alls.  operation,  we  multii)ly   by   the   2000   and   divide   by 

100;  or  wo  may    reason  thus:    When  gold  is   15  per 
cent  premium,  $1  is  equal  to  $1.15  currency,  and  $2000  is  equal  tu  2000  times  as  much. 

2.  Gold  is  12J  per  cent  premium;    wliat  is  $0810  wortli  in  currency  or  in 
silver?  Ans.  $7005. 

3.  Gold  is  185  per  cent  premium;  wliat  amount  of  currency  ■will  it  require 
to  pay  a  gold  debt  of  $18500  ?  Ans.  $52725. 


TO  FIND  THE  VALUE  OF  GOLD  IN  CURRENCY  OR  IN  SILVER  WHEN 
THE  DISCOUNT  ON  CURRENCY  OR  ON  SILVER  IS  GIVEN. 

943.     1.     Currency  is  25  per  cent  discount.     AVbat  is  $2000  gold  worth  in 
currency?  Ans.  $26GG2. 

OPERATION. 


75 


$        ■                              $100  currency  assumed. 
100    currency.                 25  =  25%  discount. 
2000  


$75  =  gold  value  of  $100  currency. 


$266C|  Ans. 

Currency  is  12J  iier  cent  discount;  Avhat  is  $3072  gold  Avorth  in  currency? 

Ans.  $4196.571. 


TO  FIND  THE  VALUE  OF  CURRENCY  OR  OF  SILVER  IN  GOLD  WHEN 
THE  DISCOUNT  ON  CURRENCY  OR  ON  SILVER  IS  GIVEN. 

944.     1.     Currency  is  25  per  cent  discount  below  gcjkl ;  what  is  $2000  wortb 
in  gold  ?  Ans,  $1500. 

OPERATION. 

25%  of  $2000  is  $500     $2000  —  $500  =  ^1500. 

2.     Silver  is  10  per  cent  discount  below  gold ;  what  is  $1755  worth  in  gold  T 

Ans.  $1579.50. 


GOLD,    CURRENCY,    SILVER    AND    UNCURRENT    MONEY. 


497 


TO   riKD  WHAT  AMOUNT  OF  GOLD  IT   WILL  EEQUIEE   TO  PAY  A 

CERTAIN  DEBT  TS  CUEEENCY,  OE  IN  SILVEE  WHEN  THE 

GOLD  IS  AT  A  CEETAIN  PEE   CENT  PEEMIUM. 

915.     1.    I  owe  a  cui-reucy  debt  of  $2000.    Gold  is  15  per  cent  premium. 
What  amount  of  gold  -will  it  require  to  pay  the  debt?  Ans.  $1739.13/3. 


OPERATION, 

$100  gold  assumed. 
15  =  15%  premium. 


$115  currency. 


115 


$  Gold, 
100 

2000 

$1739,13-1  Aus. 


Explanation. — In  the  solution  of  this  problem,  Tve  first 
assume  $100  gold,  aiul  then  add  thereto  15  per  cent  of  the 
same  and  thus  obtain  $1115  as  tlie  value  in  curreucy  of  $100 
gold;  or,  in  other  words,  we  find  by  these  equivalent  values 
that  $100  gold  will  pay  |113  currency  debt.  With  these 
ratio  figuri'S  we  make  the  sohitioii  statement,  and  placing 
the  §100  gold  ou  the  line,  we  reason  thus:  If  $115  curreucy 
is  equal  to  $100  gold,  $1  currency  is  equal  to  the  115th  part, 
and  $2000  currency  is  equal  to  2U00  tinu's  as  much  ;  or  thus : 
If  $100  gold  will  pay  a  currency  debt  of  $115,  then  by 
trausjiosition  $1  currency  debt  will  require  the  115th  part  of 
$100  gold,  aud  $2000  currency  debt  will  require  2000  times  as 
much.     The  result  is  $1739.13-/j  gold. 


2,  Gold  is  12 J  i)er  cent  premium  ;  what  amount  will  it  require   to   pay   a 
silver  debt  of  $.581.15?  Ans.  $519.21i. 

3,  A  merchant  owes  iu  curreucy  a  debt  of  $21294.10,  which  he  pays  in  gold 
at  1083.    What  amount  of  gold  will  pay  the  debt?  Ans,  $195S0.7Si4. 

4,  A  currency  debt  of  $5200  is  i)aid  iu  gold  at  IGO  per  cent  premium.     How 
much  gold  is  required  ?  Ans,  $2000. 


MISCELLANEOUS  PEOBLEMS  IN  GOLD,  CUEEENCY,  SILVEE  AND 

UNCUEEENT  MONEY. 


916.     1.     Gold  is  33J  per  cent  ijremium ;  what  is  the  corresponding  discount 
on  currency?  Ans,  25%. 

2.  Currency  is  33J  per  cent  discount;  what  is  the  corresi>onding  premium 
on  gold?  Aus,  50%. 

3.  The  discount  on  uncurrent  paper  money  is  12J  per  cent;    Avhat   is  the 
coiTesi)ouding  premium  on  currency  ?  Ans.  14f%. 

4.  Gold  is  14f  per  cent  premium ;  what  is  $10000  worth  in  currency  ? 

Ans.  $18340. 

5.  Gold  is  14f  i)er  cent  premium;  what  amount  will  it  require   to   pay   a 
currency  debt  of  $18340  ?  Ans.  $10000, 

6.  Gold  is  50  per  cent  jiremium;   how  much  silver  can  I  buy  with   $1000 
gold?  Ans.  $1500, 

7.  Gold  is  50  per  cent  premium;   how  much  gold  can   I   buy   with   $1000 
silver  T  Ans,  $CC6.6G§, 


498 


SOULE  S    rillLOSOPIIIC    TRACTICAL    MATHEMATICS. 


8.     UiiciuTeiit  money  is  1(1  per  cent  diseoui.t,  ;u.<l  gold  is  10  per  eent  ])ieiniuui. 
"Wnat  anunmt  of  gold  eau  1  buy  lor  $ll.'000  uneunenr,  ui'niey  'i      Aus.  $l)81SJ8/j. 


$12000  iincurrent  inonej', 
1200  =  lU'/p  discount. 


$10800  currency. 


110 


opi;ration. 

100 
10800 

$9818.18  Sf 


or  tlms : 


100 
110 


90 

100 

12000 


$981S.18-iV 


9.  now  mucli  uncun-ent  money  at  37i  i)er  cent  discount  can  I  buy  with  $00 
currency?  Ans.  $S0. 

10.  How  much,  uncurrent  money  at  12.J  per  cent  discoujit  can  be  bought  for 
$4500  gold  at  OJ  per  cent  premium  ?  Ans.  $561S.57J. 

11.  Uncurrent  money  is  5  per  cent  discount,  and  a  young  dealer  supposiug- 
that  currency  was  the  same  rate  per  cent  prenuum,  sold  $10000  currency  at  5  per 
cent  ijremium.     What  was  his  loss,  and  his  loss  per  cent  ? 

Aus.  $2C.31ii,  uncurrent  money  loss.     -0)%,  loss  ou  current  money. 


OPERATION. 


95 


100 
100 

.$105^^      value  $100  cnrroTit  money. 
100 


$10000 


5 

100 


^A'o  premium  ou  current  money. 
5    ^Q  discount. 


>^ft"ii  premium. 
5    ^^  discount. 


1.31}^  loss  in  un- 
current 
money. 


l^a  "o  ^'^^^  "^  cuiTeut  money. 


Or  thus, 


$10000    currency. 
5 


95 


$    500.00  premium. 
10000.00 


100 
10000 


$10526. 31|^  amount  of  uncurrent  money  he  should  have  received. 
10500.00 


$10500.00  uncurrent  money 
received. 


26.31J4  loss  in  uncurrent  money. 


■  12.  A  broker  purchased  from  one  class  of  customers  uncurrent  money  at  5  f^ 
discount,  and  ])ai(l  current  money  ;  and  from  anotlier  class  of  customers  lie  jmrcliasecL 
current  money  at  5'/,  i)reniiuiii,  siiid  ]iaid  uiiciurenL  money.  During  the  day  he 
jmrchased  $l(i()(iO()  of  each,  current  nnd  uncurrent  money.  What  was  his  gain, 
and  gain  %  t'"  tli<-'  purchase  of  uncurrent  money  I  Wlint  was  his  loss,  and  loss  fp 
on  the  purchase  of  current  money  f  and  liow  i  mli  did  t!:e  ]iarties  lose,  who  sold  the 
current  money,  by  reason  of  their  ignorance  <if  t.o  <'(iii-eit  jninciplesof  ])ercentage'? 

Arts.  1.  The  brolvcr  gained  15000  in  uiicui-ii-nt  money  ]>nrih;ise(l,  luid  gained  S,'''^,",,'  on  the 
current  money  jiuid  out. 

Note. — When  uncurrent  money  is  S^n"  discount,  currciit  money  is  5-i^y  5o  premium. 

Ans.  Z.  The  liroUcr  lost  §5000  in  uncurrent  iiioiiey  p;iiil  I'm-  currency,  and  lost  i^i^^i  on  (he 
uncurrent  money  ]iaid  out. 

Note. — When  uncurrent  money  is  41 'f  ,"„  discount,  current  numey  is  v>%  premium. 

Ans.  3.  The  jmrties  who  sold  the  current  money  lost  $2G3.15i§  in  uncurrent  money,  or  $250  in. 
current  money  by  reason  of  their  ignorance  of  jicrccutage  calculations. 

Note. — These  parties  should  have  received  $105000  uncurrent  money  for  $00750  current  money,, 
or  $105263. 15 1  §  for  §100000  current  money. 


GOLD,    CURRENCY,    SILVER    AND     UNCUKRENT    MONEY. 


499 


13.  A  broker  purchased  $100000  of  uucurreut  money  at  10  per  cent  discount, 
for  which  he  i)aid  current  money.  Then  with  the  $100000  uncurrent  money  he 
bought  current  money  at  10  per  cent  premium.  "What  were  his  net  profits  and  his 
gain  per  cent  ?  Ans.  $909.09  i-  gain  in  current  money. 

Ws)%  gain  on  currency  invested. 


OPERATION. 


$100000  unciirreut  money. 

10000  =  10,%'  discount  deducted. 

$90000  cost  in  current  money. 


UNCUR.    MONEY. 

110 


11 

90000.00 


9  GAIN. 

10000.00 

100 

^9%  gai".  Ans. 


$   CUR.    MONEY. 

100 

UNCUR.   MONEY. 

100000 

.S90909.09i\  current  money. 
90000.00      current  investment. 


$909.09i'i  gain  on  current  investment. 


125 


100 
.20 


16/    gold  price,  or  when  gold  is  at  par. 
2f/  =  15%  premium. 


li.  The  selling  price  of  coffee  is  20/?  when  gold  is  25  per  cent  premium; 
what  should  it  sell  for  when  the  jiremium  on  gold  is  15  per  cent  ?  Aus.  ISf/. 

OPEEATION. 

$  Explanation. — In    this     solu- 

tion, we  first  tiud  tlie  gold 
value  of  tbe  20e.  currency, 
which  is  16c.  The  reasoning 
for  this  work  was  fully  explain- 
ed in  Article  945,  iiage  497  ; 
then  having  the  value  when 
gold  i.i  at  ]iar,  to  liud  the 
value  in  currency  when  gold  is 
15  ]icr  cent  jircmiura  we  simply 
add  the  15  per  cent,  as  shown 
in  the  operation,  and  have  in 
the  result  the  selling  price 
desired. 

15.  Wlien  gold  is  10  per  cent  premium,  goods  sell  for  $2.75;  what  should 
they  sell  for  when  gohl  is  worth  132  1  Aus.  $3.30. 

IC.  If  goods  are  worth  $4.50  when  gold  is  150,  Avhat  are  they  worth  when 
gold  is  at  par?  Aus.  $3. 

17.  If  goods  are  worth  .^i  in  curreucy,  what  are  they  worth  in  uncurrent 
money  which  is  10  per  cent  discount?  Ans.  $4.41^. 

18.  The  gold  price  of  goods  is  $2 ;  what  are  they  worth  in  uncurrent  money, 
the  iiremium  on  gokl  being  12^  per  cent,  and  the  discount  on   uncurrent  money 


ISf/  selling  price  when  gold  is  15%  premium. 


being  12J  per  cent  ? 

FIRST   OPERATION. 

$ 

2.00 
87J 


Aus. 


112J 


JSJL'.O  ( , 


Or, 


$2.00 
25 


ii.IiO 


SECOND   OPERATION. 

$ 

100 
2.25 


S7J 


Ans. 


$2,574,  Ans. 


aistomhouse  Business. 


917.  Custom  Houses  are  buildings  or  offices  erected  or  established  by  the 
national  goveninient,  wlieie  the  collection  of  duties  or  customs  on  goods  imported 
is  made,  and  where  vessels  are  entered  and  cleared,  etc. 

948.  Duties  or  Customs  are  taxes  levied  on  certain  imported  goods  for  tlio 
benefit  of  the  general  government  and  the  protection  of  certain  kinds  of  home 
industry. 

Note. — In  some  countries,  customs  or  taxes  are  charged  liotli  on   imported  and  exported 
goods.     The  govcrunient  of  the  United  States  charges  customs  only  on  goods  imi>orted. 

There  are  two  kinds  of  duties,  specific  and  ad  valorem. 

919.  Specific  Duty  is  a  tax  levied  on  the  -weight  or  measure  per  ton,  bale,  i 
pound,  gallon,  or  yard,  without  regard  to  its  cost  or  value.  ; 

950.  Ad  Talorem  Duty  is  a  certain  per  cent  levied  or  assessed  on  the  j 
actual  cost  of  the  goods  in  the  country  from  which  they  were  imported,  as  shown  i 
by  the  invoice  or  the  appraised  value  in  the  absence  of  an  invoice.  \ 

Note. — On  certain  kinds  of  goods  hoth  a  specific  and  an  ad  valorem  duty  is  assessed.  ' 

951.  Au  Invoice  or  Manifest  is  an  itemized  account  of  the  goods  imported  i 
or  shipped  to  a  purchaser,  agent,  factor,  etc.,  with  the  actual  cost  or  value.    Eegard- 
ing  foreign  invoices,  the  law  requires : 

1.  That  all  invoices  of  imported  merchandise  shall  be  made  out  in  the  cur. 
rency  of  the  place  or  country  from  whence  the  importations  shall  be  made,  or  if  pur- 
chased in  the  currency  actually  paid  therefor,  shall  contain  a  correct  descrijitiou 
of  such  merchandise,  etc. 

2.  That  all  such  invoices,  shall  at  or  before  the  shipment  of  the  merchandise  , 
be  proditced  to  the  consul,  vice-consul,  or  commercial  agent  of  the  United  States  of 
the  consular  district  in  which  the  merchandise  was  manufactured  or  purchased  as  , 
the  case  may  be,  for  expoit  to  the  United  States,  and  shall  have  indorsed  thereon, 
when  so  produced,  a  declaration  signed  by  the  purchaser,  manufacturer,  owner,  or  : 
agent,  setting  forth  that  the  invoice  is  in  all  respects  correct  and  true,  and  was 
made  at  the  place  from  which  the  merchandise  is  to  be  exported  to  the   United 
States;  that  it  contains,  if  the  merchandise  is  to  be  obtained  by  purchase,  a  true 
and  full  statement  of  the  time  when,  the  place  where,  the  person  from  whom  the 
same  was  purchased,  and  the  actual  cost  thereof,  and  of  all  charges  thereon,  as 
provided  by  this  act;  and  that  no  discounts,  bounties,  or  drawbacks  are  contained 
in  the  invoice,  but  such  as  have  been  actually  allowed  thereon,  etc. 

(500) 


*  CUSTOMHOUSE    BUSINESS.  5oi 

3.  That  no  different  invoice  of  the  merchandise  mentioned  in  the  iuvoice  so 
produced  has  been  or  will  be  furnished  to  any  one. 

The  most  of  invoices  are  made  out  iu  the  weights  and  measures  of  the 
country  from  wliicli  the  goods  are  imported. 

952.  Tariff  is  a  list  of  goods  alphabetically  arranged,  with  the  rates  of 
duties  and  drawbacks,  established  by  the  laws  of  the  United  States. 

953.  The  Free  List  comprises  such  goods  as  are  imported  free  of  duty. 

954.  Tonnage  Duty  is  a  tax  levied  per  toTi  on  vessels  for  the  privilege 
of  entering  ports. 

955.  Ports  of  Entry  are  places  where  customhouses  are  established,  and 
it  is  lawful  to  leceive  goods  into  a  country  only  at  ports  of  entry. 

NoTK  1. — There  are  several  kinds  (if  entries,  such  as,  entry  of  a  vessel  from  a  foreign  port 
■with  passengers,  entry  of  merchandise  for  immediate  use,  bonded  warehouse  entry,  re-warehouse 
entry,  withdrawal  entry  for  consumption,  or  exportation,  etc.,  etc. 

Note  2. — The  forms  for  the  various  entries  are  generally  furnished  hy  the  customhouse  or 
the  custom-house  broker. 

956.  Bonded  Warehouses  are  buildings  in  which  imported  goods  on  which 
duties  have  not  been  jJaid  are  placed  under  bond  to  the  general  government  by  the 
importer,  until  he  may  wish  to  put  them  on  the  market  or  export  them. 

Note  1. — Goods  may  he  withdrawn  from  bonded  warehouses  for  export,  without  the  payment 
of  duties. 

Note  2, — Goods  n^maining  in  a  bonded  warehouse  more  than  one  year,  are  liable  to  an  .addi- 
tional duty.  And  goods  remaining  over  three  yeiirs  are  considered  as  abandoned  to  the  govern- 
ment. 

957.  Smuggling  is  the  act  of  bringing  foreign  goods  into  this  country 
without  paying  duties  on  them.  This  is  done  1,  By  not  entering  them  at  the 
Customhouse,  and  2,  By  showing  less  than  tlieir  real  value  in  the  invoice.  It  is  a 
crime  for  the  jireveiition  and  iiunishment  of  which  stringent  laws  have  been  enacted 
and  a  strict  supervision  is  exercised  by  the  general  government. 

958.  A  Clearance  is  a  certificate  given  by  the  collector  of  a  port  after  the 
requirements  of  law  have  been  complied  with,  that  the  vessel  has  been  properly 
entered. 

959.  Duties  are  collected  at  the  port  of  entry  by  a  Customhouse  officer 
appointed  by  the  United  States  government,  under  the  title  of  collector  of  the  port. 

Note.- — Duties  are  not  computed  on  the  fractions  of  a  dollar ;  when  less  than  50c.  they  are 
rejected;  if  50c.  or  more,  they  are  counted  as  a  dollar. 

The  collector  of  the  port  appoints  deputy  collectors,  appraisers,  weighers, 
gaugers,  inspectors,  etc.,  and  supervises  all  entries  and  papers,  estimates,  duties,  and 
receives  all  moneys  aiid  securities. 

960.  A  Naval  Officer  is  appointed  at  the  more  important  posts,  whose  duty 
it  is  to  receive  copies  of  all  manifests,  to  countersign  all  documents  issued  by  the 
collector,  to  certify  his  estimates  and  accounts,  etc.,  etc. 

961.  The  Surveyor  superintends  the  employees  of  the  collector,  inspects 
vessels  and  cargoes,  and  revises  all  entries  and  permits.  He  is  personally  respon^ 
sible  to  the  collector  of  the  port. 


502  SOULe's    rHILOSOPIIIC    PRACTICAL    MATHEMATICS.  * 

962.  The  Appraiser  is  an  ofQcer  whose  duty  it  is  to  examine  all  imported. 
goods  and  to  determine  tlieir  dutiable  value  so  tbat  ad  valorem  duty  may  be  charged. 

9(53.     The  Store-keeper  is  an  ofBcer  who  has  charge  of  the  warehouse. 

964:.  A  Customhouse  Broker  is  a  person  who  is  thoroughly  acquainted 
with  Customhouse  laws,  and  acting  under  power  of  attorney,  makes  entries,  secures 
permits  and  transacts  business  for  importers,  for  which  service  he  makes  a  charge. 

965.  Internal  Revenue  is  the  revenue  derived  from  the  sale  of  public 
lands  and  postage  stamps,  and  from  taxes  on  certain  home  manufactured  articles, 
such  as  distilled  and  malt  liquors. 

The  laws  prescribing  the  Internal  Eevenue  are  called  excise  laws,  iu 
contradistinction  to  the  Tariff  Laws,  which  prescribe  Import  Duties. 

966.  Excise  Duties  are  taxes  or  licenses  for  the  manufacture  or  sale  of 
certain  articles  consumed  at  home  or  in  their  transfer  from  one  person  to  another. 

967. .  A  Drawback  is  money  refunded  for  imjjort  duties  which  has  been 
previously  paid.  When  goods  on  which  duty  has  been  paid  are  exported,  the 
amount  of  duty  thus  paid  is  refunded. 

968.  Allowances  are  deductions  made  in  accordance  with  law  before 
estimating  the  duties. 

969.  Gross  Weight  is  the  whole  weight  of  the  goods,  including  the  weight 
of  the  hogshead,  barrel,  box,  bag,  etc.,  that  contains  them. 

970.  Net  AVeight  is  what  remains  after  all  allowances  have  been  deducted. 

971.  Leakage  and  Breakage.  By  Act  of  Congress  of  1894,  all  allowance 
for  leakage  and  breakage  was  repealed. 

972.  Draft  is  the  allowance  made  for  waste  in  weighing  goods. 

973.  Tare  is  the  allowance  made  for  the  weight  of  the  box,  crate,  barrel,  or 
bags  containing  the  goods. 

The  long  ton  of  2240  pounds  and  112  lbs.  to  the  cwt.  is  invariably  enqiloyed 
iu  the  Customhouses  of  the  United  States. 

The  United  States  Kevenue  Laws  have  been  often  changed  during  the  past 
few  years  and  are  still  agitating  Congress  and  the  people  of  the  country  more  than 
almost  any  other  question  of  government.  Tariff  or  Free  Trade,  High  or  Low 
Tariff,  are  questions  demanding  the  serious  attention  of  our  best  law-makers, 
statesmen,  and  financiers. 

TROBLEM. 

974.  1.  What  is  the  specific  duty  on  1.^0  casks  of  alchohol,  each  cask  CO 
gallons  capacity,  but  which  when  gauged,  are  found  to  be  240  gallons  short  of  full 
capacity  ?    What  is  the  duty  at  15/  per  gallon  ?  Ans.  §1314. 

OPERATION. 
150  X   60  =  9000  gallons,  gross  quantity.  Explanation.— in     all      oases 

9000  gals.  —  240  gals,  short  measure  =  '"  «■.''''''  !'";  ^^I'^cific  duty  is 

„„„,.  "^    ,  .  ^   ...  reqniri'd,  find  the  net  nuaiititv 

Si60gals.net  quantity.  .,„j   c^ujpme   the   duty    there- 

8760  gals.  X  15/  duty  =  $1314  specific  duty.  on. 

Note  1. — To  find  the  net  quantity  of  liquids  in  casks,  the  law  requires  that  they  shall  he 
ganged  hy  a  C'ustomhnu.se  gauper. 

Note  2. — To  tind  the  net  quantity  of  liquids  in  bottles,  the  hottles  must  he  counted  and 
certified  hy  a  Customhouse  appraiser. 


CUSTOMHOUSE    BUSINESS. 


503 


TO  FLSD  AD  VALOEEM  DUTY. 

PEOBLEM. 

975.     1.    Wbat  is  the  ad  ralortni  duty  on  an  invoice  of  12  dozen  Geneva 
watches,  amounting  to  21S00  francs,  if  the  duty  is  25  pei  ceut  T        Aus.  $1051.75. 

OPERATION. 

21S00  X   10. oi*   r=   84207.40  ETplanaiion. — In  nllcases  in  wbicb  niJ  rnJorfm  tlnty  is 

4^0Qy   w   •ijr'    ^  *1051  75  required,  tiuil  the  uet  value  and  multiply  the  same  by 

''    ~  -    /C  V  •  ^jj^  ^.^^^  ^f  duty. 

Note  1. — The  40c.,  in  the  net  valne,  bemjr  less  than  50c.,  are  not  used  in  xhe  Custombonse, 
■when  computiu<i  ihity. 

XoTE  2. — The  iutrinsic  and  Customhouse  value  of  a  franc  is  19.3  cents.  See  Table  of  Foreign 
Monev. 


TO  FIND  THE  EATE  OF  DUTY  WHEN  THE  NET  VALUE  AND  AMOUNT 

OF  DUTY  AEE  GIYEX. 

PEOBLEMS. 

976.     1.    Tbe  net  value  of  an  invoice  is  §1260  and  the  duty  is  $441.    What 
was  the  rate  ?  Aus.  35%. 

OPEEATIOXS   INDICATED. 


12G0 


100%  rate. 

441  or. 


1260 


441 
100 


2.  A  merchant  imported  from  London,  England,  a  case  of  ^voolen  goods. 
The  cost  of  the  goods  was  £310  5s.  6d;  commission  2J  per  cent.  Freight  £3  4s. 
Consul  fees  16s.  Sd.  The  weight  of  the  case  was  640  pounds  gross;  allowing  10 per 
cent  for  tare,  what  is  the  amount  of  duty,  the  specific  being  50 f  per  pound  and  the 
adra/ore»M  35per  cent?  Aus.  $S29.S0. 


OPERATION. 

£       s.       d. 
310        5        6      cost  of  goods. 

7      15        2  =  2i%  commission  on  cost  of  goods. 


EipJanatioti. — Commis- 
sion is  by  I.-1W  a  part  of 
the  dutiable  cost,  but 
freight  and  Consul  fees 
are  "not.  The  intrinsic 
and  the  Customhouse 
value  of  the  i"  is  $4.86t>5. 
See  Table  of  Foreign 
Money 

The  71  cents  in  the 
$1547.71  beinsi  more  than 
50,  $1  is  added  to  the 
dollars.  niakin<r  $1548 
dutiable  cost.  To  tiuil 
the  commission  on  £,  s.  d.  and  to  reduce  £.  s.  and  d.  to  the  decimal  of  a  £,  see  page  484. 

Note. — The  Tariff  Law  of  1894,  reduces  the  duty  on  woolen  goods  to  40  per  cent  on  goods 
valued  at  not  over  50c.  per  poand,  and  to  50  per  cent  on  goods  valued  at  more  than  50c,  per  pound. 


31 S        0        8     total  duti.able  co.<t. 

£318.0333  X  S4.8665  =  81547.71  =  dutiable  cost  in  dollars. 

81548  X  35%  =  8541.80  ad  ralorem  duty. 

640  lbs.  —  64  lbs.  tare  =  576  lbs.  weight  of  goods. 

576  X  50 f  =  8288  specific  duty. 

f541.S0  +  8288  =  8829.80  total  duty. 


5o4  soule's  riiiLOsopHic  practical  mathematics.  *■ 

3.  Iiuported  from  Autwcip  o(J00  meters  of  Brussels  carpet,  27  iuclies  wide  at 
4.50  francs  per  meter.  The  specific  duty  is  44/  per  sq.  yd.  and  the  ad  valorem  35 
per  cent.     Allowing  39.37  in.  to  a  meter,  wliat  was  the  duty"? 

Ans.  $2393.66. 

PARTIAL    OPERATION. 

3600m  X  4.50  francs  =  10200  francs. 

10200  fr.  X  19.3  =  .$3120.00  dutiable  value. 

$3127  X  35%  =  $1094.45  ad  valorem  duty. 

3C00m  =  3937  yds.     3937  yds.  27  inches  wide  =  29523  sq.  yjg. 

29523  sq.  yds.  ^  44/  =  $1299.21  specific  duty. 

$1094.45  +  1299.21  =  $2393.06  total  duty. 

Note. — The  specific  duty  ou  Brussels  carpets  w.is  repealed  liy  the  Tariff  Hiil  of  1894.  aud  tbe 
ad  valorem  duty  increased  to  40  per  cent, 

4.  Imported  from  Geneva,  Italy,  20  blocks  of  marble  each  block  measuring 
8x4x2  ft.  The  duty  is  50/  per  cubic  foot.  The  total  invoice  is  33700  lire.  What 
is  the  duty?  Ans.  $640. 

Note. — A  lire  is  tlie  s.ame  in  value  as  a  franc,  19.3  cents. 

5.  Imported  from  Hamburg,  Germany,  an  invoice  of  toys  amounting  in 
dutiable  value  to  12200  marks.  What  is  the  duty  at  45  per  cent  ou  6000  marks  and 
30per  cent  ou  0200  marks?  Ans.  $1085.28. 

Note. — The  intrinsic  and  cnstDUiliouse  value  of  a  mark  is  23.8  cents.     See  Table  of  Foreign 
Money. 

6.  The  dutiable  value  of  an  invoice  of  leather  from  Vienna,  Austria,  is  5850 
crowns.     What  is  the  duty  at  20  per  cent  rtrf  rrt/ore)»?  Ans.  $237.51. 

Note. — The  value  of  a  crown  is  20.3  cents. 

7.  What  is  the  duty  on  449650  pounds  of  \l.  II.  iron  at  -/y/  per  pound  ? 

Ans.  $1573.77J. 

8.  Imported  from  Havana  COm  cigars,  weighing  900  lbs.,  total  invoice  in 
American  money  is  $1500.  Under  tlie  Revenue  Law  of  1894,  the  specific  duty  is  $4 
per  lb.  gross,  and  the  ad  valorem  duty  is  25  per  cent.  What  is  the  duty,  what  was 
the  cost  of  the  cigars  per  m,  in  Havana,  and  what  was  the  importing  cost,  not 
counting  freight  and  insurance? 

Ans,  $3975.00  duty.     $25  per  m,  Havana  cost.     $91.25  per  m,  importing  cost. 


CUSTOMHOUSE    BUSINESS. 


5o5 


977.  AN  INVOICE  OF  COFFEE  IMPORTED  FROxM  KIO  JANEIRO,  BRAZIL. 

KoTE  — This  iuvoice  IS  made  in  tUe  mouetary  unit  aixl  iii  the  iiuit  of  weight  of  the  Uuited 
States,  which  has  been  the  custom  since  March,  1890 

Invoice  of  1997  bags  of  cofl'ee  shipped  by  Hernandez  Brothers  &  Co.,  on 
board  the  Briti.sh  Str.  Hassell,  for  New  Orleans. 

For  accoiuit  and  risk  of  whom  it  may  concern  and  consigned  to  order. 


i-  B 

L  1/9 

N  10/21 
O  21/27 
P  28/43 
K  44/49 
S  50/54 
T  55/60 


38,645 

"  1686c. 

20,043 

•'  1631c. 

91,962 

••  1594c. 

25,152 

"  1558c. 

28,034 

"  1533c. 

31,571 

"  1496c. 

200  bags  cofiee  \vg.  26,200  lbs.  ®  1741c.  p.  lb.,  cost  ami  freight, 

295 

153 

702 

192 

214 

241 


Less  freight  to  New  Orleans,  on  1997  bags  ®  60c.  and  5  jut  cent  i>er  bag. 

£8388.17  0  Exchange,  4.85        ...  « 

By  onr  draft  ®  90  d/st  on  London,  £8388.17,0 

_^__-_  E.  4-  O.   E 

Rio  UK  .Ia.neiuo,  28  .Iniie,   1894. 
HEIINAXDEZ  BROTH EIIS  .J-  CO. 


4561  42 
6515,55 
326901 

14658  74 
391868 
429761 
4723  02 


4 1944 
1258 

40685 


03 
U 


92 


Verify  all  the  computations  in  the  above  bill.  See  Index,  Engli.sh  Exchange, 
for  the  method  of  reducing  the  $-10,685.92  to  the  English  money  of  account. 

For  the  entries  resulting  from  the  importation  of  goods  from  Foreign 
countries,  and  from  the  remittance  of  exchange  in  payment  therefor,  See  Soul6's 
new  Science  ami  Practice  of  Accounts. 


5o6 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


978.    INVOICE  OF  COFFEE  IMPORTED  FROM  RIO  JANEIRO,  BRAZIL. 

Note. — This  Invoice  is  made  in  the  monetary  unit,  and  in  the  unit  of  weights,  of  Brazil, 
■which  were  used  in  Brazil,  up  to  March,  1890. 

Invoice  of  4000  bags  Coffee  shipped  by  C.  SINNOTT  &  CO., 
on  board  the  British  Steamer  "Plato,  " 

for  Nkw  Oijleans, for  account  and  risk  of 

Messrs.  J.  M.  BUTCHEE  &  CO., and  consigned  to 

the  order  of  Messrs.  K.  SOULE  &.  CO. 


Ji 

F 

400 

IS 

(r 

(M. 

K 

H 

48.1 

li 

,r 

450 

K 

K 

522 

11 

N 

.54.i 

Ji 

4) 

.■iOO 

14 

P 

500 

4000  ba^s  of  60  kiIo3  =  IG337.G4  arra.  ■S)  3  $  840  p.  arroba. 


Charges:  ■ 


Duty  on  Custoruboiifte  vabiation  of  290  reis  p.  kilo 

on  240.000  kilos  =  Rs.  CO  :  600.000  ®  11  per  cent.      - 
Difference  of  70  ra.  per  Kilo  =  Rs.  18,960,000  -S)  4  per  cent 

CapataziaaCO  ra.  p.  bag  

Erokerage  50  ra.     " 

4000  baj;s  "ai  700  rs  p.  bag 

Mending  bags,  &e.  -       - 

Fire  Insurance 

Conaular  Certificate 

Negro  liiro  --.-..■.-. 


Commission  for  purcbasing  ■&  2^  per  cent. 

Shipped  against  cable  order  dated  2d  Feb.,  1890,  from  New  York. 
^^^-—  £.  <£  O.  E.  ^^-^_^ 

Kid  de  Janeiko,  27  February,  1890. 
C.  SINNNOTT  it-   CO. 


Note  1.— A  Kilo  is  Ibe  French  unit  of  weight  and  equals  2.2046  Avoinlnpois  pounds. 
Note  '2.— Tbo  Arroba  is  tbo  llrazilian  unit  of  weight  and  ia  equal  to  :i2.38  Avoirdupois  pounds. 
Note  3.—  14.09  Kilos  are  equal  to  one  x\rroba.     Hence  to  rediue  Kilos  to  Arrnbas  (iiviile  the  Kilos  l>v  14.69. 
Note  4. — Seo  Index,  liraziJiau  Exchange,  lor  the  Brazilian  Money  of  Account  and  the  manner  of  reducing  Rs.  to 
£,.  s.  aud  d. 


ns. 

7650  000 

758^400 

240  000 

200  000 

2800 

000 

23 

800 

25 

090 

9 

000 

840 

710 

Es. 

Rs. 

02,730  540 


12.558 


75,294 
1.882 


000 


540 
360 


979.  STATEMENT  OF  FULL  COST  OF  INVOICE  OF  COFFEE  PER  "  PLATO." 

J.    21.    DUTCH EE  c^-    CO., 

In  account  with  C.   SINNOTT  <|-   CO. 


l«w 

I\.Liy. 


To  iTivniro  Coat,  of  4000  baps'of  Coffee,  per  "  Plato,"  to  New  Orleans, 

"  I'Ttipo.  nf  riist  of  (',ablL';,n.'inis 

"  Slnrups  un  (li;it».i  -        ■        - 

"  Uill  iilokelayo.  3  lOpercenton  Ks.  77  :426$100 


Ks. 
J5y  our  dr.Tft  w  00  d/st.  ou  Siuuolt  &  Soule,  London,  exchange  21  gd.  =  .£0970.8.3 

Es. 


Cost  on  board,  including  freight  of  40c.  and  5  per  cent  per  hag  =  0.78g.  per  16. 

E.  a  O.  E. 

ElO  DE  JA.VEIRO,  Feby.  27,  1890. 

C.   SJNXOTT  <t   CO. 

Verify  the  calculations  in  the  Invoice  and  in  the  Account   Current  and   Interest  Account 
which  follows. 

Note  1 .—  It  is  the  custom  to  charge  brokerage  on  the  amount  invested  pins  the  brokerage. 
UoTE  2.  — Sometiuies  Ibe  stalemeutof  the  full  coat  ia  made  at  the  end  of  the  Invoice. 


77:170 

noo 

18 

(lUO 

86 

ii:io 

145 

170 

77  ;  426 

1110 

lis. 

77  :  426 

77  :  420 

100 

77  :  426 

100  ' 
100 


CUSTOMHOUSE    BUSINESS. 


5o7 


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nsurance. 


981.     Insurance  is  a  contract  by  wliicb  one  party,  in  consideration  of  a 
certain  sum  calleil  a j^twhoh,  and  whicli  is  proportioned  to  the  risk  involved,  under 
takes  to  conii>ensate  tbe  otlier  party  for  loss  on  a  specified  thing,  from  specified 
causes,  within  a  sj)ecified  time. 

9S2r  Insurance  is  divided  into  two  general  classes:  1.  Property  Insurance. 
2.  Personal  Ixsuraiice. 

983.  Property  Insurance  is  security  against  loss  by  fire  or  damage  by 
transportation,  or  the  injury  or  death  of  cattle,  etc.,  and  includes  Fire  Insurance^ 
River  Insurance,  Marine  Insurance,  Transit  Insurance  and  Live  Stock  Insurance. 

984.  Fire  Insurance  is  the  insurance  against  loss  or  damage  by  fire  of  all 
kinds  of  property  on  land,  such  as  liouses,  merchandise,  factories,  etc. 

985.  River  Insurance  is  the  insurance  of  river  steamers,  barges,  and 
their  cargoes. 

986.  Marine  Insurance  is  the  insurance  of  ocean  vessels  and  their  cargoes. 

Note. — In  river  or  uiariiie  insurance,  when  tbe  steamer  or  vessel  onlj'  is  insured,  H  is  called 
Sull  Insurance  ;  and  when  the  cargo  only  is  insured,  it  is  called  Cargo  Insurance. 

987.  Transit  Insurance  is  tlie  insurance  of  goods  or  property  during 
transportation  by  land,  or  by  both  land  and  water. 

988.  Live  Stock  Insurance  is  the  insurance  of  live  stock,  such  as  horses, 
mules,  etc.,  against  accident  or  death  within  a  specified  time. 

989.  Personal  Insurance  includes  Life  Insurance,  Health  Insurance  and 
Accident  Insurance. 

990.  Life  Insurance  is  the  insurance  on  the  lives  of  persons  whereby  a 
certain  sum  of  money  is  to  be  paid  to  them  when  they  attain  a  certain  age,  or  at 
their  death,  the  amount  is  to  be  paid  to  their  heirs,  or  to  some  one  named  in  the 
policy. 

991.  Health  Insurance  is  the  insurance  by  which  the  insured  parties 
receive  a  weekly  allowance  in  case  of  sickness. 

992.  Accident  Insurance  is  the  insurance  of  persons  against  death  or 
certain  injuries  received  under  certain  circumstances  while  at  home,  or  when 
traveling. 

Note. — The  following  conditions  are  emljraced  in  the  policies  of  most  Accident  Insurance 
Companies,  giving  .such  companies  an  unjust  advantage  over  their  patrons. 

"The  insurance  under  this  Certificate  shall  not  extend  to  or  cover  dis.appe.arances,  or  injurie."* 
whether  fatal,  or  disabling,  of  which  there  is  no  visible  mark  on  the  body  of  the  insured;  nor 

(508) 


*  INSURANCE.  509 

extend  to,  or  cover  ACCIDENTaI;  injuries  or  death,  resulting  from,  or  caused  directly  or  indirectly, 
wholly  or  in  part,  by  hernia,  fits,  vertigo,  somnambulism  or  disease  in  any  form,  gas  or  poison  iu 
any  I'orm  or  manner,  contact  with  ruisoNOUS  sudstanxks,  surgical  operations  or  medical  treat- 
ment, dueling,  lighting  or  wrestling,  war  or  riot,  lilting  or  over  exertion,  suicide,  felonious  or 
otherwise,  sane  or  insane,  sunstroke  or  freezing,  riding  or  driving  races,  voluntary  exposure  to 
unnecessary  danger,  nor  extend  to  or  cover  intentional  injuries  iuliicted  by  the  insured  or  any  other 
])ersou,  or  where  the  accident,  injury  or  death  happens  while  the  insured  is  under  the  iudueuce  of 
intoxicating  drinks  or  narcotics,  or  iu  consequence  thereof;  or  while  or  iu  consequence  of.  violating 
the  law  or  the  rules  of  any  company  or  corporation;  or  while  employed  iu  mining,  blasting  or 
wrecking,  or  iu  the  manufacture,  sale  or  transportation  of  gun-powder,  or  any  other  explosive 
compound,  (unless  insured  to  cover  such  occupation).  Standing  or  riding  upon  the  platl'orms  of  any 
conveyance  using  steam  or  electricity  as  a  motive  power,  or  entering  or  artemjiting  to  leave  such 
conveyance  while  the  same  is  in  motion,  or  walking  on  the  road-bed  or  bridge  of  any  railway,  are 
hazards  not  counted  or  covered  by  this  insurance,  and  no  sura  will  be  paid  for  injuries  or  death  in 
conse(|uence  of  such  exposure,  or  while  thus  exposed,  liap])eniiig  to  any  person  other  than  railroad 
employes,  who  shall  have  been  accepted  under  such  occupatiou." 

993.  There  are  also  other  forms  or  kinds  of  insurauce  under  different  names 
according  to  the  nature  of  the  subject  of  insurance,  or  the  objects  or  interests 
insured :  as  Guarantee  or  Fidelity  Insurance,  Co-operative  Insurance,  or  Benevolent 
Insurance,  Boiler  Insurance,  and  Plate  Glass  Insurance. 

994.  Guarantee  or  Fidelity  Insurance  is  the  insurance  against  loss  that 
may  occur  through  fraud  or  unworthy  agents  or  employes  who  occupy  positions  of 
trust.  Guarantee  or  Fidelity  Companies  issue  bonds  of  suretyship  for  a  specified 
amount  to  employers,  guaranteeing  the  faithful  conduct  and  tlie  honesty  of  the 
party  employed  for  a  given  time,  for  which  they  charge  the  employe  a  given  per 
cent  on  the  face  of  tlie  bond. 

Employes  must  possess  good  characters  and  good  business  capacity  before 
they  can  be  thus  guaranteed. 

Some  guarantee  companies  also  guarantee  titles  to  lands. 

995.  Co-operative  Insurance  or  Benevolent  Insurance  is  that  in  which  the 
members  of  some  Company,  Association  or  Society,  contribute  weekly  or  monthly, 
or  at  stated  times,  to  a  fund  from  which  a  specified  sum  is  paid  to  the  heirs  of 
deceased  members,  or  to  some  persou  named  in  the  Benefit  Certificate  or  Policy  of 
tlie  deceased  member.  The  sum  contributed  depends  upon  tlie  age  of  tlie  person  at 
tlie  time  lie  becomes  a  member,  and  the  amount  of  his  benefit  certificate  or  jiolicy. 
Of  this  class  the  Knights  of  Uonor,  the  Legion  of  Uonor,  the  Catholic  Knights  of 
America  and  the  Royal  Arcanum,  are  leading  examples. 

996.  Tbe  Insurer,  generally  a  stock  or  mutual  company  doing  business 
under  the  title  of  an  Insurance  Company,  is  the  party  that  contracts  to  indemnify 
the  other.  The  Insurer,  after  having  signed  the  policy,  is  called  the  under irr iter,  so 
called  from  the  fact  of  having  signed  under  the  conditions  of  the  policy. 

997.  The  Insured  or  Assured  is  the  party  that  is  protected  by  the  under- 
taking or  contract  of  the  insurer. 

998.  The  Policy  is  the  written  or  printed  instrument  containing  the  numer- 
ous conditions  of  the  contract  between  the  insurer  and  the  insured.  In  marine  and 
river  insurance,  there  are  two  kinds  of  policies,  valued  and  open. 

999.  A  A'^alued  Policy  is  one  in  which  a  value  has  been  set  upon  the 
property  or  interest  insured,  and  inserted  in  the  policy.  A  valued  policy  is  often 
called  a  special  policy. 


5io  soule's  riiiLObOPUic  practical  mathematics.  * 

On  coniitiy  risks,  most  Iiisuiaiice  C()iii])iiiiii's  liiuit  tlie  amount  of  insurance 
to  tbree-t'ourtlis  or  two-thirds  of  tin-  appraised  value  of  the  jiroperty  insured. 

Ill  many  policies,  especially  on  buildinj;s,  the  Insurance  Com])anies  reserve 
the  right  to  repair,  rebuild  or  replace  the  property  destroyed  or  damaged  with 
other  of  like  kind  aud  quality,  witliin  a  reasonable  time. 

1000.  An  Open  Policy  is  one  in  which  the  amount  of  interest  or  property 
is  not  tixed  by  the  policy,  but  is  left  to  be  ascertained  in  case  of  loss  or  damage. 

In  OPEN  POLICIES  for  Kiver  and  Marine  Insurance,  additional  insurance 
may  be  entered,  from  time  to  time,  as  goods  are  shii)ped,  either  on  the  xiolicy  or  in  a 
pass  book,  which  is  regarded  as  having  the  force  of  the  jiolicy. 

Premiums  on  open  policies  are  jtaid  at  the  end  of  a  voyage,  or  monthly,  or 
quarterly,  or  otherwise,  as  may  be  agreed  upon. 

lOOOi.  Open  Endorsement  Policy  is  a  policy  issued  to  the  assured  in 
blank,  to  cover  on  i)roi)erty  held  by  the  assured,  as  described  in  the  several  entries 
to  book  attached,  while  in  such  places  and  insured  for  such  periods  of  time,  and 
at  such  rates  of  premium,  and  for  such  amounts  as  shall  be  written.  Hence,  no 
liability  on  the  part  of  the  Company  exists,  until  application  is  accepted.  These 
entries  are  hence  called  endorsements. 

1001.  Perpetual  Policies  are  sometimes  issued,  the  rate  being  usually  equal 
to  that  of  TEN  annual  premiums. 

In  PERPETUAL  policies,  the  premium  is  considered  merely  a  deposit  with 
the  Insurance  Company ;  for  at  any  time,  at  the  instance  of  either  party,  the  policy 
may  be  cancelled,  and  00  per  cent  of  the  premium  or  deposit  must  be  returned  to 
the  policy  holder. 

Policies  are  not  transferable  without  the  consent  of  the  insurers. 

1002.  The  Subject  of  the  Insurance  is  the  property  itself,  and  the  title 
or  interest  that  the  assured  has,  is  called  his  Insurable  Interest. 

1003.  The  premium  is  the  price  paid  for  the  insurance. 

1004.  The  Rate  of  Premium  is  either  a  certain  per  cent  on  the  amount 
insured,  as  3  per  cent  or  li  i)er  cent,  or  it  is  a  specified  number  of  cents  on  one 
hundred  dollars,  as  a  rate  of  65;',  or  $1.25,  which  means  respectively,  G5t'  and  $1.25 
on  each  $100  insured. 

1005.  The  Eate  of  Premium  depends  in  a  great  measure  ui)on  the  time 
for  which  the  policy  is  issued  ami  the  degree  of  risk,  which  is  so  varied,  that  the 
calculation  of  the  rate  becomes  one  of  probabilities. 

1006.  Short  Rates.  Rates  for  less  than  a  year  are  called  short  rates, 
and  are  ])roporti()iialIy  higher  than  yearly  rates. 

1007.  Retui'u  Premium.  When  a  policy  is  taken  for  a  year  and  is  can- 
celled at  the  re(pu-st  of  the  insured  ]irior  to  the  close  of  the  year,  a  Eeturn  Premium 
is  paid  to  the  party  insured.  In  such  cases,  the  company  has  the  right  to  charge 
Short  Hates  for  the  time  the  policy  has  been  in  force. 

lOOS.  A  Risk  is  the  probability  that  the  subject  of  the  insurance  may  be 
lost  or  damaged,  or  it  is  the  danger  to  which  the  thing  insured  is  exposed. 

Risks  are  denominated  '"'hazardous,"  "extra  hazardous,"  and  "specially 
hazardous,"  according  to  the  probability  of  loss,  which  depends  very  much  upon 
the  combustibility  or  perishable  nature  of  the  sul)je<'t  of  insurance  and  its  sur- 
roundings. For  example,  a  hardware  store  would  be  less  liable  to  burn  than  a  dry 
goods  store,  and  a  store  in  which  inflammable  substances  or  explosive  materials 


*  INSURANCE.  5ll 

were  kept  would  be  more  liable  tluiu  either;    and  a  liotel  is   more   liable   tbaii    a 
private  residence. 

1009.  Keiusurauce  is  a  contract  which  the  first  insurer  enters  into,  in 
order  to  relieve  himself  from  some  of  the  risks  that  he  has  undertaken,  by  dividing 
them  witli  other  insurers. 

1010.  Caucellation  of  Policies.  A  policy  is  cancelled  when  the  agreement 
between  the  party  insured  and  the  insurers  is  annulled. 

1011.  Trausferring  Policies.  A  policy  is  transferred  when  the  insured, 
with  the  consent  and  acceptance  of  the  insurers,  conveys  or  transfers  his  rights 
under  the  terms  and  conditions  of  the  policy  to  some  third  person. 

1012.  All  Insurance  Agent  is  a  person  who  is  duly  authorized  to  act  for 
an  insurance  comjiany  in  soliciting  business,  contracting  risks,  collecting  pre- 
miums, and  adjusting  losses. 

1013.  An  Insurance  Broker  is  a  person  who  eflects  insurance  and  negoti- 
ates policies,  for  which  service  he  is  paid  a  commission  or  brokerage  by  the  com- 
pany taking  the  risk. 

Brokers  are  regarded  as  agents  of  the  parties  insured,  and  not  of  the  insur- 
ance company. 

101-1.  Insurance  Companies  are  of  several  kinds  according  to  the  principle 
or  form  of  their  organization :  1.  Stock  Companies.  2.  Mutual  Comjianies.  3. 
Mixed  or  Stock  and  Mutual  Companies. 

1015.  A  Stock  Insurance  Company  is  one  in  which  the  Capital  is  divided 
into  shares,  which  are  subscribed,  paid  for  and  owned  by  persons  or  firms  called 
stockholders.  The  stockholders  share  all  the  profits  and  are  liable  for  the  losses, 
to  the  extent  of  the  value  of  the  shares  owned. 

1016.  A  Mutual  Insurance  Company  is  one  in  which  the  profits  and  losses 
are  shared  by  all  who  are  insured,  who  are  the  policy  holders. 

In  this  form  of  company,  there  are  no  stockholder^  or  capital  stock.  The 
caj)ital  consists  of  unused  premiums  of  the  reserved  earnings  and  of  the  invest- 
ments of  the  company. 

1017.  A  Mixed  or  Stock  and  Mutual  Insurance  Company  is  one  which 
combines  the  principles  of  both  the  Stock  and  Mutual  Companies,  (.  e.  they  have  a 
capital  stock  and  shareholders,  and  divide  a  portion  of  the  net  profits  among  share- 
holders and  policy  holders.  Generally  in  mixed  companies  a  limited  dividend  of  3 
to  10  per  cent  is  first  paid  to  the  stockholders,  a  portion  of  the  profits  is  carried  to 
surplus  or  reserve  ftind,  and  the  surplus  profits,  if  any,  are  divided  among  the 
particijiating  jiolicy  holders. 

1018.  Non-participating  policies  are  sometimes  issued  at  reduced  rates,  by 
mutual  and  by  mixed  comxjanies,  and  such  policies  do  not  share  in  the  losses  or 
gains  of  the  company. 

In  some  mixed  companies,  the  capital  is  unlimited,  provided  it  is  not  less 
than  a  certain  sum;  and  in  the  organization  of  the  company,  it  is  represented  by 
deposit  notes ;  and  subsequently  by  the  installments  collected,  premiums  paid  on 


512  soule's  philosophic  practical  mathematics.  * 

insurances,  aiul  by  the  sums  earned  in  tlie  business.  lu  these  companies,  the 
insured  becomes,  by  the  mere  fact  of  insurance,  a  member  of  the  company,  and  is 
obliged,  therefore,  to  observe  its  rules  and  recognize  the  authority  of  its  agents. 

1(U9.  Insurance  Scrip  consists  of  tlie  certificates  given  by  mutual  and 
mixed  companies  to  the  insured,  in  lieu  of  cash  dividends.  They  are  equivalent  to 
receipts  or  certificates  for  money  loaned.  They  genjerally  bear  interest,  are  nego- 
tiable by  delivery  and  trnnsfer  on  the  books  of  the  company,  and  are  payable  at 
such  time  as  the  company  may  subseipiently  specify. 

10'20.  The  Surplus  of  an  insurance  company  is  the  excess  of  the  assets 
over  the  liabilities.  Capital  stock  and  unearned  premiums  are  considered  a  part 
of  the  liabilities  when  estimating  the  surplus  or  when  determining  gains  and  losses. 

1021.  Misrepresentation.  Any  material  misrepresentation  jjrejudicial  to 
the  insurers,  whether  intentional  or  not,  and  whether  made  by  the  principal  or  his 
agent,  vitiates  the  policy. 

1022.  Bottomry  and  Respondentia.  Bottomry  is  a  contract  by  which  the 
sliip  owner,  or  tlie  master  on  his  belialf,  pledges  the  keel,  tackle,  and  apparel  of  his 
shij)  as  security  for  money  which  he  borrows  for  the  use  of  the  ship,  in  contempla- 
tion of  a  particular  voyage  or  for  a  particular  and  lixed  period  of  time.  It  is  a 
condition  of  a  bottomry  contract  tliat  if  the  ship  should  lie  lost  during  the  voyage, 
or  l)y  any  of  the  perils  enirmerated  in  the  contract,  the  lender  shall  lose  his  money  ; 
but  if  the  ship  arrives  in  safety,  then  he  shall  receive  his  principal  and  also  the 
interest  agreed  u|ion,  however  much  it  may  exceed  the  irsnal  or  legal  interest  for  the 
use  of  money.  When  the  money  is  loaned  on  other  property  exposed  to  miiritime 
risks,  the  contract  is  called  Remixnulentut. 

1023.  Salvage  is  a  compensation  allowed  parties  for  their  voluntary  services 
in  saving  a  ship,  steamboat,  cargo,  or  other  property  from  the  perils  of  tire  or  water. 

ADJUSTMENT  OF  LOSSES. 

1024.  In  Marine  Insurance  the  policies  contain  an  "Average  Clause"  by 
which,  "  if  the  vessel  is  valued  in  the  statement  thereof,  beyond  the  amount 
insured,  the  insurers  are  to  pay  such  part  only  of  the  sum  insured  (or  of  the  partial 
loss)  as  the  amount  insured  bears  to  such  valuation."  And  in  case  of  the  cargo, 
if  that  "  is  value<l  beyond  the  amount  insured,  the  insurers  are  to  pay  such  part 
only  of  the  sum  insured  (or  of  the  partial  loss)  jis  the  amount  insured  bears  to  the 
full  valuation."  The  effect  of  this  principle  is  that  the  insured  is  himself  the  insurer 
for  that  part  or  portion  of  tlie  property  insured. 

Forexamiile:  If  a  vessel  valued  at  $40000  were  insured  for  J  value,  or 
$30000,  and  were  totally  destroyed,  the  coni]>any  or  the  insurers  would  jiay  the 
$30000;  but  if  the  vessel  were  damaged  to  the  amount  of  $10000,  the  (•omi)any 
would  i>ay  only  three-fourths  of  the  partial  loss,  or  $7r)00.  Again,  if  a  cargo  valued 
at  $40000  were  insured  for  .$10000,  and  sustained  a  loss  of  $10000  or  more,  the 
insurers  would  pay  only  $2500. 

1025.  In  Fire  Insurance,  under  the  ordinary  Fire  Insurance  Policy,  which 
does  not  contain  the  average  or  the  To  per  cent  clause,  the  company  will  i)ay  the 
full  amount  of  the  loss  or  damage,  provided  it  does  not  exceed  the  sum  covered  by 
insurance.  Thus,  if  a  house  or  stock  of  goods  were  insured  for  $10000  and  a 
partial  loss  of  say  $7000  be  sustained,  the  conqiany  would  pay  the  full  loss  or 
$7000.  But  if  the  loss  should  reach  $11000  or  more,  the  company  would  pay  only 
$10000,  the  amount  specified  in  the  policy. 

Sometimes  FIRE  INSURANCE  policies  contain  the  "  Average  Clause,"  and 

then  the  company  issuing  such  policies  would  pay  only  such  portion  of  the  loss  as  the 
amount  insured  bears  to  the  total  value  of  the  property. 

Note. — Tlio  usual  form  of  the  "  Average  Clause,"  to  be  used  wheti  jiolicy  covers  any  mer- 
chaudise  in  two  or  more  locations,  is  as  follows  : 

"  It  is  a  part  of  the  cousideratiou  of  this  policy  and  the  basis  upon  which  the  rate  of  premium 


*  INSURANCE.  513 

18  fixed,  that  the  assured  shall  maintain  insurance  on  each  item  of  property  insured  by  this  policy 
equal  to  the  actual  cash  value  thereof,  and  that  failing  so  to  do,  the  assured  shall  he  au  iusurer 
to  the  extent  of  such  deficit,  and  in  that  event  shall  bear  his,  her  or  their  proportion  of  auy  loss.'' 

1026.  A(ynstmeutg  of  Losses,  under  policies  containing  the  "Average  Clause,"  would  be 
made  as  follows  :  Suppose  property  valued  at  $20000  was  stored  in  two  warehouses,  and  insured 
for  $15000,  and  that  the  goods  in  one  warehouse  were  destroyed  by  fire  making  a  partial  loss  of 
say  $8000,  the  company  would  pay  $6000,  i.  e.  such  proportion  of  the  loss  as  the  amount  insured 
bears  to  the  full  value  of  the  property. 

1027a  To  Adjust  the  loss  when  several  difl'erent  insurance  companies  have  risks  on  the 
eame  property,  the  loss  is  divided  among  all  the  companies  in  proportion  to  the  amount  of  their 
respective  risks,  whether  all  the  companies  be  solvent  or  not.  Thus,  if  goods  valued  at  $25000  be 
insured  in  Co.  A,  fur  $3000,  in  Co.  B,  for  $5000,  and  in  Co.  C,  for  $10000,  making  $18000,  and  a  loss 
of  $9000  should  be  sustained,  Co.'s  A,  B,  and  C,  would  pay  respectively,  VVj  Ai  ^""i  it  "^  ^^^  l'^^*- 

1028.  The  75  per  cent  Co-insnrance  Clause,  which   is  now  generally  inserted  in  fire 

policies,  reads  as  follows  :  "  It  is  a  part  of  the  consideration  of  this  policy  and  the  basis  uponl 
which  the  rate  of  premium  is  fixed,  that  the  assured  shall  at  all  times  maintain  a  total  iusurance 
on  the  property,  insured  by  this  policy,  of  not  less  than  75  per  cent  of  the  total  cash  value  thereof 
(as  covered  uuder  the  several  items  of  this  policy,)  aud  that,  failing  so  to  do,  the  assured  shal 
"become  co-insurers  to  the  extent  of  the  deficiency,  aud  in  that  event,  shall  bear  their  proportion 
of  auy  loss  occurring  under  this  policy." 

1029.  Tlie  Lightning  Clause  reads  as  follows:  "  This  policy  shall  cover  any  direct  loss  or 
damage  caused  by  lightuiug,  (meaning  thereby  the  commonly  accepted  use  of  the  term  Lightning; 
and  in  no  case  to  include  loss  or  damage  by  cyclone,  toruado,  or  windstorm,)  not  exceeding  the 
sum  iusured,  nor  the  interest  of  the  insured  in  the  property,  aud  subject  in  all  other  respects  to  the 
terms  aud  conditions  of  this  policy.  I'rovided,  however,  if  there  shall  be  any  other  insurance  on 
said  property,  this  company  shall  be  liable  only  pro  rata  with  such  other  iusurance,  for  any  direct 
loss  by  Lightning,  whether  such  other  insurance  be  against  direct  loss  by  lightning  or  not." 

1029a*  Iron  Safe  Clause. — 1st.  The  assured  will  take  a  complete  itemized  inventory  of 
stock  on  hand  at  least  once  in  each  calendar  year,  aud  unless  such  inventory  has  been  taken  withiu 
twelve  calendar  muulbs  prior  to  the  date  of  this  policy,  one  shall  be  taken  in  detail  withiu  thirty 
days  of  issuance  of  this  policy,  or  this  policy  shall  be  null  aud  void  from  such  date. 

2ud.  The  assured  will  keep  a  set  of  books,  which  shall  clearly  aud  plainly  present  a  com- 
plete record  of  business  transacted,  including  all  purchases,  sales  and  shipments,  both  for  cash 
and  credit,  from  date  of  inventory  as  provided  for  in  first  section  of  this  clause  and  also  from  date 
of  last  preceding  inveyitory,  if  such  has  been  taken,  and  during  the  continuance  of  this  policy. 

3rd.  The  assured  will  keep  such  books  and  inventory,  and  also  the  last  preceding  inventory,  if 
such  has  been  taken,  securely  locked  in  a  fire-proof  safe  at  night,  aud  at  all  times  when  the  build- 
ing mentioned  in  this  policy  is  not  actuallj-  opeu  for  business  ;  or  failing  in  this,  the  assured  will 
keep  such  Ijooks  and  inventories  in  some  secure  place  uot  exposed  to  a  fire  which  would  destroy  the 
aforesaid  Ijuilding;  and  unless  such  books  and  inventories  are  produced  and  delivered  to  this 
company  for  examination  after  loss  or  damage  by  fire  to  the  personal  property  insured  hereunder, 
this  policy  shall  be  null  aud  void  and  no  suit  or  action  shall  be  maintained  hereon.  It  is  further 
agreed  that  the  receipt  of  such  books  and  inventories  and  the  examination  of  the  same  shall  not 
be  an  admission  of  auy  liability  under  the  policy,  nor  a  waiver  of  auy  defense  to  same. 

1029b.  The  Foundation  Clause.  It  is  understood  that  the  foundations  and  piling  on 
which  the  within  described  buildiug  rests,  are  not  included  in  this  Insurance,  aud  in  no  event 
shall  they  be  taken  as  part  of  the  valuation  of  the  building  for  the  purpose  of  arriving  at  the 
assured's  contributory  proportion  in  the  application  of  the  Co-Insurance  Clause. 

The  foundations  being  exempted  from  the  operation  of  the  coinsurance  clause  of  this  policy 
and  not  beiug  coTered  by  this  insurance,  it  is  hereby  understood  and  agreed  that  in  the  event  of  loss, 
if  the  insuring  company  should  elect  to  rebuild,  and  it  be  found  necessary  on  account  of  statutory 
or  other  causes  to  replace  the  foundations  in  whole  or  in  part,  the  cost  of  replacing  or  repairing 
same  will  be  at  the  expense  of  the  assured. 


514  SOULES    PHILOSOPHIC    TRACTICAL    MATHEMATICS.  * 

1030.  The  Three-quarter  Loss  Clause  or  Country  Clause  inserted  in  tli© 
country  policies  issued  by  many  Insurance  Companies,  reads  as  follows: 

"It  is  agreed  and  understood  to  be  a  condition  of  this  insurance,  tliat  in  case  of  any  loss  or 
damage  under  this  policy,  this  company  shall  not  be  liable  for  an  amount  greater  than  thrkk- 
FOURTHS  of  said  loss,  not  exceeding  the  sum  herein  insured,  the  other  one-foukth  to  be  borne  by 
the  assured,  and  in  the  event  of  other  insurance  on  property  covered  under  this  policy,  this 
company  shall  not  be  liable  for  more  than  its  proportion  of  three-fourths  of  such  loss  or  damage." 

1030«.    The  Three-fourths  Value  Clause  is  as  follows : 

"It  is  understood  and  agreed  to  be  a  condition  of  this  insurance,  that  in  the  event  of  loss  or 
damage  by  fire  to  the  property  insured  under  this  policy,  this  company  shall  not  be  liable  for  an 
amount  greater  than  three-fourths  of  the  actual  cash  value  of  each  item  of  property  insured  by 
this  policy  (not  exceeding  the  amount  insured  on  each  such  item)  at  the  time  immediately  preceding 
such  loss  or  damage;  and  in  the  event  of  additional  insurance — if  any  is  permitted  hereon — then 
this  company  shall  be  liable  for  its  proportion  only  of  three-fourths  such  cash  value  of  each  item 
insured  at  the  time  of  the  fire,  not  exceeding  the  amount  insured  on  each  such  item." 

1031.  A  Floating  Policy  sometimes  termed  a  "  blanket  policy"  covers 
goods  in  a  number  of  diflereut  warehouses,  docks,  etc.,  with  the  "  Average  Clause" 
added. 

IMPORTANCE  OR  VALUE  OF  INSURANCE. 

1032.  Insurance,  considered  in  all  of  its  various  relations  and  different 
applications,  is  one  of  the  most  important  and  valuable  institutions  of  the  age, 
without  which  trade  and  commerce  would  languish,  and  the  poor  and  medium 
classes  of  society  would  be  deprived  of  many  blessings  and  opportunities  of  success, 
which  through  it  they  now  enjoy. 

Considering  the  hazards  which  threaten  property  and  the  uncertainty  of 
human  life,  prudence,  wisdom,  regard  for  tinancial  obligations,  and  affections  for 
family,  demand  that  every  man  should  insure  both  his  property  and  liis  life. 

For  a  business  man  not  to  insure  his  jtroperty  is  evidence  of  his  incapacity, 
and  unless  his  goods  or  other  property  is  fully  paid  for,  it  is  i)roof  of  his  want  of 
fidelity  to  his  creditors.  The  only  exception  to  this  is  where  the  party  has  sufficient 
capital,  outside  of  his  business,  to  enable  him  to  continue  business,  pay  his  financial 
obligations  and  to  support  those  dependent  upon  him,  in  case  of  loss. 

In  regard  to  Life  Insurance  Companies,  it  must  be  said  that  while  they  are 
doing  a  grandly  humane  work  and  merit  the  favorable  consideration  and  patronage 
of  all  men,  some  of  the  companies  are  characterized  for  extravagance  beyond 
prudence  or  reason  ;  and  some  have  violated  the  principles  of  ethics  m  their  prom- 
ises and  statements  of  facilities,  advantages  and  resources. 

These  facts  make  it  necessary  for  the  applicant  for  Life  Insurance  to  investi- 
gate and  exercise  Judgment  and  discretion  in  selecting  a  company. 

The  various  Social,  Co-operative  or  Benevolent  Insurance  Associations  with 
their  low  rate  of  assessments,  and  their  economical  system  of  management,  offer  to 
the  poor  man  facilities  for  insuring,  which  as  he  loves  his  mother,  his  wife,  his 
children,  or  his  fellow-man,  he  should  accept. 

These  Co-operative  Insurance  Associations  are  the  children  of  this  progressive 
age;  they  are  supplementing  the  work  of  the  regular  Life  Companies,  and  are 
proving  a  blessing  to  mankind. 

1033.  Bead  the  Policy  of  Insurance.  To  all  who  insure,  whatever  may  be 
the  subject  of  insurance,  or  the  character  of  the  company  in  which  you  insure,  we 
say,  read  your  policies  and  fully  understand  your  rights  and  duties. 

/m  case  of  loss  of  furniture,  books,  pictures,  wearing  apparel,  jewelry,  etc., 
the  insured  is  required  to  give  a  full  itemized  statement  of  the  same  before  the 


*  INSURANCE.  5l5 

insurance  will  be  paid.  The  importance  therefore  of  having  a  list  or  an  inventory 
of  them  safely  preserved  is  easily  seen. 

The  PARTY  INSURED  should  also  fully  understand  that  if  he  insures  his  house 
and  then  sustains  a  loss  on  his  kitchen,  or  his  furniture,  he  has  no  chiims  on  the 
insurers  for  loss  or  damage.  Mis  policy liroiccts  onhj  such  property  as  is  specified 
therein.  He  should  also  understand  that  where  his  house,  when  insured,  is  damaged, 
and  said  damage  paid  for  or  repaired  by  the  insurers,  that  his  policy  is  reduced  to 
the  extent  of  the  damage  sustained.  A  want  of  this  knowledge  frequently  results 
in  unpleasant  conti'oversies  between  the  insured  and  insurers. 

The  problems  iii  insurance  involve  some  of  the  principles  of  per  cent,  and 
sometimes,  especially  in  life  insurance,  the  calculation  of  the  average  length  of 
human  life,  and  the  worth  of  money  invested  at  various  rates  of  interest,  and  in 
many  different  ways. 

TO  FIND  THE  RATE  OF  PREMIUM. 

1034.  The  first  thing  to  be  determined  by  an  insurance  company  after  its 
organization,  is  the  rate  per  cent  ])remium  to  charge.  To  determine  this  for  river 
insurance,  iiidepeudeut  of  the  charges  and  statistics  of  other  companies,  we  present 
the  following  figures: 

From  the  statistics  of  river  commerce,  we  will  suppose  that  during  the  past 
season  there  were  40  steamers  plying  between  the  ports  of  New  Orleans  and  St. 
Louis,  the  aggregate  value  of  the  same  being  $2,400,000,  and  which,  in  accordance 
with  the  regulations  of  insurance  companies  in  New  Orleans,  was  insured  for  two- 
thirds  of  its  value,  $1,600,000;  that  during  the  season 

1  steamer  was  lost,  the  value  of  which  was  $45,000,  §  of  which  is      -      $30,000 
And  1  was  damaged  to  the  amount  of  $18,000,  §  of  which  is      -     -     -        12,000 


Making  a  total  loss  of        -        -  ...        $42,000 

STATEMENT  TO  OBTAIN  THE  PER  CENT  LOSS. 


IGOOOOO 


42000 
100 


2§7(,  loss,  Ans. 

By  these  figures,  we  see  that  to  insure  a  steamer  in  this  trade,  the  probabili- 
ties of  loss  are  2§  per  cent,  and  hence  the  rate  per  cent  of  premium  must  be  2|  per 
cent,  plus  such  increase  as  may  be  required  to  remunerate  the  insurer  for  his 
trouble,  cover  expenses,  and  leave  a  fair  margin  for  profit. 

To  show  approximately  the  per  cent  to  charge  for  fire  risks,  we  present  the 


5i6 


SOULE  S    I'HILOSOPHIC    PRACTICAL    MATHEMATICS. 


following  figures,  ■which  are  very  nearly  tlie  actual   figures  of  oue  of  the  New 
Orleans  Insurance  Companies  for  one  year's  unusually  disastrous  business : 


Amount  of  Fire  Risks,  $8,500,000 
<'  "  Marine  "  2,800,000 
«        "   Eiver     "         1,750,000 


Losses  on  Fire  Eisks,  $75,540 
51,000 
20,210 


it 
u 


"   Marine  " 
"   Kiver     " 


$13,050,000 

The  expenses  of  the  Company  during  the  year  were 

"    taxes            "             "             '<          "      "        " 
By  these  figures,  we  see  that  $8,500,000  have  lost 
To  which  we  add  J  of  the  expense 
And  ,W5-o-o-o"o  of  the  taxes 


$36,000 
12,000 


$146,750 


$75,540 

12,000 

7,810 


Total  loss  and  expense  on  fire  risks 


$95,356 


Having  the  total  losses  of  the  fire  risks,  we  then  find  the  per  cent  loss  by  the 
following  statement* 


8500000 


95356 
100 


8500000 


or, 


1.1218  +  %  (nearly  1J%).    Ans. 


100% 
95356 


1.1218  +  % 


The  result  of  this  work  shows  that  the  probabilities  of  loss,  including 
expenses  and  taxes  are  1.1218  +  %,  and  hence  the  rate  per  cent  of  premium  must 
be  1.1218  +  %,  plus  such  iucrease  as  may  be  necessary  to  produce  such  profit  as 
equity  would  warrant  and  the  wisdom  of  the  Directors  of  the  Company  may  deter- 
mine. To  fix  the  premium  at  1^  per  cent,  the  Company  would  realize  on  the  fire 
risks,  as  shown  by  the  above  figures,  a  profit  of  $10,894. 

In  actual  practice,  many  other  small  items  of  expenditure  and  allowance 
might  be  made,  and  thus  bring  the  result  still  nearer  the  exact  premium  to  charge. 
The  supply  of  water  and  the  efficiency  of  the  fire  department  would  also  be  con- 
sidered. Our  subject  in  presenting  this  work  is  more  to  show  the  general  manner 
of  arriving  at  a  very  imr»crtant  result  than  the  minute  details  of  the  operation, 
which  must  necessarily  xitiar  in  every  case,  and  can  only  be  fully  understood  as  they 
occur  in  the  actual  business  of  a  company ;  in  actual  operations,  the  figures  showing 
the  amount  insured  and  the  amount  of  loss  sustained  for  several  years  would  be 
taken,  instead  of  the  result  of  one  year's  business. 


J 


INSURANCE. 


5i7 


TO  FIND    THE    AMOUNT    OF    PKEMIUM,    OR    COST    OF    INSURANCE, 

WHEN  THE  AMOUNT  INSURED  AND  THE  RATE  OF 

PREMIUM  ARE  GIVEN. 

PROBLEMS. 

1035.  1.  A  real  estate  owner  insures  for  one  year  two  houses,  valued  at 
$7000  each,  at  1 J  per  cent.     What  is  the  premium  ?  Ans.  $157.50. 

2.  What  is  the  premium  for  au  insurance  of  $120000  ou  a  hotel,  at  $2.25  per 
$100?  Ans.  $2700. 

3.  A  ship  was  insured  for  $60000  at  1 J  per  cent,  and  her  cargo  for  $25000  at 
$1.30  on  $100.     What  was  the  cost  of  the  insurance  ?  Ans.  $1225. 

4.  What  is  the  premium  pro-rata  on  a  $3500  policy,  dated  Sept.  12, 1894,  aud 
expiring  Jan.  30, 1895,  at  the  annual  rate  of  75/  per  $100  ?  Ans.  $10.06^. 

Note. — 30  days  to  a  month  is  allowed  in  this  problem. 

5.  What  is  the  premium  on  a  $3500  policy  dated  as  in  the  above  problem  at 
the  short  rate  of  95/  per  $100 1  Ans.  $12.74  ,i. 

6.  Insured  my  house  for  $8000,  furniture  for  $4500,  library  for  $1500,  paint- 
ings for  $1000,  piano  for  $500  and  stable  for  $200.  The  policy  costs  $1.50 ;  the  rate 
of  premium  is  1|  per  cent,  with  20  per  cent  discount  or  rebate;  what  is  the  cost  of 
the  insurance  and  what  is  the  actual  rate  of  premium  charged? 

Ans.  $158.50  cost  of  insurance.    1%  actual  premium. 

NoTK. — The  law  of  Louisiana  does  not  allow  insurance  companies  to  give  a  rebate.     The  net 
rate  of  insurance  must  be  stated. 


OPERATION  TO  OBTAIN  COST 
OF   INSURANCE. 

$15700        sum  insured. 
14%   rateof  prem. 


$196.25 
39.25 


$157.00 
1.50 


amt.  of  prem. 
=  20%  dis. 

net  prem. 
policy. 


$158.50      cost  of  ins. 


OPERATION  TO  OBTAIN  THE  ACTUAL 
RATE  PER  <;ENT. 

1J%  premium  =  |% 

20%  discount  =  J  of  f  =  ^  percent, 

which  deducted  from  J  leaves  1% 

actual  premium. 

or  thus : 

1J%  =  $1.25 

20%  of    which    is    .25,    and    which 
deducted  leaves  $1,  which  =  1%. 


or  thus: 


15700 


$  PREMIUM. 
157 

1P0 

1%     Ans. 


7.  A  merchant  insured  for  one  year  his  house  for  $12000,  his  furniture  for 
$5000,  library  for  $3000,  paintings  for  $2000,  statuary  for  $2500,  silver-plate  for 
$1500,  piano  for  $500,  mirrors  for  $2000,  cisterns  for  $400,  stables  for  $1000,  fences 
for  $500,  and  wearing  apparel  for  $3000.  The  policy  cost  $1.50;  the  rate  of  pre- 
mium was  1|  per  cent,  with  15  per  cent  rebate.  What  was  the  cost  of  the  insur-. 
ance,  and  what  was  the  actual  per  cent  paid  ? 

Ans.  $427.35,  cost  of  the  insurance.    111%,  actually  paid. 

8.  What  will  be  the  cost  to  insure  a  stock  of  extra  extra  hazardous  goods 
valued  at  $28500,  at  $3.25  on  $100  ?  Ans.  $926.25. 


5i8 


SOULE  S    rilU.OSOPHIC    I'RACTICAT^    MATHEMATICS. 


9.  A  real  estate  owuer  insures  2  dwelliugs  valued  respectively  at  $8500  and 
$7000;  and  also  the  rent  of  same,  -wliich  is  $1800  for  the  $8500  dwelling  and  $1500 
for  the  $7000  dwelling.  The  rate  was  IJ  per  cent.  What  did  the  insurance  cost 
liim?  Ans.  $282.00. 

10.  The  following  entries  are  made  on  an  "oj>en  ])olivy^^  or  in  the  pass  book 
nsed  with  an  open  policy :  Make  the  extensions  and  find  the  total  amount  due. 


Dates 


Names  ok  Steamer 


I'EOM 


To 


Subject       Amt.  Ins.  Eatk     Pekm. 


1895 
Nov. 


Garlaufl, 

l';iri;<>U(l, 

T.  r.  Leatln'Ts, 

Queen  City, 

City  of  New  Orleans. 


New  Orleans. 


Shreveport, 
Vicksbiirjj, 
St.  Louis, 
Cincinnati, 
Momphis, 


Mdse. 
Mdse. 
Mdse. 
Mdse. 
Mdse. 


4100 

00 

2+ 

(1500 

00 

i 

9800 

00 

n 

2300 

00 

2 

1685 

00 

1 

$ 

373 

35 


11.  What  will  it  cost  to  renew  a  policy  of  insurance  on  property  valued  at 
$24200  at  the  rate  I  per  cent  premium?  Ans.  $211.75. 

12.  An  insurance  company  insured  a  block  of  buildings  valued  at  $400000 
at  2  per  cent,  allowing  a  rebate  of  15  per  cent.  The  company  subsequently  re-insured 
$100000  at  IJ  per  cent,  $100000  at  1:^  per  cent,  and  $100000  at  13  per  cent.  How 
much  premium  did  the  company  receive?  How  much  did  it  pay  for  re-insuraucCj 
and  at  what  rate  per  cent  were  its  receipts  on  the  amount  that  it  protects  ? 

Ans.  $G800,  premium  received.    $4125,  premium  paid. 


^lU  JO' 


PARTIAL    OPERATION. 


$400000  at  2%  =  $8000  —  15%  ^  $6800.  $100000  at  1^%  =  $1125.  $100000  at  1:1% 
=  $1250.  $100000  at  1^%  =  $1750.  $1125  +  $1250  +  $17.50  =  $4125.  $CS00 — 
$4125  =  $2075.  $400000  —  $300000  =  $100000.  ($2675  x  100)  -^  $100000  =  2.675% 

=  225(«,. 

13.  You  insure  your  house  for  $9000 ;  furniture  for  $4500 ;  jewelry  for  $6000 ; 
books  for  $1800;  paintings  for  $2500;  statuary  for  $1400;  silver-i»late  for  $1200; 
piano  for  $400;  cisterns  and  outhouses  for  $1500;  wearing  apjiarel  for  $2500.  The 
rate  is  1^  per  cent  with  15  per  cent  rebate.  What  is  the  cost  of  the  insurance,  and 
what,  111  case  of  a  fire  and  a  total  loss,  would  be  your  various  duties  in  order  to  col- 
lect the  insurance?  Ans.     To  the  first,  $.327.25. 

Answer  to  the  second  :  Your  first  duty  would  be  to  read  your  policy  and  act 
in  accordance  with  its  conditions.  2.  To  make  an  itemized  statement  or  bill  of  the 
furniture,  jewelry,  books,  paintings,  statues,  silver-i)Iate  and  wearing  apparel,  and 
present  the  same  with  your  account  to  the  insurance  company. 

14.  If  you  keejj  an  insurance  on  an  extra  hazardous  stock  of  goods  for  20 
years  at  5  per  cent,  and  they  then  burn  and  you  collect  the  full  insurance,  have  you 
gained  or  lost?  Ans.     Making  no  allowance  for  interest,  you  have  Iosti)remiuia 

paid  to  the  amount  of  the  value  of  the  goods. 


INSURANCE. 


5i9 


TO  FIXD  WHAT  AMOUNT  TO  INSURE    SO   AS   TO    COVER   BOTH   THE 

PROPERTY  INSURED  AND  PREMIUM  PAID,  WHEN  WE 

HAVE  THE  AMOUNT  OF  PROPERTY  AND  THE 

RATE  PER  CENT  PREMIUM  GIVEN. 

PROBLEMS. 

1036.  1.  I  wish  to  insure  a  stock  of  goods  valued  at  64:4100,  for  svtch  a  sum 
that  ill  case  of  loss,  I  Mill  receive  the  value  of  tlie  merchandise  and  the  premium 
paid  for  insurance,  or  in  other  words  I  wish  to  insure  the  merchandise  and  the 
premium  jiaid  for  the  insurance.  The  rate  is  2  i)er  ceut.  For  what  amount  must  I 
iusui-e,  and  what  is  the  premium  ?  Aus.  S45000,  amount  to  insure. 

$900,  luemium. 

FIRST   OPERATION. 


$100  assumed. 

2  =  2(fc  premium. 


98 


S'JS  value  of  goods. 


Explntialinn. — Wc  first 
assuiiio  $100  to  represent 
the  viihie  of  the  goods, 
jihis  the  2"n  premium  for 
insuring  tlie  goods.  We 
then  deduct  therefrom  the 
2  per  ceut  premium  and 
obtiiin  $98,  as  the  value 
of  the  goods,  which,  with 
the  preniinmpaid  for  in- 
suring the  goods  reifuired 
an  insurance  of  $100.  We 
then  reason  thus  :  Since  $98  v.ilue  of  goods  require  an  insurance  of  .$100,  SI  will  reijuire  the  98th 
part,  and  .$44100  will  require  441UO  times  as  much,  the  result  of  which  is  |4o000.  We  then  calculate 
the  2  per  ceut  on  the  ^45000,  and  tiud  the  premium  to  be  $900. 


$ 

100 

44100 

$45000      amount  to  insure. 
2%  premium. 


$900.(10     premium. 


SECOND   OPERATION. 


$44100      value  of  goods. 
2fo  premium. 


98 


582.00     premium  on  goods. 


9 
100 

882 

$900  premium  on  goods  ami  premium, 
44100  value  of  jroods  added. 


$45000  amount  to  insure. 

2.  My  house,  situated  iu  a  country  village,  is  valued  at  $12000,  and  the 
fm-niture  at  $5000.  I  wish  to  insure  f  of  the  value  of  the  house  and  furniture,  so 
that  iu  case  of  loss  I  may  receive  the  ^  value  of  both  and  the  iiremium  paid  for  the 
insurance.  The  rate  is  1^  per  cent;  policy  cost  $1.50;  the  rebate  or  discount  is  10 
per  cent.    What  amount  of  insurance  must  I  effect  and  what  will  be  the  cost? 

Ans.  $12954.03,  insurance.     $205.53,  cost. 

PARTIAL    OPERATION. 

$12000  +  $5000  =  $17000.  f  of  $17000  r=  $12750.  If  %  =  $1.75.  10%  of  $1.75  = 
.175/.  $1.75— .175  =  $1,575.  $100— $1,575  =  $98,425.  ($12750  x  100) -4- $98,425 
=  $12954.03  to  be  insui-ed.  1|%  on  $12954.03  =  $226.70  ^- 10%  rebate  =  $204.03 
+  $1.50  =  $205.53. 


520 


SOULE  S    rniLOSOPHIC    PRACTICAL    MATHEMATICS. 


3.  A  carg:o  of  cotton  is  valued  at  $200000;  for  what  sum  must  a  policy  be 
taken  to  cover  both  cotton  and  iiremiuni,  the  rate  being  $1.02^  on  $100? 

Ans.  $2G.4294.70+. 

4.  For  what  amount  must  I  insure  my  dwelling,  valued  at  $0000,  so  that  in 
case  of  loss  I  may  receive  both  the  value  of  the  house  and  the  premium  ^wiVZ  ?  The 
rate  of  insurance  is  If  per  cent,  rebate  20  jier  cent  ?  Ans.  $9736.30+. 


PARTIAL    OPERATION. 


13%  =  l    20%  =  1.     J  X  i  =  A 


=  $983 


I  —  iro  =  IB  =  1|%  net  premium.    $100  — 11% 
($9000  X  100)  -^  9S|  =  $9730.30+. 


TO  FIND  THE   AMOUNT  INSURED,   WHEN  WE  HAVE  THE  AMOUNT 
OP  PEEMIUM  PAID  AND  THE  RATE  PER  CENT  PREMIUM  GIVEN. 


PROBLEMS. 

1037.     1.     The  cost  of  insuring  a  house  was  $114.00;   the  cost  of  policy  wa» 
$1.50 ;  the  rate  of  insurance  was  1^  per  cent.     What  was  the  amount  insured  ? 

Ans.  $9000. 

]<:r]>Ianalion. — We  first 
deduct  the  cost  of  the 
policy  from  the  total  cost 
of  insurance,  and  have  iu 
the  remainder  $112.50  tho 
amount  of  premium  which 
is  H  per  cent  of  tho  sum 
insured.  The  reasoning 
for  the  statement  to  linil 
this  sum  is  the  same  as 
given  on  page  446. 

2.     A  merchant  paid  $121.50  for  insuring  his  store.    The  rate  of  premium 

was  $1.50  per  $100;  the  rebate  was  20  per  cent;  the  policy  cost  $1.50.    For  what 


OPERATION. 

$114.00  cost  of  insurance. 
1.50  cost  of  jiolicy. 


Statement  to  find  the  sum  insured. 


100 

4 

112.50 


$112.50  amount  of  premium. 


$9000.00  Ans. 


amount  did  he  insure  his  store  ? 


$121.50  premium  and  policy. 
1.50  policy. 


Ans.  $10000. 


OPERATION. 

^.50  rate  of  premium. 
.30  20%  rebate. 


1.20 


$120.00  premium. 


$1.20  net  rate  ot  premium. 


100 
120.00 

$10000. 


3.  The  premium  for  insuring  J  of  a  stock  of  goods  at  J  per  cent  was  $137.81J. 
What  was  the  value  of  the  goods?  Ans.  $24500. 

4.  The  cost  of  insuring  an  invoice  of  merchandise  was  $180.10^.  The  rate 
of  insurance  was  IJ  per  cent;  the  rebate  was  10  per  cent;  the  cost  of  the  policy 
$1.50.     What  was  the  amount  of  the  invoice?  Ans.  $17040. 


OPERATION  INDICATED. 


1J%  =  1.125—  10%  rebate.     (.1125)  =  1.0125%.     $180,105  — 
(178.605  X  100)  -  $1.0125  =  $17040. 


.50  =  $178,605. 


INSURANCE. 


521 


TO   FIND   THE   EATE   PER   CENT   PEEMIUM,   WHEN    WE    HAVE    THE 
AMOUNT  INSUEED  AND  THE   AMOUNT  OP  PEEMIUM  GIVEN. 


PKOBLEMS. 


1038.  1.  The  cost  m'  insuring'  a  stock  of  extra  liazartlous  goods,  vahied  at 
$38400.  was  $1153.50  iucludiug  the  cost  of  policy,  $1.50.  What  was  tlie  rate  per 
cent  premium  ?  Ans.  3%. 


OPERATION. 

$1153.50  cost  of  ins. 
1.50  cost  of  i^olicy. 


Statement  to  fiml  tlie  rate  per  cent 
l>remium. 

100 

38400  1153 


$1152.00  amt.  of  prem. 


3%    Ans. 


Explanation. — We  iirst  de- 
duct from  the  total  cost,  tbe 
cost  of  the  policy,  and  thus 
obtain  $1152,  ■which  is  the 
amount  of  ]iicniinni  .Tt  a  certain 
per  cent  on  tbe  value  of  the 
goods.  We  then  make  the 
statement  to  find  the  rate  per 
cent.  The  reasoning  for  this 
statement  is  the  same  as  that 
given  on  pages,  439  and  410. 


2.  A  merchant  insures  his  goods  for  $12000.  The  insurers  allow  a  rebate  of 
25  per  cent  on  the  premium,  and  charge  $1.50  for  the  policy.  The  ccst  of  insurance 
and  policy  was  $163.50.    What  was  the  rate  of  premium  on  $100 ! 

Ans.  $1.80,  or  1.8%. 


OPERATION. 


f  163.50  cost  of  insurance. 
1.50  cost  of  jiolicy. 


Statement  to  find  the  amount  of 

premium  before  tbe  25  per 

cent  rebate  was  made. 


100 
162 


Statement  to  find  the  rate  per 
cent  of  premium. 


$162.00  net  prem.  paid. 


12000 


$216  amt.  of  prem.  chg'd 


216 
100 


Sl,%  =  1.8%. 


3.  The  premium  paid  for  insuring  a  house  valued  at  $9400,  was  $70.50. 
"What  was  the  rate  per  cent  ?  Ans.  f  %. 

4.  I  paid  $193.50  for  insuring  my  house  for  $12000.  The  policy  cost  $1;50. 
The  rebate,  in  lieu  of  participation  in  the  profits  of  the  company,  was  20  per  cent. 
What  was  the  rate  per  cent  premium  charged,  and  what  was  the  actual  rate  jjaid? 

Ans.  2  «;^  premium  charged.    lf%  actual  premium  jmid. 


522 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


PEACTICAL  LABOR  SAYING  SYSTEJI  OF  CASTING  UNEARNED 

PREMIUMS. 

1039.  The  following  iisefnl  talilc  is  tlie  resiiit  of  tlie  experience  and  genius 
of  Mr.  II.  Carpenter,  an  expert  insurance  agent  of  New  Orleans,  througli  wliose 
kindness  we  are  permitted  to  use  it. 

TABLE  FOE  CASTING  THE  UNEARNED  PREMIUMS,  THE   YEAR  COMMENXTNG  ON  THE 

FIRST  OF  JANUARY. 


(To  find  the  unearntd  premium,  multiply  the  premium  charged  by  the  amount  opposite  the  date  of 

the  policy). 


DATK. 

.lANTAlCY. 

DATE. 

i'i;ui!Lai:y. 

DATE, 

MARCH. 

DATE. 

AI'lilL. 

1 

.0027 

•   1 

.0876 

1 

.1643 

1 

.2492 

2 

.0055 

2 

.0904 

2 

.1671 

2 

.2519 

3 

.0082 

3 

.0931 

3 

.1698 

3 

.2547 

4 

.0109 

4 

.0958 

4 

.1725 

4 

.2574 

5 

.01.37 

5 

.0986 

5 

.17.53 

5 

.2601 

6 

.0164 

6 

.1013 

6 

.1780 

6 

.2629 

7 

.0192 

7 

.  104 1 

7 

.1807 

7 

.2656 

8 

.0219 

8 

.1068 

8 

.1835 

8 

.2683 

9 

.02^6 

9 

.1095 

9 

.1862 

9 

.2711 

10 

.0274 

10 

.1123 

10 

.1889 

10 

.2739 

n 

.0301 

U 

.1150 

11 

.1918 

11 

.2766 

12 

.0328 

12 

.1177 

12 

.1944 

12 

.2793 

13 

035C 

13 

.1205 

13 

.1972 

13 

.2820 

14 

.03S3 

14 

.1232 

14 

.1999 

14 

.2848 

15 

.0411 

15 

.1260 

15 

.2027 

15 

.2876 

16 

.0438 

16 

.12S7 

16 

.2054 

16 

.2903 

17 

.0466 

17 

.1314 

17 

.2081 

17 

.2930 

1« 

.0493 

18 

.1342 

IS 

.2109 

18 

.29.58 

19 

.0520 

19 

.1369 

19 

.2136 

19 

.2985 

20 

.0548 

20 

.1397 

20 

.2164 

20 

.3013 

21 

.0575 

21 

.1424 

21 

.2191 

21 

.3040 

22 

.0602 

22 

.1451 

22 

.2218 

22 

.3067 

23 

.0630 

23 

.1479 

23 

.2246 

23 

.3095 

24 

.(657 

24 

.1506 

24 

.2273 

24 

.3122 

25 

.0085 

25 

.1.534 

25 

.2300 

25 

.3150 

26 

.0712 

26 

.1.561 

26 

.2328 

26 

.3177 

27 

.0739 

27 

.15S« 

27 

.2355 

27 

..3204 

28 

.0767 

28 

.1610 

28 

.2383 

28 

.3232 

29 

.0794 

29 

.1643 

29 

.2410 

29 

.32,59 

30 

.0822 

30 

.2437 

30 

.3287 

31 

.0849 

31 

.2465 

The  nnit  of  this  table  is  based  ou  the  premium  of  §1.  for  1  year,  Tvhich  divided  by  365,  the 
number  of  days  in  a  year,  gives  the  jiremium  ou  $1  for  1  day,  which,  carried  to  4  ))laces  of  decimals, 
i.s  .0027 ;  this  being  multijilied  by  the  number  of  the  days  of  the  year,  gives  the  tabular  multi- 
iilieis,  -which  in  the  table  presented  is  only  extended  to  4  months,  or  119  days  of  the  year.  Having 
*liC'Kc  imiltipliers,  by  simply  multiplying  the  amount  of  the  premium  by  the  multiplier  in  the  table 
corresponding  -Bith  the  date  of  the  policy,  we  produce  the  unexpired  j)remium.  For  example,  the 
premium  on  a  jiolicy  issued  for  1  year  from  March  17th,  amounts  to  $80.  To  find  the  unearned 
premium  ou  the  Z\st  day  of  December,  vre  look  for  the  multiplier  opposite  March  the  17th,  in  the 


I 


INSURANCE. 


523 


table,  and  -n-ith  it  multiply  the  $80  thus:  $80  x  -2081  ='|;16.6480,  or  practically  SI";  or  iu  practice 
the  premium  may  be  multiiilied  by  the  ceuts  ouly  increased  by  1,  when  the  decimal  is  in  excess  of 
one-half,  thus : 

S80  X  .21  =  $16.80  ;  or  practically  $17. 

The  above  table  Tvas  prepared  for  a  company,  ivbose  fiscal  year  begins  on  the  1st  of  January; 
but  it  can  easily  be  adapted  to  the  work  of  other  offices  \Yhose  fiscal  year  begins  at  a  different 
date,  by  substituting  the  first  day  of  their  business  year  for  January  1st. 

Want  of  space  prevents  the  insertion  of  the  table  for  the  entire  year,  but  the  4  mouths 
presented  is  sufficient  to  demonstrate  the  simiilicity,  convenience  and  value  of  it. 

ADJUSTMENT  OF  LOSSES. 

1040.  1.  Jones  &  Smith  hold  three  policies  on  a  stock  of  good.s;  one  policy 
in  Co.  A,  for  $5000,  one  iu  Co.  B,  for  $8000,  and  one  in  Co.  C,  for  $3000.  A  fire 
occnrs  and  damages  the  goods  to  the  amount  of  $9800.  "What  is  due  from  each 
company  ?  Ans.  $3081.25  from  Co.  A. 

$4930.00  from  Co.  B. 


$1848.75  from  Co.  C. 


$5000 
8000 
3000 


OPERATIONS  INDICATED. 
Co.  A.  Co.  B. 


IGOOO 


9800 
5000 


ICOOO 


9800 
8000 


16000 


Co.  C. 
9800 
3000 


$10000 

2.  A  dwelling  is  insured  in  Company  X,  for  $5500,  and  in  company  Y,  for 
$3500.  The  house  is  destroyed  by  fire,  and  the  companies  find  that  they  can  re-build 
it  for  $7000;  this  amount,  in  accordance  with  tlie  conditions  of  the  policies,  they 
ofier  to  the  owner,  who  accepts  it.     What  sum  will  each  company  pay? 

Ans.     Co.  X,  $4277.78.     Co.  Y,  $2722.22. 

3.  J.  M.  Butchee  &  Co.,  carry  policies  on  goods  in  16  companies  for  $150000, 
Company  A  carries  a  risk  of  $10000.  The  whole  stock  of  goods  is  destroyed  by 
fire.  ]\Ierchandise  Account  iu  tlie  books  of  J.  M.  Butchee  &  Co.  is  debited  with 
$875000  and  credited  with  $862500.  By  comparing  the  cost  and  sales  of  100  arti- 
cles of  goods,  it  is  determined  and  agreed  between  the  insurers  and  insured  that 
15  per  cent  gain  has  been  realized  on  sales,  and  a  settlement  is  made  on  this  basis. 
Make  the  bill  for  the  amount  due  from  Company  A, 


OPERATION. 


115 


100 
802500 

$750000  cost  of  sales. 


$87.5000  total  cost. 
750000  cost  of  sales. 


(BILL.) 


INSURANCE  CO.  A., 


$125000  cost  of  goods  lost. 


New  Orleans,  July  18,  1895. 


To  J.  M.  Butchee  &  Co.,  Dr. 


For  r'sWo'^rof  $125000,  Mdse.  destroyed  by  fire  at  No.  108  Magazine  Street, 
New  Orleans,  June  2,  1895.        ----..... 


$8333.33 


Received  payment, 


524 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Adjustment  ioidcr  the  Average  Clause. 

1.  A  inercliiiiit  has  goods  valued  at  S20000  in  two  warehouses,  on  which  he  has 
a  policy,  with  the  "Average  Clause,"  for  $<S(){)().  The  goods  iu  the  warehouse  in 
•which  he  had  $ll.'()(K)  was  damaged  by  lire  to  the  aniouut  ot  $0000.  What  is  due  by 
the  Insurance  Company  ?  Ans.  $13400. 

OPKRATION. 


$12000  insured  by  owner. 
8000   insured  by  company. 

$L'0000   total  insurance. 


20000 


16000 
8000 


or,  20000  :  8000  :  :  GOOD  :  2400 


$2400. 


Adjustments  under  the  Three-fourth  Loss  or  Coventry  Clause. 

1.  A  country  merchant  who  had  a  $10000  policy  on  his  goods,  with  the  three- 
quarter  loss  clause,  sustained  a  loss  by  tire  amounting  to  $8000.  How  much  do  the 
insurers  owe  the  merchant  ■?  Ans.  $0000. 

Note. — In  policies  contiiininL;  the  ihrcr-qnarier  loss  chiiise,  tbe  insurers,  in  case  of  a  tot.il  loss 
pay  only  three-fourtba  of  tbe  value  of  tlic  property  destroyed,  and  if  tbe  loss  is  less  than  the  sum 
insured,  tbo  insurers  pay  only  throe-fourths  of  the  loss. 

2.  If  in  Problem  1,  with  the  three-quarter  loss  clause,  how  much  ■would  the 
merchant  have  leceived :  1.  If  the  loss  had  been  $10000;  2.  If  the  loss  had  been 
f  12000  ;  3.  1  f  the  loss  had  been  $10000  ?  Ans.  1.  $  7500  on  a  $10000  loss. 

2.  $  9000  oil  a  $12000  loss. 

3.  $10000  on  a  $10000  loss. 
Remarks. — In  case  of  a  $10000  or  a  $12000  loss,  as  previously  stated  in  note 

to  Problem  1,  the  underwriters  are  responsible  for  ^  of  the  loss,  because  it  does  not 
exceed  the  face  of  the  policy  and  because  of  tlie  ^  condition  clause  of  the  ])olicy. 

In  case  of  the  $10000  loss,  ^  of  the  same  being  $12000,  the  underwriters  or 
insurers  carrying  the  $10000,  are  responsible  for  the  full  face  of  the  policy,  $10000. 

Note.— flOOOO  is  f  of  ,f  133.33};  hence,  for  a  loss  of  f;13333J  or  more,  the  insurers  are  respon- 
sible for  the  face  of  the  $10000  policy. 

Marine  Adjustments. 

1.  The  owner  of  |  of  a  vessel  insured  f  of  his  interest  at  2.J  per  cent;  the 
sum  paid  for  premium  was  $.502.50.  The  vessel  sustained  a  loss  by  tire  of  $5000. 
What  sum  do  the  insurers  owe  him?  Ans.  $1400.25. 

OPERATIOX    INDICATED. 


$100 
2 

562..50 
$22i,vJ0.00  =  i  of  his  | 


22500 
4 

.$30000  =  I  of  value  of  vessel. 


80000 


5ono 

22500 
$1406.25 . 


$80000  =  value  of  vessel. 

Note. — In  M.arino  Insur.ance,  according  to  the  "Average  Clause,"  tbo  insurers  pay  for  iv 
PARTIAL  T.oss  of  property  pautiai.i.y  insured,  only  such  a  jiroportion  of  the  loss,  as  the  sum  insured 
bears  to  the  whole  value  of  the  property. 

2.  A  cargo  of  fruit  valued  at  $8500,  and  insured  for  $8000  was  damaged  at 
sea  and  sold  at  auction  on  arrival  in  port  for  $7500.  What  was  paid  by  the  insurers 
on  the  adjustment  of  the  loss?  Ans.  $041.17+. 

Adjustment  under  the  Seventy-five  per  cent  Co-insurance  Clause. 

1.  A  merchant  effected  $40000  insurance  ou  property  worth  $00000.  His 
policies  contain   the    75  per  cent  co-insurance  clause.      A   loss  of    $30000   was 


*  INSURANCE.  525 

snstaiiR'd  ;  what  sum  will  the  mercliaiit  receive  from  tLe  underwriters  iu  a  correct 
udjustmeut  of  the  loss,  and  what  will  be  his  loss  as  a  co-insurer? 

Ans.  $2UG(J0|  due  from  underwriters. 

3333^  merchant's  loss  as  a  co-insurer. 
Note. — See  Article  1028,  page  513,  for  the  73  per  ceut  lusurauce  Clause. 

FIRST    OPERATION. 

$fiOOOO  X  "io^o  =  $-(5000  aiiuinnt  to  be,  or  which  is  really  insured.  $-15000  —  40000  =  $5000  the 
sum  tliat  tlie  iiierchaut  is  cu-iusurer.  Thus  making,  under  the  75%  clause  of  the  polic}',  $45000 
iusurauce  on  the  property 

45000  I  $30000  Explanation. — The  reasoning  for  the  line  statement  is   as 

I    40000  follow.s:    Since  $45000  insurance  lose  $30000,  $1  will  lose  the 

45000dth  part  and  40000  (the  sum  insured  by  the  uuderwntcis) 

I  $266ti6j    Ans.  will  lose  40000  times  as  much. 

$30000  —  2666Gf  =  $3333^  merchant's  loss  as  co-insurer. 

SECOND    OPERATION. 

I  $30000  $40000  X  66|  =  $26G66|,  ETplanaiion.— Since  $45000  lose  $30000,  $1 

45000  I     100  which  is  the  sum  lost  and         will  lose  the  4.5000dth   part,    and   $100   will 

I to  be  paid   by   the   insur-         lose  100  times  as  much.     The  loss  tliiis  ]iro- 

I    66j5o' loss.  ance  companies.  duced,  being  on  $100,  dollars,  is  hence   the 

loss  per  cent. 
$5000  X  66|°(|'  =  $3333J-  which  is  the  sum  lost  and  to  be  borne  by  the  merchant  under  the 
75,"j^  co-insurance  clause. 

2.  Suppose  in  Problem  1,  tliat  the  loss  Lad  been  $45000,  liow  much  would 
the  Insurance  Companies  owe?  Ans.  $40000. 

Kemarks. — When  the  loss  is  equal  to  or  greater  than  the  full  amount  insured 
by  the  underwriters  and  owner  as  co-insurer,  then  the  underwriters  will  owe  the 
face  of  the  policies. 

3.  Suppose  in  Problem  1,  that  a  total  loss  of  850000  or  $G0000  had  been 
sustained,  how  much  would  the  Insurance  Comjjanies  owe?  Ans.  $40000. 

Remarks. — See  remark  of  Problem  2.  In  no  case  on  fire  insurance  can  the 
underwriters  owe  to  the  insured  more  than  the  face  of  the  policies. 

Adjustment  under  the  Three-fonrths  Vahie  Clause. 

Under  this  clause,  which  is  inserted  iu  some  policies,  the  insured  will  receive  the  full 
amount  of  his  loss,  provided  it  does  not  exceed  three-fourths  of  the  value  of  the  property. 
Thus :  1st.  Suppose  property  valued  at  $1000,  is  insured  for  $750,  and  a  loss  of  $600  is  sustained, 
the  insured  will  receive  $600  from  the  underwriters.  2d.  Suppose  property  valued  at  $1000  is 
insured  for  $750,  and  a  loss  of  $850  or  more  is  sustained,  the  insured  will  receive  $750  from  the 
underwriters. 

Remarks. — This  three-fourths  value  clause  differs  from  the  three-fourths  loss  or  countri/  clause 
explained  iu  Articles  1030  and  1040,  pages  514  and  524,  inasmuch  as  under  this  three-fourths  value 
clause  the  insured  will,  iu  case  of  a  partial  loss  not  exceeding  three-fourths  of  the  sound  value  of 
the  property  insured,  receive  the  full  amount  of  loss.  But  under  the  three-fourths  loss  clause  the 
insured  will  receive  only  three-fourths  of  the  amount  of  his  loss  whether  it  be  more  or  less,  uot 
exceeding  the  face  of  the  policy.     See  Article  1030a. 

FINDING  THE  LOSS  OF  MERCHANDISE  DESTROYED  BY  FIRE    AND    ADJUSTING   THE 
LOSS  WITH  THE  INSURANCE  COMPANIES  CARRYING  THE  RISKS. 

PROBLEMS. 

1041.     To  find  the  loss  of  Mdse.  by  fire  is  a  far  more  difficult  question  than  is  supposed  by 
one  man  iu  a  thousand. 

To  elucidate  this  matter,  let  us  assume  the  following  conditions: 

1.  That  Jones  &  Smith  carry  $90,000  iusurauce  on  Mdse.  in  12  Companies. 

2.  That  their  entire  stock  is  destroyed  by  fire. 

3.  That  their  Double  Entry  Ledger,  with  all  purchases  and  sales  posted  to  date  of  fire, 
shows  Mdse.  Account  debited  $486,500  and  credited  $467,450. 

From  these  assumed  facts,  what  has  been  the  loss  and  what  does  Insurance  Co.  A  owe, 
which  carried  a  policy  of  $8000  on  the  stock  f 

The  first  step  in  a  case  of  this  kind  is  to  read  your  policies  and  notify  the  underwriters  of 
all  the  known  facts  of  the  fire.  Then  consult  with  them  regarding  your  loss,  exhibiting  all  books 
^and  papers  pertinent  thereto.  When  the  underwriters  are  satisfied  that  all  is  fair,  then  the 
iquestion  of  GAIN  or  LOSS  per  cent  on  the  sales  of  goods  must  be  determined.     And  the  solution  to 


526  soule's  philosophic  practical  mathematics.  * 

this  question  ia  the  golden  kry  thnt  unlocks  the  treasury  of  the  nnder-nriters  carrying  the  risks. 
No  correct  adjustment  of  loss  can  be  nuide  bel'ore  tliis  question  is  solved. 

It  is  to  the  interest  of  J(nies  &  .Smith  to  show  a  large  gain  per  cent  on  sales;  and  it  is  to 
the  interest  of  the  Insurance  Cnnipanies  to  show  a  small  gain  per  cent  on  sales. 

If  an  agreement  eaunot  ho  reached  by  tlie  two  jiartics  in  interest,  from  the  general  accounts, 
then  .an  average  gain  per  cent  must  bo  determined  by  listing  the  cost  and  sales  of  a  number  of 
the  leading  lines  of  goods,  thus  : 

Statement  to  fliul  average  gain  Statement  to  find  coat  of  sales  of  Mdse.  based  ou 

per  cent.  an  average  gain  of  16.19  per  cent. 

4.31  I     100 

116.19      467450.00 


Co.st. 

Sales. 

«;4. 

4.80 

.22 

.30 

1.50 

1.65 

2.75 

3.00 

.15 

,18 

18. 

21.00 

Si26.62 

$30.93 

Net  Gain 

,   $4.31 

26.62 


100 


16.19  ^"o  gain. 


$402315.17  cost  of  sales. 


Total  cost  of  Mdse.,    $486500.00 
Cost  of  sales       "  402315.17 


Loss  by  fire,  $  84184.83 

In  the  above  statement  of  cost  and  sales,  to  .ibridge  the  work,  we  limited  the  statement  to 
Bis  items.  In  priictice,  fifty  to  one  hundred  should  be  used.  In  the  selection  of  the  items  lies 
the  great  difBculty.  The  underwriters  insisting  on  items  ujicui  which  a  suuill  gain  was  realized, 
and  Jones  &  .Smith  insisting  on  items  upon  which  a  large  gain  was  realized.  A  nice  discrimina- 
tion is  recjuired  in  making  this  statement  to  render  justice  to  both  parties.  The  author  hereof,  in 
behalf  of  his  client,  had  nine  sittings  of  two  hours  each,  -with  Insurance  Adjusters,  in  a  vain 
etiort  to  make  a  statement  of  costs  and  sales,  from  which  the  average  gain  per  cent  should  be 
accepted. 

Where  the  parties  in  interest  fail  to  agree  upon  this  point,  arbitration  of  the  question  seems 
to  our  mind  to  be  tlie  most  ethical  and  reasonable. 

The  following  statement  shows  the  .amount  of  loss  due  from  the  Insurance  Co.  having  a  risk 
of  $8000. 

Total  amount  insured,  $90000.  Total  amount  lost,  $84184.83.  Hence  the  Co.  owes  ^^^^ 
of  $8000  =  $7483.10. 

INSURING   GOODS. 

1042.  Business  prudence  demands  that  merchiints  and  manufacturers  should  give  careful 
attention  to  their  stock  of  goods  and  to  the  amount  of  insurance  covering  tlio  same,  adding  or 
cancelling  jiolicics  so  as  to  keep  fully  protected  and  losing  nothing  by  over-insurance.  A  careful 
record  should  l>o  kejit  of  the  date  the  policies  expire,  and  daily  inspection  should  he  m.ade  thereof. 
The  insured  slionld  also  note  with  care  all  the  special  clauses  and  conditions  imjiosed  by  his 
policy.  The  Lightning,  the  Average,  the  Found.aticm  .and  the  75%  or  more  Co-Insurance  clauses 
should  be  thoroughly  uuderstood  when  the  policy  is  written. 

Points  to  be  Specially    Observed  in  the  Mdse.  Insurance. 

When  considering  merchandise  account  and  the  statement  of  costs  and  sales,  due  regard 
must  be  given  to  the  fullowing  elements  on  purchases  and  sales : 

(1).  Trade  Discounts.  (2).  Cash  Discounts.  (3).  Rebates.  (4).  Time  Purchases  and  Time 
Sales. 

The  merchandise  returned  to  sellers,  and  the  merchandise  re-purchased  from  purch-asers,  if 
entered  in  merchandise  acc<mnt,  must  be  eliminated  therefrom.      After  cimsidering  all  these  ele- 
ments, an<l  finding  the  loss,  then  the  question  of  d<-preciation  and  apjireciation  of    the  goods   lost 
is  to  be  considered,  in  order  to  lind  the  cash  value  of  the  goods,   which  the  policy  of  insurance  _ 
covers. 

It  is  the  custom  of  Insurance  Companies,  when  adjusting  fire  losses,  to  allow  i%  discount 
for  each  30  days,  on  time  purchiises  and  on  time  sales.     This  is  equal  to  6°,,  interest. 

VALUED  POLICIES  ON  IMMOVABLE  PROPERTY. 

The  Valued  Policy  Insurance  Law  of  Louisiana,  enacted  in  1900,  requires  that  the  insur.ance 
on  imniov.able  property  shall  be  the  \'alue  assessed  or  ajqiraised  "l)y  tlie  insurer,  or  as  by  him  per- 
mitted to  be  assessed  at  the  time  of  the  issuance  of  the  policy,"  and  that  this  value  "shall  be 
conclusively  taken  to  be  the  true  value  of  the  property  at  tiio  time  of  the  issuance  of  the  policy, 
and  tlie  true  value  of  the  pro]ierty  at  the  time  of  damage  or  destruction.'' 

On  tbis  value.  Insurers  are  obliged  to  adjust  the  loss.  The  law,  however,  gives  the  insurers 
the  right  to  restore  the  damage  or  to  replace  the  property  destroyed  if  they  prefer  to  do  so, 
instead  of  making  payment  therefor. 

By  this  law  the  adjustments  of  loss  are  easily  and  expeditiously  made,  and  the  vexatious 
contentions,  so  often  made  by  the  insurers  under  the  former  law,  are  obviated. 


GENERAL  AND  PARTICULAR  AVERAGE.  52/ 

TAKING  STOCK  OR    INVENTORYING  GOODS. 
'When  "talcing  stock,"  or  inventorying  floods  and  other  property  preparatory  to  closing  the 
books  and  declaring  the   gains  and  losses  at  the  close  of  the  iiscal  year,  or  in  case  of  a  corpora- 
tion, at  a  dividend  period,  the  cnrrent  market  valne  shonld  be  placed  upon  the  goods  and  all  other 
property,  stocks,  bonds,  etc. 

Note. — In  the  caao  of  storks  aud  lionds  Tvliich  Iiave  an  nnstatilo  value  and  a  wido  raar2:in  of  flachiation,  tlio  avf  rape 
Tveekly  i>r  iiiontlily  value  is  somelimes  placed  upon  them.  Circumstances,  jud^pneut  and  etliies  are  the  proper  factors  to 
govern  such  casesi 

In  the  case  of  merchandise,  instead  of  listing  every  article  at  the  current  market  value 
Tvitli  allowances  for  depreciation  ou  account  of  damage,  out  of  fashion  or  style,  the  inventory  is 
often  taken  ou  ditferent  lines  of  goods,  at  the  orignal  invoice  price,  and  then  a  specified  per  cent, 
reduction  or  addition  is  made  thereto,  according  to  the  present  market  price  of  such  lines  of  goods; 
allowances  being  made  also  for  damage,  old  stock,  etc. 

In  some  cases  of  unimproved  real  property,  it  is  often  difficult  to  determine  its  market  value. 
Whether  it  is  worth  the  original  cost,  more  or  less,  plus  taxes,  is  a  question  to  he  decided  by  the 
facts  attending  the  case,  and  the  judgment  of  the  appraisers. 

General  and  Papticular  Average. 

1043.  General  Average  is  the  proportional  coutributiou  by  all  parties  con- 
cerned in  a  vessel,  cargo  or  freight,  towards  a  loss  or  damage  sustained  or  incurred 
for  the  common  safety  or  benefit. 

1044.  Particular  Average  is  the  contribution  for  loss  or  damage  incurred  or 
sustained  by  or  for  one  or  more  of  the  interests,  vessel,  ctirgo  or  freight;  such  loss 
much  be  borne  by  the  interest  or  interests  which  sustained  the  loss. 

1045.  An  Average  Adjuster  is  one  whose  profession  it  is  to  average  and  adjust 
losses  in  whicli  general  and  particular  average  cases  occur.  He  is  required  to  be 
familiar  with  insurance  laws  governing  the  subject,  and  to  deteimine  which  loss 
shoidd  b(>  charged  to  General  Average  and  which  to  Particidar  Average,  and  then 
to  apportion  the  loss  and  expense  among  the  several  contributory  interests. 

To  determine,  in  all  cases,  what  kind  and  character  of  losses  and  damages  are 
subject  to  general  average  and  what  to  particular  average  is  a  very  difHcult  matter 
and  often  gives  rise  to  prolonged  litigation.  Arnould's  Maritime  Law  and  Abbott, 
on  Shipping,  give  valuable  information  on  this  topic. 

The  following  are  some  of  the  losses  and  damages  subject  to  general  average : 
1st.  The  damage  sustained  by  a  ship  purposely  run  ashore  to  prevent  her  foundering 
at  sea  or  driving  on  the  rocks.  2d.  All  dannige  jturposely  done  to  the  vessel  or 
cargo,  to  preserve  the  whole  from  impending  danger,  such  as  the  cutting  away  of 
masts,  riggings,  etc.,  when  the  ship  is  in  distress.  3d.  All  charges  incurred  for  the 
general  good,  by  entering  a  foreign  port  in  distress.  4th.  The  ex])ense  for  dis- 
charging the  cargo  for  the  purpose  of  repairing  the  vessel,  or  for  floating  her  when 
she  accidentally  gets  aground.  5th.  The  expense  of  i)ilotage  when  putting  into  a 
port  in  distress.  Cth.  The  expense  of  labor  to  pump  the  sliip  after  having  sprung  a 
leak.  In  some  cases,  seamen's  wages  and  the  provisions  of  the  ship's  company  are 
subjects  of  general  average. 

The  following  are  a  few  of  the  los.ses  and  damages  subject  to  particular 
average : 

1st.  Damage  by  fire  from  accidental  causes,  unattended  by  a  sacrifice  to 
extinguish  the  fire;  also  the  damage  to  goods  by  water  thrown  upon  the  burning 
goods  to  extinguish  the  fire,  which  is  regarded  not  a  sacrifice  of  goods,  but  a  real 
advantage  to  them. 

2d.  The  damage  done  to  a  vessel  and  cargo  by  being  run  foul  of  accidentally 
and  without  fault  on  either  side. 

3d.  The  springing  of  masts,  breaking  of  the  upper  works  or  timbers,  parting 
of  cables,  and  all  other  loss  and  expense  upon  the  ship  occasioned  by  the  perils, 
insured  against  not  purposely  incurred,  and  not  amounting  to  a  total  loss. 


528  soule's  philosophic  practical  mathematics.  * 

1046.  Jettison.  The  goods  thrown  overboard,  the  cutting  away  of  masts  or 
cables,  and  whatever  the  master  of  a  vessel  does  in  distress  for  the  safety  or 
preservation  of  the  whole,  is  called  jettison. 

In  making  an  adjustment  of  general  average,  it  is  necessary  to  first  take  an 
account  of  the  various  losses  which  are  to  be  paid  by  contribution,  and  secondly,  to 
take  another  account  of  tlio  value  of  all  the  articles  that  are  to  contribute.  In  the 
list  of  contributory  articles — vessel,  cargo  and  freight — must  be  included  the  goods, 
etc.,  thrown  overboard ;  for  otherwise  the  owners  of  these  goods  would  receive  full 
value  for  them  and  yiay  nothing  towai'ds  the  loss. 

In  estimating  the  amount  of  loss  to  the  vessel  for  masts,  rigging,  etc.,  it  is 
the  custom  to  deduct  ^  of  the  cost  of  the  new  articles ;  for  being  new  they  will  be 
of  greater  value  than  the  articles  lost.  In  estimating  the  value  of  the  contributory 
interests,  the  general  custom  is  as  follows:  To  value  the  s/oj;  at  what  it  is  worth  at 
the  end  of  the  voyage ;  to  value  the  cargo  at  what  it  would  have  brought  at  its  port 
of  destination,  but  it  is  sometimes  valued  at  its  invoice  price  at  the  j)ort  of  lading; 
and  to  value  the  freight  at  §  the  amount  due  at  the  time  of  the  jettison  or  other 
Bacriflce.  But  in  some  ports  the  value  of  the  freight  is  determined  by  deducting 
from  the  gross  amount  the  wages  for  seamen,  pilotage,  etc.  In  New  York,  J  of  the 
freight  is  deducted. 

As  remarked  in  the  introduction  of  this  subject,  it  is  so  vast  and  complicated 
that  it  cannot  be  satisfactorily  treated  in  a  work  of  this  kind,  and  it  is  questionable 
whether  or  not  it  is  proper  to  present  it  in  works  of  this  kind.  It  is  a  subject  that 
merits  and  lias  received  the  attention  of  the  most  profound  jurists  of  both  hemi- 
Bpheres.  We  liave  briefly  introduced  it,  in  conformity  to  custom,  and  present  the 
following,  to  show  the  operation  for  a  general  and  i^articular  average  computation ; 

PROBLEMS. 

1.  The  steamer  City  of  Baltimore  left  New  Orleans  for  Philadelphia  with  a 
general  cargo  amounting  to  $150000.  During  the  voyage,  the  vessel  encountered 
very  severe  weather,  and  goods  to  the  amount  of  $4o00  were  thrown  overboard.  Masts 
and  rigging  were  cut  away  which  cost  $S10  to  replace;  one  anchor  cost  $250.  The 
expenses  of  the  vessel  in  port  while  being  repaired  were  $540.  The  charges  for 
protest  and  adjustment  of  average  were  $250.  The  gross  amount  of  freight  amounted 
to  $12420.    The  value  of  the  vessel  is  $90000. 

There  were  300  hhds.  of  sugar  on  board,  158  of  which  were  damaged  by 
water,  according  to  the  report  of  the  surveyor,  to  the  extent  of  34  hhds.,  which,  at 
$100  amounts  to  $3400. 

The  goods  and  vessel  were  fully  insured. 

What  is  the  general  and  particular  average  loss  per  cent.,  and  the  amount  of 
loss  sustained  by  each  contributing  interest,  estimating  the  cargo  at  the  port  of 
lading  invoice?  Ans.  2.44884  +  %  general  average  loss. 

11 J  %  particular  average  loss. 
$2203.97  to  be  contributed  by  vessel. 
$3673.26     " 
$  202.77     " 
$3400.00     " 


u 

"  cargo. 

a 

"  freight 

u 

"  sugar. 

GENERAL    AND    PARTICULAR   AVERAGE. 


529 


OPEKATION. 
General  Average  Loss. 

Masts,  rigging,  etc., 

Less  ^  deducted  because  uew 

Goods  thrown  overboard  .... 

Anchor,  cost  of  new 

Expenses  in  port  

Protest  and  average  adjustment 

Total  amount  of  general  average  loss 

Particular  Average  Loss. 

34  hhds.  sugar  ®  $100 


$810 
270 


i  540 

4500 

250 

640 

250 


$3400 


Contributory  Interests  to  General  Average  Loss. 


Value  of  steamer  (when  rei^aired) 

Cargo  (invoice  value) 

Freight 

Less  J,  allowed  for  expenses  .        .        .        . 

Total  amount  of  contributory  interests 


Loss. 

$  90000 

150000 

$12420 

4140 

8280 

Contributory  Interests  to  Particular  Average  Loss. 
300  hhds.  sugar  ®  $100 


$248280 


$30000 


Statement  to  find  the  General  Average  per 
cent  Loss. 


248280 


$  LOSS. 

6080 
100 


Statement  to  find  the  Particular  Average 
per  cent  Loss. 


30000 


$  LOSS. 

3400 

100 


2iU^%  =  2.44884  +  %.  11 J  % 

Statement  showing  the  amount  of  loss  that  each  contrihuting  interest  must  pay. 


Yessel,  $  90000  ®  2.44884% 

Cargo,  150000  ®  2.44884% 

Freight,  8280  ©  2.44884% 

Sugar,  30000®  11 J  % 


$2203.97 

3673.26 

202.77 

3400.00 


Eemarks. — In  actual  practice,  average  adjusters  individualize  every  con- 
signee's invoice,  and  specify  the  loss  on  each  article  of  goods.  They  also  itemize 
every  expense  incurred  in  the  settlement  of  the  various  losses. 

When  the  rate  per  cent  loss  contains  an  inconvenient  fraction,  as  is  the  case 
in  the  above  example,  it  is  extended  to  three,  four  or  more  places  of  decimals,  and 
a  table  prepared  therefrom  to  facilitate  the  operation  of  multiplying  the  amounts  of 
the  different  invoices  composing  the  cargo.  If  the  contnbuting  interests  were  few 
in  number,  time  might  be  saved  by  making  proportional  statements  to  find  the  loss 
that  «ach  interest  must  sustain. 


530  soule's  philosophic  practical  mathematics.  * 

2.  In  tbe  above  problem,  suppose  that  Jones  &  Co.  Lad  an  invoice  of  general 
mercliandise  valued  at  $2000,  fully  insured,  ■wbicli  sufi'ered  no  damage,  -what  -would 
be  the  sum  due  by  Jones  &  Co.  or  by  their  insurers?  Ans.  $48.97+. 

operation. 
$2000  X  2.44884+%  =  $48.97+. 

Note. — Since  tlie  cargo  is  ono  of  tlio  contribiitory  interests  to  the  general  average,  each 
owner  of  goods  in  tbe  cargo  (or  bis  insurer)  must  pay  on  his  invoice  the  general  average  per  cent 
of  loss. 

3.  In  the  same  problem,  suppose  that  Smith  &  Co.  had  an  invoice  of  sugar 
valued  at  $1000,  which  sufi'ered  no  loss,  what  would  be  the  sum  due  by  Smith  &  Co. 
or  by  his  insurers  ?  Ans.  $113,334. 

operation. 

$1000  X  11J%  =  $113.33J. 

Note. — The  loea  on  sngar  is  a  Particnlar  Average  Loss  inasmuch  as  it  was  not  incurred  for  the 
common  safety  of  the  vessel,  cargo  and  freight,  therefore  each  owner  of  sugar  (or  his  insurer)  must 
pay  on  his  invoice  the  particular  average  per  cent  loss. 

4.  The  ship  (jucen  of  the  Ocean,  on  her  voyage  from  Bombay  to  New  Orleans, 
was  struck  by  a  severe  gale  in  tbe  Gulf  of  Mexico,  and  her  sails,  rigging  and  upper 
works  were  torn  and  damaged  to  the  amount  of  $1500.  To  relieve  her  sufferings, 
goods  were  thrown  overboard  to  the  amount  of  $9400.  And  one  mast  was  cut 
away,  the  cost  of  which,  to  replace  it,  was  $3C0.  The  cost  of  the  surveys,  protest 
and  adjustment  of  average  amounted  to  $850. 

The  vessel,  when  repaired,  was  worth  $50000.  The  value  of  the  cargo  in 
New  Orleans  was  $110000.  The  freight  amounted  to  $9840.  The  cargo  and  vessel 
were  fully  insured. 

What  was  the  general  and  particular  average  loss  per  cent  sustained  by  each, 
of  the  contributory  interests  ? 

Ans.  6H44%  or  0.2980+%  general  average  loss. 
2%  particular  average  loss. 

OPERATION. 

General  Average  Loss. 

Value  of  goods  thrown  overboard         ....  $  9400 

Cost  of  mast $360 

Less  J  for  new 120                 .240 

Cost  of  protest,  surveys  and  adjustment  of  average  850 

Total  amount  of  general  average  loss         -  $10490 


GENERAL    AND    PARTICULAR    AVERAGE. 

Particular  Average  Loss. 

Cost  to  repair  damage  to  sails,  rigging  and  upper  works 
Less  J  deducted  for  being  new  .... 


Net  amount  of  particular  average  loss 
Contrihiitory  Interests  to  General  Average  Loss. 


Value  of  vessel  (after  being  repaired) 
Value  of  cargo  in  iS'ew  Orleans 

Freight 

Less  i,  allowed  for  expenses 


$9840 
3280 


Total  amount  of  contributory  interests 
Contributory  Interests  to  the  Particular  Average  Loss. 
Value  of  vessel 


531 


$1500 
500 

$1000 


$  5000(> 
110000 

6560 
$160560 


$50000 


Statement  to  iind  tlie  general  average  loss  per 
cent  that  each  interest  must  sustain. 


16G500 


$  LOSS 

10490 
100 

6HH%  Aus.     or  6.2980+%, 


Statement  to  fiutl  tlio  particular  average 

per  cent  loss  that  the  insurers  of 

the  vessel  must  sustain. 


50000 


1000 
100 


2%  Ans. 


Kife  Insurance. 


J  ^/-<-<-'r<-'/-W^,-M-'/-H-<-'H-!-W: 


1047.  Life  Insurance  is  a  contract  by  which  a  life  insurance  company,  in  con- 
sideration of  certain  payments  made  by  the  insured,  agrees  to  pay  a  specified  sum 
of  money  to  his  heirs  at  his  death,  or  to  himself  if  living,  at  a  specified  age. 

1048.  Life  Insurance  Companies  are  organized  on  the  plan  of  Stock  Com- 
panies, Mutual  Companies,  Mixed  Companies,  and  on  the  Co-operative  and  Benevo- 
lent Association  plan. 

Note, — See  Fire  Insurance  for  definition  of  each  plan,  page  511. 

IMPORTANCE  OF   LIFE  INSURANCE. 

1049.  The  subject  of  Life  Insurance  is  of  vast  importance  to  all  classes  of 
people,  the  rich  as  well  as  the  poor.  For  in  the  ever-evolving  drama  of  life,  fortunes 
often  disappear,  and  in  the  contests  for  wealth  and  fame  small  gains  to  the  many 
are  the  rule,  while  to  some,  cold  faced  poverty  is  the  life  and  death  companion. 
Hence  the  rich  should  insure  to  guard  against  misfortune;  those  who  are  not  rich 
should  insure  so  that  in  case  of  early  death,  their  families  and  those  dependent  upon 
them  for  support,  would  be  i)rotected  from  want  and  sufi'ering ;  or  in  case  of  old  age, 
they  would  possess  the  means  through  an  Endowment,  or  an  Annuity  Policy,  or  a 
Trust  Certificate,  of  self-support. 

From  the  policy  treasuries  of  the  various  forms  of  Life  and  Co-operative 
Insurance,  come  millions  of  dollars  annually  to  relieve  the  necessities  of  distressed 
widows,  of  orphan  children,  and  of  decrepit  people.  And  for  this,  thanks  and 
gratitude  are  due  to  the  wisdom,  to  the  humanity,  and  to  the  practical  sense  of  those 
who  iirovided  the  insurance. 

The  world  is  most  liberal  and  gracious  in  bestowing  upon  needy,  afflicted,  and 
suffering  humanity,  words  of  sympathy,  tears  of  pity,  and  prayers  of  supplication. 
But  these  gifts,  however  bounteous,  do  not  pny  rent  bills;  they  do  not  snpi)ly 
covering  for  naked  bodies;  they  do  not  provide  sustenance  for  hungry  stomachs,  or 
medicine  for  fevered  brains.  Therefore,  by  the  light  of  these  facts,  and  since  by 
law  of  nature  all  must  live  before  they  die,  we  conclude  that  it  is  the  first  and  higli- 
«st  duty  of  man  to  provide  for  himself  and  those  dependent  upon  him,  and  to  assist 
in  the  financial  charities  of  the  world.  And  all  this  we  submit  can  be  best  accom- 
plished through  Life  or  Co-operative  Insurance. 

(532) 


*  LIFE    INSURANCE.  533 

The  Policies  of  Life  Insurance  yarj'  in  their  conditions  and  are  named  as 
follows : 

1050.  Ordinaey  Life,  Limited  Payment  Life,  Teem  Endowment, 
Reserve  Endowment,  Annuity,  Tontine  Savings  ok  lNVF:sTiMENT,  Joint 
Life  Policy,  and  Trust  Certificate  Policy. 

There  are  also  some  other  special  policies  issued  by  some  companies. 

1051.  An  Ordiiiary  Life  Policy  is  one  on  -which  the  premium  is  paid  annually 
during'  tlie  life  of  tlie  insured,  and  at  his  death,  the  amount  of  the  policy  is  jtaid  to 
the  party  named  therein. 

1052.  A  Limited  Payment  Life  Policy  is  one  on  which  the  premium  is  paid 
annually  for  a  certain  number  of  years,  spccilied  at  the  time  the  jiolicy  is  issued  or 
until  the  death  of  the  insured,  should  that  occur  before  the  elapse  of  the  time  speci- 
fied. 

This  class  of  policies  is  also  issued  on  single  payments,  or  on  5,  10  or  more 
annual  jiayments.  Wlien  issued  on  one  jiayment,  the  insured  receives  annual  cash, 
dividends  from  the  company. 

1053.  A  Term  Policy  is  one  payable  at  the  death  of  the  insured  if  it  occur 
during  a  given  number  of  years,  the  j^ayment  of  the  annual  jiremiums  to  continue 
till  the  ])olicy  expires. 

1054.  An  Endowment  Policy  is  one  that  is  paid  to  the  insured  at  the  end  of 
a  specified  number  of  years,  or  to  his  heirs  in  case  he  should  die  within  the  time. 
The  premium  is  payable  annually,  or  in  cash,  in  5  or  10  annual  pajTuents. 

1055.  A  Reserve  Endowment  Policy  is  one  combining  the  Life  and  Endow- 
ment Plans,  the  premium  being  the  same  as  in  a  Life  Policy,  but  the  insurance 
terminates  at  such  a  time  as  the  insurer  names,  when  an  endowment  will  be  paid 
equal  to  the  legal  reserve  of  the  policy. 

1056.  An  Annuity  Policy  is  one  which  secures  to  the  insured  the  payment 
of  a  specified  sum  annually,  as  long  as  he  lives.  The  premium  is  jjaid  in  one  cash 
payment. 

1057.  A  Tontine  Savings  or  Investment  Policy  is  one  payable  at  the  death 
of  the  insured,  but  in  which  the  excess  of  premiums  received  over  the  claims  by 
death  and  expenses,  is  divided  at  the  end  of  a  specified  jieriod  of  time,  among  the 
survivors  of  those  insured  for  that  specified  time,  and  at  this  time  these  survivors 
may  either  accept  the  money  as  in  an  Endowment  Policy,  or  change  the  form  of 
their  jjolicy  as  may  be  agreed. 

1058.  A  Joint  Life  Policy  is  one  payable  at  the  death  of  the  first,  of  two  or 
more  persons,  to  the  survivor.  The  ijremium  is  payable  in  the  same  manner  as  in 
Life  Policies. 

1059.  A  Trust  Certificate  Policy  is  a  policy  issued  at  a  rate  of  premium  much 
lower  than  the  ordinary  policies,  and  by  whicli  tlie  beneficiary  named  therein  receives 
a  specified  number,  more  or  less,  as  may  be  desired,  of  annual  payments  after  the 
death  of  the  insured. 


534  soule's  philosophic  practical  mathematics.  * 

For  the  heads  of  families,  this  is  oue  of  tlie  best  methods  of  insurance,  as  by 
it  the  money  can  neither  be  lost  nor  squandered,  and  the  family  receives  a  certain 
annual  income  to  i^rotect  them  from  care,  to  educate  the  children,  etc. 

1060.  A  Non-forfeiting  Policy  is  one  which  does  not  become  void  because  of 
the  non-payment  of  i)remiums;  provided  that  two  or  three  annual  payments  have 
been  made  thereon. 

Note. — The  limited  forfeit  law  of  life  insurance  of  Massachusetts,  makes  the  policy  non- 
forfeiting -svheu  two  full  annual  payments  have  been  made;  the  laws  of  the  State  of  New  York, 
make  the  policy  non-forfeiting  wheu  three  full  annual  payments  have  been  made. 

1061.  A  Dividend  in  Life  Insurance  Companies  is  a  share  of  the  proiits  or 
suri^lus  paid  to  the  stockholders  in  stock  companies,  to  policy  holders  in  mutual 
companies,  and  to  stockholders  and  policy  holders  iu  mixed  companies,  after  the 
liabilities  of  the  company  are  reserved. 

1062.  The  Reserve  Fund  of  a  Life  Insurance  Company  is  a  sum  which, 
invested  at  a  given  rate  of  interest,  together  with  the  maturing  premiums  on  exist- 
ing policies  will  x^roduce  a  sufficient  revenue  to  pay  all  obligations  as  they  mature 
or  arise. 

1063.  Surplus  is  the  sum  left,  after  providing  for  the  liabilities,  claims  and 
exi^enses  of  the  comi^any. 

1064.  Script  is  a  Certificate  entitling  the  policy  holder  to  certain  profits  or 
surplus,  when  payable. 

1065.  Loading  or  Margin,  is  a  percentage  added  to  the  net  premium  to  defray 
ex])enses  and  provide  for  au  excess  of  mortality. 

1066.  The  Reserve  of  Life  Insurance  Companies  is  that  part  of  the  premiums 

of  all  policies,  M-ith  the  interest  thereon,  which  is  reserved  or  set  aside  as  a  fund  for 

the  payment  of  policies  when  they  become  due. 

The  law  of  New  York  State  requires  Life  Insurance  Companies  to  reserve  such  a  part  of  the 
premiums  as  will,  at  4  per  cent  interest,  amount  to  a  sufficient  sum  to  pay  all  policies.  The  State 
of  Massachusetts  specifies  4i  per  cent  interest  on  the  reserve. 

1067o  "Value  of  a  Policy.  "The  net  value  of  a  policy  is  the  difference 
between  the  net  single  i)remium  for  the  sum  insured  at  the  age  of  the  policy  holder 
when  the  policy  is  valued,  and  the  present  value  of  all  future  net  premiums  calcula- 
ted to  be  received  on  the  life  of  the  party  insured.  The  gross  value  of  a  policy  is 
the  difference  between  the  net  single  premium,  as  given  above,  and  the  present 
value  of  all  future  gross  premiums  to  be  received  on  the  policy." 

1068.  The  Surrendered  Value  or  Market  Value  of  a  policy  is  commonly 
determined  by  deducting  from  the  reserve  25  to  50  per  cent,  as  a  surrender  charge, 
and  to  i)ay  the  remainder  as  an  equitable  surrender  value. 

Much  discussion  has  followed  this  method  of  finding  the  surrendered  value 
of  policies,  to  which  we  refer  those  interested  to  the  "  Principles  and  Practice  of 
Life  Insurance." 

1069.  The  Rate  of  Premium  is  based  mainlj^  upon  the  probability  of  life  as 
shown  in  the  following  tables;  but,  specifically,  it  involves  six  elements:  1.  The 
reserve  fund,  as  above  defined.    2.  The  probable  rate  of  interest  on  the  reserve 


LIFE    INSURANCE.  S.'^S 


O' 


ftiiid.  3.  The  cost  of  mortality^  wlilcli  is  the  estimated  amount  of  each  insurer's 
share  of  the  losses  by  death  each  year.  4.  The  loading,  which  is  the  part  of  the 
premium  required  for  expenses.  5.  The  age  of  the  insured.  6.  The  period  of  insur- 
ance and  the  kind  of  policy. 

CO-OPEKATIVE  OR  BEXEVOLENT  INSURANCE  ASSOCIATIOXS. 

1070.  Associations  of  this  kind,  of  which  there  are  several,  as  described  on 
page  509,  difi'er  materially  in  their  general  features  and  plan  of  work,  from  the 
regular  Life  Insurance  Companies.  They  also  differ  from  one  another  in  some 
respects,  while  on  many  points  there  is  a  close  similarity. 

The  Knights  of  Honor,  which  is  one  of  the  leading  Co-operative  Associations 
or  orders  of  this  class,  works  on  the  following  plan  : 

Members  are  admitted  by  vote  into  social  societies  called  Lodges,  and  on  the 
payment  of  an  assessment  of  from  one  to  two  dollars,  according  to  age,  receive  a 
Benefit  Certificate  for  $2000.  These  certificates  are  the  equivalent  of  the  policies 
in  Life  Insirrance  Companies  and  are  payable  at  the  death  of  the  member  to  the 
beneficiary  named  therein.  The  continuance  of  the  Benefit  Certificate  is  conditioned 
on  the  prompt  payment  of  all  subsequent  assessments  upon  the  holder  or  member. 

To  replenish  the  fund  for  the  payment  of  the  Benefit  Certificates,  assessments 
are  made  at  different  times  on  a  graded  })lau  according  to  the  age  of  the  members, 
which  vary  in  amount  from  $1  to  $1.50.  The  sum  thus  raised  is  paid  into  the 
General  Fund  called  the  Widows'  and  Orj)hans'  Benefit  Fund,  from  which  is  paid 
the  Benefit  Certificates  of  all  deceased  members  until  the  fund  is  reduced  below 
$2000.  Then  another  assessment  is  made  and  paid,  and  thus  the  fund  is  maintained 
and  the  certificates  of  deceased  members  are  jiaid. 

Very  similar  principles  and  methods  govern  nearly  all  the  Co-operative  Insur- 
ance Associations,  and  thus  they  ofl'er  economical  advantages  to  all  who  desire  to 
insure,  but  more  especially  do  they  commend  themselves  to  persons  of  small  or  limi- 
ted means. 

By  the  co-operative  plan  of  insurance,  the  assessments  are  collected  and  paid 
over  to  the  general  fund  by  the  officers  of  the  societies  or  lodges,  without  charge  or 
commission,  and  the  reserve  and  surplus,  it  is  claimed,  are  in  the  pockets  of  the 
members. 

The  Knights  of  Honor  issue  half  rate  Benefit  Certificates  for  $1000.  Some 
of  the  other  Co-operative  Associations  issue  Benefit  Certificates  ranging  from  $500 
to  $5000. 

The  amount  of  the  assessments  is  determined  upon  the  expectancy  of  life, 
the  age  of  the  member  when  he  joins  the  order,  and  some  other  elements. 

1071.  Debenture  Insurance  or  Investment,  is  an  investment  evidenced  by 
a  written  instrument  siiecity.ng  the  amount  invested,  the  revenue  to  be  derived 
therefrom,  the  conditions  upon  which  it  is  received,  the  manner  of  payment,  etc. 

While  some  Debenture  Companies  are  conducted  ethically,  and  are  based  upon 
correct  financial  and  mathematical  principles  to  ensure  their  solvency,  many  are 
founded  ui)on  a  chimerical  and  delusive  basis,  and  are  conducted  through  fraud  and 


5^,6  soule's  philosophic  practical  mathematics.  * 

false  pretense.  Hence  all  -who  would  patronize  Debenture  Companies  sLould 
critically  investigate  their  Charter,  the  financial  princi])les  upon  which  they  work, 
the  management,  promises,  etc.  of  the  company,  before  they  invest  in  Debenture 
Bonds. 

1072.  From  the  Manual  of  the  Life  Association  of  America,  and  the  Principles 
and  Practice  of  Life  Insurance,  we  collate  the  following  Statistics  and  the  Tables  of 
Comparative  Expectancy  of  Life  and  of  Mortality. 

BASIS  OF  LIFE  mSURANCB. 

1073.  The  duration  of  human  life  is  governed  by  a  law  as  regular  in  its  opera- 
tion as  any  other  law.  While  the  actual  continuance  of  any  individual  life  is  wholly 
uncertain,  the  average  duration  of  a  large  number  of  lives  can  be  accurately  fore- 
told. 

Ten  thousand  persons  of  ordinary  health,  at  the  age  of  35  years,  will  survive 
31  years  longer,  on  an  average.  Consequently,  the  exi)ectation  of  life  for  a  person 
35  years  of  age  is  31  years. 

The  basis  on  which  the  whole  structure  of  life  insurance  rests  is  the  laic  of 
mortality  among  the  human  race.  Strange  as  it  may  seem,  war,  pestilence,  and 
famine  cause  but  slight  variations  in  the  operation  of  this  law.  In  spite  of  these 
destructive  agencies,  happening  as  it  were  by  chance,  :n  spite  of  all  the  unforeseen 
accidents  and  catastrophies  to  which  human  life  is  exposed  ;  in  spite  of  the  absolute 
irncertainty  of  prolonged  life  in  any  individual  case,  yet  the  average  duration  of 
existence  among  a  large  number  of  persons,  say  10,000  or  100,000  is  a  mathematical 
verity — a  certainty  as  fixed  and  invariable  as  any  other  law  of  nature. 

HOW  HAS  this  law  BEEN  DISCOVERED  AND  WHAT  ARE  THE  PROOFS  OF  IT?— 

1074.  Within  the  last  200  years,  close  observation  and  actual  measurement  of 
the  duration  of  lil'e  liave  been  made  among  different  classes  of  persons  in  different 
countries,  and  under  entirely  different  sets  of  circumstances.  These  tests,  made 
independently  of  one  another,  extended  to  embrace  large  iiumbers  of  persons — in 
some  instances,  entire  populations — so  fully  corroborate  and  verify  one  another,  that 
there  is  left  no  reason  to  doubt  that  the  law  of  mortality  is  accurately  determined, 
and  that  it  is  in  fact  an  absolute  law  of  nature. 

The  separate  and  distinct  observations  commonly  adduced  as  proof  or 
authority  for  tliis  law  are  thirteen  in  number.  The  following  are  those  in  general 
use  in  America  and  England:  The  American  Experience,  the  Carlisle,  the  Farr  or 
English  Ijife  No.  3,  the  Actuaries  or  Combined  Experience,  and  the  ISTorthamiJton. 

1075.  The  Am,erican  Experience  is  made  from  actual  experience  of  several 
leading  American  Companies,  extending  over  a  period  of  25  years. 

107G.  The  Carlisle  table  is  a  record  of  the  mortality  of  the  town  of  Carlisle, 
England,  from  1778  to  1787. 

1077.  The  Farr  or  English  Life  No.  3,  is  a  record  of  mortality  of  all  ages 
throughout  Great  Britain,  from  1838  to  1844. 

1078.  The  Actuaries  or  Comhincd  Experience  is  a  record  of  mortality  based  on 
the  record  of  02,537  assured  lives  in  17  Life  Companies  in  England. 

The  Northampton  Table  was  prepared  by  Dr.  Price  of  Northampton,  England. 


LIEE    INSURANCE. 


537 


1079. 


Table  No.  1. 
COMPAEATIVE  EXPECTANCY  OF  LIFE. 


A  Tal(!e  ebowing  the  expectancy  of  life  in  years  and  ImnilietUlis  of  a  year,  deduced  from  four 
English  and  the  American  Experience  Tables. 


Farr  or 

Actuaries 

Ammoau 

Farr  or 

Actuaries  A 

iieiicaa 

Age. 

North- 

Carlisle. 

JSnslisU 

or  Com- 

Experi- 

Age. 

Xorth. 

Carlisle, 

EndisU 

vr  Com-      i 

xperi- 

ampton. 

Life, 

bined  JEx- 

euce. 

amptou. 

IJle, 

bined  Ex- 

eace. 

Ko.  3. 

perience. 

No.  3. 

pDiicncu. 

1 

32.74 

44.67 

46.65 

53 

16.54 

18.97 

17.67 

18.16 

18.79 

2 

37.79 

47.55 

48.83 

.... 

» -  . » 

54 

16.06 

18.28 

17.06 

17.50 

18.09 

3 

39.55 

49.81 

49.61 

.... 

• .  -  • 

55 

15.. 58 

17..58 

16.45 

16.86 

17.40 

4 

40.58 

50.76 

49.81 

•  •   □  ■ 

•  •  •  • 

56 

15.10 

16  89 

15.86 

16  22 

16.72 

5 

40.84 

51.24 

49.71 

•  •  •  • 

■  •  >  ■ 

57 

14.63 

16.21 

15.26 

15. .59 

16.05 

6 

41.07 

51.16 

49  39 

3   •    •• 

•  >  >  • 

58 

14.15 

15. .55 

14.68 

14.97 

15.39 

7 

41.03 

50.79 

48.92 

•   ••• 

■ . . . 

59 

13.68 

14.92 

14.10 

14.37 

14.74 

8 

40.79 

50.24 

48.37 

•    •     •   ■ 

■  •  •  • 

60 

13.21 

14.34 

13.53 

13  77 

14.09 

9 

40.39 

49.57 

47.74 

•    •     •    • 

• .  . . 

61 

12.74 

13.82 

12.96 

13  18 

13.47 

10 

39,78 

48.82 

47.05 

48.36 

48.72 

62 

12.28 

13.31 

12.41 

12.61 

12.86 

11 

39.14 

48.04 

46.31 

47.68 

48  08 

63 

11.81 

12.81 

11.87 

12.05 

12.26 

12 

38.49 

47.27 

45.54 

47.01 

47.45 

64 

11.35 

12.30 

11.34 

11.51 

11.68 

13 

37  83 

46.50 

44.76 

46.33 

46.82 

65 

10.88 

11.79 

10.82 

10.97 

11.10 

14 

37.17 

45.74 

43.97 

45.64 

40.16 

66 

10.42 

11.27 

10.32 

10.46 

10..54 

15 

36.51 

44.99 

43.18 

44.96 

45.50 

67 

9.95 

10.75 

9.83 

9.96 

10.00 

16 

35.85 

44.27 

42.40 

44  27 

44.85 

68 

9.. 50 

10.23 

9-36 

9.47 

9.48 

17 

35.20 

43  57 

41.64 

43.58 

44.19 

69 

9.05 

9.70 

8.90 

9.00 

8.98 

18 

34  58 

42.87 

40  90 

42.88 

43.53 

70 

8.60 

9.15 

8.45 

8.54 

8.48 

19 

33.99 

42.16 

40.17 

42.19 

42.87 

71 

8.17 

8.65 

8.03 

8.10 

8.00 

20 

33.43 

41.46 

39  48 

41.49 

42.20 

72 

7.74 

8.16 

7.62 

7.67 

7..54 

21 

32.90 

40.75 

38.80 

40.79 

41.53 

73 

7.32 

7.72 

7.22 

7.26 

7.10 

22 

32.39 

40.03 

38,13 

40.09 

40.85 

74 

6.92 

7.33 

6'85 

6.80 

6.68 

23 

31.87 

39.31 

37.46 

39.39 

40.17 

75 

6.54 

7.00 

6.49 

6  48 

6.28 

24 

31.36 

38.58 

36.79 

38.68 

39.49 

76 

6.18 

6.99 

6.15 

6.11 

5.88 

25 

30.85 

37.86 

36.12 

37.98 

38.81 

77 

5.83 

6.40 

5.82 

5.V6 

5.48 

26 

30.33 

37.13 

35.44 

37.27 

38.11 

78 

5.48 

6.11 

5.51 

5.42 

5.10 

27 

29.82 

36.40 

34.77 

36..56 

37.43 

79 

5.11 

5.80 

5.21 

5.09 

4.74 

28 

29.30 

35.68 

34.10 

35  86 

36.73 

80 

4.75 

5.51 

4.93 

4.78 

4.38 

29 

28.79 

34.99 

33.43 

35.15 

36.03 

81 

4.41 

5.20 

4.66 

4.48 

4.04 

30 

2H.27 

34  34 

32.76 

34.43 

35.33 

82 

4.09 

4.93 

4.41 

4.18 

3.71 

31 

27.75 

33.68 

32.09 

33  72 

34.62 

83 

3  80 

4.65 

4.17 

3.90 

3.. 39 

32 

27.24 

33.02 

31.42 

3301 

33.92 

84 

3..5S 

4.39 

3.95 

3.03 

3.08 

33 

26.72 

32  36 

30.74 

32.30 

33.21 

85 

3.. 37 

4.13 

3.73 

3.. 36 

2.77 

34 

26.20 

31.68 

30.07 

31.58 

32.50 

86 

3.18 

3.90 

3..53 

3.10 

2.47 

35 

25.68 

31  00 

29.40 

30.87 

31.78 

87 

3.01 

3.71 

3.34 

2.84 

2.19 

36 

25.16 

30.32 

28.73 

30.15 

31.07 

88 

2.86 

3.60 

3.16 

2.. 59 

1.91 

37 

24.64 

29.63 

28.06 

29.44 

30.35 

89 

2.66 

3.47 

3.00 

2.35 

1.66 

38 

24  12 

28  65 

27  39 

28-72 

29.63 

90 

2.41 

3.28 

2.04 

2.11 

1.42 

39 

23.60 

28.27 

26.72 

28.00 

28.90 

91 

2.08 

3.26 

2.69 

1.89 

1.19 

40 

23.07 

27.61 

26.06 

27  28 

28.18 

92 

1.75 

3.37 

2.55 

1.67 

.98 

41 

22.56 

26.97 

25.39 

26.56 

27.45 

93 

1.37 

3.48 

2.41 

1.47 

.80 

42 

22.04 

26.34 

24.73 

25.84 

26.72 

94 

1.05 

3.53 

2.29 

1.28 

.64 

43 

21  54 

25  71 

24.07 

25.12 

25.99 

95 

.75 

3.53 

2J7 

1.12 

.50 

44 

21,03 

25.09 

23.41 

24.40 

25.27 

96 

.; 

)0 

3.46 

2.06 

.99 

45 

20.52 

24.45 

22.76 

23.69 

24.54 

97 

... 

3.28 

1.95 

.89 

46 

20.02 

23.81 

22.11 

22.97 

23.81 

98 

■  >  . 

3.07 

1.85 

.75 

47 

19.51 

23.17 

21.46 

22.27 

23.08 

99 

... 

2.77 

1.7G 

.50    ■ 

,  . 

48 

19.00 

22.50 

20,82 

21.56 

22.36 

100 

.  .  . 

2  28 

1.68 

49 

18.49 

21.81 

20.17 

20.87 

21.63 

101 

1.79 

.... 

50 

17.99 

21.11 

19.54 

20.18 

20.91 

102 

... 

1..30 

51 

17.50 

20.39 

18.90 

19..50 

20.20 

103 

.  .  . 

.83 

•  .  .  • 

52 

17.02 

19.68 

18.28 

18.82 

19.49 

104 

.50 

.... 

.. 

538  soule's  philosophic  tractical  mathematics.  * 

[Table  Ko.  2\. 
1080.         COMPARATIVE     MORTALITY    TABLE,— Showing  (lie  number  expected  «o  die  out  of  1000 


persona  entering  each  year,  aeeordiiigto  three 

[English  and  the  American  E.\perienco  Tahles. 

Age. 

Aiuerieau  Experience. 

Carhsle. 

Actuaries. 

English  Life  Ko.  3. 

Age. 

10 

7.49 

«... 

• .  . . 

15 

7.63 

.... 

.... 

•M 

7.80 

.... 

.... 

21 

7.85 

.... 

.... 

.... 

22 

7  90 

.... 

23 

7.95 

2.t 

8.01 

25 

8.06 

Y.Sl 

7!77 

9^20 

25 

2(j 

8.13 

7.36 

7.88 

9.38 

26 

27 

8.19 

7.76 

8.00 

9  55 

27 

28 

8.26 

8.69 

8.13 

9.74 

28 

29 

8.34 

9.82 

8.27 

9.93 

29 

30 

8.42 

10.10 

8,42 

10.13 

30 

31 

8.51 

10.20 

8., 57 

10. .34 

31 

32 

8,60 

10.13 

8.74 

10.. 56 

32 

33 

8-71 

10.05 

8.91 

10.80 

33 

34 

8.83 

10.15 

9.09 

11.05 

34 

35 

8.94 

10.25 

9,28 

11.33 

35 

3lj 

9  08 

10.55 

9.48 

11.62 

36 

37 

9^23 

10.85 

9.68 

11.94 

37 

38 

9.40 

11.16 

9.90 

12.29 

38 

39 

9.58 

11.87 

10.13 

12.65 

39 

40 

9.79 

13.05 

10.36 

13  06 

40 

41 

10.00 

13.77 

10,61 

13.18 

41 

42 

10.25 

14.37 

10,89 

13,94 

42 

43 

10.51 

14.. 58 

1 1 ,  25 

14.44 

43 

44 

10.81 

14.79 

11,69 

14,97 

44 

45 

11.16 

14,80 

12.21 

15,. 54 

45 

4r> 

11. 56 

14.81 

12.83 

16.15 

46 

47 

12.00 

14.60 

13  51 

16.18 

47 

48 

12.. 50 

13,93 

14.25 

17.49 

48 

49 

13.10 

13.68 

15.06 

18  23 

49 

50 

13.78 

13,41 

15.93 

19.02 

50 

51 

14., 54 

14.29 

16.89 

20.42 

51 

52 

15  38 

15.20 

17.94 

21.45 

.52 

53 

16.33 

16,14 

19.09 

22,51 

53 

51 

17.39 

16.89 

20.31 

23,64 

54 

55 

18.57 

17.92 

21.66 

24.85 

.55 

56 

19.88 

19.00 

23.12 

26.17 

56 

57 

21.33 

20.89 

24.67 

27.63 

57 

58 

23,10 

24.20 

26,38 

29.25 

58 

59 

24.72 

28.27 

28.24 

31.05 

59 

60 

26.69 

33.48 

30,33 

33.05 

60 

61 

28.88 

35.78 

32,61 

.35.29 

61 

62 

31.29 

37.40 

35  12 

37.77 

62 

63 

33.94 

38.25 

37.83 

40.53 

63 

64 

36.87 

39.77 

40.82 

43.60 

64 

65 

40.12 

41.08 

44.08 

46.98 

65 

GG 

43.76 

42.50 

47,01 

50.71 

66 

67 

47.64 

44.38 

51.47 

.51.83 

67 

68 

52.00 

46.45 

55.63 

.59.33 

68 

69 

56.76 

49.10 

60.08 

64 .  25 

69 

70 

61  99 

51.64 

61.93 

69.62 

70 

71 

67.66 

58.85 

70.01 

72.70 

71 

72 

73.73 

68.12 

75,80 

78.54 

72 

75 

94,36 

80 

144,46 

85 

235.55 

90 

4.54.55 

94 

8,57.14 

The  above  table  sboTva  how  many  of  lOOD  persons  will  die  in  the  year,  for  each  age  given. 

Tlie  foregoing  tables  and  the  interest  ami  annnitv  tables,  given  farther  on,  furiiisli  llie  principal  data  for  the  calcula- 
tions of  Life  Insurance  (Companies.  It  is  not  claimed  by  life  insurers  or  others,  that  tlie  duration  of  any  one  lile  can  lie 
detenuin.'d  hut  that  of  a  ninltitude  of  lives  is  easily  ascertained.  Ily  reason  of  this,  the  iiltim.ate  results  ot  the  calcula- 
tions o,"  lile  insurers  are  always  correct,  for  if  one  person  lives  longer  than  those  insui'ed  vnth  hmi,  he  helps  those  who  Ul9 
early;  ami  if  ho  dies  early,  they  help  him. 


LIFE    INSURANCE. 


539 


Table  No.  3. 
1081.  SHOWING  THE  ANNUAL  TEEMIUM  BATES 

For  an  Insurance  of  $1000. 


LIFE  POLICIES.-PAUBLE  AT  BEATH  DULY. 

Enflowient  Policies.— PayaMe  as  inilicated,  or  at  Eealli  if  prior,  u 

PaymcDts  tc 

contiDne  for 

Ten  Tear.s. 

Fifteen 
Tears. 

Twenty 
Tears. 

Twenty 
Yeai-3.    Ten 
Paymenta. 

Age. 

Life 

Teu  Tears 

Fifteen 
Tears. 

Twenty 
Tears. 

Age. 

21 

$1860 

$40.50 

$30.80 

$26.10 

$105.40 

$66.90 

$48.20 

$76.30 

21 

22 

19  10 

41.20 

31.30 

26.60 

105.50 

67.00 

48.30 

76.50 

22 

23 

19.50 

41.90 

31  90 

27  00 

105.70 

67.10 

48.40 

76.60 

23 

24 

20.00 

42.70 

32.50 

27,60 

105  80 

67.20 

48.60 

76.70 

24 

25 

20  50 

43.50 

33.10 

28.10 

105.90 

67.40 

48.70 

76.90 

25 

26 

21  00 

44.30 

33  80 

28  60 

106  00 

67.50 

48.90 

77.00 

•26 

27 

21.50 

4.5.20 

34  40 

29.20 

106.10 

67.60 

49.00 

77.20 

27 

28 

22-10 

46  10 

35  10 

29.80 

106.30 

67.80 

49.20 

77.40 

28 

29 

22,70 

47.00 

35  90 

30  50 

10640 

68.00 

49.40 

77.60 

29 

30 

23  30 

48,00 

36.00 

31  10 

106.60 

68.20 

49.60 

77.80 

30 

31 

24,00 

49.10 

37  40 

31  80 

106.80 

68.30 

49.80 

78.10 

31 

32 

24,70 

50.10 

38.30 

32  60 

107.00 

t)8  60 

50.10 

78.30 

32 

33 

25.50 

51.20 

39.10 

33  30 

107.20 

68  80 

50.30 

78.00 

33 

34 

26  30 

52,40 

40.00 

34  10 

107.40 

69.00 

50.60 

78.90 

34 

35 

-27.10 

53,60 

41  00 

35.00 

107.60 

69.30 

50.90 

79.20 

35 

36 

28  00 

54.80 

42,00 

.35,80 

107  80 

69  60 

51.30 

79.60 

36 

36 

29,00 

56  20 

4300 

36  80 

108  10 

69  90 

51.70 

80.00 

37 

38 

30  00 

.57.50 

44,10 

37.70 

lOS  10 

70,20 

52.10 

80.40 

38 

3;t 

31.10 

.59,00 

45  30 

38  80 

108,70 

70,60 

52.50 

80.90 

39 

40 

32.20 

60  40 

46.50 

39  80 

109  10 

71.00 

53.00 

81.40 

40 

41 

3340 

62,00 

47,70 

41  00 

10940 

71.50 

53.60 

81.90 

41 

42 

34.70 

63.60 

49  00 

42.20 

109,80 

72,00 

54.20 

82.60 

42 

43 

36.10 

65.30 

50,40 

43..50 

110  30 

72.50 

54.80 

83.30 

43 

44 

37.50 

67.10 

51  90 

44.80 

110.80 

73.10 

55.60 

84.00 

44 

45 

39.10 

69.00 

53  40 

46.20 

111  30 

73  80 

56.40 

84.90 

45 

46 

40.70 

70.90 

55  10 

47.80 

11200 

74.60 

57.30 

85.80 

46 

47 

42.50 

72.90 

56  80 

49.40 

112  60 

75.40 

58.30 

86.80 

47 

48 

44.40 

75.10 

58,60 

51.10 

113  4(1 

76  30 

B9.40 

88  00 

48 

49 

46.40 

77.30 

60  50 

52.90 

114.20 

77.30 

60.70 

89.20 

49 

50 

48. .50 

79.60 

62.50 

54,80 

115.10 

78.40 

62.00 

90.60 

50 

51 

.50  80 

82.10 

64.60 

56  90 

116.10 

79.70 

63.50 

92.10 

51 

52 

.53  30 

84.60 

06  90 

59.10 

117.20 

81.00 

65.20 

93.70 

52 

53 

.55.90 

87.30 

69  20 

61.40 

118.40 

82.50 

67.00 

95.50 

53 

54 

58  70 

90.10 

71,80 

63  90 

119.80 

84.10 

68.90 

97.50 

54 

55 

61  60 

93.00 

74.40 

66.00 

121.20 

85.90 

71.10 

99.60 

55 

56 

64.80 

96.10 

77-30 

69.50 

122.80 

87.90 

73.50 

101.90 

56 

57 

68.20 

99,:i0 

80  30 

72.60 

124.60 

90.10 

76.10 

104.40 

57 

58 

71.80 

102,70 

83  .50 

75.80 

126.50 

92.50 

78.90 

107.10 

58 

59 

75.70 

106  .SO 

86.90 

79.40 

128.70 

95.10 

82.10 

110.10 

59 

CO 

79,90 

110.10 

OOfiO 

83.20 

131.00 

98.00 

85.50 

113.20 

60 

61 

84.30 

114.10 

94  .50 

87.30 

133.60 

101.20 

61 

62 

89,10 

118  30 

98  70 

91.70 

136.40 

104  70 

62 

63 

94.20 

122  70 

103  20 

96.40 

139.60 

108.50 

63 

64 

99  60 

127  50 

108  (10 

101.50 

143.00 

112.70 

64 

65 

105.. 50 

132.50 

113  20 

107.10 

146.80 

117  30 

65 

Note  1.  — The  above  table  is  taken  by  permission  from  The  Mutual  Life  Insurance  Com- 
pany of  New  York. 

Note  2.— Policies  which  do  not  participate  in  the  dividends  of  a  company  are  issued  at  a 
lower  rate  of  preiimim,  than  given  in  the  above  table. 

NoTK  3.— TLe  above  rates  <if  preminm  are  for  annual  payments.  In  some  cases  the  payments 
are  made  semi-annually  or  quarterly,  when  a  small  i>er  cent  is  added  to  the  annual  payment  before 
dividing  bv  2  or  4. 


540  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

PROBLEMS. 

1.     What  is  the  aiiuual  premium  for  a  mau,  aged  21  years,  on  a  Life  Policy 
for  $oOO0 1  Ans.  $93.00. 

OPERATION. 

*^*'"^  Explanation.— The  premium  on  $1000  at  the  age  of  21  is,  as- 
shown  in  the  table,  $18.60;  hence,  for  $5000,  it   is  5  times- 


$93.00    Ans.  as  much. 

2.  What  is  the  anuual  premium  on  a  lO-j-ear  Endowment  Policy  for  $5000, 
the  age  being  21  years  ?  "  Ans,  $527.00. 

OPERATION. 

$105.40 

5  Explanation. — Since  the  premium  on  $1000  for  the   cou- 

'  ditions  named  is  $105.40,  for  $5000  it  is  5  times  as  much. 

$527.00 

3.  A  man  is  32  years  old  and  takes  out  a  20year  Endowment  Policy  for 
$25000.     What  is  his  annual  premium  ?  Ans.  $1252.50, 

4.  A  person,  28  years  old,  took  out  a  15-year  Endowment  Policy  for  $2500. 
What  is  the  anuual  premium  1  Ans.  $109.50. 

5.  What  is  the  annual  premium  on  a  Life  Policy  for  $8500,  payable  in  ten 
annual  payments,  the  insured  being  44  years  of  age?  Ans.  $570.35. 

C.  A  man,  21  years  old.  took  out  a  10-year  Endowment  Policy,  and  at  its 
maturity,  the  premiums  paid  amounted  to  $5270.  What  was  the  amount  of  the 
policy?'  Ans.  $5000. 

OPERATION.  Explanation. — Since  the  premium  for  10  years  Teas  $5270, 

for  one  year  it  is  the  -,^„-  p.art,  $527      Then  since  tlie  annual 

^^(,„„    ,     ..„   *-■"•)'"  payment   on   a  $1000   ten-year   Endowment    Policy   is,    per 

<tic>^/U  -:-  W  —  <l>0-/  taijle,  $105.40  for  a  person  aged  21  years,  there  were  as  many 

527  -^  105.40  =  5  thousands       thousands  insured  as  $527  is  equal  to  $105.40,    which  is  5 
5  X   $1000  :=  $5000  times.     Hence  there  was  $5000  insurance, 

7.  Mr.  X,  aged  22  years,  took  out  a  Life  Policy  payable  in  15  years.  After 
making  eight  annual  payments  amounting  to  $870.40,  he  died.  What  was  the  face 
of  his  policy  ?  Ans.  $3500. 

8..  A  man,  21  years  old,  paid  $93.00  i>remium  annually,  on  a  policy  of  $5000, 
What  was  the  rate  per  cent?  Ans.  1.86%. 

OPERATION    INDICATED. 


5000 


93 

100  or,  5000 


100% 
93 


9.  The  annual  premium  on  a  $25000  Endowment  Policy,  payable  in  twenty 
years,  for  a  person  aged  32,  is  $1252.50.     What  is  the  rate  per  cent?     Ans.  5.01%. 

10.  A  man,  28  years  old,  paid  an  annual  premium  of  6.78  per  cent  on  a  15- 
year  Endowment  Policy  of  $2500.  What  was  the  annual  premium,  and  how  much 
would  he  pay  in  all  ?  Ans.  $109.50  annual  premium. 

$2542.50  total  premium. 

11.  A  man,  25  years  old,  took  out  a  Life  Policy  for  $100000  and  paid  the 
regular  annual  premiums  until  his  death,  at  the  age  of  09  years.  How  much  did  he 
pay  in  premiums?  Ans.  $90200. 


LIFE    INSURANCE. 


541 


Table  No.  4. 


10S2. 


AMERICAN  EXPERIENCE  TABLE  OF  MORTALITY. 


Age. 

Number 
Living. 

Number 
Dying. 

Especation 
of  Lile. 

Age. 

Number 
Living. 

Number 

Uy.ug. 

Expectation 
ot  Lite. 

10 

100000 

749 

48.72 

53 

66797 

1091 

18.79 

11 

99251 

746 

48.08 

54 

65706 

1143 

18.09 

12 

98505 

743 

47.45 

55 

64563 

1199 

17.40 

13 

97762 

740 

46.80 

56 

63364 

1260 

16.72 

14 

97022 

737 

46.16 

57 

62104 

1325 

16.05 

15 

96285 

735 

45.50 

58 

60779 

1394 

15.39 

16 

95550 

732 

44.85 

59 

59385 

1468 

14.74 

17 

94818 

729 

44.19 

60 

57917 

1546 

14.10 

18 

94089 

727 

43.53 

61 

56371 

1628 

13.47 

19 

93362 

725 

42.87 

62 

54743 

1713 

12.86 

20 

92637 

723 

42.20 

63 

53030 

1800 

12.26 

21 

91914 

722 

41.53 

64 

51230 

1889 

11.67 

22 

91192 

721 

40.85 

65 

49341 

1980 

11.10 

23 

90471 

720 

40.17 

66 

47361 

2070 

10  54 

24 

89751 

719 

39.49 

67 

45291 

2158 

1000 

25 

89032 

718 

38.81 

68 

43133 

2243 

9.47 

26 

88314 

718 

38.12 

69 

40890 

2321 

8.97 

27 

87596 

718 

37.43 

70 

38569 

2391 

8.48 

28 

86878 

718 

36.73 

71 

36178 

2448 

8.00 

29 

86160 

719 

36.03 

72 

33730 

2487 

7.55 

30 

85441 

720 

35.33 

73 

31243 

2505 

7  11 

31 

84721 

721 

34.63 

74 

2S73S 

2501 

6.68 

32 

84000 

723 

33.92 

75 

26237 

2476 

6.27 

33 

83277 

726 

33.21 

76 

23761 

2431 

5.88 

34 

82551 

729 

32.50 

77 

21330 

2369 

5.49 

35 

81822 

732 

31.78 

78 

18961 

2291 

5.11 

36 

81090 

737 

31.07 

79 

16670 

2196 

4.74 

37 

80353 

742 

30.35 

80 

14474 

2091 

4.39 

38 

79611 

749 

29.62 

81 

12383 

1964 

4.05 

39 

78862 

756 

28.90 

82 

10419 

1816 

3  71 

40 

78106 

76.5 

28.18 

83 

8603 

1648 

3.39 

41 

77341 

774 

27.45 

84 

6955 

1470 

3.08 

42 

76567 

785 

26.72 

85 

5485 

1292 

2.77 

43 

75782 

797 

26.00 

86 

4193 

1114 

2.47 

44 

74985 

812 

25.27 

87 

3079 

933 

2.18 

45 

74173 

828 

24  54 

88 

2146 

744 

3.91 

46 

73345 

848 

23.81 

89 

1402 

555 

1.66 

47 

72497 

870 

23.08 

90 

847 

385 

1.42 

48 

71627 

896 

22.36 

91 

462 

246 

1.19 

49 

70731 

927 

21.63 

92 

216 

137 

.98 

50 

69804 

962 

2091 

93 

79 

58 

.80 

51 

68842 

1001 

20.20 

94 

21 

18 

.64 

52 

67841 

1044 

19.49 

95 

3 

3 

50 

Note  1.— This  table  is  adopted  by  the  State  of  New  York  and  several  States,  in  estimatiDg 
life  endowments. 

Note  2. — Massachusetts  has   adopted   the  tables  prepared   by  Dr.    Digglesworth,    for 
estimating  life  estates. 

Note  3. — The  Carlisle  Tables  prepared  by  Milne  are  generally  used  in  England, 


542  SOULe's    riilLOSOPHlC    PRACTICAL    MATHEMATICS.  * 

THE  PROBABILITY  OF  HUMAN  LIFE. 

1083.  The  probaliility  of  life  is  the  likelihood  which  a  person  has,  according  to  the  table  of 
jnortality,  of  living  a  certain  nnnilier  of  years.  The  probability  of  death,  is  the  likelihood  a 
person  has  of  dying  "within  a  given  jieriod.  As  it  is  ab.solntely  certain  that  a  person  "will  always 
be  dead  or  alive,  the  probability  of  death  is  found  by  subtracting  the  probability  of  life  from 
unity,  and  vice  versa. 

PROBLEM. 

1.  What  is  the  probability  that  a  person,  aged  30  years,  will  live  to  be  50  years  old,  according 
to  the  figures  of  the  American  Experience  Table  ?  Aus.  .81698. 

OPERATION. 

According  to  the  table,  the  number  living,  of  the  100000,  at  the  age  of  30,  is  85441;  and  the 
number  living,  at  the  age  of  50,  is  69804;  consequently,  15637  Lave  died. 

The  probability  of  any  member  surviving  for  20  years,  will  be  expressed  by  the  fraction 
of  which  69804  is  the  numerator,  and  85441  is  the  denominator;  and  that  the  probability  of  dying 
will  be  expressed  by  the  fraction  in  which  15637  is  the  numerator  and  85441  the  denominator. 

Thus  f?§5t  =  .81698,  the  probability  of  living. 
And  ii^i\  =  .18302,  the  probability  of  dying. 

It  will  be  observed  that  these  amounts  stand  to  each  other  as  complement  and  supplement. 
Hence,  when  we  have  one  of  the  amounts  we  can  find  the  other  by  subtracting  it  from  the  unit. 

Thus  1  —  .81698  =  .18302;  and  1  —  .18302  =  .81698. 

^OTE. — As  there  are  85441  chances  in  all  of  the  persona  agetl  30  attaining  to  50  years,  and  as  the  probahility  is  that 
only  60804  will  Uve,  the  contingencj'  that  each  individual  -^-ill  Uve,  is  .81698,  and  hence  the  contingency  of  his  dying 
is  .18302 

2.  A  man  25  years  old  pays  $140.50  annual  premium  on  a  20  year  policy  of  $5000.  What 
would  this  annual  payment  be  worth  to  the  company  if  continued  for  20  years,  and  the  company 
put  the  payments  at  i%  compound  interest  ?  Ans.  $4183.82. 

Operation  by  the  Annuity  Table,  page  925. 

$140.50  X  29.778079  =  $4183.82. 

3.  Suppose  in  the  above  problem  that  the  payment  of  $140.50  was  continued  for  20  years 
and  was  annually  invested  at  4/'^  coni])oniid  interest,  what  would  be  the  value  of  the  investments 
at  the  expectancy  of  the  policy  holder's  life,  based  upon  the  American  Experience  Table,  (38.81 
years)  ? 

OPERATION. 

First  step  by  the  Annuity  Table,  page  925. 
$140.50  X  29.778079  =  $4183.82  =  value  of  the  annual  payments  or  annuity  for  20  years. 

38.81  years  —  20  years  =  18.81  years,  which  is  the  time  the  value  of  the  Annuity  $4183.82, 
was  at  compound  interest  at  i%. 

Second  step  by  C.  I.  Table,  page  611. 

$4183.82  X  2.02581652  =  $8475.65  =  compound  amount  for  18  years. 
Int.  on  $8475.65  at  4%,  .81  years    274.61 

D.26  Ans. 


For  further  information  regarding  Life  Insurance,  see  "The  Principles  and  Practice  of  Life 
Insurance,"  or  consult  the  officers  of  some  Life  Insurance  Company.  For  Co-operative  Insurance, 
consult  some  member  of  a  Co-operative  Association. 


nte^rest. 


1084.  Interest  is  defined  by  many  political  economists  to  be  the  sum  the 
borrower  pays  for  the  use  of  money.  But  this  definition  is  opposed  by  the  more 
acute  and  critical  political  economists,  who  define  interest  to  be  the  share  of  the 
product  obtained  by  the  owner  of  capital. 

1085.  The  Rate  of  Interest  is  the  per  cent  paid  for  the  capital  or  money 
borrowed,  for  a  specified  time. 

It  was  formerly  believed  that  the  rate  depended  upon  the  abundance  or  scar- 
city of  capital  in  the  money  market  at  any  given  time.  But  this  opinion  has  been 
proven  to  be  erroneous.  The  rate  is  now  shown,  by  our  best  economists,  to  be 
governed  by  the  productiveness  or  profits  of  capital  employed  in  the  various  lines 
of  business. 

Mill  says,  after  thoroughly  arguing  the  case,  "it  is  an  error  to  imagine  that 
the  rate  of  interest  bears  any  necessary  relation  to  the  quantity  or  value  of  money 
in  circulation." 

EiCARDO  says,  that  "  the  rate  of  interest  is  not  regulated  by  the  rate  at  which 
the  bank  will  lend,  but  by  the  rate  of  profit  which  can  be  made  by  the  employment 
of  capital,  and  which  is  totally  independent  of  the  quantity  and  of  the  value  of 
money." 

Note. — The  value  of  money  is  its  purchasing  power. 

But  independent  of  this  general  law  based  upon  profits,  the  rate  is  sometimes 
affected  by  the  greater  hazard  that  is  incurred  by  loaning  to  less  responsible  parties 
or  on  less  certain  security,  and  also  by  reason  of  the  difficulties  that  sometimes 
attend  the  transactions  of  loan,  because  of  the  legal  limit  placed  upon  the  rate, 
called  Usury  Laws. 

Again,  the  rate  is  influenced  by  the  duration  of  the  loan.  Money  loaned  for  a 
short  period,  commands  a  higher  rate  than  money  loaned  for  a  long  period.  And 
call  loans  bear  a  lower  rate  than  loans  made  for  a  definite  period  of  time. 

HISTORY  OF  INTEREST. 

1086.  For  upward  of  thirty  centuries,  the  mind  of  man  has  been  engrossed 
with  this  subject,  and  yet  no  unanimity  of  opinion  has  been  reached,  as  regards  the 
individual  rights  of  men  who  loan  and  borrow  money,  to  fix  the  rate  of  compensa- 
tion to  be  paid  to  the  owner  or  lender  thereof.  In  the  early  ages,  the  charging  and 
receiving  of  compensation  for  the  use  of  money  was  called  usury,  and  was  denounced 

(543) 


544  soule's  philosophic  practical  mathematics.  * 

as  unjust.  This  mistakeu  and  prejudicial  view  undoubtedly  arose  from  a  misunder- 
standing or  misinterjiretatiou  of  some  of  the  enactments  of  the  Mosaic  laws,  which 
read  as  follows  :  "  Thou  shalt  not  lend  upon  usury  to  thy  brother ;  usury  of  money, 
usury  of  victuals,  usury  of  anything  that  is  lent  upon  usury ;  unto  a  stranger  thou 
niayest  lend  upon  usury ;  but  unto  thy  brother  thou  shalt  not  lend  upon  usury." 
(Deuteronomy,  XXiii,  19,  20.)  And,  '-If  thou  lend  money  to  aiiy  of  my  people  that 
is  jjoor  by  thee,  thou  shalt  not  be  to  him  a  usurer,  neither  shalt  thuu  lay  upon  him 
usury."  (Exodus,  xxii,  25.)  And  again,  in  speaking  of  a  poor  brother,  "  Thou 
shalt  not  give  him  thy  money  upon  usury,  nor  lend  him  thy  victuals  for  increase." 
(Leviticus,  xxv,  37.) 

These  laws  are  specific  and  direct  in  their  application,  and  do  not  forbid  the 
charging  of  usury  except  to  poor  brothers.  For  ages,  the  followers  of  Moses  and 
the  disciples  of  his  teachings  were  the  money  lenders  of  Europe  for  usury;  and  by 
reason  of  this,  and  the  fact  that  they  were  as  a  class  shrewd,  trafficking,  and  a 
money  dealing  people,  is  to  be  attributed  the  unjust  prejudice,  that  in  a  measure  to 
this  day  exists  against  them  and  others  who  follow  the  profession  of  money  lending. 

Aristotle,  the  Grecian  jihilosopher,  gravely  contended  that,  as  money  could 
not  beget  money  it  was  barren,  and  usury  should  not  be  charged  for  its  use. 

In  this  regard,  as  in  many  others,  the  old  philosopher  did  not  philosophize 
aright.  He  did  not  see  that  with  money  the  borrower  could  add  to  his  flocks,  his 
fields  and  his  merchandise,  and  thus  profit  by  the  increase  of  all. 

The  Greeks  were  not,  however,  governed  by  the  opinion  of  Aristotle,  for  in 
his  age,  they  paid  an  average  rate  of  about  30  per  cent  on  the  capital  employed  in 
the  shipping  trade,  and  about  12  per  cent  for  capital  employed  in  business  on  land. 

The  Eoman  Law  allowed  interest  or  usury,  and  during  the  days  jf  the  repub- 
lic, the  rate  of  interest  or  usury  was  very  high.  This  resulted  mainly  from  the  fact 
that  the  debtors,  the  plebeians,  repeatedly  threatened  and  sometimes  made  eiforts 
to  deprive  their  creditors,  generally  the  patricians,  not  only  of  the  interest  on  their 
capital,  but  also  of  the  capital  or  princij^al  itself. 

But  from  the  time  of  the  4th  century,  the  writings  of  the  Catholic  Church 
Fathers  and  the  Canon  Law  prohibit  interest.  This  prohibition  was  based  upon  the 
Mosaic  law  as  above  quoted  and  upon  the  sophistry  of  Aristotle. 

The  council  of  Vienna,  Austria,  in  1311,  declared  any  defense  of  interest  to 
be  heresy. 

In  England,  from  the  earliest  period  of  history  to  the  reign  of  Henry  VIII, 
interest  -was  forbidden  to  all  persons  except  to  Jews  and  foreigners,  and  they  were 
often  jilundered  to  enrich  the  crown,  on  the  miserable  pretext  of  jmnishment,  for 
what  the  government  called  their  "hellish  extortions." 

But  the  opposition  of  the  people  to  this  ruinous  interdiction  against  interest, 
by  the  government,  resulted  in  a  statute  passed  in  1546,  allowing  interest  to  the 
extent  of  10  per  cent  per  annum.  Yet,  in  the  reign  of  Edward  VI,  in  1552,  the 
government  again  prohibited  the  taking  or  charging  of  any  interest,  and  passed  an 


INTEREST.  545 

act  declaring  that,  tatiug  interest  vas  ''a  rice  most  odions  and  detestable"  and 
•'  contrary  to  the  "word  of  God." 

On  the  passage  of  this  act,  interest  rose  to  14  per  cent  and  continued  at  this 
rate  until  1571,  -vrhen  an  act  was  passed  repealing  the  act  of  Edvard  YI,  and  re-en- 
acting that  of  Henry  Till,  allowing  10  per  cent  interest. 

The  word  intercut  was  now  substituted  for  the  word  usury,  and  the  meaning 
if  usury  was  changed  to  mean  the  rate  x>er  cent  charged  in  excess  of  the  legal  rate. 

By  the  middle  of  the  17th  century,  nearly  all  the  governments  of  Europe 
followed  England  in  abolishing  the  Usury  Laws,  and  establishing  fixed  rates.  But 
as  the  fixed  rates  were  evaded  just  as  the  prohibition  had  been  evaded  before,  and 
as  the  legally  fixed  rates  resulted  in  mating  the  rate  of  interest  higher  than  it 
would  have  been  without  such  laws,  all  European  States  have  since  abolished  Usury 
Laws. 

In  1S39.  England  abolished  Usury  Laws  on  commercial  paper  running  less 
than  twelve  months;  in  1S50,  on  all  capital  except  loans  on  real  estate;  and  in 
1S54,  they  were  abolished  altogether. 

The  United  States  stands  almost  alone  in  retaining  Usury  Laws.  The  legal 
rate  of  interest  in  the  2>rational  Banking  Law  is  7  per  cent,  in  States  where  no  rate 
is  fixed.  In  States  where  the  State  Laws  fix  the  rate  per  cent,  the  2>rational  Bants 
are  to  be  governed  accordingly. 

See  Table  of  Interest  Laws  of  the  diEFerent  States  and  territories,  page  548 
for  the  various  rates  per  cent  allowed,  penalties  for  usury,  and  days  of  grace. 

EVIL  EFFECTS  OF  USUBY  LAWS. 

10S7.  Usury  Laws,  like  other  laws  fixing  prices,  are  not  only  ineffectual,  but 
they  prevent  all  persons,  whose  credit  is  not  first  class,  from  obtaining  loans  from 
capitalists  of  the  highest  character,  and  force  borrowers  to  secure  loans  from  those 
who  are  less  scrupulous  and  who  are  also  compelled  to  charge  above  the  current 
rate  a  premium  sufBcient  to  comj>en5ate  the  lender  for  the  extra  rist  and  odium 
incorred  in  violating  the  Usury  Law. 

Usury  Laws  undertate  to  prevent  what  is  not  an  evil,  and  hence  what  should 
not  be  interfered  with  ;  consequently,  they  are  productive  of  injury  instead  of  bene- 
fit, Who  would  pretend  that  it  is  just  for  the  legislature  to  compel  underwriters  to 
insure  extra  hazardous  or  specially  hazardous  properties,  on  the  same  terms  or  rate 
per  cent  as  they  write  simply  hazardous  rists.  And  yet  such  a  law  would  be  no 
more  absurd  and  unjust  than  the  law  of  usury  which  requires  that  the  rate  per 
cent  of  interest  shall  be  the  same  on  capital  loaned  to  those  whose  security  or 
character  are  widely  different. 

Again,  it  is  impossible  to  make  the  legal  rate  a  fair  rate ;  for  the  martet  rate 
is  continually  changing,  and  independent  of  the  hazard  of  the  rist,  a  rate  that 
would  have  be*n  fair  at  one  time  would  be  too  low  at  another  time,  and  perhaps 
extortionate  at  another  time.    A  loan  for  a  short  period  commands  a  higher  rate 


S46  soule's  philosophic  practical  mathematics.  * 

than  for  a  long  period  ;  and  a  rate  that  is  fair  for  a  short  period  would  be  unfair  for 
a  long  period.  The  only  way  a  fair  rate  can  be  determined  is  by  competition  in  the 
money  market,  when  all  the  elements  of  supply  and  demand,  of  risk,  of  time,  and 
of  profit,  are  duly  considered  by  the  borrower  and  the  lender. 

BOEEOWEES  AND  LEJSTDEES. 

1088.  When  the  Usury  Laws  were  passed,  the  poorer  classes  were  the  bor- 
rowers and  the  rich  were  the  lenders,  and  hence  the  Usury  Laws,  were  enacted 
mainly  with  a  view  to  jirotect  the  poor,  against  high  rates  of  interest. 

But  at  the  present  time,  the  borrowers  are  largely  the  rich  men,  and  the  rich 
corporations,  and  a  large  part  of  the  lenders,  are  the  poorer  or  medium  classes. 
Hence,  considering  the  reason  for  the  passage  of  the  Usury  Laws,  it  follows  logic- 
ally that  they  should  be  now  abolished. 

One  evidence  of  this  change  of  borrowers  and  lenders  is  the  fact  that  in  1894^ 
three  million  depositors,  mostly  poor  people,  had  nearly  1,300,000,000  dollars  ou 
deposit  in  the  savings  banks.  Another  evidence  is  the  fact  that  hundreds  of  mill- 
ions of  dollars  of  bonds  of  various  kinds  of  corporations,  of  States,  and  of  nations^ 
are  held  by  the  poorer  classes. 

Considering  the  Usury  Laws  in  all  their  bearings,  the  conclusions  are  as 
follows : 

1.  They  are  unwise  and  unjust.  2.  They  are  practically  inoi>erative.  3.  They 
are  evidence  of  financial  iuexjierience  on  the  part  of  the  law  makers.  4.  They  are 
violative  of  the  rights  of  borrowers  and  lenders,  and  of  the  ethical  iirinciples  of 
business.    5.  They  should  be  repealed. 

That  a  law  regulating  interest  in  cases  of  judgments,  successions,  and  trans- 
actions in  which  no  contracts  of  interest  are  made,  is  necessary,  there  is  no  question 
of  doubt.  But  further  than  this,  the  question  of  interest  should  be  left  to  the  judg- 
ment of  the  contracting  parties. 

DEFINITIONS. 

1089.  Interest  is  the  sum  jiaid  for  the  productiveness  of  capital. 

Interest  is  calculated  on  a  scale  of  a  certain  number  of  units  on  every  $100 
for  1  year,  and  hence  it  combines  j;er  cent  and.  per  annum,  from  the  hsitm  per  centum, 
on  a  hundred,  andjjjer  annum,  for  a  year. 

1090.  The  Principal  is  the  sum  of  money  or  capital  on  which  interest  is 
charged  or  paid. 

1091.  The  Rate  of  Interest  is  the  per  cent  paid  on  the  principal,  for  its  use, 
for  a  specified  time. 

1092.  Time  is  the  iieriod  for  which  the  principal  is  loaned  or  bears  interest. 


*  INTEREST.  547 

1093.     The  Amount  is  the  sum  of  the  principal  aud  the  interest. 

109-i.     The  Proceeds  is  the  difl'erence  between  the  principal  and  the  interest. 

The  subject  of  interest  is  classified  into  several  divisions,  viz:  Simple  Interest, 
Annual,  Semi-Annual,  and  Quarterly  Interest,  aud  Compound  Interest,  which  we 
will  present  in  the  order  here  given. 

1095.  Simple  Interest  is  the  interest  on  the  principal  uuincreased  by  inter- 
est, however  long  overdue. 

Annual,  Semi-Annual  and  Quarterly  Interest,  is  Simple  Interest,  and  the  interest 
on  the  Simple  Interest  from  the  time  it  becomes  due  until  i)aid. 

Note. — Anmial,  Semi-Animal  and  Quarterly  Interest  is  allo-netl  \>y  some  States  on  notes  and 
contracts  where  the  interest  is  Payable  Annually,  Senii-Annually  or  Quarterly. 

Compound  Interest  is  interest  on  both  the  principal  and  the  interest,  i.  e.  it  is  interest  Tvhcre 
the  principal  is  increased  at  the  expiration  of  each  period  of  interest  payment  by  the  interest  on 
the  principal  for  such  period,  -whether  for  12,  6,  3,  or  1  month  ;  or  for  one  day. 

Note. — Compound  Interest  is  allowed  in  some  States,  under  certain  conditions. 

1096.  Legal  Interest  is  the  rate  per  cent  fixed  by  the  law  of  each  state,  to 
apply  when  no  agreement  is  made.    In  Louisiana,  it  is  5  per  cent. 

1097.  Conventional  Interest  is  the  rate  per  cent  agreed  upon  by  the  parties 
concerned.  The  law  of  many  of  the  States  places  a  limit  to  this  interest.  In 
Louisiana,  the  limit  is  8  per  cent. 

1098.  Usury  is  a  higher  rate  per  cent  than  the  law  allows.  The  law  of  dif- 
ferent States  prescribes  different  penalties  for  usury.  In  Louisiana,  the  penalty  is 
the  forfeiture  of  all  interest. 

The  Principal,  the  Interest,  the  Bate,  the  Time,  and  the  Amount  or  Proceeds, 
constitute  the  five  quantities  involved  in  interest  questions  ;  and  when  any  three  of 
these  are  given,  the  others  may  be  found.  Hence  there  are  five  classes  of  interest 
questions. 


548 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


1099. 


TABLE  OF  INTEREST  LAWS. 


The  followiug  table  shows  the  Interest  Laws  of  the  different  States  and  Territories  compiled  from 
the  State  and  Territorial  Statutes  of  1891. 


States  aud  Territories. 


Alabama 

Arizona 

Akka.nsas, 

•California, 

colouado 

Connecticut 

Dakota,  North 

"        South 

Delawark, 

1)ist.  of  Columbia,. 

Florida, 

Georgia, 

tlDAIlO 

Illinois, 

Indiana, 

Indian  Territory,. 

Iowa, 

Kansas, 

Kentucky, 

JLouisiana, 

Maine, 

Maryland, 

Massachusetts, 

Michigan, 

Minnesota, 

Mississippi, 

Missouri, 

Montana, 

Nebraska, 

Nevada, 

New  Hampshire,... 

New  Jersey, 

New  Mexico, 

IINew  York, 


North  Carolina, 

Ohio 

Oklahoma  Territory,.. 

Oregon, 

§Pennsy-lvania 

Rhode  Island, 

South  Carolina, 

Tennessee 

UTexas, 

Utah, 

Vermont, 

Virginia, 

Washington, 

West  Virginia 

Wisconsin, 

Wyoming 


Rateallowec 

Legal  rate 

by  contract 

of  juterest. 

generally  lu 

writing. 

PER    CENT 

PER    CENT 

8 

8 

7 

Any  rate 

6 

10 

7 

Any  rate 

8 

Any  rate 

0 

6 

7 

12 

7 

12 

() 

G 

13 

10 

8 

Any  rate 

7 

8 

10 

18 

6 

8 

6 

8 

6 

6 

6 

8 

G 

10 

0 

6 

5 

8 

6 

Any  rate 

G 

6 

6 

Any  rate 

G 

10 

7 

10 

G 

10 

G 

10 

10 

Any  rate 

7 

10 

7 

Any  rate 

6 

G 

G 

6 

G 

12 

C 

6 

G 

8 

G 

8 

7 

10 

8 

10 

G 

6 

6 

Any  rate 

7 

8 

6 

6 

8 

12 

8 

Any  rate 

G 

G 

G 

6 

10 

Any  rate 

6 

G 

7 

10 

12 

Any  rate 

Penalties  for  Usury. 


Forfeiture  of  all  the  interest 

None 

Forfeiture  of  principal  aud  interest. . . 

None c 

None 

None 

Forfeiture  of  excess 

Forfeiture  of  all  tlie  interest 

Forfeiture  of  principal  and  interest. ., 

Forfeiture  of  all  the  interest. , 

None 

Forl'eiture  of  excess  of  interest 

Forfeiture  of  thrice  the  excess  of  int. 

Forfeiture  of  all  tbe  interest 

Forfeiture  of  excess  of  interest 

Forfeiture  of  principal  and  interest... 

Forfeiture  of  10  per  cent  on  amount... 

Forleiture  of  excess  of  interest 

Forfeiture  of  excess 

Forfeiture  of  all  the  interest 

None 

Forfeiture  of  excess  of  interest 

None 

Forfeiture  of  excess  of  interest 

Forfeiture  of  excess  over  10  per  cent. . 

Forfeiture  of  all  the  interest 

Forfeiture  of  all  the  interest 

None 

Forfeiture  of  interest  and  cost 

None 

Forfeiture  of  thrice  the  excess 

Forfeiture  of  all  the  interest 

$100  fine 

Forfeiture  of  principal  and  int.  and 
fine  not  exceed  ing$1000,  or  imi)rison- 
men  t  not  exceed  i  ng  6  mon  ths  or  both 

Forfeiture  of  all  the  interest,  etc 

Forfeiture  of  excess  above  6  per  cent. . 

Forfeiture  of  all  the  interest 

Forfeiture  of  principal  and  interest 

Forfeiture  of  excess  of  juterest 

None 

Forfeiture  of  all  the  interest 

Forfeiture  of  excess  of  interest 

Forfeiture  of  all  the  interest 

None 

Forfeiture  of  excess  of  interest 

Forfeiture  of  excess  over  6  per  cent 

None 

Forfeiture  of  excess  of  interest 

Forfeiture  of  all  the  interest 

None 


Grace  or  no 
Grace, 


Grace. 

Grace. 

No  statute. 

No  grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace, 

Grace. 

No  statute. 

Grace. 

No  grace, 

Grace. 

Grace. 

Grace, 

Grace. 

Grace, 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

No  grace. 

Grace. 

Grace. 

Grace. 

Grace. 

Grace. 

No  grace. 

Grace. 
Grace. 
Gr,ace. 
Grace. 
Grace. 
Grace. 
Grace, 
Grace. 
Grace. 
Grace. 
Grace. 
Grace. 
Grace. 
Grace. 
Grace. 
Grace. 

*In  California,  compound  interest  is  allowed  by  written  contract. 

Iln  Idaho,  compound  interest  is  allowed  by  written  contract.  _ 

iln  Louisiana  interest  njion  inteiest  may  be  collected  provided  it  be  added  to  the  principal  and  by  a  new  contract  made! 

■  a  new  debt.    On  Eottomry  and  Respoiidenlia  Loans,  a  higher  rate  than  8  may  he  coUected.    AKso  a  higher  rate  may  be  col-f 

looted  provideil  it  be  embodied  in  the  instrument  as  dlscoiint.  L 

||In  New  York,  on  demand  loans  of  $DOUO  and  upward,  which  are  secured  by  collaterals,  any  rate  of  interest  may  DOi 

agreed  upon  in  writing. 

^In  Pennsylvania.  Commission  Merchants  may  contract  for  7  per  cent. 

Illn  Texas,  open  accounts  draw  8  per  cent  interest  I'rom  the  1st  day  of  January  after  the  same  are  made. 


PROMISSORY  NOTES  AND  NEGOTIABLE  PAPER.  549 

PROMISSORY  NOTES  AND  NEGOTIABLE  PAPER. 


1100.  A  Negotiable  Promissory  \ote  is  au  imcouditional  promise  in  writing, 
to  pay  to  the  order  ot  a  certain  party,  or  to  bearer,  at  a  certain  time,  a  specified  sum 
of  money. 

Tlie  following  is  the  usual  form  of  a  negotiable  promissory  note: 


New  Orleans,  Marcli  21,  1895. 

Thirty  days  after  date,  for  value  received,  I  promise  to  pay  to  the  order 
of  P.  W.  Sherwood,  Five  Hundred  Dollars. 

H.   JONES. 
Due  April  20/23,  1895. 


1101c  Parties  to  a  Note.  The  original  parties  to  a  note  are  those  whose 
names  appear  in  the  paper,  when  it  is  made. 

Subsequent  parties  are  those  who  become  interested  in  the  paper  after  it  is 
made.  In  the  above  note,  H.  Jones  is  the  promissor,  maker  and  payor.  P.  W.  Sher- 
wood is  the  promissee,  payee,  bolder,  and  first  indorser.  The  holder  of  a  note  is  the 
party  who  owns  it.  The  first  indorser  of  a  note,  is  he  to  whom  it  is  made  payable, 
whether  his  name  appears  first  on  the  back  of  the  note  or  not.  Parties  whose 
names  are  written  above  that  of  the  first  indorser  are  sureties,  and  bke  the  maker, 
are  responsible  without  protest. 

If,  in  the  above  note,  the  words  "the  order  of,"  before  the  name  of  the  payee,  had  been 
omitted,  it  would  have  been  uuiiegotiable. 

No  particular  form  of  words  is  necessary  to  form  a  valid  promissory  note.  The  omission  of 
the  words  "value  received"  does  not  render  the  note  invalid,  but  in  the  hands  of  the  original 
holder  it  would  be  necessary,  if  reijuired  by  the  maker,  to  prove  that  value  had  been  received. 

A  promissory  note  bears  no  interest  until  maturity  unless  specified,  and  in  either  case  unless 
the  per  cent  is  specified,  it  will  bear  only  the  legal  rate. 

1102.  Negotiable  Paper  is  that  which  may  be  transferred  from  one  owner  to 
another  by  assignment  or  indorsement. 

There  are  several  kinds  of  negotiable  paper,  namely:  Promissory  Notes,  Bills  of  Exchange, 
Due  Bills,  Bank  Notes,  Checks  on  Banks  or  iSankers,  Coupon  Bonds,  Certificates  of  Deposit,  Let- 
ters of  Credit,  and  Bills  of  Lading. 

Negotiable  paper  that  does  not  specify  the  time  of  payment,  is  payable  immediately,  and  if 
the  place  of  payment  is  not  specified,  it  is  payable  wherever  it  is  held  at  maturity.  When  the 
place  of  payment  is  specified,  it  must  then  be  presented  at  such  place  the  day  that  it  hgally 
matures.  The  customs  and  laws  of  the  place  m  which  negotiable  instruments  are  made  payable, 
always  apply. 

1103.  The  Face  of  a  note,  draft,  etc.,  is  the  sum  specified  or  named  therein 
and  promised  to  be  paid. 

110-1.    The  Maturity  of  a  note  is  the  day  that  the  note  becomes  due. 

1105.  Days  of  Grace  are  days  allowed  for  the  payment  of  a  note,  draft,  etc., 
after  the  expiration  of  the  time  specified  in  the  instrument.    By  custom  in  most  of 


55o  soule's  philosophic  practical  mathematics.  ■* 

the  states  3  days  of  grace  are  allowed  on  all  notes,  drafts,  etc.,  that  are  not  drawn 
without  grace,  at  sight  or  on  demand. 

Note. — In  some  States  sight  drafts  are  accepted,  and  bear  three  days  of  grace. 

1106.  Dishonoring  a  note  is  the  failure  to  pay  it  when  due. 

1107.  The  Protesting  of  a  note  is  a  notice  to  the  indorsers  that  it  is  due,  that 
payment  has  been  demanded  and  refused,  and  that  the  holder  looks  to  them  for  the 
payment.  The  object  of  protesting  a  note  is  to  secure  the  responsibility  of  the 
indorsers.    Protests  are  generally  made  by  a  notary. 

1108.  Discount  Day  is  the  day  that  a  note,  draft,  etc.,  is  discounted.  Many 
bankers  and  business  men  when  discounting  notes,  etc.,  charge  interest  for  this  day. 

1109.  Discounting  Notes,  etc.,  consists,  according  to  the  custom  of  bankers, 
in  calculating  the  interest  on  their  face  for  the  unexpired  time  including  3  days  of 
grace,  and  in  some  communities  discount  day,  at  some  specified  rate  per  cent,  and 
deducting  the  same  from  the  face. 

1110.  The  Proceeds  or  Cash  Yalue  of  Notes,  etc.,  is  what  remains  after 
the  interest  or  discount  is  deducted. 


INDORSEMENTS  ON  PROMISSORY  NOTES  AND  BILLS. 


nil.  An  Indorsement  is  anything  written  on  the  back  of  a  note  or  bill, 
which  pertains  to  the  payment  thereof. 

1112.  An  Indorser  is  the  party  who  writes  his  name  on  the  back  of  a  note 
or  bill.  The  first  indorse?-  on  a  note,  is  he  to  whom  the  note  is  made  payable.  The 
first  indorser  on  an  accepted  bill,  is  the  Draicer. 

A  note  or  bill  is  not  negotiable  unless  it  is  indorsed  by  the  payee,  except  in  the  case  of  a 
note  ^hich  is  uiailo  payable  to  the  bearer. 

Indorsements  vary  in  form  as  follows:  1.  Blanlc.  2.  In  full.  S.  Qualified.  4.  Reslriciive. 
5.   Conditional.     €.  Protest  and  Notice  of  I'rotesl  Waited. 

The  following  elucidates  the  above  indorsements,  J.  B.  Anderson  being  the 
indorser: 

1.  (Indorsement  in  Blank),  {Second  form  of  Qualified  5.         {Conditiniinl  Indorse- 

J.  B.  Anderson,  Indorsement).  meni). 

Pay  to  the  order  of  A.  B. 

2.  (Indorsement  in  Full).  J.  B.  Anderson,  Pay  to  A.  B.,    -nhen   he   shall 

Pay  to  the  order  of  A.  B.  Agent.  have  comj)leted    the    house 

J.  B.  Anderson,  he  is  now  building  for  me. 
4.     {Eeslrictire  Indorsement).  J.  B.  Anderson. 

3.  {Qualified  Indorsement).  Pay  to  A.  B.  only. 

P.ay  tothe  orderof  A.  B.,  J.  B.  Anderson.            6.           {Protest  Waived) . 

■without  recourse  to  me.  or  thus: 

J.  B.  Anderson.  Pay  to  the  Fourth  National            Protest  and  Notice  of  Pro- 

or  thus:  Bank  of  St.  Louis,  for  my                test  waived. 

Without  recourse.  account.                                                               J.  B.  Anderson. 

J.  B.  Anderson.  J.  B.  Anderson. 
Note. — When  indorsing  notes,  bills,  or  checks,  write  your  name  on  the  back  in  the  same 

way  as  it  is  written  on  the  face.  If  your  name  is  wrongly  spelled  on  the  face,  then  you  should  also 

mis-spell  it  in  the  indorsement.  Or  you  may  both  mis-spell  it  as  it  is  on  the  face  of  the  paper,  and 
then  write  it  correctly. 


*  MISCELLANEOUS  COMMERCIAL   INSTRUMENTS  OF  WRITING.  55 1 

DISCOUNTING  NOTES  AND  ACCEPTANCES. 

1113.  It  is  tlie  custom  of  the  banks  and  the  business  men  of  New  Orleans, 
■when  discounting  notes,  to  count  the  actual  unexpired  days,  including  one 
DISCOUNT  DAY;  i.  e.,  the  day  the  note  is  discounted. 

Remarks.  In  1904,  the  Legislature  of  Louisiana  adopted  the  New  York 
Commercial  Instrument  Law,  thus  making  nearly  half  the  States  of  the  Union 
which  have  adopted  this  Law,  up  to  December  190-4. 

This  Law  does  not  allow  days  of  grace  on  Commercial  paper.  Hence  when 
maturing  notes  and  bills,  and  wliea  discounting  or  computing  interest  on  them  in 
this  book,  and  in  states  which  have  adojited  this  Law,  consider  the  notes  to  be 
drawn  for  3  days  more  than  the  time  specified  therein,  escei^t  such  notes  as  may 
be  drawn  without  grace. 

Note. — For  a  full  discussion  anil  elucidation  of  the  foregoiuj;  iudorsements  of  accommoda- 
tion paper,  acceiitauce  or  payment  for  honor,  treatment  of  accommodation  pajier,  renewals  or 
extension  of  notes,  and  an  extended  presentation  of  commercial  instrumeuta  of  writing,  see  Soule's 
New  Science  and  Practice  of  Accounts,  pages  25  to  39  and  199  to  201. 

Miscellaneous  Commercial  Instruments  of  Writing. 


KECEIPTS. 


1114,  1.     lieceijit  in  full  where  payvienis  hm'e  been  prerioushj  made. 


New  Orleans,  November  6,  1895. 

Received  from  H.  J.  Calvert,  Three  Thousand  Dollars,  Two   Thousand 
having  been  previously  paid,  making  in  all.  Five  Thousand  Dollars. 

R.  W.  ABBOTT. 


Seeeipt  for  Rent. 


New  Orleans,  December  4,  1895. 

Received  from  W.  A.  Beer,   Fifty  Dollars,    for  rent  of  house  No.   457 

Baronne  Street,  for  November,  1895. 

E.  R.  GURLEV. 


3.     Receipt  through  a  Third  Party. 


New  Orleans,  January  3,  1895. 

Received  from  R.  B.  Montgomery,  through  F.  A.  Golden,  Twenty-three 
Hundred,  Forty-seven  and  -,"„"(,-  Dollars,  in  full  of  all  demands  to  date. 

GEO.  B.  MUSE. 


Note. — The  words  in  italics  have  no  money  value  whatever.  A  receipt  is  evidence  for  the 
amount  named  therein,  and  no  words  can  make  it  worth  any  more.  Hence,  the  words  in  italics 
are  useless,  save  as  general  iufoimation. 


552 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


4.     Receipt  to  apply  as  an  Indorsement  on  a  Note. 


New  Orleans, 

November  16,  1895 

Received 

from 

Geo 

.  B. 

Brackett, 

Three 

Hundred  Doll 

ITS, 

tho  same 

to 

apply 

on 

his  no 

te  of 

One 

Thousauil  Doll 

ars. 

T. 

C. 

HENRY. 

Note. — Money  paid  on  account  of  notes  should  be,  when  practicable,  indorsed  on  the  back 
of  the  note.  The  above  payment  thus  indorsed  would  read  as  follows :  Received  on  the  within  not© 
Three  Hundred  Dollars. 


New  Okleaxs,  November  16,  1895. 


T.  C.  HENRY. 


NOTES 


S.     An  Individual  Note. 


New  Orleans 

July  16,  1895. 

Sixty  days 

after  date,  for  value 

received,  I  promise 

to 

pay  to  the  order 

of  H. 

A.  Spencer 

One  Thousand  Six 

Hundred  Dollars. 

Due  September  14/17,  1895. 

R. 

P.  HENRY. 

A  Joint  Note. 


New 

Orleans 

,  July  16,  1895. 

Two 

months  after  date, 

for 

value 

received, 

we 

promise  to  pay  to  the  | 

order 

of  S. 

S.  Packard, 

Sixteen  Thousand  Dollars. 

Due 

September  16/19, 

1895. 

A. 

G.  MAYLIE, 

C. 

S.  CLARKE. 

note. 


N.  B. — In  this  note,  the  joint  makers  are  each  liable  for  one-half  of  the  sum  named  in  the 


7.     A  Joint  and 

Separat 

3  Note. 

New 

Orleans,  February  4,  1895. 

Six  months  after 

date,  for  value  received, 

we  or  either  of  us  promise  to 

P'ly 

to  the  order  o 

f  H. 

B.  Stevens  &  C 

3,,  Nine 

Thousand  Two  Hundred  Dol- 

lars 

payable  at  th 

e  Canal  Bank  in  New 

Orleans 

Due  August  4/7, 

1895 

0.  H. 

KARSTENDIEK, 

A.  V. 

DkGRUY, 

H.  C, 

YOUNG. 

N.  B. — In  this  note,  in  place  of  the  words,  tee  or  either  of  «s,   the   words,   we  jointly  and 
aeeerallij,  may  be  used. 


NOTES. 


553 


S,     Collateral  Note. 


$5000. 


New  Oklkans,  May  14,  1895. 


Ninety  days  after  date,  I  promise  to  pay  to  the  order  of  J.  G.  Anderson, 

Five  Thousand  Dollars,  for  value  received,  with  interest  at  the  rate  of  six  per 

cent  per  annum  from  date  hereof  uutil  paid. 

This  note  is  secured  by  pledge  of  the  secnritieg  mentioned  on  the  reverse  hereof,  and  in 
case  of  its  non-paynient  jit  iiiatui  ity,  or  should  the  drawer  hereof,  when  called  ou.  refuse  or  lail  to 
keep  the  uiarjiiu  heienu  gooil,  the  holder  is  hereby  authorized  to  sell  tlie  said  securities  at  public 
or  pi-ivate  sale,  "witliout  recourse  to  lesal  pi-oeeedings,  and  to  make  any  ti-an-slers  that  may  be 
required,  applying  proceeds  of  sale  towards  the  payment  of  within  note. 


Due  Aug.  12/15,  1895. 


L.  V.  WELLS. 


Note. — On  the  back  of  this  note  must  be  itemized  the  securities  pledged  for  its  protection. 


9.     A  Note  with  a  certain  Party  as  First  Iiiclorser. 

If  Henry  Levy  wishes  to  give  Lis  note  to  Louis  "Winn,  and  to  give  J.   D. 
Purcell,  as  first  indorser,  the  note  would  be  drawn  as  follows : 


$1621 

New 

Orleans 

February 

4,  1895. 

Thirty  days  after  date,  for  va 

ue  received, 

I  promise 

to  pay  to 

the  order 

of  J. 

D.  Purcell, 

One  Thousand  Six 

Hundred  Twenty-one 

Dollars. 

Due  March  6/9, 

1895. 

HENEY  LEVY. 

This  note  must  now  be  indorsed  by  J.  D.  Purcell,  aud  delivered  to  Louis  Winn. 

10.     A  Note  with  Surety. 


New  Orleans,  March  21,  1895. 

Four  months  after  date,  for  value  received,  I  promise  to  pay  to  the  order 
of  N.  E.  Ball,  Eight  Thousand  Dollars. 

B.  TURNER,  Jr. 

I  hereby,  for  value  received,  agree  to  become  security  on  the  above  note. 

C.  W.  BUTLER. 


IfoTE. — The  contract  of  surety  may  be  indorsed  on  the  back  of  the  note. 

11.     A  Note  with  a  Guarantor. 


New  Orleans,  March  21,  1895. 

Sixty  days  after  date,  for  value  received,  I  promise  to  pay  to  the  order 
of  L.  Powell,  Twenty-four  Hundred  Five  and  -f^"^  Dollars. 

J.  W.  LUCE,  Jr. 

For  value  received,   I  guarantee  the  payment  of   the  above  note  at 
maturity.  l^  P-  BOLTON. 


XoTE.— The  contract  of  guarantee  may  be  indorsed  on  the  back  of  the  note  with  the  word 
"within"  in  place  of  the  word  "above." 


554 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


12.     A  Note  of  a  Married  Woman  in  Louisiana. 


New  Orleans,  March  21,  1895. 

Ninety  days  after  date,  for  value  received,  I  promise  to  pay  to  the  order 
of  J.  H.  McNeely,  Forty-six  Hundred  Seventy-three  and  ,',r'',i  Dollars. 

MRS.  KATIE  HOPKINS. 
I  hereby  authorize  my  wife  to  sign  the  above  note. 

A.   15.   HOPKINS. 


13. 

J  Note  signed 

with  a 

cross 

6.'/ 

a  party  who  cannot  write. 

New  Orleans, 

May  23, 

1895. 

One 

yenT 

after  date,  for 

value 

receiveti 

,  I  promise  to  p 

ly  to  the 

order 

of 

s. 

S. 

Levy, 

Twenty-five  Hundred  Dollars. 

HENRY   X 
mark 

SCHNEIDER. 

Witnesses : 

E. 

P.  Delanne. 

H. 

C.  Aycock. 

BILLS  OF  EXCHANGE,  DRAFTS  AND  CHECKS, 


1115.  A  draft,  check  or  bill  of  cscbaiigc  i.s  a  written  oi'der  for  the  payment 
of  money  nnconditionally.  The  bill  may  be  printed,  except  the  signatui-e  of  the 
drawer — that  must  be  written. 


14.     A  Time  Bill  or  Draft,  Time  reckoned  from  Acceptance. 


|5623i',fti  New  Orleans,  January  1,  1895. 

Thirty  days  after  sight,  pay  to  the  order  of  V.  Wintz,  Five  Thousand 
Six  Hundred  Twenty-three  and  -,\Po  Dollars. 
To  A.  Greenwood,  I  W.  C.  KEEVER. 


New  Orleans. 


s 


PARTIES  TO  BILLS  AND  DRAFTS. 


There  are  three  ori£;inal  parties  to  a  bill  of  exchange,  viz:  the  drawer, 
drawee  and  payee.  The  subsequent  jjarties  to  a  bill  are  those  to  whom  it  may 
be  successively  transferred.  In  the  above  bill,  W.  C.  Keever  is  the  drawer,  A. 
Greenwood  is  the  drawee,  and  when  he  accepts  the  bill,  he  will  become  the  acceptor, 
maker,  and  payor,  and  the  drawer  then  becomes  first  indorser.      V.  Wintz  is  the 


BILLS    OF    EXCHANGE,    DRAFTS    AND    CHECKS, 


555 


payee.  Sight  bills  are  paid  at  sight  in  Louisiana;  but  in  some  States,  they  are 
aceepted  and  bear  three  days  of  grace.  Drafts  and  bills  should  be  presented  for 
acceptance  without  delay,  and  if  acceptance  is  refused,  they  should  be  protested  for 
non-acceptance  in  order  to  hold  the  drawer. 

Accepted  bills  bear  grace,  are  protested  for  non-payment  the  same  as  notes, 
and  are  subject  to  all  the  laws  governing  promissory  negotiable  notes. 


15.     A  Sight  Bill. 


$1500.  New  Orleans,  February  3,   1895. 

At  sight,  pay  to  the  order  of  C.  B.  Modinger,  Fifteen  Hundred  Dollars, 
and  charge  the  same  to  account  of 


To  W.  n.  TxeitnoMs, 

Alexandria,  La. 


A.  L.  SOULfi. 


IG.     Bank  Check. 


New  Ohleaxs,  May  1,  1895. 

SOULfi  COLLEGE  liAXK. 

Pay  to  the  order  of  H.  C.  Brown,  Two  Thousand  Seventeen  and  •,<,"„  Dol- 

lars. 

$2017,1,'^;. 

J.  J.  BELL  &  CO. 

17.     Order,  Payahle  in  Merchandise. 


New 

Orleans,  April 

11,  1895. 

Mr. 

J.  Vega 

Pay   to 

Mr. 

J. 

N. 

Harrison   or 

order 

Forty 

-four   and  -f^ff 

Dollars 

in 

merchandise, 

and 

charge 

the  same  to  iii 

y  account. 

C.  W.  BUTLER. 

IS.     Certificate  of  Deposit. 


No.  400.  Soul6  College  Bank, 

New  Orleans,  July  5,  1895. 

N.  L.  Barfield  has  deposited  in  this  hauk  One  Thousand  Dollars,  payable 
to  his  order  ou  return  of  this  certificate  properly  indorsed. 
$1000.  M.  GREEN, 

Cash  ier. 


19.     Due  Bill. 


New 

Orleans, 

April  1,  1895. 

D 

ue  to  Wm. 

Hardie 

or  order,  on  demand 

,  One  Thousand 

Seven  Hundred 

Dollars, 

with 

interest  from  date  at  8 

per  cent. 

GEO. 

C.  FAHEY. 

556 


SOULES    rniLOSOPHlC    PRACTICAL    MATHEMATICS. 
20.     Due  Bill  Payable  in  Merchandise. 


Nkw  Orleans,  March  16,  1895. 

Due  R. 

C. 

Spencer 

or  order, 

Se 

centy 

-five 

Dollars, 

payable 

in  merchan- 

disc 

from  my 

store,  May  1 

1895. 

C.  W. 

GLYNN. 

St.     A  Set  of  Domestic  Bills. 


CD. 


SouLF,  College  Bank, 

New  Orleans,  January  3,  1895. 


At  siglit  of  this  First  of  Exchange,  (second  unpaid)  pay  to  the  order  of 
G.  Loomts,  Two  Thousand  Dollars,  value  received. 

EDWARD  E.  SOULfi. 


To  the  City  National  Bank, 


of  New  York. 


Cashier. 


(2).  SoulA  College  Bank, 

New  Orleans,  January  3,  1895. 

At  sight  of  this  Second  of  Exchange,  (first   unpaid)  pay  to  the  order  of 
G.  Looniis,  Two  Thousand  Dollars,  value  received. 


To  the  City  National  Bank, 

of  New  York. 


EDWARD  E.  SOULfi. 

Cashier. 


SPECIAL  LAWS  AND  BUSINESS  CUSTOMS 


TO    BE    OBSERVED    IN    INTEREST    AND    DISCOUNT    CALCULATIONS. 


1116.  1.  In  performing  the  operations  of  interest  calculations,  we  must 
observe  certain  special  laws,  and  also  certain  business  customs  whicli,  by  reason  of 
being  in  general  use  or  of  long  standing,  have  become  law  without  any  legislative 
enactments. 

The  following  are  some  of  these  laws  and  customs  governing  the  most  of  the 
mercantile  communities  of  America: 

2.    To  consider  the  year  as  consisting  of  12  mouths  of  30  days  each,  or  360 
days,  for  a  divisor  in  all  interest  or  discount  calculations. 

Note. — The  American,  English,  and  French  governments,  and  the  State  of  New  York  con- 
eider  the  year  as  consisting  of  365  days,  hence  in  calculating  interest  on  United  States,  English  or 
French  l)onds,  or  on  notes  maturing  and  payable  in  New  York,  the  divisor  would  be  365  instead  of 
360.     See  Interest  Divisor,  pages  560  and  561. 


*  SPECIAL    LAWS    AND    BUSINESS    CUSTOMS.  557 

3.  To   allow  3   days  of  grace  on   all   commercial  paper,  not  payable  on 

demand,  at  siglit,  or  without  grace,  except  in  States  in  wliicli  days  of  grace  are 

not  allowed. 

Note. — Also  to  allow  three  days  of  grace  on  sight  drafts  or  hills  of  exchange  in  States  in 
which  sight  bills  are  accepted. 

4.  In  maturing  notes,  drafts,  etc.,  tliat  are  drawn  in  years  or  montlis,  it  is  the 
custom  to  count  from  the  day  of  the  month  tliat  the  instrument  is  dated  to  the  same 
day  of  the  month  in  which  it  matures.  Thus  a  note  dated  July  15,  1895,  and  made 
payable  3  months  after  date,  matures  October  15/18,  1895. 

5.  In  maturing  notes,  drafts,  etc.,  that  are  drawn  in  days,  it  is  the  custom  to 
mature  them  in  days.  Thus,  if  a  note  was  dated  July  15, 1895,  the  same  as  above, 
but  m;w;le  payable  in  90  days  after  date,  it  would  mature  October  13/lG,  1895. 

C.  Kotes,  acceptances,  etc.,  maturing  on  the  29th,  3()th  or  31st  days  of 
months  having  less  than  29,  30  or  31  days,  are  considered  due  on  the  last  day  of  the 
month,  and  allowing  for  grace  are  payable  on  the  3d  of  the  following  month. 

Thus,  a  note  dated  December  28,  29,  30  or  31,  1895,  and  j^ayable  2  months 
after  date,  matures  February  28th  and  March  3d,  1896. 

7.  Kotes,  drafts,  etc.,  maturing  on  Sunday  or  a  legal  holiday,  are  payable  in 
Louisiana,  the  first  business  day  following.  In  most  of  the  States,  commercial 
paper  maturing  on  Sunday  or  a  legal  holiday,  is  i)ayable  on  the  preceding  business 
day. 

8.  When  discounting  notes  or  drafts  drawn  in  years  or  months,  it  is  the 
custom  to  count  the  actual  days  of  unexpired  time,  including  3  days  grace  and  also 
discount  day,  in  cities  or  places  where  interest  is  computed  for  the  day  the  note  is 
discounted.  Thus,  a  note  dated  July  15,  1895,  and  made  i)ayable  3  months  after 
date,  would  mature  October  15/18, 1895,  and  if  discounted  the  day  it  was  drawn,  the 
discount  time  would  be  90  days,  including  grace  and  discount  day. 

If  this  same  note  had  been  made  payable  90  days  after  date,  it  would  have 
matured  October  13/16, 1895,  and  if  discounted  the  day  it  was  drawn,  the  discount 
time  would  have  been  94  days,  allowing  grace  and  discount  day. 

And  in  like  manner  a  note  made  payable  in  one  year  would  be  discounted  for 
369  days,  if  discounted  the  day  it  was  drawn,  including  grace  and  discount  day. 
In  discount  comjiutations,  the  time  that  has  elapsed  is  deducted  from  the  actual 
time  the  note  has  to  I'un. 

Note  1. — In  cities  or  places  where  no  interest  is  charged  for  the  discount  day,  but  where  3 
days  of  grace  are  .allowed,  the  discount  time  on  the  above  notes  would  be  1  day  less  on  each,  making 
them  respectively  95,  93  and  368  days. 

Note  2. — In  States  where  days  of  grace  are  not  allowed,  and  the  discountday  is  not  included, 
the  discount  time  would  be  4  days  less  on  each  of  the  above  notes,  making  respectively  92,  90  and 
365  days. 

TO  COLLECT  USTTEREST-BEARING  NOTES. 

1117.  In  collecting  a  note  that  is  drawn  in  months  or  years  and  which  bears 
interest,  it  is  customary  to  charge  interest  for  the  months  at  30,  or  years  at  360 
days  each,  instead  of  the  actual  days. 


558 


SOULE  S   PHILOSOPHIC   PRACTICAL   MATHEMATICS. 


If  a  note  is  dated  March  10,  1895,  and  made  payable  1  year  after  date  with 
interest,  it  would  mature  March  10/13,  1800,  and  if  held  by  the  payee  till  maturity, 
lie  would  collect  the  face  of  the  note  and  the  interest  for  one  year  or  3G0  or  363 
days. 

In  case  the  note  was  not  paid  until  the  expiration  of  the  3  days  of  grace,  then 
the  holder  would  be  entitled,  in  equity  at  least,  to  interest  for  3  days  of  grace, 
making  3G3  days. 

Remarks. — The  question  of  charging  interest  for  the  3  days  of  grace,  as  here 
stated,  when  the  payor  avails  himself  of  the  3  days  of  grace,  is  not  as  far  as  we  can 
learn,  covered  by  any  statute  or  legal  decision ;  and  our  financial  men  are  not  agreed 
upon  this  point.  Some  argue  that  in  States  where  the  law  allows  the  maker  of 
notes  3  days  of  grace,  he  should  not  be  charged  interest  for  those  days.  Others 
argue  that  while  the  law  allows  3  days  of  grace  in  which  to  pay  the  note  or  bill,  it 
does  not  stop  the  interest  on  those  days  any  more  than  it  does  on  any  of  the  other 
days  the  instrument  has  to  run. 

The  equity  of  the  question  is  clearly  in  favor  of  charging  interest  for  the 
days  of  grace,  and  we  shall  so  present  the  matter  in  this  work.  It  is  an  ethical 
principle  in  business  that  value  must  be  given  for  value  received.  That  no  i)ersou 
shall  use  the  pr()i)erty  of  another  without  rendering  compensation  therefor.  Hence, 
if  the  payor  of  a  note  retains  the  use  of  the  payee's  money  for  the  3  days  of  grace, 
he,  the  ])ayor,  should  pay  for  such  use.  And  moreover,  the  3  days  of  grace  in  States 
allowing  grace,  is  as  much  a  part  of  the  specified  time  a  note  has  to  run  as  the  time 
written  in  the  body  of  the  note. 

ELUCIDATION  OF  THIS  TEINCIPLE. 

1118.     A  note  is  given  March  26,  1895,  for  $5000,  payable  one  year  after  date 
with  interest  at  8  per  cent. 

What  is  the  interest  due  in  States  allowing  grace,  if  collected  March  26  or  27 
or  28  or  29,  1896  ">.  Ans.  See  statements  following. 


Statement  to  find  inter- 
est due  March  26. 

$50.00 
360       8 

300 


Statement  to  find  inter- 
est due  March  27. 

.$50.00 
8 
360       361 


Statement  to  find  inter- 
est due  March  28. 

$50.00 
8 
362 


360 


Statement  to  find  inter- 
est due  March  29. 

$50.00 
8 
360       303 


«400.00  Ans.      $401,111  Ans.      $402.22|  Ans.      $403.33J  Ans. 

TO  DISCOUNT  NOTES  BEARING  INTEREST. 

1119.  In  discounting  a  note  drawn  in  months  or  years  and  which  bears  inter- 
est, it  is  the  custom  to  add  to  the  face  of  the  note  the  interest  for  the  months  at  30, 
or  years  at  360  days  each,  i)lus  three  days  of  grace,  which  gives  the  maturity  value 
of  the  note,  and  then  to  discount  the  sum  thus  produced,  also  for  the  actual  number 
of  days  including  3  days  grace  and  also  1  discount  day,  in  places  where  the  discount 
day  is  included.  See  page  580,  for  the  operations  of  discounting  notes  that  bear 
interest. 


*  SPECIAL    LAWS    AND     BUSINESS    CUSTOMS.  SSq 

POINTS  IN  DRAWING  NOTES. 

1120.  The  followiug  may  be  useful  to  those  who  have  frequent  occasion  to 
draw  notes : 

1.  33  days  =  4  weeks  aud  5  days;  03  days  =  9  weeks;  93  days  =  13  weeks 
and  2  days. 

Therefore  a  note  at  30  days  will  fall  due  five  days  later  than  the  day  of  the 
■week  on  which  it  was  given. 

A  note  at  60  days  will  fall  due  on  the  same  day  of  the  week. 
A  note  at  90  days  will  fall  due  two  days  later. 

2.  In  drawing  notes  or  accepting  drafts,  it  should  be  borne  in  mind  that 
often  bills  of  different  dates,  having  the  same  time  to  run,  will  mature  on  the  same 
day  and  thus  occasion  unexpected  inconvenience.  Thus  a  note  drawn  Januaiy  31st 
for  3  months,  and  another  drawn  January  30th  for  3  months,  both  mature  Ajiril 
30th  and  May  3d.  Sometimes  one  day's  diflercnce  in  the  date  or  acce])tance  will 
make  2,  3  or  4  days'  difference  in  the  maturing  of  bills  having  the  same  time  to 
run, thus : 

Notes  drawn  or  drafts  accepted  April  tlie  30th  at  3  months,  mature  July 
30th  and  August  2(1,  hut  if  dated  or  accepted  ]May  1st,  they  will  mature  August 
1/4,  2  days  later.  W  a  note  was  drawn  February  the  2Sth  at  6  months,  it  would 
mature  August  28th  and  31st;  but  if  dated  IMarch  1st,  it  would  mature  September 
1st  and  4th,  4  days  later.  An  observance  of  these  principles  by  accountants  and 
business  men  will  frequently  save  much  inconvenience  aud  sometimes  result  in 
preserving  the  right  of  action  against  indorsers. 

3.  In  discounting  notes,  the  most  of  bankers  reject  the  cents  if  less  than 
50,  aud  if  in  excess  of  50,  they  increase  the  sum  by  $1  and  reject  the  cents. 

In  the  operations  of  interest  in  this  book,  we  will  use  the  exact  figures. 
For  Time  Tables,  see  page  298  of  this  book. 

Table  of  Interest  Laws  of  the  States  aud  territories,  as  existing^  in  1891,  see 
page  548. 

INTEREST   DIVISORS. 

1121.  An  Interest  Divisor  is  a  number  which  used  as  a  divisor  of  any  prin- 
cipal will  give  Interest  in  the  quotient.  The  Interest  IHvisor  differs  according  to 
the  rate  per  cent,  and  also  according  to  the  number  of  days,  360  or  305,  used  in  a 
year  as  a  basis  for  interest  computations. 

1122.  Interest,  as  shown  in  the  preceding  remarks  and  definitions,  is  a  com- 
pound of  per  ceut,  on  a  100,  and  per  annum,  for  a  year,  300  days.  Hence  to  obtain 
1  per  cent  interest /or  i  2/mr,  on  any  principal,  we  simply  divide  by  100.  And  to 
obtain  1  per  cent  interest  for  1  day,  on  any  principal,  we  first  divide  by  100  to  get 
1  per  cent  interest  for  1  year  and  then  divide  that  quotient  by  3G0  to  get  the  interest 
for  one  day.  Or  we  may  divide  the  principal  at  once  by  30000  which  is  the  protluct 
of  100,  the  basis  of  per  ceut,  aud  300  days,  the  basis  of  a  year. 


56o 


soule's  philosophic  tractical  mathematics. 


The  quotient  arising  by  this  division,  being  interest,  we  therefore  name  the 
3CO0O  the  Interest  Divisor  for  1  per  cent. 

Note.— When  365  <Iays  are  used  as  the  basis  of  interest  computations,  then  the  365  -would  be 
used  where  360  is  used  herein,  and  the  1  per  cent  Interest  Divisor  ivoiild  he  SGOOO. 

Having  thus  produced  the  interest  at  1  per  cent  for  one  year  or  one  day,  to 
find  the  interest  at  any  desired  rate  per  cent  and  for  any  desired  number  of  years 
or  days,  'sve  bave  but  to  multiply  this  interest  by  the  desired  rate  per  cent  and  the 
desired  number  of  years  or  days. 

TABLE  OF  INTEREST  DIVISORS. 

1123.  To  aid  in  understanding  the  interest  divisor  and  the  use  of  the  same, 
we  present  the  following  table  which  gives  the  interest  divisors  at  1,  2,  3,  4,  4J,  6, 
6,  8,  9, 10,  12,  15,  IS,  20,  24,  25,  30  and  36  per  cent : 


-rf  Interest 

/O  Divisois. 

36000  -^  1  =  36000 

36000  -^  2  =  18000 

36000  -^  3  =  12000 

36000  -^  4  =  9000 

36000  4-  4J  =  8000 

36000  ->  5  =  7200 


^  Interest 

fo        Divisors. 

36000  -^  C  =  6000 


30000 


4500 


36000  4-  9  =  4000 
36000  4-  10  =  3600 
36000  -^  12  =  3000 
36000  -^  15  =  2400 


fjf         Interest 
/O        Diviaora. 

36000  -^  18  =  2000 
36000  -^  20  =  1800 
36000  4-  24  =  1500 
36000  -^  25  =  1440 
36000  ■-  30  =  1200 


36000 


36  =  1000 


When  the  rate  of  interest  will  not  cancel  the  36000  without  a  remainder,  then 
we  iiroceed  as  shown  in  the  statement  for  the  second  operation  following : 

Note. — In  many  compntations,  Equation  of  Accounts,  Accounts  Current  and  Interest 
Accounts  by  the  product  or  etpiation  method.  Interest  on  Daily  Cash  Balances,  Cash  Notes,  True 
Discount,  etc.,  the  Interest  Divisor  is  of  great  service,  and  should  be  well  understood  by  all 
accouutants. 

BASIS  OF  ALL  INTEREST  COMPUTATIONS. 


1124.  The  foregoing  is  the  &as/s  o/ fflZJ  interest  computations,  and  by  working 
in  accordance  therewith  we  avoid  all  the  arbitrary  rules  which  confuse  and  confound 
the  millions. 

To  perform  the  operations  of  interest  in  detail,  as  above  indicated,  would 
require  considerable  time  and  labor.  Hence,  with  a  view  to  economize  both,  and 
Btill  work  from  the  foundation  principles  of  interest,  we  combine  reason  and  can- 
cellation with  the  foregoing  principles  and  evolve  a  brief,  a  simple,  and  a  universal 
formula,  applicable  to  all  interest  computations. 


INTEREST  EVOLUTION. 


56l 


A  UKIVEESAL  FORMULA  FOR  COMPUTING  INTEREST. 

1125.  The  solution  of  the  following  problem  sLows  tLe  application  of  the 
foregoing-  principles  of  interest,  and  tbe  evolution  iu  the  oi^erations  by  wliicli  the 
brief,  simple,  and  universal  formula  is  obtained : 

What  is  the  interest  on  $72000  at  8  per  cent  for  11  days  ?  Ans.  $176. 

First  Operation,  in  detail. 

36000)  $72000  ($2  =  interest  at  1%  for  1  day. 

8  =  8%. 

$16  =  interest  at  8%  for  1  day. 
11  =  11  days. 

$176  =  interest  at  8%  for  11  days. 


Second  Operation 
in  Interest  Evolution. 


Third  Operation  in 
Interest  Evolution. 


Fourth  Operation  in 
Interest  Evolution. 


$ 


100 
360 


72000    or, 

8  36000 

11 


.16,  Ans. 


72000 

8 

11 

$176,  Ans. 


4500 


72000 
11 

$176,  Ans. 


45 


720.00 
11 

$176.00  Ans. 


Explanation. — In  the  first  operation,  we  divided  by  36000,  the  1  per  cent  interest  dirisor  and 
obtained  $2,  tbe  interest  at  1  per  cent  for  1  day  ;  this  we  multiplii-d  by  the  rate,  8  per  cent,  and 
obtained  $16,  the  interest  at  8  per  cent  for  one  day;  this  we  nmltiplied  by  the  days,  11,  and 
obtained  $176,  the  interest  at  8  per  cent  for  11  days. 

In  the  SECOND  operation,  we  indicated,  on  "the  statement  line,  the  work  of  the  first  operation, 
and  then  cancelled. 

In  the  THIRD  operation,  we  mentally  divided  the  36000  by  the  rate  per  cent,  8,  and  produced 
4500,  the  8  per  cent  interest  divisor.  By  this  mental  cancellation,  we  very  much  shortened  the 
operation. 

In  the  FOURTH  operation,  we  first  produce  the  4500,  the  8  per  cent  interest  divisor,  then  cancel 
tbe  two  O's  and  in  compensation  therefor  point  off  two  jilaces  in  the  principal.  By  this  mental 
cancellation,  we  shorten  the  operation  to  the  greatest  practical  limit,  and  present  a  universal 
formula  for  interest  computations,  far  superior  to  the  arbitrary  rules  given  lu  the  arithmetics  and 
calculators  now  before  the  public. 

TABLE  OF  CONTRACTED  INTEREST  DIVISORS. 

1126.     The  following  table  shows  the  Contracted  Interest  Divisors,  for  the  most 
usual  rates  per  cent  : 

TABLE   uV   CONTRACTED   INTEREST   DIVISORS. 

C/^     Interest 
Ds.  /''    Divisors. 

360  -^  15  =  24 
360  -r  18  =  20 
360  -^  20  =  18 
360  4-  24  =  15 
360  -V  30  =  12 
360  -1-  36  =  10 

Note. — When  the  rate  per  ceut  is  not  a  factor  of  360,  such  as  7  per  cent,  11  per  cent,  etc., 
then  360  will  be  the  Interest  Divisor,  and  the  rate  per  cent  will  be  used  as  a  multiplier  as  shown  in 
the  first  operation  of  the  first  problem  following  this  table. 


Ct,        Interest 

yri      Interest 

Ds.             '''       Divi.iuis. 

Ds.            7"     Divisors. 

360  4-  1    =  360 

360  - 

-  5    =  72 

360  4-  2    =  180 

360  - 

-    6  =  60 

360  4-  2i  =  144 

360  - 

-    8  =  45 

360  4-  3"  =  120 

360  - 

-    9  =  40 

360  4-  4    =    90 

360  - 

-  10  =  36 

360  4-  4J  =    80 

360  - 

-  12  =  30 

720.00 

8 

11 


SECOND 

45 

OPERATION. 

720.00 
11 

$170.00,  Ans. 

562  soule's  ruiLosopiiic  practical  mathematics.      '  * 

A  PniLOSOPUIC  METHOD  OP  USING  THE  PACTOKS  OP  THE  DTTEK- 

EST  DIVISOR. 

1127.  In  tlio  pr.actical  computation  of  interest,  we  prefer  to  use  the  factors  of 
the  Interest  Divisor  (100  and  SCO,  or  SOo)  in  a  slightly  (lifl'erent  manner  from  that 
shown  by  the  four  operations  above,  yet  strictly  in  accordance  with  reason  and  logic. 

In  order  to  fully  elucidate  the  work,  we  will  re-state  and  re-work  the  above 
problem. 

What  is  the  interest  on  $72000  at  8  per  cent  for  11  days  1  Ans.  $17G. 

FIRST   operation.  SECOND   OPERATION.  Ji:xplanation.—la  the  first  opera- 

tion, our  statemeut  conforms  to  th« 
statement  made  in  the  second  oper- 
ation preceding,  except  that  wa 
first  divide  the  principal,   $72000, 

3G0      11  — .    by  100,  hy  pointing  oflf  two  places, 

and  then  by  360,  instead  of  using 
ftl7rnn     i-,  o  '  the  36000  as  a  single  divisor. 

ifi^o.uu,  AUS.  The  reasoning  for  the  work  based 

upon  the  foregoing  elucidations,  i» 
as  follows:  1  per  cent  interest  on  any  principal  for  1  year  is  the  ji^o  part  of  it,  which  we  produce 
by  pointing  off  two  places.  Tlien,  at  8  per  cent  it  is  8  times  as  much,  which  is  indicated  by 
writing  the  8  on  the  increasing  side  of  the  statement  line.  Then,  since  the  interest,  as  indicated 
by  the  statement  is  for  1  year,  for  one  day  it  is  the  360th  part,  which  is  indicated  by  writing  the- 
360  on  the  decreasing  side  of  tbe  statement  line  ;  and  for  11  days,  it  is  11  times  as  much  as  for  1 
day,  which  is  indicated  by  writing  the  11  ou  the  increasing  side  of  the  statement  line. 

Note. — The  reasoning  for  the  second  operation,  is  as  follows:  The  interest  on  $72000  for  1 
year  at  1  per  cent  is  the  100th  part,  which  is  $720;  and  for  1  day  at  8  per  cent  it  is  the  45th 
part,  and  for  11  days  it  is  11  times  as  much  as  it  is  for  1  day. 

This  is  a  philosophical  and  logical  method  of  work,  step  by  step,  and  it  is 
universal  in  its  application.  ' 

In  the  second  operation,  our  statement  is  the  same  as  that  in  the  fourth  oper- 
ation preceding  and  constitutes  the  most  valuable  method  iwssible,  of  computing 
interest. 

The  reasoning  is  the  same  as  in  the  first  operation,  except,  instead  of  multi- 
plying by  8  and  dividing  by  300,  we  mentally  divide  the  300  by  8,  and  then  use  the 
quotient,  45,  as  a  contracted  interest  divisor.  In  this  manner,  we  contract  the  opera- 
tion to  the  greatest  practical  limit  and  use  reason  and  logic  throughout  the  solution. 

PEOBLEMS  IN  INTEREST 

WORKED  BY   THE  PHILOSOPHIC   SYSTEM. 

1128.  The  Principal,  Rate  Per  Cent,  and  Time  given  to  find  the  Interest,  the 
Amount,  or  the  Proceeds : 

FOR  YEARS. 
1.    What  is  the  interest  on  $560  at  8  jjer  cent  for  3  years !      Ans.  $134.40. 

OPERATION. 

$5.00  =  1  %  of  principal.  Explanfilhn. — Considering  that  interest  invol- 

g  __  g  (V  ves  ]ier  cent  and  ])er  annum,  as   elucidated   in 

'    *  the  foregoing  work,  and  in  consonance  with  the 

.  foregoing  logical  method  of  operation,  we  hero 

$44.80  =  mt.  for  1  year.  reason  as  follows:  The  interest  on  $560  at  1  per 

3  =  years.  cent   for  l  year  is  the   hundredlh  part,    $5.60, 

which  we  express  by  pointing  off  two  places; 

tiQ/i   in        ;,,f    r..r  9  ..r..c      a-^o  and  at  8  per  cent  it  is  8  times  as  mnch,  which  is 

$134.40  =  mt.  for  3  yrs.,  Ans.  ^^44  go ;  and  for  3  years  it  is  3  times  as  much  as 

for  1  year,  which  is  $134.40. 


INTEREST. 


563 


FOR  YEARS  AND  MONTHS. 

2.    What  sum  must  be  paid  for  the  interest  on 
and  9  months. 

OPERATION. 


320  at  8  per  cent  for  1  year 
Ans.  $114.80. 


8.20  =  1%  of  principal. 
8       =  8%. 
21     =  months. 


$114.80  =  int.  for  1  yr.  9  mos.,  Ans. 


Erplanaiion. — In  this  and  in  all  proljlenis 
■where  there  are  months  in  the  time,  we  reason 
as  follows:  The  interest  on  $820  at  1  per  cent 
for  1  year,  or  12  months,  is  the  Imndredlh  part, 
$8.20;  and  at  8  per  cent  it  is  8  times  as  much; 
and  for  1  month  instead  of  12,  it  is  the  12tli 
part;  and  for  21  months  (1  year  and  9  months 
rednced  to  months)  it  is  21  times  as  much  as  it 
is  for  1  month. 


FOE  TEARS,  MONTHS  AND  DAYS. 
3.    What  is  the  interest  on  $1230.40  at  9  per  cent  for  2  years,  5  months,  and 


24  days  T 

FIRST   OPERATION. 


Ans.  $274.99+. 


3G0 


12.3040  =  1%  of  principal. 
9  =  9%. 

894         =  days. 


SECOND   OPERATION. 

$ 

12.3040 
40    894 


$274.9944  =  int.  for  2  yrs.  5  mos. 
and  24  ds. 


$274.9944 


THIRD   OPERATION. 


$12.30.40 
9 


$110.73.00  =  interest  for  1  year, 
o 


360 


$221.47.20  =  interest  for  2  years. 
53.52.24  =  interest  for  174  days. 


12.3040 

9 

174 

53.5224 


$274.99.44 

Explanation. — In  this  and  in  all  problems  where  there  are  days  in  the  time,  in  accordance 
■with  the  foregoing  elucidated  principles,  ■vs-e  reason  as  follows  in  the  first  operation:  The  interest 
on  $1230.40  at  1  percent  for  1  year  is  the  hundredth  part,  $12.3040:  and  at  9  per  cent  it  is  9  times  as 
much  ;  and  for  1  day,  instead  of  1  year,  it  is  the  360th  part ;  and  for  894  days  (2  years,  5  months 
and  24  days  reduced  to  days)  it  is  894  times  as  much. 

In  the  second  operation,  the  same  reasnniuf;  governed  the  statement,  but  instead  of  ■writing 
the  9  per  cent  and  the  360,  we  used  the  Contracted  Interest  Divisor,  as  elucidated  in  the  foregoing 
work. 

GENERAL  DIEECTIONS  FOR  CALCULATING  INTEREST. 


1129.     From  the  foregoing  elucidations,  ■we  derive  the  following  general  direc- 
tions for  calculating  interest: 

1.    For  years,  first  find  1  ;per  cent  of  the  principal,  hy  dividing  hy  100  (pointing 


564  soule's  philosophic  practical  mathematics. 


off  two  placesj  ;  then  multiply  hy  the  rate  per  cent,  and  thin  pirudnct  by  the  mnnhcr  of 
yearn.     See  problem  1,  page  502. 

2.  For  months,  write  the  principal  on  the  statement  line  and  find  1  per  cent,  by 
dividing  by  100,  or  pointing  off'  two  places ;  then  indicate  on  the  statement   line  the 
division  by  12,  and  the  midtiplieation  by  the  rate  per  cent  and  the  number  of  monthSf,^ 
and  tvork  out  the  statement.    Sec  problem  2,  page  5G3.  ■■ 

3.  For  days,  write  the  principal  on  the  statement  line  and  find  1  per  cent  by 
pointing  off'  two  places,  or  by  indicating  the  division  by  100;  then  indicate  on  the  state- 
ment line,  the  divison  by  3G0,  for  365  when  that  is  used  J  and  the  mnltipiication  by  the 
rate  per  cent  and  the  number  of  days,  and  tcork  out  the  statement,  cancelling  as  much 
as  possible.     See  j)roblem  3,  page  5G3. 

Note. — Instead  of  indicating  the  division  by'360  and  the  multiplication  by  the  rate  per 
cent,  we  may  simply  divide  by  the  quotient  (the  contracted  Interest  Divisor)  of  360  divided  by 
the  rate  x)er  cent,  as  explained  on  page  502. 

4.  To  find  the  amount,  add  the  interest  to  the  principal. 

5.  To  find  the  proceeds,  subtract  the  interest  from  the  principal. 

Note. — There  are  always  as  many  decimals  in  the  answer  as  there  are  on  the  statement  line, 
and  no  more. 

PROBLEMS. 

6.  What  is  the  interest  on  $13o0  at  8  per  cent  for  64  days  ?     Ans.  $19.20. 

7.  What  is  the  interest  on  $550  at  7  per  cent  for  72  days  ?  Ans.  $7.70. 

8.  What  is  the  interest  on  $727.20  at  5  per  cent  for  11  days!  Ans.  $1,111. 

9.  What  is  the  interest  on  $7200  at  10  per  cent  for  121  days  ?     Ans.  $242. 

10.  What  is  the  interest  on  $3155.16  at  6  per  cent  for  5  years  T 

Ans.  $946,548. 

11.  What  is  the  interest  on  $2344.80  at  8  per  cent  for  3  months  ? 

Ans.  $46,890. 

12.  What  are  the  interest  and  the  amount  of  $5000  at  8  per  cent  for  2  years, 
5  months,  and  15  days  ?  Ans.  $983.33^  interest;  $5983.33J  amount. 

Note. — To  obtain  the  amount,  add  the  interest  to  the  principal. 

13.  What  are  the  interest  and  the  proceeds  of  $45000  at  4  per  cent  for  S 
months  and  20  days  ?  Ans.  $1300  interest;  $43700  proceeds. 

Note.— To  obtain  the  proceeds,  subtract  the  interest  from  the  principal. 

14.  What  is  the  interest  on  $1440  at  7  per  cent  for  123  days  1      Ans.  $34.44. 

15.  What  is  the  interest  on  $1711.90  for  34  days  at  15  per  cent  ? 

Ans.  $24.25+. 

16.  What  is  the  interest  on  $3240  for  94  days  at  9  per  cent  ?      Ans.  $76.14. 

17.  What  is  the  interest  on  $21636.72  for  63  days  at  10  per  cent? 

Ans.  $378.64+. 

18.  What  is  the  interest  on  $0840  for  5  years,  8  months,  and  14  days  at  4J 
percent?  Ans.  $1756.17. 

19.  What  is  the  amount  of  $2700  for  1  year,  2  months,  and  5  days  at  5  per 
cent?  Ans.  $2859.37 J. 


INTEREST. 


565 


20.  What  are  tbe  proceeds  of  $5284.50  for  4  montlis,  and  12  days  at  7  per 
cent?  Alls.  $ol4S.S0+. 

21.  What  is  the  interest  ou  $1500000  for  3  days  at  ^  per  cent  per  day  ? 

Aus.  $11250.00. 

OPERATION   INDICATED. 

;  $15000.00 

15000.00  i% 

90  =  A^o  per  day  =  90%  per  year. 

3G0    3  '  or  thus:  $  3750.00  =  interest  for  1  day. 

3  —  days. 

$11250.00,  Aus. 


$11250.00  =  interest  for  3  days. 

22.     What  is  the  interest  on  $30000  at  4  per  cent  per  mouth  for  28  days  f. 

Aus.  $1120.00. 

OPERATION    INDICATED. 


3C0 


300.00 

48  =  4%  per  month  =  48%  per  year. 
28 


or, 


30 


300.00 

4 

28 

$1120.00,  Aus. 


23. 


$1120.00,  Ans. 
What  is  the  interest  on  $260000  for  11  days  at  |  per  cent  per  day  ? 

Ans.  $10725.00. 

24.  What  are  the  interest  and  the  amount  of  $1804500  from  July  12  to  Sep- 
tember 13,  at  4.^  per  cent  per  auiiuiu,  last  day  inclusiye? 

Ans.  $14(382.94  interest.     $1879182.94  amount. 

25.  W^hat  is  the  interest  on  $2700  for  50  days  at  2  per  cent  per  month  1 

Ans.  $90. 

26.  A  note  for  $4100,  having  80  days  to  run,  is  discounted  at  IJ  per  cent  per 
month.     What  are  the  proceeds'?  Ans.  $3903.33. 

27.  W' hat  are  the  interest  and  proceeds  of  $10000  for  3000  days  at  10  per 
cent?  Ans.  $10000,  interest.     No  proceeds. 

28.  The  cash  price  of  a  lot  of  goods  is  $300.  A  buyer  purchases  the  goods 
on  a  credit  of  90  days  when  money  is  worth  8  per  cent.  What  ought  to  be  the 
credit  price?  Ans.  $306.00. 

OPERATION  INDICATED. 


45 


3.00 
90 

$6.00  int. 


29.  The  terms  for  an  invoice  of  goods  amounting  to  $2350  are  "4  months,  or 
5  per  cent  discount  lor  cash  in  10  days."  The  buyer  paid  cash  in  10  days.  What 
did  he  gaiu  or  lose,  money  being  worth  0  per  cent?  Ans.  $76.57  gain. 

OPERATION   INDICATED. 


$2350  —  5%  ($117.50)  =  $2232.50.  00 


2232.50 
110 

$40.9291+. 


$117.50  —  $40.93  =  $76.57. 


566  soule's  philosophic  practical  mathematics.  * 

THE  PHILOSOPHIC  SYSTEM  OF  CONTEACTING  INTEEBST  QUESTIONS. 


1130.  There  are  a  great  many  methods  of  contracting  interest  calculations, 
but  the  greater  number  of  them  are  applicable  only  to  sjiecial  combinations  of 
numbers,  while  others  require  the  memorizing  of  arbitrary  rules  and  exceptions 
instead  of  the  exercise  of  the  reasoning  faculties.  The  philosophic  system  which 
■we  here  present  is  general  in  its  application,  and  is  based  upon  one  beautiful  system 
of  work  by  which  we  solve  every  interest  problem,  and  give  a  reason  for  every  figure 
of  the  work  without  the  aid  of  rules,  no  matter  what  may  be  the  sum,  the  rate  per 
cent,  or  the  time. 

The  following  problems  will  elucidate  this  beautiful  system  of  contraction: 

PROBLEMS  TO  BE  SOLVED  MENTALLY  WITHOUT  THE  AID  OF  PEN  OR  PENCIL. 

1.  What  is  the  interest  on  $1200  for  G4  days  at  5  per  cent  ? 

2.  What  is  the  interest  on  $1400  for  78  days  at  6  per  cent  ? 

3.  What  is  the  interest  on  $1500  for  34  days  at  8  per  cent? 

4.  What  is  the  interest  on  $2000  for  124  days  at  9  per  cent? 

5.  What  is  the  interest  on  $1528.60  for  51  days  at  12  per  cent  ? 


FIRST  QUESTION. 


72 


OPERATION     OF 
SECOND   QUESTION. 


12.00 
64 

$10.66§,  Ans. 


FOURTH  QUESTION. 
9% 


60 


14.00 

78 


$18.20,  Ans. 
OPERATION    OF 


THIRD   QUESTION. 

8% 


45 


15.00 
34 

$11.33J,  Ans. 


40 


20.00 
124 


$62.00,  Ans. 


FIFTH    QUESTION. 
12% 


30 


15.28.00 
51 

$25.9802,  Ans. 


lu  worliuig  out  these  statements,  'we  make  no  figures  Imt  the  result,  or  answer;  thus  iu 
■working  the  lirst  <iuestioii  wo  see  that  the  1200  and  the  72  may  he  canceled  by  12,  and  that  1200  is 
e(iual  to  12,  100  times,  and  that  72  is  equal  to  12,  6  times;  wo  have  therefore  hut  to  multiply  the 
64  by  100  and  divide  by  6 ;  all  of  ■which  is  mentally  performed. 

Iu  the  second  example,  Tve  cancel  mentally,  one  0  on  the  1400  and  the  0  on  60,  then  the  78  by 
6,  ■which  gives  us  a  quotient  of  13,  ■with  ■which  we  multijily  110,  and  produce  the  answer. 

In  the  third  example,  'we  cancel,  mentally,  the  45  and  the  1500  by  15;  we  then  annex  the  two 
Os  to  the  34  and  divide  by  3 ;  the  result  is  the  answer. 

In  the  fourth  example,  we  cancel  a  0  on  each  side  of  our  statement  line,  th«n  the  200  or  the 
124  by  4,  and  multiply  the  two  factors  together,  which  gives  us  the  correct  result  or  answer. 

In  the  fifth  example,  wo  cancel  a  0  on  each  side  of  our  statement  line,  then  the  51  by  3,  which 
gives  us  a  quotient  of  17,  with  which  we  multiply  the  15286,  and  produce  the  correct  result. 


INTEREST. 


56; 


In  all  interest  calculations,  we  must  be  careful  to  point  off  1  per  cent  in 
commencing  the  work,  and  in  the  final  result  point  oft'  the  same  number  of  figures. 

We  connnend  this  system  of  interest  calculations  and  this  method  of  con- 
traction over  all  others,  because  of  their  simplicity  and  their  easy  application  to  all 
problems  in  interest  ■without  referring  to  rules,  and  because,  taking  the  problems  as 
they  naturally  occur  in  buisness,  they  are  the  shortest  and  by  far  the  easiest. 

Note. — In  the  following  problems,  place  the  figures  on  the  st.itemeut  line  and  solve  mentally, 
writing  no  figures  but  the  answer. 

C.  Find  the  interest  on  $1440  for  G7  days  at  2.^%  ? 

7.  "  "  "      8400  for  28  days  at  3 7«? 

8.  "  "  "      ICOO  for  54  days  at  3.}%? 

9.  '•  '•  '•      2817  for  48  days  at  4i%  ? 

10.  "  "  "  21G0forC4daysat5%? 

11.  "  "  "  3000  for  13J  months  at  C%  ? 

12.  "  "  "  450for40d"aysat7%? 

13.  "  "  "  15000  for  33  days  at  8%  ? 

14.  "  "  "  2408  for  10  days  at  9%  ? 

15.  "  '•  "  1137.G0  for  72  days  at  10%  ? 

16.  "  "  "  25000  for  IS  days  at  12%? 

17.  "  "  "  728.27  for  9G  days  at  15%  1 


Ans. 

$6.70. 

Ans. 

19.60. 

Ans. 

8.40. 

Ans. 

10.90 

Ans. 

19.20. 

Ans. 

202.50 

Ans. 

3.50. 

Ans. 

110.00. 

Ans. 

0.02. 

Ans. 

22.752. 

Ans. 

150.00. 

Ans. 

29.1308 

PECULIAR  INTEREST  QUESTIONS. 
1131.     1.     "What  is  the  interest  on  10/  for  10  days  at  10  per  cent? 


STATEMENT. 


100 
360 


10 
10 
10 


Ans.  -aij/. 

Explanation. — In  problems  in  which  there  are  not  two 
figures  to  rei)resent  dollars,  instead  of  indicating  the 
divison  by  prefixing  and  pointing  off  two  O's  we  divide 
the  sum  by  100  and  place  the  result  or  quotient  on  the 
statement  line,  and  then  proceed  as  iisual.  In  this  prob- 
lem, the  interest  on  10c.  for  1  year  at  1  per  cent  is  -^'-Xc. 


3^/  Ans. 

2.  What  is  the  interest  on  $1.15  for  15  days  at  0  per  cent?  Ans.  ||/. 

3.  What  is  the  interest  on  12^/  at  4 J  per  cent  for  lOJ  days?     Ans.  ilio/* 

OPERATION    INDICATED. 


/ 


/^ 


2 

25 

100 

2 

9 

3G0 

O 

21 

or 


3G000 


or 


9 

21 


2 

8000 
o 


/ 


21 


4.  What  is  the  interest  on  1G|/  at  5.J  per  cent  for  8  days,  6  hours,  and  24 
minutes?  *"«    ---- 

5.  What  is  the  interest  on  12^^  for  5.i  days  at  6^  per  cent? 

6.  What  is  the  interest  on  6^/  for  10  days,  4  hours,  and  : 
per  cent  ?  Ans, 


Ans. 


-5  5_e' 

ieosr • 


5  minutes  at  4| 


6B3  5  6-JO/ 


568 


SOULE  S    PHILOSOPHIC    PRACTICAL   MATHEMATICS. 


DIFFERENCE  OF  INTEREST  BY  USING  THREE  HUNDRED  AND  SIXTY 
AND  THREE  HUNDRED  AND  SIXTY-FIVE  DAYS  TO  THE  YEAR. 

1132.  Since  tbere  are  3G5  days  in  a  common  year,  it  follows  tliat  when  360 
days  are  used  as  an  interest  divisor  on  tlie  basis  of  a  year,  tlio  interest  is  3I5  =  -,^3 
greater  tbaii  would  be  produced  by  using  305  as  tlie  interest  divisor,  and  wlien  305 
days  are  used  as  the  basis  of  a  year  ortlie  interest  divisor,  the  result  is  ^ao  =  ri  l^ss 
thau  would  be  ijroduced  by  dividing  by  300. 

The  only  exception  to  the  above  is  when  the  time  is  in  even  years;  then 
there  is  no  difference. 

The  following  problem  will  more  clearly  elucidate  this  difference : 

1.  What  is  the  interest  on  $7300  for  81  days  at  8  per  cent  when  using  3G0 
days  for  a  year,  and  also  when  using  365  days  for  a  year? 

Ans.  See  operations  following. 


Oiieratiou  for  360  days  to  tbe  year. 
7300    ' 
360     8 
81 


Operation  for  3G.5  days  to  the  year. 
7300 
8 
81 


305 


$131.40  int. 


$129.00  int. 


Operations  to  sliow  that  tlio  interest  for  3fiO  days  is  v'j  more  than  the  interest  for  365  days,  and  that 
the  interest  for  305  days  is  7"^  less  than  for  360  days. 

$131.40  4-  73  =  $1.80.  $129.60  ->  73  =  $1.80. 

$131.40  —  $1.80  =  $129.60.  $129.60  +  $1.80  =  $131.40 

INTEREST  AT  THItEE  HUNDRED  AND  SIXTY-FIVE  DAYS  TO  THE  YEAR. 


1133.     1.     What  is  the  interest  on  $5000  for  93  days  at  7  per  cent,  using  365 
days  for  a  divisor  ?  Ans.  $89.18. 


OPEKATION. 


30.J 


50.00 

7 
93 

$89.18  Ans. 


Explanation. — The  reasoning  in  this  problem  is  as 
follows:  The  interest  on  J^oOOO  for  1  year  at  1  per  cent  is 
$50,  and  at  7  per  cent  7  times  as  ninch;  and  for  a  day, 
instead  of  a  year,  it  is  the  365th  part,  and  for  93  days  it 
is  93  times  as  much  as  for  1  day. 


2.  What  is  the  interest  ou  $2500  at  6  per  cent  for  146  days,  counting  365 
days  as  the  interest  year  ?  Ans.  $00.00. 

3.  What  is  the  interest  ou  $1111.11  at  11  per  cent  for  11  times   11   days, 
counting  305  days  to  the  year?  Ans.  $40.51+. 


INTEREST. 
INTEREST  ON  UNITED  STATES  BONDS. 


569 


1134.     1.     What  interest  lias  accrued  on  $10000  United   States  6  per  cent 


registed  bonds  from  July  1,  1885,  to  August  1,  18901 

OPERATION. 


Ans.  $3049.32. 


365 


10000 

6 

1855 

$3019.32  Ans. 


$10000 

0  % 


or 


$000.00 

5  years. 


365 


$3000.00  int.  for  5  years. 
49.32  int.  lor  30  days. 


10000 

6 

30 

$49.32 


$3049.32  Ans. 

Note.— The  United  States  Government  computes  interest  on  tbe  L.-isis  of  3G5  (Lays  to  .a  year, 
and  for  any  number  of  days  less  than  a  year  a  iiroportional  part  is  taken:  Thus,  for  19  days,  jVf 
would  bo  taken. 

2.     What  interest  has  accrued  oti  $250000  United  States  4J  per  cent  funded 
loan  from  September  1st,  1882,  to  December  1st,  1890?  Ans.  $92773.97  gold. 

OPERATION. 


365 


250000 

9 

3010 

$92773.97  Ans. 


$250000 


or 


$11250.00 

8  j'ears. 


2 
365 


$90000.00  int.  for  8  yrs. 
2773.97  int.  for  90  ds. 


250000 

9 

90 

$2773.97 


92773.97  Ans. 


INTEREST  ON    ENGLISH  MONEY. 


1135.     1.     What  interest  has  accrued  on  £100000  English  consols  from  Jan- 


uary 1,  1894,  to  July  1,  18951 


STATEMENT. 
£ 
1000.00 

3 
540 


365 


Ans.  £4487.07. 


Explanation. — Consols  are  English  bonds  that  bear  3 
per  cent  interest.  See  Stocks  and  Bonds  for  full  explan- 
ation. 


£4487.67  Ans. 
Note. — 365  days  are  used  la  Interest  Computations  in  England. 


570 


SOULE  S    rillLOSOPIlIC    rUACTICAL    MATHEMATICS. 


2.    What  is  the  interest  on  £1420  12s.  'Jd.  for  G3  days  at  5  per  cent? 

Aus.  £12.  5s.  2d.  + 

OPERATION. 


£,      s,    d. 
1420    12    9 
5    41 


305 


.60  +  .037i  =  £.C3S. 


£  £12.26+  Ext>laiialion.  —  \Ya 

1420.638  20  Ijcre    iiist    reduce    the 

5  Bliilliiii^s  and  iience  to 

G3  5.20s.         the  decimal  of  a  ])Ound, 

12  and    then    i)roceed    in 

the    usnal    manner    of 
computing  interest. 


£12.26030+ 

2.40d. 

Note. — For  an  explanation  of  the  method  of  reducing  shillings  and  pence  to  the  decimal  of 
B  pound,  see  Article  869,  page  450. 

3.    What  is  the  iuterest  on  £3199  4s.  4id.,  at  G  per  cent  for  G3  days  allowing 
360  days  to  the  year?  Aiis.  £33  lis.  lOd. 

OPERATION. 

4s.    4Jd. 
5       4i 


60 

.20  +  .0183  =  £.219.  — 


£ 

31.99.219 

63 

£33.59179 
20 

11.  S3.580S, 
12 

£33.59179 

10.     0296d 

4.  What  is  the  interest  on  14s.  7d.,  for  50  days  at  8  per  cent,  allowing  365 
days  to  the  year?  Ans.  1.9176d.  practically  2d. 

5.  What  is  the  interest  on  £8  3s.  lOd.,  for  34  days  at  5  i)er  cent,  allowing  360 
days  to  the  year  ?  Ans.  9.2S4d.;  practically  9d. 

6.  What  is  the  iuterest  on  £15940  12s.  Cd.  for  125  days  at  8  per  cent,  allow- 
ing, as  is  the  English  custom,  365  days  to  the  year?  Ans.  £436  14s.  7d. 

Note. — See  Index,  English  Exchange,  for  operations  in  English  Exchange. 


INTEREST  ON  FRENCH  CURRENCY. 


1 136.    1.    What  is  the  interest  on  fr.  55620.25,  for  73  days  at  4  per  cent  ? 

Aus.  Tr.  444.96. 
STATEMENT. 
Fr. 

556.20.25 
4 
365     73 


Fr.  444.9620  Ans. 

Note. — 305  days  are  used  in  interest  computations  in  France. 

2.    What  is  the  iuterest  on  fr.  4190.80  for  165  days  at  3  per  cent? 

Ans.  Fr.  56.83+. 


INTEREST.  571 

INTEREST  ON  GERMAN  CURRENCY. 

1 137.     1.    What  is  tlie  interest  on  6870.60  marks  for  64  days  at  4 J  per  cent  ? 

Ans.  54.21+  marks. 

OPERATION    INDICATED. 


2 
365 


68.70.60 

9 

64 


Note. — 365  days  are  used  in  interest  computatious  in  Germany. 

2.    What  is  the  interest  on  128400  marks  for  235  days  at  5  per  cent  ? 

Ans.  4133.42+  marks. 

DIFFERENT  METHODS  OF  COMPUTING  INTEREST. 

1138.  We  present  the  following  different  methods,  not  because  tliey  possess 
any  great  merit,  but  that  they  may  be  contrasted  with  our  philosophic  system,  and 
also  tliat  they  may  be  used  for  reference  by  business  men  who  have  previously 
learned  one  of  these  methods,  and  who  may  not  have  the  time  or  desire  to  learn  a 
new,  though  better  method. 


BANKERS'  METHOD,  OR  THE  SIXTY-DAY  OR  SIX  PER  CENT  METHOD. 

1139.    The  following  problems  will  show  the  method  of  calculating  interest 
as  used  by  many  bankers. 

Note. — It  is  the  nniform  custom  of  bankers  in  tbeir  daily  operations  of  banking,  to  calculate 
interest  from  interest  tables. 

1.     What  is  the  interest  on  $2485  for  75  days  at  6  per  cent?      Ans.  $31.06. 

OPERATION. 

^  of  $24     85  =  interest  for  60  days.  Explanation.— The  basis  of  tbia  work   is  that  1 

=        6      21  =  "  "    15      "  1'*'"  '^^"t  '°''  60  days  is  equal  to  1  per  cent  li)r  3G0 

"^  days,  and  hence  to  obtain  6  per  cent  for  GO  days  we 

~~  simply  divide  the  principal,  or  sum  on  ■which  we 

$31      06  Ans.     "  "    75       '<  wish  to  find  the  interest,  by  lOO;  this  we  do  in  this 

case  by  drawing  a  vertical  line  through  the  sum, 
so  as  to  cut  off  the  two  right  hand  figures  of  the  dollars.  Then,  ns  75  days  is  J  more  than  60,  we 
add  i  of  the  interest  for  OO  days  to  itself,  and  thus  produce  S31.06,  the  correct  interest.  In  c.nse 
we  had  but  one  figure  representing  dollars,  then  we  would  prefix  a  0  before  drawing  the  line,  and 
if  no  dollar  figures,  then  two  Os  would  be  prefixed. 

If  we  wish  the  interest  for  a  different  per  cent  than  6,  we  would  add  to  or  subtract  from  6 
per  ceut  interest  such  a  part  of  itself  as  the  rate  per  cent  was  more  or  less  than  6. 

Note.— The  interest  for  years  and  months  is  found  by  multiplying  the  interest  for  60  days,  ot 
two  mouths,  by  one  hall'  the  number  of  mouths. 


572 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


What  is  tbe  interest  on  $543  for  63  days  at  5  per  cent?  Ans.  $4.75. 


OPERATION  BY  THE  BANKERS'  METHOD. 


of  $." 


^  Of 


4;?         =  interest  for  60  days  at  G% 
1'715     =         "         "3     ""      "  6% 


7015 
0502 


"   03     "      "  6% 
"    03     "      "  1% 


OPERATION  BY   OUR 
PHILOSOPHIC    METHOD. 


8  ;^ 


5.43 
3801 


7513  Ans.       "         "   03     "      "  5% 

$4.75J  Ans. 

Exjtlaiin(io)i.  —  }]ere  we  ileiluct  I  of  tbe  iuterest  at  6  per  cent,  and  thus  obtain  in  Ibe  remain- 
der 5  per  cent.  • 


3.     What  is  the  interest  on  $1428  for  35  days  at  8  per  cent?     Ans.  $11.10§. 


OPERATION  BY  THE  BANKERS'  METHOD. 


i    ) 

■6^  ) 


h     ) 


$11 


28  interest  for  CO  days  at  6% 
14         "         "   30     "      "   6% 


19 


II 


a 


0 


n 


"    6% 


.33         "         "   35     "      "   0% 
77§  is  i  of  0%  interest  added. 

10§  Ans. 


OPERATION   BY  OUR  PHILOSOPHIO 
SYSTEM    OR  METHOD. 


14.28 

?^  7 


9996 


L1.10§  Ans. 


4.     What  is  the  interest  of  $1260  for  106  days  at  6  per  cent  ?    Ans.  $22.26. 


Operation  by  tbe  Bankers'  or  60-day  or  6  per  cent  Metbod.  . 

$12.60  =  interest  for  60  d.  Cut  off  two  places  to  the  right. 

6.30  =  interest  for  30  d.  ^  of  $12.60  =  $6.30 

2.10  =  interest  for  10  d.  |  of      6.30  =    2.10 

1.26  =  interest  for    6  d.  ^  of      6.30  =     1.26 


Operation  b.y  our   Philo- 
sopbic  Metbod. 


00 


^22.26  =  interest  for  106  days. 


1-l^p  21 
106 

$22.26 


5.     What  is  the  interest  of  $2340  for  3  months,  27  days  at  7  per  cent? 

Ans.  $53.23J. 


Operation  by  tbe  Bankers'  or  60-day  Metbod. 

$23.40  =  interest  for  60  days,  or  2  months. 
11.70  =  interest  for  30  days,  or  1  month. 
7.80  =  interest  for  20  days,  4  of  23.40  =  $7.80 
2.34  =  interest  for    6  days,  I  of  11.70  =     2.34 
39  =  interest  for    1  day,    i  of    2.34  =       39 

$45,63  =  interest  for  given  time  at  6% 

7,ei  =  interest  for  given  time  at  1  %  i  of  $45.63  =  $7.61 

$53.24  =  interest  for  given  time  at  7% 

Note. — Add  one  cent  wben  tbe  fractioQ  is  i  or  more. 


Operation  by   onr  Philo- 
sopbic   System. 


0(30 
6 


$       39 

7 
117 

819 


31941 


$53, 23 J  Ans. 


INTEREST. 


573 


What  is  tbe  interest  of  $3000  for  5  montlis,  19  days,  at  4  per  cent? 

Aus.  $07.60. 


Operation  by  the  Bankers'  or  60  day  Method. 


Operation  Ijy   onr  Philo- 
sophic System. 


$30.00  =  interest  for  60  days,  or  2  montlis. 
18.00  =  interest  for  1  mouth. 


9(1.00  =  interest  for  5  months,  $18.00  x  5 
9.00  =  nit.  for  la  d. 

1.80  =  int.  for    3d.       3  =  I  of  15.  i  of 
60  =  int.  for    Id.       1  =  |  of    3.  |  of 


$90.00 


15  =  i  of  00.  i  of  $30.00  =  $9.00 

"      9.00  =     l.SO 

1.80  =       .60 


?P 


36.00  40 
169 

$07.60  Ans, 


$101.40  =  interest  for  given  time  at  6% 
33.80  =  interest  for  given  time  at  2% 

67.00  =  interest  for  given  time  at  4% 

Take  i  of  $101.40  from  itself. 

7.    Which  of  the  above  methods  has  the  greater  merit  ? 

Ans.  The  Philosophic. 

Taking  all  principals,  all  rates  per  cent  and  all  jieriods  of  time,  the  Philo- 
sophic System  is  far  superior  to  all  others. 


THE  THIRTY-SIX  PER  CENT  METHOD  OF  COMPUTING  INTEREST. 


1140.  This  method  is  based  upon  the  Interest  Divisor  for  36  per  cent  which  is, 
as  shown  on  page  500,  1000.  Hence,  as  elucidated  on  page  501,  to  find  the  interest 
on  any  amount  for  1  day  at  36  per  cent,  divide  by  1000;  then  compute  it  for  the 
required  time  and  rate  per  cent. 


PROBLEM. 

.    What  is  the  interest  on  $2758.40  for  75  days  at  8  per  cent? 

Ans. 

OPERATION    INDICATED. 


15.97+. 


36 


75840  =  interest  for  1  day  at  36%. 


$45.9733J  Ans. 


Explanafion, — .Since  the  3G  per  cent  Tniereaf 
Dirisor  is  1000,  we  divide  the  auionut  by  1000 
and  thns  obtain  J.li. 75840  interest  for  1  day  at  36 
per  cent.  Then,  as  indicated  in  the  statement, 
we  divide  the  interest  for  1  day  by  36,  to  obtain 
the  interest  at  1  per  cent ;  this  result  we  multi- 
ply by  8  to  obtain  the  interest  at  8  per  cent,  and 
the  resnlt  thus  jiroduced,  by  75  to  obtain  the 
interest  for  75  days. 


Various  methods  are  used  by  different  parties  to  conclude  the  operation  after 
dividing  by  1000.  But  as  none  of  the  methods  are  equal  to  the  above  for  ease  and 
simplicity,  we  will  not  occupy  sjiace  with  them. 

The  100  day  method,  of  computing  interest  is  also  based  upon  the  36  per  cent 
Interest  Divisor,  and  like  the  above  method,  is  far  inferior  to  the  unrivaled  Philo- 


574 


SOULE -S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Sophie  System,  as  sbowu  in  previous  problems,  pages  572  and  573  aud  by  tbe  fol- 
lowiug  oi)eratioii  of  the  above  problem. 

OPERATION   INDICATED   BY   THE   PHILOSOPHIC   SYSTEM. 


45 


27.5840 
75 


or. 


15.9733^  Ans. 


27.5840 
5 


$45.9733i  Ans. 


THE  S.  L.  BLACK  METHOD  OF  CALCULATING  INTEREST. 


1141.  The  following  method  is  generally  known  as  the  S.  L.  Black  Method,  it 
having  been  taught  by  him  some  40  years  ago.  It  is  one  of  the  best  of  the  rule 
methods  aud  yet  it  surrenders  without  a  condition  to  our  Philosophic  System. 

NoTK. — See  tlie  operations  of  the  same  problems  by  the  two  systems. 

1.    What  is  the  interest  on  $2400  for  1  year,  4  months  and  15  days,  at  6  per 
cent!  Ans.  $198. 

OPERATION. 


$24.00 


16.5 
1200 


$198,000  Ans. 
of  12,  which  is  i,  and  proiUice  the  answer  in  mills. 


Explanation. — In  this  problem,  we  first  rednco 
the  1  year  to  months  and  add  to  it  the  4  months, 
thus  makiuj;  16;  we  then  divide  the  days  by  3 
and  annex  the  f|notient  5  to  the  16  months.  We 
then  multiply  the  number  thus  obtained  by  such 
a  part  of  the  sum  $2400  as  the  6  per  cent  is  part 


By  this  method  of  computing  interest,  when  the  rate  per  cent  is  less  than  12, 
reduce  the  time  to  months  and  annex  thereto  J  0/  the  days,  if  any,  and  then  multiply 
the  sum  or  principal  by  such  a  part  of  the  time,  or  the  time  by  such  a  part  of  the  sum 
or  principal  as  the  rate  per  cent  is  part  of  12,  and  jtoint  off  three,  as  the  answer  is  in 
mills,  when  there  are  days.  When  there  are  cents  in  the  principal,  point  off  five 
places.  When  the  rate  is  12,  multiply  all  of  the  time  by  all  of  the  principal.  When 
the  rate  is  more  than  12,  first  find  the  interest  at  12  per  cent  and  then  add  such  a 
part  of  the  interest  to  itself  as  the  excess  of  rate  is  part  of  12. 

The  basis  of  this  system  is  that  1  per  cent  for  one  year  is  12  per  cent  for  one 
month,  and  hence  to  find  the  interest  for  months  at  12  per  cent,  we  Lave  but  to 
multiply  the  sum  by  the  number  of  months,  or  the  number  of  months  by  the  sum^ 
and  the  answer  is  in  cents. 


*  INTEREST.  S75 

2.    What  is  the  interest  ou  $1230  for  4S  days  at  S  per  cent  ?     AiU.  $13.12. 


OPERATION. 

%)  $1230 

4T0  X  2  =  820 
5248      =  16 


OPERATION 
BY  OUR  PHILOSOPHIC  SYSTEM. 


$13,120  Ans. 


0 


1230  82 
^^       16 

$13.12    Ans. 


Explanation. — In  this  example  we  see  that  8  per  cent  is  |  of  12,  and  that  f  of  $1230  is 
which  we  multij)ly  by  16,  which  is  i  of  the  Jays,  and  we  have  the  answer  in  mills. 


3.    What  is  the  interest  on  $0400  for  94  days  at  9  per  cent  ?     Ans.  $150.40. 


OPERATION 
BY   THE   S.   L.   BLACK  METHOD. 

I    of  $6400 


=  $4800 
31^ 

148800 
1000 

$150,400  Ans. 


OPERATION 
BY   OUR   PHILOSOPHIC   SYSTEM. 


/P 


0400  160 
94 

$150.40     Ans. 


4.     What  is  the  interest  on  $1512  for  35  days  at  10  per  cent !    Ans.  $14.70. 


OPERATION  BY   THE   S.   L.   BLACK  SYSTEM. 


25"' 

1   $1512 

or, 

i 

504 

$1512 
llg 

$16,632 
l.OOS 

1   420 

1200 
11§ 

13860 

840 

)  $17,640 
2.94 

$14,700  Ans. 

SI  4.70 

$14.70      interest  at  10%. 


OPERATION 
BY   OUR    PHILOSOPHIC   SYSTEM. 


t^J-i  42 
35 


$14.70    Ans. 


By  the  contrasts  here  made,  it  is  clear  that  our  philosophic  system  possesses 
advantages  over  the  very  best  rule  work  koowu. 


576 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


A  ^"EW  TABLE. 

1142.     Tbe  following  table  is  used  by  many  accountants,  with  advantage,  in 
interest  computations. 

6%  jier  annum  =    J 7c  of  the  face  of  a  note  or  debt  for  30  days. 


9% 
10% 
12% 
15% 
18% 
20% 
24% 


a 
u 

a 
a 


=     Wc 
=     lie 

=  \% 
=  1  % 

=  1^% 
=  li% 

=  !§% 
=  2  % 


(( 


" 

30 

30 

30 

30 

30 

30 

30 

.-^n 

Tbe  following  problem  will  illustrate  tbe  metbod  of  work  by  tbe  use  of  this 


table : 


PROBLEMS. 


1.     What  is  tbe  interest  on  $843. CO  for  15  days  at  S  per  cent!     Ans.  $2.S1. 


OPERATION. 


fS.43.60 
$2.81.20    Ans. 


Erplanaiion. — As  8  per  cent  per  annum  is  equal  to  |  per 
cent  of  tbe  face  for  30  days,  and  as  15  days  is  i  of  30,  it  is 
plain  that  the  interest  is  i  of  |  per  cent  (=  i)  of  the  face 
of  the  S843.60.  which  is  S2.81.  Had  the  time  been  18 
days,  we  would  have  first  found  it  for  15  days,  then  added 
}  of  the  interest  to  itself,  as  18  is  i  or  20  per  cent  more 
than  15.  Had  it  been  for  20  days,  we  could  have  found  it 
for  15.  and  then  added  J  of  the  interest  to  itself,  or  we  could  have  found  it  for  30  days  and  then 
deducted  J  of  tbe  interest,  or  we  could  have  reasoned  thus :  20  days  being  |  of  30,  and  8  per  cent 
per  annum  being  j  per  cent  of  the  face  for  30  days,  then  §  of  |  (=  J  of  the  face)  is  the  interest. 

By  a  little  reflection,  it  is  readily  seen  that  all  of  these  contracted  methods  are  but  the 
canceled  results  of  our  philosophic  system,  and  to  the  ready  calculator  they  can  often  be  used 
mentally  to  great  advantage, 

2,  Wbat  is  tbe  interest  on  $1624.50  for  22  days  at  6  per  cent? 

3.  ^\'bat  is  tbe  interest  on  $420.80  for  34  days  at  9  per  centt 

BANKEES'  A^'D  MERCHANTS*  DlSCOU:ST. 


1143.  In  the  preceding  work,  we  have  defined  simple  interest  without  making 
mention  of  Bank  Discount ;  and  as  there  has  been  and  still  is  much  discussion  on 
tbe  subject  of  bank  discount,  we  will  therefore  appropriate  a  small  space  to  a  con- 
sideration of  tbe  subject  before  giving  esauijiles.  Bank  Discount,  or  tbe  more  gen- 
eral term.  Discount,  is  interest  paid  in  advance.  Hence,  when  bankers  and  business 
men  discount  notes  or  debts,  where  tbe  element  of  time  is  considered  as  well  as  the 
rate  per  cent,  they  get  the  interest  on  the  face  of  tbe  note  or  debt  for  tbe  actual 
time,  and  at  tbe  specified  rate  per  cent.  The  sum  thus  obtained,  and  called  interest 
or  discount,  is  then  deducted  from  tbe  face  of  tbe  note  or  debt  discounted,  and  the 
remainder  or  diflereuce  is  the  cash  value  or  proceeds. 


■*  bankers'  discount.  577 

Many  persons  iniinfornied  on  the  subject  of  business  customs  pertaining  to 
the  operations  of  interest,  earnestly  contend  tbat  bank  discount  is  the  same  as 
jHiother  divrsion  of  interest  called  True  Discount.  True  discount  diiiers  materially 
from  bank  discount  as  ■we  sball  elucidate  a  few  pages  furtber  on,  and  coHsts/s  in  Jind- 
iiig  tchat  sum  of  money  loaned  for  a  certain  time  at  a  certain  rate  per  cent  icill  amount 
to  a  given  sum  ;  or,  to  define  it  differently,  it  is  the  finding  and  deducting  of  such  a 
.v»»i  called  true  discount,  from  the  face  of  a  note  or  debt,  as  icill  leave  a  remainder 
which,  if  the  simple  interest  be  added  thereto  for  the  same  time  and  rate,  icill  reproduce 
the  original  sum. 

Tbe  real  difference  between  bank  and  true  discount  arises  from  tbe  fact  tbat  in 
bank  discount  we  operate  ou  tbe  face  of  tbe  note  or  debt,  wbicli.  of  course,  includes 
the  interest  cbarjred  :  and  in  true  discount  we  operate  ou  tbe  tnie  present  value  of 
the  note  or  debt,  and  hence  the  difference  atnounts  to  the  simple  interest  on  the  true 
discount. 

Tbe  term  true  discount  does  not  imply  tbat  bank  discount  is  false  or  untme. 
It  is  thus  named  to  distinguish  it  from  bank  discount,  and  because  tbe  true  discount 
of  a  note  or  debt  is  just  equal  to  the  interest  ou  tbe  proceeds  of  tbe  sum  from 
-which  the  true  discount  was  deducted. 

1144.      PROBLEMS  IN  BAN'KERS'  AND  MERCHANTS'  DISCOUNT. 

Note. — The  following  examples  are  all  of  a  practical  character,  and  each  involves  some  point 
of  law  or  of  linsiness  custom  necessary  to  he  known  in  dealing  in  notes  and  securities,  or  in  per- 
forming the  daily  discount  transactions  of  the  merchant  and  the  banker. 

TO  MATURE  AND  DISCOUNT  NOTES,  WHEN  DRA^^■N  IN   MONTHS   AND   WHEN   DRAWN 

IN  DAYS. 

Note. — Observe  carefully  the  difference  in  the  maturity  and  discount  of  the  two  following 
notes : 

$2540.80  New  Orleans,  December  18,  1895. 

1.  Two  months  after  date,  for  value  received.  I  jiromise  to  pay  to  tbe  order 
of  Frank  Draxler,  Two  Thousand  Eive  Hundred  Forty  and  iVd  Dollars. 

A.   D.   HOFELINE. 

When  does  this  note  mature  ?  What  are  the  proceeds,  if  discounted  the  day 
it  was  drawn  at  9  per  cent  f  Ans.     It  matures  February  IS  21,  1S96. 

Tbe  net  proceeds  are  $2498.88. 

OPERATION. 

*  ^  iTjrpZanad'on.— In  maturing  this  note  in  accordance  with 

law,  we  count  the  months,  but  in  discounting  in  aceord- 
_  ance  with  business  custom  in  New  Orleans,  we  count  the 


40 


66 


$41 .9232  discount.  actual  number  of  days  in  the  two  months  and  add  thereto 


I249S.S8      proceeds.  3  ^^ys  of  g^a^e  and  discount  day. 


578 


SOULES    nilLOSOrHIC    PRACTICAL    MATHEMATICS. 


$2540.80  New  Orleans,  December  18,  1S05. 

'2.     Sixty  days  after  date,  for  value  received,  I  promise  to  pay  to  the  order  of 
G.  Reynolds,  Two  Thousand  Five  Hundred  Forty  and  ,%%  Dollars. 

J.  B.  Anderson. 

When  does  this  note  mature?     What  are  the  ]>roceeds,  if  discounted  the  day 
it  was  drawn,  at  9  per  cent?  Ans.     February  10/10,  189G,  it  matures. 

$2500.15,  proceeds. 

OPERATION. 


40 


25.40.80 
64 


$40.6528  discount. 
$2500.15      proceeds. 


Explaiialion.—lii  this  problem,  according  to  law,  w© 
mature  in  days,  and  according  to  custom  we  discount  ii> 
days,  counting  grace  and  discount  day. 


GENERAL  DIRECTIONS  FOR  BANKERS'  AND  MERCHANTS'  DISCOUNT. 


1145.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  Bankers'  and  Merchants'  Discount: 

1.  Calculate  the  interest  on  the  note  at  the  yiven  rate  for  the  actual  number  of 
days  that  the  note  has  to  run,  plus  three  days  of  (/race  and  discount  day. 

2.  Subtract  the  interest  thus  found  from  the  face  of  the  note  ;    the  remainder 

will  be  the  proceeds. 

Note  1. — Wbcn  notes  bear  interest,  find  tbe  amount  or  value  of  tbe  same  at  maturity,  and 
calculate  tbe  discount  on  siicb  maturity  value. 

Note  2  —In  many  cities  and  States,  interest  is  not  cbarged  for  discount  day. 

Note  3,  — In  all  problems  following,  discount  day  is  included  unless  otberwise  stated. 


PROBLEMS. 

3.     January  3,  1895,  a  note  was  drawn  for 


1200  and  made  i)ayable  one  year 
after  date,  without  interest.  When  does  this  note  matuie?  If  discounted  on  th& 
day  that  it  was  drawn  at  8  per  cent,  what  was  the  discount  and  proceeds? 

Ans.    January  3/G,  1896,  it  matures.     $98.40,  discouut.    $1101.60,  proceeds. 

OPERATION. 

$  Explanation. — This  note  being  drawn  in  years,  we  therefore,  in 

/^.00   10  conformity  to  custom,  mature  it  in  years,  and  when  we  discounts 

i'^  it,  in  conformity  to  a  diflerent  business  custom,  we  count  the 

0   300      303   123  actual  days  of  unexpired  time,  including  3  days  grace  and  dis- 

connt  day.     In  making  tbe  statement,  we  use  the  same  reasoning 

$98.40  as  in  the  third  problem  of  interest. 

$6231.50.  New  Orleans,  November  1,  1895. 

4.  Ninety  days  after  date,  for  A-alue  received,  1  promise  to  pay  to  the  order  of 
R.  Cole,  Six  Thousand  Two  Hundred  Thirty-one  and  ^o%  Dollars;  payable  at  th«* 
Germauia  National  Bank,  New  Orleans. 

J.  P.   WiLKIKSON. 


BANKERS      DISCOUNT, 


5/9 


When  does  this  note  become  due? 
December  -!3,  1S'J5,  at  6  per  cent  ? 

OPERATION. 


What  are  the  proceeds,  if  disconiited 
Aus.     January  '30j'2  189G,  it  is  due. 
$6187.88,  proceeds. 


62.31.00 

0 

^-i  7 

$43.62().n 


$0187.88 


discount, 
net  proceeds 


Explanation. — Had  this  note  lieen  discounted 
the  day  that  it  was  drawn,  it  would  have  been 
discounted  for  04  days;  but  as  it  was  not  dis- 
counted until  December  i!3d,  we  see  that  52  days 
have  already  expired,  and  as  discount  is  only 
charged  for  the  time  that  the  note  or  debt  has 
to  run,  wo  therefore  deduct  the  52  days  from  the 
94,  and  obtain  42  days  unexpired  time  for  which 
we  discount  the  note. 


5.  A  note  for  $7281.27,  bearing  date  September  14, 1805,  and  payable  90  days 
after  date,  was  discounted  at  the  Germania  National  Bank,  September  28,  1895,  at 
8  per  cent.  What  were  the  interest  and  net  proceeds,  and  when  does  the  note 
mature  1  Aus.    $129.44,  interest. 

$7151.83,  net  proceeds. 
December  13/16, 1895,  it  matures. 

6.  A  draft,  bearing  date  July  18,  1895,  and  accepted  July  tbe  21st,  1895,  for 
$10000,  payable  30  days  after  sight,  was  discounted  August  7th,  at  8  per  cent.  What 
were  the  net  iiroceeds,  and  when  does  the  draft  mature  ? 

Ans.    $9962.22,  net  proceeds.    August  20/23, 1895,  it  matures. 

7.  A  note  for  $5800,  bearing  date  February  12,  1895,  and  payable  6  months 
after  date,  was  discounted  by  a  note  broker  Jlarcli  1,  1895,  at  12  i^er  cent.  What 
was  the  maturity  of  the  note  ?  What  was  the  discount,  and  what  the  cash  value  or 
proceeds?  Aus.    August  12/15, 1895,  it  matured. 

$324.80,  interest. 
$5475.20,  cash  value. 

8.  How  much  is  the  interest  on  $20000,  for  04  days  at  8  per  cent? 

Ans.  $284.44. 

9.  A  note  for  $75000  dated  August  19,  1895,  and  payable  60  days  after  date, 
was  discounted  August  20,  1895,  at  5|  per  cent.  What  were  the  proceeds  counting 
days  of  grace  and  discount  day?  Ans.  $74245.31. 

10.  In  the  above  problem,  what  would  have  been  the  proceeds,  allowing  365 
days  to  the  year,  and  not  counting  discount  day  ?  Ans.  $74267.47. 

OPERATION    INDICATED. 


365 
4 


750.00 

62 

23 


732.53  discount. 

11.  A  note  for  $385000,  bearing  date  May  7,  1895,  payable  15  days  after 
date,  was  discounted  May  8,  1895,  at  6J  per  cent.  What  was  the  discount,  not 
counting  discount  day !  Ans.  $1181.74. 


58o  soule's  philosophic  practical  mathematics.  * 

TO  COLLECT  NOTES  THAT  BEAR  INTEREST  AND  EVENTUAL  INTEREST. 

1146.  1.  Januarys,  1895,  a  note  is  drawn  for  $1200  and  made  payable  one 
year  after  date,  with  interest  at  8  per  cent.  What  is  the  interest  and  amount  due 
the  holder  at  the  maturity  of  the  note.     When  does  the  note  fall  due? 

Ans.  $90,  interest.     $1296,  amount.     Due,  January  3/G,  1896. 

QpEnATioN  IF  PAID  January  3,  1896.  Operation  iv  paid  January  G,  1896. 


51200      face  of  note. 
8% 

or,        45 


$96.00      interest  for  1  year. 


$1296       amount. 


1200        45 
300        — 


1200 
363 

$96.80  int.  for  1  yr.  and  3  ds. 


$  96  $1296.80  amount. 


Note.— Had  tbe  note  been  paul  January  4,  interest  would  Lave  been  charsed  for  361  days; 
if  paid  January  5,  interest  would  bave  been  computed  lor  362  days.  See  pages  557  and  558,  for  an 
explanation  for  thus  counting  the  days. 

2.  I  receive  for  the  sale  of  real  estate  a  note  for  $9000,  bearing  date  January 
5,  1895,  payable  two  years  after  date  with  G  per  cent  current,  and  8  per  cent  event- 
ual interest.    The  note  was  paid  August  10,  1897 ;  what  was  the  amount  received  ? 

Ans.  $10512.50. 
Note. — Eventual  Interest  is  the  interest  tbe  note  bears  after  maturity,  in  case  it  is  not  paid 
at  maturity. 

OPERATION. 

Face  of  note $9000 

Interest  for  2  years  and  3  days,  at  6  per  cent       -        -        $1084.50 
Interest  on  face  of  note  from  Jan.  8,  '97,  to  Aug.  10,  '97, 

at  8  per  cent  (214  days) 428.00       1512.50 


Amount  due  August  10,  1897 $10512,50 

3.  Received  a  note  dated  October  2, 1895,  for  $6006.67,  payable  one  year  after 
date,  without  grace  with  6  ])er  cent  interest.  When  does  it  mature,  and  if  I  hold 
the  note  until  it  matures  what  will  be  the  interest  and  what  the  amount  due? 

Ans.    October  2d,  1896,  it  matures. 
$400,  interest. 
$7066.67,  amount  due. 

TO  DISCOUNT  NOTES  THAT  BEAR  INTEREST. 

$4500.  New  Orleans,  June  4,  1895. 

1147.  1.  Four  months  after  date,  for  value  received,  I  promise  to  pay  to  the! 
order  of  O'Neil,  Sullivan  &  Co.,  Four  Thousand  Five  Hundred  Dollars,  with  6  perj 
cent  interest. 

G.  Pharr. 


J 


BANKERS     DISCOUNT. 


58  c 


When  does  this  note  mature  1    If  discounted  the  day  it  was  drawn  by  a  note 
1      broker  at  8  i^er  cent,  what  proceeds  would  the  holder  receive  ? 

Ans.    It  matures  October  4/7, 1S95.    The  proceeds  are  $4463.67. 


60 


OPERATION 

To  find  the  amount  or  value  of  tbe  note 

at  maturity. 

4500 

123 


$92.25  =  int.  for  123  ds.  at  6%. 
4500.      —  face  of  note  added. 


OPERATION 

To  discount  tbe  maturity  value  of  the  iiote 
and  find  the  proceeds. 

4592.25 
45         126 


$4592.25  =  amount  or  value  of  note 
at  maturity. 


$  128.5830  =  interest  or  discount. 
4592.25      =  value  of  note  at  matu- 

rity. 

$4463.67      =  proceeds  of  note. 


Explanation— Tn  this  problem,  we  first  find  the  amonnt  or  value  of  the  note  at  maturity. 
This  \fe  do  by  calculating  and  adding  to  the  face  of  the  note,  the  6  per  cent  interest  that  it  bears 
for  4  months  and  3  days.  This  work  gives  lis  §4592  25  as  the  value  of  the  note  when  it  matures; 
hence  it  is  clear  that  this  is  the  amount  to  be  discounted.  We  then  discount  the  $4592.25  according 
to  business  custom  for  the  actual  unexpired  time,  including  3  days  of  grace  and  discount  day,  at 
the  specified  8  per  cent. 

$9730.  New  Orleans,  November  24,  1895. 

2.  One  year  after  date,  for  value  received,  I  promise  to  pay  to  the  order  of 
E.  B.  Oope,  Nine  Thousand  Seven  Hundred  and  Thirty  Dollars,  with  interest  at  8 
per  cent. 

Louis  Bush. 

What  is  the  above  note  worth  on  the  1st  June,  1896,  if  discounted  at  12  per 
cent?  Ans.  $9884. 


OPERATION 

To  find  the  worth  of  the  note  at  maturity. 

$  9730 
45         303 


OPERATION 

To  discount  or  find  the  cash  value  of  the  note 
June  1,  1896. 
$10514.89 
30       ISO 


784.89  =  int.  for  1  yr.  and  3  ds. 
9730.      =  face  of  note. 


$10514.89  =  value  of  note  at. matu- 
rity. 


$    630.8934  =  interest  or  discount. 

10514.89  =  princii)al  or  value  of 
note  at  maturity. 

$9884.00  =  proceeds  or  value  of 
note  June  1,  1896. 


Explanation. — In  this  as  in  the  preceding  example,  we  first  find  the  value  of  the  note  at 
maturity  by  adding  to  the  face  of  the  same  the  8  per  cent  interest  that  it  bears  for  one  year  and  3 
days;  and  then  we  discount  the  maturity  value  at  12  per  cent  for  the  unexpired  time,  counting 
actual  days,  with  grace  and  discount  day.  In  finding  the  time  we  count  thus:  June  29,  July  31, 
August  31,  September  30,  October  31  and  November  24  days,  3  days  grace  and  discount  day  (June 
the  1st)  =  180  days. 


582 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


3.  A  mortgage  note  for  $20000,  dated  December  16, 1895,  payable  18  months 
after  date,  with  8  per  cent  interest,  was  discounted  January  G,  1800,  at  7  per  cent. 
What  was  the  maturity  value  and  the  proceeds  ? 

Ans.  $22413.33  —  maturity  value.     $20103.51  =  proceeds. 

Note.— The  inortgaKo  matured  June  16il9,  1897.  Interest  added  at  8  per  cent  for  IS  months 
and  3  days  (543  days)  ecpials  $2413.33.  Maturity  value  discounted  at  7  per  cent,  for  tlie  tune 
from  January  6,  189G,  to  June  16/19,  1897,  including  three  days  of  grace  and  discount  day,  eiiuals 
530  days. 

4.  A  note  drawn  April  2, 1895,  at  3  months,  for  $1635.15,  bearing  8  per  cent 
interest,  was  discounted  June  1,  ]895,  at  8  per  cent.  When  did  the  note  mature? 
W.  at  is  it  worth  at  maturity,  and  what  was  the  cash  value  when  discounted? 

Ans.     July  2/5,  1895,  it  matures. 

$1668.94,  worth  of  note  July  8,  1895. 
$1655.96,  cash  value  June  1,  1895. 

Note. — Interest  is  added  for  93  days.  The  maturity  value  is  discounted  for  35  days,  which 
includes  discount  day. 

"CASH  NOTES" 

OR    DRAFTS   WHICH,    WHEN    DISCOUNTED,    WILL   rRODCCE   A   SPECIFIED   SUM. 


$4500  note  assumed. 

64  interest  for  64  days  at  8%. 

$4436  proceeds,  cash. 


$ 


4436 


4500 

8872 


1148.     1.     For  what  sum  must  a  60  day  note  be  drawn,  so  that  when  discounted 

at  8  per  cent,  the  proceeds  will  be  $8872  ?  Ans.  $9000. 

FIRST  OPERATION.  Explanation. — As  it  is  the  custom  of  hankers  to  calcu- 

late their  discount  on  the  face  of  notes,  drafts,  etc,  it  is 
plain  that  if  we  were  to  add  the  simple  interest  of  the 
$8872  to  itself  for  the  time  and  rate  given,  and  draw  the 
note  fur  the  anioiiiit  thus  produced,  it  would  not,  when 
discounted,  produce  the  required  sum  for  the  reason  that 
the  interest  on  this  increased  amount  would  bo  more  than 
the  interest  on  the  first  sum.  The  dclicit  would  be  the 
interest  on  the  interest  first  obtained,  jilus  the  interest  on 
each  succeeding  sum  of  interest,  interminably.  Conse- 
quently, to  produce  exact  results,  we  cannot  work  on  the 
face  of  the  sum  that  we  desire  to  obtain  for  the  note 
when  discounted.  We  are  therefore  constrained  to  assume 
some  number  to  represent  the  face  of  tlie  note  to  be  drawn, 
and  to  facilitate  the  operation  we  assume  $4500  the  8  per 
cent  Interest  Divisor,  to  represent  the  face  of  the  note. 
We  assume  the  Interest  Divisor  for  the  reason  that  the  interest  thereon  is  always  as  many  dolliirs 
as  there  are  days  in  the  time;  and  knowing  this  we  are  saved  the  labor  of  comjinting  the  interest 
on  the  sum  assumed.  Accordingly,  in  this  jiroblem,  the  time  being  04  days,  the  interest  or  discount 
on  $i4500  for  64  days  at  8  per  cent  is  $61,  which  subtracted  from  $4500,  the  assumed  note,  gives 
$4436  ])roceeds.  We  now  observe  that  as  we  discounted  the  $4500  for  64  days  at  8  per  cent,  the 
same  ratio  exists  between  the  $4436  proceeds  and  the  $4500  assumed  note  as  exists  between  the 
$8872  proceeds  and  the  face  of  the  note  required  to  produce  the  same  when  discounted  for  the  given 
time  and  rate  pr.  ct.  Hence  we  have  but  to  find  the  proportional  result  of  these  two  ratios.  To  do 
this  we  place  the  $4500  assumed  note  on  the  statement  line  aud  reason  thus;  Since  $4436  cash 
require  $4500  note,  $1  cash  will  require  the  4436th  part  and  $8872  will  require  8872  times  as  much. 

SECOND   OPERATION 
By  assuming  $100  as  the  face  of  the  note. 


$9000,  Ans. 


45 


$100     note  assumed. 
64 

$1.42=  interest. 


lA'. 


887.20 


$98.57  J  proceeds. 


100 

9 

8872.00 

$9000    =  face  of  note. 


i 


•*  CASH    NOTES.  583 

GENERAL   DIEECTIONS. 

1149.     From  tlie  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  finding  the  face  of  "  Cash  Notes : " 

1.  Assume  as  the  face  of  the  note  the  Interest  Divisor  for  the  rate  per  cent 
given  or  $100,  and  find  the  proceeds  of  the  same  for  the  given  time  and  rate  per  cent. 

2.  Then  divide  the  assumed  note  multiplied  by  the  required  proceeds,  hy   the 
proceeds  of  the  assumed  note. 

PROBLEMS. 

2.     What  must  be  the  face  of  a  note,  so  that,  when  discounted  for  04  days  at 
12  per  cent,  it  will  produce  $10000  proceeds'?  Ans.  $10323.47= 

OPERATION 

By  the  use  of  the  Interest  Divisor. 


^3000  note  assumed. 

94  interest  for  94  days  at  12  per  cent.  2906 


3000    note  assumed. 
10000 


$10323.47,  Ans. 


^2900  cash  proceeds. 

3.  A  creditor  owed  me  a  balance  of  $3212.65  and  settled  the  same  with  his 
"Cash  Note,"  payable  4  months  after  date.  The  note  was  dated  June  17,  1S95. 
Allowing  8  jier  cent  iuterest,  what  was  the  face  of  the  note,  and  when  does  it 
mature  ?  Ans.     $3305.19+  face  of  note. 

October  17/20,  1895,  it  matures. 

4.  What  must  be  the  face  of  a  note  to  net  or  produce  $1777.95,  when  dis- 
counted at  7  per  cent  for  63  days  ?  Ans. 


OPERATION. 


$36000  =  note  assumed. 

441  =  interest  for  63  days  at  7%.  35559 


$355.59  =  proceeds. 


36000 
1777.95 

$1800.00,   Ans. 


Explanation. — There  l)eing  no  7  per  cent  Interest  Divisor,  we  .issnnie  as  Ihe  face  of  the  note 
tlie  1  per  cent  Interest  Divisor  and  then  multiply  the  interest  at  1  per  cent,  which  is  $63  {as  many 
<l()llars  as  there  are  days)  by  7,  the  rate  per  cent  and  thus  obtain  iiiil  interest.  The  statement  is 
then  made  as  in  the  preceding  examples. 

Note. — Whenever  there  is  no  Interest  Divisor  for  the  rate  per  cent  given,  the  Interest  Divisor 
for  1  per  cent  should  bo  assumed  and  the  interest  found  thereon  as  above,  or,  if  i>referred,  by 
assuming  and  discounting  a  |100  note,  as  shown  in  the  second  operation,  page  582. 

5.  A  creditor  owed  me  a  balance  of  $6812.45  and  settled  the  same  with  his 
"Cash  Note,"  payable  4  months  after  date.  The  note  was  dated  June  17,  1895, 
allowing  8  per  cent  interest.  What  was  the  face  of  the  note,  and  when  does  it 
mature?  Ans.    $7008.69,  face  of  note. 

October  17/20, 1895,  it  matures. 


584  soule's  philosophic  practical  mathematics.  ♦ 

(3.  A  merchant  wishes  to  obtain  $7500  from  bank,  on  a  30  day  note.  Th© 
rate  of  discount  is  8  per  cent.  "What  size  note  must  he  ofler  so  that,  when  dis- 
counted, Lo  will  receive  the  exact  sum  1  Ans.  $7557.10. 

7.  A  creditor  owes  $1250  and  wishes  to  pay  it  with  a  draft  at  45  days  sight, 
without  grace  or  discount  day.  What  must  be  the  face  of  the  draft,  the  rate  of  dis- 
count being  8  per  cent?  Ans.  $12G2.C3. 

8.  For  what  sum  must  a  note  be  drawn  for  45  days,  without  grace  or  dis^ 
count  day,  to  settle  a  cash  balance  of  $10802.50,  interest  at  10  per  cent  1 

Ans.  $11000. 

"CASH   NOTES" 

WITH   INTEREST,    COMMISSION,   AND   BROKERAGE   COMDINED. 

1150.  1.  A  customer  desires  to  obtain  from  a  bank  $8000  on  his  90  day  note. 
In  conformity  with  bank  custom,  which  prudence  and  sai'ety  demand,  he  is  required 
to  have  one  or  more  indorsers  on  the  note,  and  as  his  correspondent  I  indorse  and 
negotiate  the  note  for  him.  The  rate  of  bank  discount  is  8  per  cent ;  I  charge  2J 
per  cent  commission  for  indorsing,  and  ^  per  cent  brokerage  for  negotiating.  What 
must  be  the  face  of  the  note  ?  Ans.  $8406.80. 

OPERATION. 

Face  of  note  assumed,  $4500. 

Interest  on  same  for  94  days  at  8  per  cent        -        •        $  94. 

Commission  on  same  at  2i  per  cent  -        -        -  112.50 

Brokerage  on  same  at  ^  per  cent       ....  11.25 217.75 


Cash  value,  or  proceeds  of  the  assumed  note.        ....        $4282.25 

$  Explanation.— In    this    solution,    for    reasons 

4500  =  note  assumed.  given  in  the   solution   ..f  tbe   first   problem    of 

/IO«o  or      crwwi  nil  Cash  Notes,"  we  assuuie  tue  »  percent  interest 

4.^»_,.-0     &UUU.UO  Divisor  as  tlie  face  of  tbe  required  note,   and 

from  it  we  deduct  the  interest,  conimissioa  and 

$8406.80,  Ans.  brokerage,    and    thns    produce    the    necessary 

relationship  numbers,  as  exjdained  in  the  first 
Bolution  with  which  we  make  tbe  proportional  statement,  the  result  of  which  gives  the  correct 
answer. 

SECOND   OPERATION. 

By  assuming  $100  as  the  face  of  tbe  note. 
$  $ 

I  100       face  of  note  assumed. 
45  !  94  856.45 


$2.08^  interest. 
li.50    =  2.^%  commission. 
25    =  i%    brokerage. 


100 

9 

8000.00 


$8406.80  Ans. 


$4.83^  amount  of  deduction. 
100.        note  assumed. 


$95.16J  proceeds. 


CASH    NOTES.,  585 

To  discount  the  above  note,  tlie  following  figures  would  be  produced : 

Face  of  note,  $8406.80 

DEDUCTIONS. 

Interest  on  face  of  note  for  94  dnys  at  8  i)er  cent,  -  $175.61 
Commission  on  face  of  note  at  2.J  per  cent,  -  •  210.17 
Brokerage  on  face  of  note  at  ^  per  cent,  -        -  21.02  406.SO 


Net  proceeds,  $8000.00 

2.  A  merchant  owes  a  cash  balance  of  $2540.37,  wLicli  lie  wishes  to  settle 
by  note  at  120  days  for  such  a  sum  as,  when  discounted  at  8  per  cent,  and  allowing 
2i  per  cent  commission  for  indorsing,  will  net  the  exact  cash  balance.  What  must 
be  the  face  of  the  note  ?  Ans.  $2081.29. 

3.  What  must  be  the  face  of  a  note  for  60  days  to  net  $1720,  discount  at  5 
per  cent  and  2i  per  cent  commission  for  indorsing?  Ans.  $1780.33+. 

4.  A  merchant  owes  a  cash  balance  of  $4000,  which  he  wishes  to  settle  by- 
note  at  120  days  for  such  a  sum  as,  when  discounted  at  8  per  cent,  allowing  2J  per 
cent  commission  for  indorsing,  will  net  the  exact  cash  balance.  What  must  be  the 
face  of  the  note  ?  Ans.  $4221.88. 


CASH  NOTE  WITH  COMMISSION  ON  FACE  AND  BROKERAGE  ON  THE 

BALANCE. 


1151.  1.  A  merchant  owes  his  correspondent  $3250  cash,  for  which  it  is 
agreed  that  be  shall  give  his  note  for  such  a  sum  as  will  net  the  amount  due,  after 
allowing  for  the  following  deductions  :  Interest  for  63  days  at  10  per  cent ;  commis- 
sion on  the  face  of  the  note  at  2^  per  cent ;  and  J  per  cent  brokerage  on  the  balance, 
Tiz.,  on  the  remainder  of  the  note,  after  the  interest  and  commission  shall  have  been 
deducted.    For  what  face  must  the  note  be  drawn  ?  Ans.  $3411.31. 


OPERATION. 

$3600  face  of  note  assumed. 

$63 

=  interest. 

90 

=  commission. 

6859.53 


$153         =  interest  and  commission. 
3600         =  note  assumed. 


3600 

2 

3250.00 
$3411.31  Ans. 


$3447    =  proceeds. 

17.23J  =  i%  brokerage  on  $3447 

$3429.76^ 

Note. — Observe  tLat  the  interest  and  commission  are  required  on  the  face  of  the  note,  and 
brokerajre  on  the  balance,  after  deducting  interest  and  commission. 


586 


n 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 
SECOND   OPERATION. 


X00  25 

n  7 

$1.75  interest. 
2.50   =  2i%  com. 


100 
4.25 


note  assumed, 
int.  and  com. 


35.75        proceeds. 
.47874  =  i7o  brok.  on  $95.75. 


190.5425 


$95.27124  net  proceeds. 


100 
2 

3250.0000 

$3411.31 

Aus. 


$4.25    =iut.&com. 

To  discount  the  above  note,  the  following  figures  would  be  produced  : 

Face  of  note,  -        -        -        , $3411.31 

DEDUCTIONS. 

Interest  on  face  of  note  for  03  days  at  10  per  cent, 


Commission  on  face  of  note  at  24  per  cent, 


$59.70 
85.28 


Balance,  .... 

Brokerage  on  balance  at  4  per  cent, 

Net  proceeds  of  note. 


144.98 

$3200.33 
16.33 

$3250.00 


CASH  NOTE  WITH  BROKERAGE  ON  THE  FACE  AND  COMMISSION  ON 

THE  BALANCE. 


1152.  1.  For  what  sum  must  a  note  be  drawn  to  produce  $742.50,  after  the 
following  deductions  are  made  :  184  days  interest  at  12  per  cent,  \  ])er  cent  broker- 
age on  the  face  of  the  note,  and  2J  per  cent  commission  on  the  balance  ? 

Ans.  $813.46  +  . 

OPERATION 
face  of  note  assumed. 


184       =  interest. 
7.50  =  brokerage. 


$2808.50 

70.2125  =  com.  on  $2808.50 


2738.2875 


g  191. .50  =  int.  and  brok.     $2738.2875 
3000.00  =  note  assumed. 


$3000 
742.5000 

$813.40 


$2808.50  =  proceeds. 

Note.— Observe  tlint  the  interest  and  brnkernge  .nro  required  on  the  face  of  tlie  note,  while 
tbe  coiumissiou  is  recjuired  only  on  the  remainder,  after  deducting  interest  and  brokerage. 


SECOND   OPERATION. 

$  100.  =  note  assumed. 

100       face  of  note  assumed.        G.38J      ==  int.  and  brok. 

30     184  182.5525 

$93.6166ij  =  proceed.s. 
2.34041  =  24%    com.    on 

$93.6166§. 

$91.27624  net  proceeds. 


3.134  interest. 
.25    =  i%  brokerage. 


100 
742.5000 


$813.46+ 
Ans. 


$6,384  ^"it.  of  deductions. 


CASH    NOTES. 


.58; 


CASH  NOTE   WITH    INTEREST  ON   THE   FACE   AND  COMMISSION   ON 

THE    llEMAINDEK. 

1153.  1.  A  merchant  wishes  to  draw  a  note  for  such  a  sum  that,  when 
discounted  for  64  days  at  9  per  cent,  and  2i  per  cent  commissiou  on  the  yeiitainder  is 
deducted,  it  will  uet'$25000.     What  must  be  the  face  of  the  note? 

Ans.  $20057.03. 


OPERATION 


$-1000       —  note  assumed. 
G4       =  interest. 


JS37.G0 


I'oG       =  proceeds. 
1)8.40  =  commissiou. 


$  4000 
25000.00 

$20057.95  Ans. 


$3837.00  =  net  proceeds. 
Note. — Observe  that  the  commission  is  required  on  the  remninder  only. 

SECOND   OPERATION. 


4p 


100     note  assumed. 
04 


95.94 


$1.00  interest  on  $100  note  assumed. 
100.00  note  assumed. 


$  N. 
100 
25000.00 

$20057.95  Ans. 


40  proceeds  of  $100  note  as.snmed. 
2.40  =r  2i%  commissiou  on  remainder  or  proceeds. 


$95.94  net  proceeds  of  the  $100  note  assumed. 

In  discounting  the  above  note,  the  following  figures  would  be  produced  : 

Face  of  note, $20057  95 

Less  interest  for  64  days  at  9  per  cent, 410.93 


Remainder, 
Less  commission  at  2i  per  cent, 

Net  proceeds, 


$25041.02 
041.02 

$25000.00 


CASH  NOTE    WITH    INTEREST  ON  THE   FACE   AND   COMMISSION  ON 

THE  NET  SUM. 

1154.  1.  A  merchant  wishes  to  draw  a  note  for  such  a  sum  that,  when  dis- 
counted for  64  days  at  9  per  cent,  and  2.i  i)er  cent  commission  on  the  ntt  siiiii  is 
deducted,  it  will  produce  $25000.     What  uiust  be  the  face  of  the  note  ? 

Ans.  $20041.07. 

OPERATION. 


$4000  =  note  assumed.  ]  Net  sum  to  be  produced,       $25000 


64 


interest. 


$3936  =  proceeds. 


Com.  on  same  at  2i%, 
Amt.  to  be  obtained  from 


62 


3936 


bank,  -         -         -     $25625 

Note.— Observe  that  the  commission  is  required  on  the  net  sum  only. 


$  4000 
25025 

$20041.07  Ans. 


588 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


40 


100     note  assumed. 
«4 


•SI. CO  interest. 
100.      face  of  assumed 

note. 

$98.40  lU'oceeds. 


SECOND   OPERATION. 

Xet  sum  to  be  pro- 
duced,        -         -         -  $2.5000 

Commission  ou  same 

at  2 J  per  cent     -        -         G25 

Amount  to  be  obtain- 
ed from  bank,     -         -  $25025 


98.40 


100 
25025.00 

$20041.67   Ans. 


In  discounting  the  above  note,  tbe  following  figures  would  be  produced  : 
Face  of  note, $20041.67 


CHARGES   OR   DEDUCTIONS. 

Interest  on  face  of  note  for  64  days  at  9  j)er  cent, 


416.67 


$25625.00 


102J 


$100 
25025 


.$25000  net  proceeds  required. 
For  further  work  of  a  kindred  nature  to  the  foregoing,  see  Domestic  Exchange. 
COMPLEX  INTEREST  PEOBLEMS. 

1155.  1.  A  planter  borrows  $3040,  and  agrees  to  pay  the  same  and  the 
interest  thereon  in  jiayments  of  $380  for  ten  consecutive  years.  What  rate  of 
interest  does  he  pay  ? 

Note.  This  problem  may  be  interpreted  in  two  ways,  viz:  1.  That  $3040  was  bor- 
rowed and  $760  interest  was  paid  for  its  use  for  10  years,  or  $76  interest  for  one  year;  thus 
making-  it  a  simple  interest  problem  with  the  pinncipal,  time  and  interest  given  to  find  the 
rate  % .  2d.  Tlie  problem  may  be  interpreted  or  classed  as  an  annuity  problem  with  $3040 
as  the  Present  Worth,  $380  as  the  annuity  and  10  years  as  the  time,  to  find  the  rate  per  cent, 
allowing'  yearly  compound  interest. 

SOLUTION  OF  THE  FIRST  INTERPRETATION. 
$380  for  10  years  =  $3800;  $3800  —  S304O  =  $760  interest  for  10  years  or  876  interest  for 
one  year.     Then  $3040  :  $76  ::  $100  :  $2.50  =■  2\%  per  annum  simple  interest. 

SOLUTION  OP  THE  SECOND  INTERPRETATION. 

By  the  use  of  the  annuity  tables  (page  926)  we  proceed  as  follows:  S3040  =  Present 
Worth  H-  $380,  annuity  ^  $8  =  Present  Worth  of  an  Annuity  of  $1  for  the  given  time,  10 
years,  and  at  the  required  rate.  Then  referring  to  the  Annuity  Table  in  the  line  of  10  years, 
we  find  7.91273  which  is  tlie  nearest  Present  Worth  to  the  $8  and  as  this  is  in  the  41%  interesS 
column,  therefore  the  annual  rate  paid  is  practically  \\fo. 

For  extended  work  of  this  character  see  Building-Loan  Interest  Tables,  and  Bond  and 
Investment  Tables,  by  J.  W.  Robinson. 

EQUAL  ANNUAL  PAYMENTS  OF  PRINCIPAL  AND  INTEREST. 


1156.  2.  Bought  a  piece  of  land  for  $5000.00  agreeing  to  pay  8  per  cent 
interest,  and  to  ]>ay  principal  and  interest  iu  live  equal  annual  jjaymeiits,  how 
much  is  the  annual  payment  ?  .  Aus.  $1252.28. 

OPERATION. 

$5000  -^  $3.9927  =  $1252.28. 

Explavafion. — By  tbe  conditions  of  tbe  problem,  we  observe  tb.it  tbe  $5000  is  tbe  present 
worth  of  an  annuity,  tbe  time  being  5  years  and  tbe  rate  |)er  cent  8,  Hence,  as  tbe  present  worth 
of  |1,  multiplied  by  the  annuity,  would  give  the  full  present  worth,  jt  is  clear  that  if  we  divide 


CASH    NOTES. 


589 

the  given  present  worth  ($5000)  by  the  present  worth  of  $1,  for  the  given  time  and  rate  per  cent,  the 
quotient  will  be  the  required  annuity. 

By  referring  to  the  annuity  table,  we  find  the  present  value  of  %X  annuity  per  annum  at 
compound  iuterest  for  5  years  at  S  per  cent  to  be  $3.9927. 

Then  by  dividing  $5(100  by  $3.9927,  we  obtain  $12.52.28  as  the  equal  annual  payment.     The 
reasoning  is,   since  $3.9927   (present  value  of  $1  annuity,  for  5  years  at  S  jier   cent)    require   $1 
ivestment,  then  $1  present  value,  etc.  will  req^uire  the  3.9927th  part  of  $1,  and  $500u  jiresent  value, 
ill  require  5000  times  as  much. 


in 
will 


To  obtain  the  i>resent  value  of  .?1  animity,  per  atiiuim,  at  compound  interest, 
for  5  years  at  S  per  cent,  witliout  access  to  annuity  tables,  tirst  tiud  the  compouud 
iimouut  of  $1  for  5  years  at  8  per  cent.     Thus  : 

i'lBfjTs  second: 


$100 
1.08 

1st,    $108.00 
1.08 


2d,     $116.6400 
1.08 


3d,    $125.971200 
1.U8 

4th,  $136.04889600 
1.08 


Then  add 

$100 
108 
116.64 
125.9712 
136.048896 

.$586.060096  =  final  valne  of  $100,  annuity  for  5 
years  at  8  ])er  cent  I'onipouud 
interest.  Divide  by  100  ^ 
15.8666  +,  final  value  of  $1. 


5th,  $146.9328076800  =  compound  amount  of  $100, 
which  divided  bj'  $100 
=  $1.4693280768. 

The  $5.8666  -^  1.4693280768  =  $3.9927  present  value  of  $1  annuity  per  annum 
at  compound  interest  for  5  years  at  8  jjer  cent. 

A  DIFFICULT  PRACTICAL  QUESTIOX  IN  EENEWING  NOTES. 


1157.  1.  Jones  has  a  note  for  $4000  in  bank  whicli  falls  dne  August  19,  1895. 
It  is  agreed  that  he  shall  pay  the  bank  §2000  casli  and  a  new  9U  day  note  for  what 
may  be  due.  The  bank  is  to  deduct  the  interest  or  discount  on  the  new  note  at  8 
l)er  cent  for  94  days,  from  the  $2000  cash  i)aid,  apply  the  remainder  on  the  $4000 
due,  and  take  the  new  note  for  the  balance. 

What  is  the  interest  on  the  new  uote,  how  much  will  the  cash  pay  on  the  old 
note,  aud  what  is  the  lace  of  the  new  note  i 

Aus.  $    42.67  is  the  interest  on  the  new  note. 

1957.33  is  the  amount  paid  on  the  old  note. 
2042.67  is  tlie  face  of  the  new  note. 

OPEKATION   TO   FIND   THE   FACE   OF   THE    NEW  NOTE. 
For  the  explanation,  see  "Cash  Notes,"  page  582. 


4406 


4500 
2000 


$2042.67  face  of  new  note. 

By  the  conditions  of  the  transaction  and  in  accordance  with  bank  custom,  the 
interest  or  discount  on  the  renewal  note  must  be  paid  in  advance,  and  the  sum 
required  to  pay  it  is  to  be  taken  out  of  the  $2000  cash  paid.    By  the  above  opera* 


5go 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


tion,  ^ylli(•ll  involves  the  principles  of  cash  notes,  we  see  tliat  the  face  of  the  renewal 
note  is  $ 2042.07 ;  and  consequently,  the  discount  is  f4l!.(!7.  This  s^ili.dT  discount 
subtracted  frmn  $2000  cash  paid,  leaves  $11)57.33  to  be  credited  on  the  old  uote. 

2.  A  man  owes  $700  now  due,  and  it  is  agreed  that  he  shall  pay  $400  cash 
and  give  his  note  for  1  year,  without  grace  or  discount  day,  for  the  balance  that 
may  be  due ;  the  dis(!ount  on  the  note  at  0  ])er  cent  to  be  paid  in  advance  out  of  the 
$400  cash  paid.  How  much  of  the  $400  cash  is  to  be  credited  on  the  $700  and  what 
is  the  face  of  the  i-euewal  note  ?  Ans.  $380.57  cash  to  credit  on  $700  debt. 

$319.43  face  of  note. 
Note. — This  note  is  discounted  for  365  days. 


TO  FIND  THE  VALUE   OF   NOTES   BEAEING   INTEREST   AND   MATUR- 
ING IN  SUCCESSIVE  AND  CONSECUTIVE  MONTHS. 

1158.  April  8,  1894,  a  party  purchased  from  a  real  estate  holder  a  house  and 
lot  for  $7500,  on  the  following  terms  and  conditions:  $300  was  paid  in  cash.  3(> 
notes  of  $200  each  bearing  7  jjcr  cent  interest  were  given  ;  three  of  the  iiotes  were 
made  payable  in  3  months,  July  8,  1894.  The  other  33  notes  were  made  payable 
monthly  after  July  8,  1894.  After  the  three  notes  payable  July  8,  and  19  notes 
payable  monthly  after  July  8  had  been  paid,  the  holder  sold  the  14  remaining  notes 
at  7  per  cent  par  annum  discount.  Counting  30  days  to  the  month  without  grace  or 
discount  day,  what  were  the  proceeds  of  the  14  notes.  Bankers'  Discount  Method? 

Ans.  $3136.74. 

SOLUTION. 

$7500  cost  of  house  and  lot,  less  $300  paid  in  cash  =  $7200  which  was  paid 
in  30  notes  of  $200  each.  The  3  notes  paid  July  8,  and  the  19  notes  i)aid  monthly 
in  successive  payments  after  July  8,  make  22  notes  which  deducted  from  the  3& 
notes  leave  14  notes  ou  hand,  19  months  after  July  8,  which  is  February  8,  1896. 

Each  of  the  14  Tiotes  of  $200  has  accrued  interest  at  7  per  cent  from  April 
8,  1894,  and  each  will  continue  to  bear  interest  until  ])ai<l.  And  as  they  fall  due 
respectively  and  successively  monthly  after  February  8,  1896,  the  following  state- 
ment and  operation  shows  the  maturity  value  of  the  14  notes,  23  to  36  inclusive : 


£ 

6 

Maturitj' 

c 

o 
a 

Cm 
O 

Face 

of 
Note. 

Date. 

Matures. 

2 

accrued 
interest 
at  7  per 

Maturity 
value." 

Date 
Discounted. 

Matures. 

1 

Dis- 
count. 

Proceeds. 

d 

1 

cent. 

. 

Mar.  8,  1896 

§ 
1 

7 

23 

$200 

April  8,  1894 

Mar.  8,  1896 

23 

$26.83i 

$226,834 

Feb 

8,  1896 

$1,32 

$225.51 

24 

200 

"     8,  1894 

Apr.    8,  1896 

24 

28.00 

228.00 

8,  1896 

AT>r.   8,  1896 

2 

7 

2.66 

225.34 

25 

200 

"     8,  1894 

May   8,  1896 

25 

29.16; 

229. 16  J 

8,  1896 

May   8,  1896 

3 

7 

4.01 

225. 1& 

26 

200 

"     8,  1894 

.lune  8,  1896I2K 

30.33i 

230.334 

8,  1896 

.June  8,  1896 

4 

7 

5.37 

224.96 

27 

200 

"     8,  lS94!,July  8,  1896 

27 

31.50 

231.50 

8,  1896 

July  8,  1896 

5 

7 

6.75 

224.75 

28 

200 

"     8,  1894  Aug.  8,  1896 

28 

32.661 

232. 66  J 

8,  1896 

Aiifj.  8,  1896 

6 

7 

8.14 

224.53 

29 

200 

"     8,  1894  .Sej).    8,  1896 

29 

33.834 

233.834 

8,  1896 

Sep.    8,  1896 

7 

7 

9.53 

224.30 

30 

200 

"     8,  1894 

Oct.    8,  1896 

30 

35.00 

235.00 

8,  1896 

Oct.    8,  1896 

8 

7 

10.97 

224.03 

31 

200 

"     8,  1894 

Nov.  8,  1896 

31 

36.161 

236.161 

8,  1896 

Nov.   8,  1896 

9 

7 

12.40 

223.77 

32 

200 

"     8,  1894  Dec.   8,  1896 

32 

37. 33  J 

237.334 

8,  1896 

Dec.    8,  1896 

10 

7 

13.84 

223.49 

33 

200 

"     8,  1894Lla,u.    8.  1897 

3;! 

38.50 

238.50 

8,  1896 

.Fan.    8,  1897 

11 

7 

15.30 

223.20 

34 

200 

"     8,  1894  Feb.   8,  189734 

39.661 

239.661 

8,  1896 

Feb.    8,  1897 

12 

7 

16.78 

222.89 

35 

200 

"     8,  1894  Mar.   8,  1897,33 

40.834 

240.834 

8,  1896 

Mar.  8,  1897 

13 

7 

18,26 

222.57 

36 

200 

"     8,  1894 

Apr.  8,  1897 

36 

42.00 

242.00 

8,  1896 

Apr.  8,  1897 

14 

7 

19.76 

222.24 

$2800 

=  face  of  the  14  notes. 

$481,834 

$3281.834 

fl45.09 

$3136.74 

VALUE    OF    STOCK. 


591 


This  statement  fiiul  operation  show  the  fonowiiiij  results:  1.  The  face  of 
the  14  notes  $-'S()(>.  2.  The  maturity  accrued  interest  on  each  and  all  the  notes. 
3.  The  maturity  value  of  each  and  all  the  notes.  4.  Tlie  discount  on  each  and  all 
the  notes.  5.  The  net  proceeds  or  i)reseut  value,  February  8,  IS'JG,  of  each  and  all 
the  notes. 


TO    FIND     THE     VALUE     OF     STOCK     WHICH     PAYS     A     SPECIFIED 

MONTHLY    DIVIDEND    OR    REVENUE,    AND     HAS     A 

SPECIFIED     TIME     TO     RUN. 


1159.  1.  A  party  holds  S2700  of  stock,  which  i)ays  $80  per  month  revenue, 
(35|%  interest  per  annum).  The  stock  lias  27  years  to  run.  What  is  the  present 
value  of  the  stock  the  current  rate  of  interest  being  G%,  allowiug  G%  interest  ou 
all  the  monthly  payments  of  the  revenue  or  income  ?  Aus.  $12G14.02 


SOLUTION. 


OPERATION 


F,ice  of  stock         ..... 

Receipts  liy  montlily  p.Tyments  of  $80  or 
monthly  interest  for  27  years  at  351% 

Interest  on  $)iO  monthly  payments  on 
Si27()0,  for  52326  terms  of  1  month,  (1 
to  323  inclusive)  =  4360^  yrs.  at  6%    - 


$2700.00 


To  find  the  sum  of  the  terms  of  the 
the  monthly  payments. 
1  +  323  =  324 
324  —  2  =  162 
162  X  323  =  52326  A'alue  of  stock  at  maturity     - 

VALUE   OP  §100   STOCK   Foil  27  YEARS   AT  6    PER    CENT. 

Face  of  Stock  assumed.  ....--. 

Interest  on  $100  for  27  years  at  6  per  cent  ... 

Interest  on  50c.,  the  monthly  interest  ou  $100  at  6  per  cent, 

for  52326  terms  of  1  mouth  =  4360^  years  at  6  per  cent, 


25920.00 


20930.40 


$49550.40 


$100.00 
162.00 

130.82 


392.82 


100 
49550.40 


I  $12614.02  Aus. 


$392.82 

2.  Suppose  in  the  above  problem,  interest  is  allowed  only  on  the  annual  pay- 
ments of  the  monthly  revenue  instead  of  the  monthly  payments,  what  would  be 
the  jjresent  value  of  the  stock  ?  Ans.  $12575.34. 

SOLUTION. 


OPERATION 
To  find  the  sum  of  the   terms 
the  annual  paj'ments. 

1  to  26 

1  or 


) 


Vilt 
26 

351 


13 
351 


Face  of  Stock  ...... 

Qf        Monthly  interest  for  27  years  at  35g  per  cent 
or  at  $80  per  month  -        -        - 

Interest  on  $960  yearly  payments  made  by 
monthly  installments  ou  $2700  for  351  terms 
of  1  year,  (1  to  26)  at  6  per  cent,  value  of 
stock  at  maturity,      ..... 


388.36 


VALUE  OF  $100  FOR  27  YEARS  AT  6  PER  CENT. 

Face  of  stock  assumed,           .        -        - $100.00 

Interest  on  $100  for  27  years  at  6  per  cent,             ....  162.00 
Interest  on  $6,  the  yearly  interest  on  $100  for  351  terms  of 

1  year  at  6  per  cent,              126.36 

Value  of  $100, -        -        $388.36 

Note. — See  Stocks  and  Bonds  in  this  work  for  extended  operations  of  a  similar  character. 


$  2700.00 
25920.00 

20217.60 
$48837.60 


100 
48837.60 

$12575.34  Ans. 


592 


soule's  philosophic  practical  mathematics. 


TO  FIND  TBE  EATE  OF  INTEREST  KECEIVED  AND  PAID  IN  THE 
SALE  OF  PROFERTY  ON  THE  INSTALLMENT  PLAN. 

1160.  A  Homestead  Association  sold  to  Levin  Cooper  Soule,  a  Louse  and  lot 
for  $5000,  on  tbe  following  conditions  of  ])aynient:  Casb  $1000;  the  remainder  in 
80  notes  of  $50  each,  bearing  7  per  cent  interest  and  payable  monthly,  as  the 
mouths  expire,  in  SO  consecutive  months. 

In  this  transaction,  what  is  tbe  rate  per  cent  interest  that  tbe  homestead 
makes  and  what  per  cent  interest  does  Levin  Cooper  Soul6  pay  on  tbe  $4000? 

Ans.  The  Homestead  makes  and  Levin  Cooper  Soule  loses, 
3.54g%  +  3.45|%  =  77o  +  .5512+  %  use  of  inter- 
est on  interest  payments,  —  7.5512+  %• 

SOLUTION. — First  Step. 

Since  tbe  80  notes  bear  7  per  cent  interest  and  are  payable  in  consecutive 
monthly  periods  as  tbe  months  successively  expire,  tbe  interest  thereon  is  obtained 
by  lindiug  tbe  interest  on  $50  at  7  per  cent  for  tbe  whole  number  of  months  that  all 
the  notes  bear  interest.  This  we  find  as  follows :  Tbe  notes  mature  successively  in  1  to 
SO  months,  thus  forming  an  arithmetical  series  of  80  terms;  hence  the  sura  of  the 
first  and  last  terms,  multiplied  by  half  the  number  of  terms,  gives  the  total  number 
of  months  that  all  the  notes  drew  interest.    Thus,  1  +  80  =  81  x  40  =  3240  months. 


FIRST   OPERATION 

To  tiud  the  interest. 


12 


50 


3240 


$  945  interest  on  all  the  notes. 


SECOND   OPERATION 

To  find  tbe  interest. 

Int.  on  first  note  for  1  mo.  is      -     $     .29^ 
Int.  on  last  note  for  80  mos.  is  -       23.33^ 

Int.  on  first  and  last  notes  is     -     $23.62i 

$23.62i  —  2  =  $11.81J  average   interest 
per  mouth. 
11.81^  X  SO  months  =  $945  interest  on 
all  tbe  notes. 

Note.— This  $945  interest  was  received  on  the  $4000  during  the  80  months. 

Second  Step. 
STATEMENT  REMARK.— The  Homestead  Association 

To  find  the  rate  per  cent  interest  on  the  $4000.        receives    Only    3.543%     interest     On     the 

but  it  had  tbe  use  of  tbe  monthly 


12 


4000 

SO 

$2G6§ 


2G6§ 


1 
945 


I 


I  3.54^%   Ans. 

See  page  599,  for  an  esplanatiou  of 
this  kind  of  work. 


])ayments  of  the  monthly  maturing  notes 
with  the  accrued  interest,  thus  making 
an  arithmetical  series  of  79  terms  of  note 
and  interest  payments. 


RATE    OF    INTEREST    RECEIVED    AND    PAID. 


593 


Third  Step. 

To  determine  the  worth  to  tbe  Homestead  Association  of  the  79  note  pay- 
ments at  7  per  cent  interest,  -we  proceed  as  follows :  The  79  note  payments  form  au 
arithmetical  series  of  79  terms ;  therefore,  as  above  explained,  the  snm  of  the  first 
and  last  terms,  multiplied  by  half  the  number  of  terms,  gives  the  total  number  of 
mouths  that  the  Homestead  received  interest  on  the  $oO  monthly  note  payments. 
Thus,  79  +  1  =  SO,  the  sum  of  the  number  of  terms.  SO  x  39J  =  3160  months  that 
the  Homestead  had  the  use  of  $50. 

Note. — 39J  is  i  of  the  number  of  terms. 
The  interest  on  $50  for  31G0  months  at  7  per  cent  is  (  m^^' )  $921§.    This 


sum  being  the  interest  worth  of  the  79  note  payments,  during  the  SO  months,  it  is 
as  justly  to  be  credited  to  the  receipts  of  the  $4000  as  is  the  $945  interest  on  tbe 
monthly  maturing  notes.  The  rate  per  cent  interest  that  the  $921g  is  of  $4000  for 
the  SO  months,  worked  as  in  the  above  problem  to  find  the  rate  per  cent  $945  was  of 
$4000,  is  3.45§  per  cent,  which  added  to3.54g  per  cent  obtained  above  by  the  interest 
received,  =  7  per  cent  the  Homestead  gains  on  the  $4000,  exclusive  of  the  interest 
on  the  interest  of  the  79  monthly  payments  of  interest  received. 

Fourth  Step. 
To  include  the  interest  on  the  interest  of  the  79  successive  monthly  payments 
of  accrued  interest  as  gain  for  the  Homestead,  we  proceed  as  follows :  The  payment 
of  interest  on  the  first  $50  note  for  one  month  at  7  per  cent  was  29^  cents  which 
was  used  by  the  Homestead  for  79  months;  the  interest  on  the  second  note  was  58J 
cents  which  was  used  by  the  Homestead  for  78  months;  tlius  the  interest  increases 
lu  au  arithmetical  ratio  of  29^  cents,  and  the  time  decreases  by  an  arithmetical 
ratio  of  1  month.  The  following  statement  shows  in  detail  the  interest  on  79 
accrued  interest  monthly  payments: 


.-'9J  X 

79  = 

23.04i 

6.4  U 

X 

58  =  372.161 

iy.54;  X 

37  =  464.04^ 

18.66S  X 

16  =  298. 66  J 

.58i  X 

78  = 

45.50 

6.70g 

X 

57  =  382.37i 

12.83i  X 

36  =  462.00 

18.95S  X 

15  =  284. 37i 

.87i  X 

77  = 

67.37i 

7.00 

X 

56  =  392. 

13.12i  X  35  =  459. 37i 

19.25  X 

14  =  269.50 

1.16i  X 

76  = 

88.66} 

7.29i 

X 

55  =  401. 04J 

13.41§  X 

34  .=  456.161 

19.54i  X 

13  =  2.54. 04i 

1.45J  X 

75  = 

10y.37i 

7.58i 

X 

54  =  409.50 

13. 70S  X 

33  =  452.37i 

19.83i  X 

12  =  238.00 

1.75  X 

74  = 

129.50 

7.87i 

X 

53  =  417.37i 

14.00  X 

32  =  448.00 

20.12J  X 

11  =  221.37^ 

2.04J  X 

73  = 

149.04i 

8.161 

X 

52  =  424. 66i 

14.29^  X 

31  =  443.04i 

20.4 IS  X 

10  =  204.165 

2. 33 J  X 

72  = 

168. 

8.455 

X 

51  =  431. 37i 

14.58i  X 

30  =  437.50 

20.70^  X 

9  =  186.37i 

2.62i  X 

71  = 

186.37i 

8.75 

X 

50  =  437.50 

14.87.J  X 

29  =  431. 37i 

21.00  X 

8  =  168.00 

2.91S  X 

70  = 

204. 16S 

9.04' 

X 

49  =  443.04i 

15.16§  X 

28  =  424.661 

21.29i  X 

7  =  149.041 

3.205  X 

69  = 

221. 37i 

9.33i 

X 

48  =  448. 

15.45S  X 

27  =  417.37i 

21.58J  X 

6  =  129.50 

3.50  X 

68  = 

238. 

9.62i 

X 

47  =  452.37i 

15.75  X 

26  =  409.50 

21.87i  X 

5  =  109. 37J 

3.79(S  X 

67  = 

254.044 

9.91§ 

X 

46  =  456.16J 

16.0JJ;  X 

25  =  401. 04J 

22.165  X 

4  =  88.661 

4.08i  X 

66  = 

269.50 

10.20s 

X 

45  =  459.37i 

16.33i  X 

24  =  392,00 

22.451  X 

3=  67.37i 

4.37J  X 

65  = 

284. 37J 

10  50 

X 

44  =  462.00 

16.62i  X 

23  =  382.371 

22.75  X 

2  =  45.50 

4.6Hi  X 

64  = 

298.661 

10.79J 

X 

43  =  464.04i 

16.91s  X 

22  =  372.16J 

23.04^;  X 

1  =  23.04J 

4.95§  X 

63  = 

312.37i 

11.08i 

X 

42  =  465.50 

17.20S  X 

21  =  361. 37i 

23.33i  X 

0  =  00.00 

5.25  X 

62  = 

325.50 

11.37J 

X 

41  =  466.37i 

17.50  X 

20  =  350.00 

5.54  i  X 

61  = 

338.04J 

11.66S 

X 

40  =  466.665 

17.79J  X 

19  =  338.04i 

Tot.  products,  $24885.00 

5.83J  X 

60  = 

350. 

11.95s 

X 

39  =  466.37i 

18.08i  X 

18  =  325.50 

6.12i  X 

59  = 

361. 37J 

12.25 

X 

38  =  465.50 

18.37*  X 

17  =  312.37i 

By  multiplying  each  interest  jiayment  by  the  time  it  has  to  run,  we  have  in 
the  product  the  sum  npon  which  to  compute  the  interest  for  07ie  month  to  equal  the 
interest  on  the  79  interest  payments  for  the  79  difl'erent  periods  of  time. 


594 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Remark. — See  Article  1161,  following  this  solution  for  an  abbreviated  method 
of  finding  the  interest  that  would  accrue  on  the  interest  payments  made  on  a 
specified  number  of  equal  installments  payable  at  successive  and  consecutive 
periods. 

Fifth  Step. 

By  the  above  statement  we  find  there  was  $945  interest  i)aid,  thus:  29g  -f 
23.33J  =  23.GL'A  x  (80  -^  2).  We  also  find  that  the  interest  on  the  interest  of  the 
79  accrued  interest  payments  amounts  to  $145. 16^,  which  is  equal  to  .5512+  per 
cent  on  $4000  for  79  mouths,  as  per  the  following  operations: 


OPERATION 

To  find  the  interest  on  $24885 
for  1  month  at  7  per  cent. 


12 


248.85 

7 

$145.16+  interest. 


OPERATION 

To  find  the  per  cent  that 

$145. 16i  is  of  S4000for 

79  months. 


40.00 
79 


or  thus: 


$145,161  4- 263  J 
=  .5512+%,       2C3,\ 
100 


1% 
145.161 


$263J 

This  .5512+%  added  to  the  7  per  cent  that  the  Homestead  gains,  as  shown 
in  the  first  part  of  the  solution,  gives  7.5512+%  that  the  Homestead  realizes  oa 
the  $4000  during  the  80  months,  the  use  of  money  being  worth  7  per  cent. 

Sixth  Step. 

The  buyer  pays  7  per  cent  interest  on  the  notes  as  they  mature.  Thus  he 
pays  29g  cents  on  the  first  note  and  $23.33J  on  last  note,  the  sum  of  which  is 
$23.G2i.  This  $23.62*  -^  2  =  $11. 81^  average  monthly  interest.  $11.81i  x  80  mos. 
=  $945  interest  paid  by  the  buyer  on  the  $4000.  This  is,  as  shown  in  the  first  part 
of  the  work,  second  step,  3.54|  per  cent. 

By  the  monthly  payments  the  buyer  loses  the  use  of  the  79  successive 
monthly  payments  of  $50  and  the  value  of  such  use  as  shown  above  in  the  thiid 
step,  is  $921§  which  =  3.45§  per  cent  on  the  $4000.  Then  3.54g  +  3.45g  =  7  per 
cent  interest  that  he  pays. 

The  buyer  also  loses  the  use  (not  the  payment)  of  the  interest  on  the  interest 
of  the  79  successive  monthly  payments,  which,  as  shown  in  the  fifth  step,  amounts 
to  $145. 16J.  This  amount,  as  shown  in  the  fifth  step,  is  equal  to  .5512+  %  on  tiie 
$4000  for  79  rr.anths.  Hence  the  buyer  loses,  altogether,  3.54g%  +  3.45g%  + 
.5512+  %  —  7.5512+  %  on  the  $4000  for  79  months,  the  use  of  money  being  worth 
7  per  cent. 


TO   FIND  THE  AMOUNT  OF  INTEREST  THAT  WOULD  ACCRUE    ON   TIIE    INTERESTS 

RECEIVED  ON  THE  SERIAL  OR  PERIODICAL   PAYMENTS  OF  A  GIVEN  AMOUNT 

DIVIDED  UP  INTO  A  CERTAIN  NUMBER  OF  EQUAL  INSTALLMENTS,  AND 

PAYABLE  AT  THE  END  OF  SUCCESSIVE  CONSECUTIVE  PERIODS, 

1161.  The  method  of  finding  the  sum  of  the  payments  and  their  regular 
interests  having  been  given  in  full  in  the  preceding  problem,  we  i)rcsent  the  fol- 
lowing shorter  method  of  finding  the  amount  of  the  interest  on  the  interest  pay- 


*  INTEREST    ON    INTEREST    PAYMENTS.  5g5 

ments,  supposing  each  original  interest  payment  to  be  invested  at  simple  interest 
from  the  time  paid  until  the  maturity  of  the  last  payment. 

The  first  interest  payment  will  bear  iuterest  for  one  month  less  than  the 
whole  number  of  months;  hence,  if  we  let  A  represent  the  interest  on  the  first  pay- 
ment for  one  mouth,  and  let  M  equal  tlie  remaining  number  of  months,  we  will  have 
the  following  series  and  their  summation  to  represent  the  successive  interests,  and 
their  total  aggregate  or  final  sum. 

Diagram  for  the  summation  of  tlie  compounding  of  the  successive  increasing  monthly  inter- 
ests, with  the  successive  decreasing  monthly  pel'iods. 

INTERESTS. 

1st  month  =    a  x  m  =    am  —  0 

2(1         "       =  2a  X  ni  —  1  =  2ain  —  2a 
3d        "       =  3a  X  m  —  2  =  3am  —  6a 
4th       "       =  4a  X  111  —  3  =  4am  —  12a 
5th       "       =  5a  X  m  —  4  =  Ram  —  20a 
6th       "       =  6a  X  m  —  5  =  6aui  —  30a 


21ani  —  70a 
Sum  of  interests  =  a  (21m  —  70) 

The  quantity  (21ni)  is  the  sum  of  the  arithmetical  series  from  1  to  6  =  (21) 
multiplied  by  the  number  of  months  m  (6)  =  (21  x  C).  And  tlie  minus  quantity  to 
be  subti'acted  (70),  the  sum  of  the  last  column,  is  the  sum  of  a  compound  series  or 
rectangular  pyramidal  series,  and  is  found  for  any  term,  (n)  by  cubing  it,  and  sub- 
tracting (n),  and  dividing  the  remainder  by  3,  hence  formula  ("^—  ")  or  as  in  above 
diagram  (6^-6)  =  (70).  '  3 

3 

Or,  it  may  be  found  by  multiplying  the  number  of  terms  (n)  by  one  less,  and 
one  more  than  itself  as  5  x  C  x  7  and  divide  their  product  by  3  =  70,  which  sub- 
tracted from  (21m)  the  remainder  is  the  equivalent  number  of  months  to  be  multi- 
plied by  one  month's  interest  (a),  which  will  give  the  amount  on  which  to  calculate 
the  interest  for  one  month,  at  the  given  rate. 

When  a  number  of  payments  have  been  made  to  find  the  final  Maturity  Value 
of  tlie  interest  on  the  remaining  payments,  proceed  as  follows:  First  find  the  value 
of  the  interests  for  the  whole  period  by  the  method  just  given  ;  then  find  the  value 
of  that  part  of  the  series  that  has  been  paid,  and  subtract  the  last  from  the  first. 

To  find  the  Present  Value  of  the  unmatured  part  of  a  series,  find  the  final 
value  as  above,  and  discount  the  amount  by  tlie  customary  methods.  The  usual 
way  would  be  by  the  principles  of  Bank  Discount.  A  more  ethical  way  would  be  to 
add  the  accrued  simple  iuterest  on  the  unmatured  notes  to  their  face  value. 


CONDENSED    METHODS    FOR    FINDING    TFTE    MATURITY    VALUE    OP 

TeE  INTEREST  ON  TDE  INTERESTS  OF  A  SERIES  OF 

PERIODICAL  NOTES  OR  PAYMENTS. 

11G2.  First. — Find  the  simple  interest  on  the  first  payment  for  one  month,  or 
period,  at  the  given  rate  per  cent,  and  call  it  (a)  or  1st  result.  Second. — Find  tiie 
sum  of  the  arithmetical  series  from  one  to  the  number  of  periods  (s)  and  multiply  it 


596 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


by  tlie  imuiber  of  iiiontbs  m  =  (sm)  or  2nd  result.  Third. — Find  tbe  sum  of  a  com- 
liouud  series  by  multiplying  tbe  number  of  terms  (n)  by  one  less  and  by  one  more 
tban  itself,  and  divide  tbe  product  by  tbree,  or  by  tbis  formula  ("'—  ")  =  3rd  result  c, 

3 

Fourth. — Subtract  tbe  tbird  result  from  tbe  second  result,  and  multiply  tbe 
remainder  by  (a)  or  first  result,  and  tbe  product  is  tbe  sum  on  wbicb  to  compute  tbe 
interest  for  one  moutb  to  equal  tbe  interest  on  tbe  interest  for  tbe  79  dilfcrent 
jieriods  of  time. 

Tbese  four  operations  may  be  expressed  by  tbe  following  formula : 

Tbis  formula  gives  tbe  number  of  dollars,  on  wbicb  to  calculate  tbe  interest 
for  one  month  at  tbe  given  rate;  (m)  represents  tbe  number  of  niontbs  tbe  first 
montbs'  interest  would  run,  and  is  tberefore  one  less  than  the  whole  number  of  terms 
(n)  of  tbe  original  series. 

To  ajtply  these  methods  to  tbe  foregoing  problem  we  proceed  as  follows: 


FIRST    STEP 

To  find  tbe  interest  on  $50  for 

1  month  at  7  per  cent. 


12 


50 

7 

p. 291  ii,t.  =  a  or  First 
result. 


SECOND     STEP 
To  find  tbe  snm  of  tbe  arith- 
metical series  and  tbe  prod- 
uct thereof  by  tbe  montbs 

1  +  80  =  81  X  40  =  3240 

=  sum. 
3240  X   79  =  255960  =  s. 

m.  or  Second  result. 


TniRD   STEP 
'io  find  tbe  sniu  of  a  compound 

series. 

SO  X  79  X  81  =  511920  - 
3  =  170040  =  c  or  Third 
result. 


FOURTn   STEP. 

Eesult  s.  m.  255900  —  result  c  170G40  =  85320  =  d  or  Fourth  result. 

FIFTH    STEP. 

Eesult  d  85320  x  result  a  29J^  =  $24885.00,  wbicb  is  the  amount  upon  which 
to  com])ute  tbe  interest  for  one  niontb,  to  equal  tbe  interest  on  tbe  79  interest  pay- 
ments for  79  different  periods  of  time,  as  shown  in  tbe  above  detailed  solution. 


TRUE  DISCOUNT. 

1163.  True  Discount  is  sucb  a  deduction  from  the  face  of  notes  or  debts,  as 
is  equal  to  tbe  siniitle  interest  on  tbe  remainder  for  tbe  same  time  and  rate  for  wliich 
tlie  deduction  was  made.  For  a  statement  of  tbe  difl'erence  between  True  and  Eank 
Discount,  see  our  remarks  on  Interest  and  Bank  Discount,  i)age  576. 

1164.  The  Present  Worth  of  a  note  or  debt  is  the  sum  that  remains  after 
tbe  true  discount  is  deducted  from  tbe  face;  or  it  is  the  principal  which,  at  the 
specified  rate  of  interest  and  time,  would  amount  to  tbe  face  of  tbe  note  or  debt 
wlien  it  becomes  due.  For  esamjile,  if  we  loan  $500  for  1  year  at  8  ])er  cent,  it  will 
amount  to  $540.  Hence  we  see  that  the  present  worth  of  $540,  due  one  year  hence 
at  8  per  cent,  is  $500. 


TRUE    DISCOUNT. 


597 


Tlie  proceeds  or  cash  value  of  tliis  $540,  by  the  bankers'  inetliod  of  discount 
is  but  $49G.S0,  which  is  $3.20  less  than  by  the  true  discount  system. 

True  discount  is  used  but  very  little  in  business.  Business  men  generally 
use  the  Merchants'  and  Bankers'  System  of  Discount. 

The  function  or  purpose  of  true  discount  is  properly  to  determine  what  sums 
are  to  be  invested  to  produce  a  specified  amount  at  a  given  rate  of  interest  and  for 
a  stated  period  of  time.  And  in  this  respect  it  is  the  same  as  interest  where  we 
have  the  time,  rate  per  cent,  and  interest  or  amount  given  to  find  the  principal. 

The  following  problems  will  fully  illustrate  the  operations  of  True  Discotint: 

PROBLEMS. 

1.  What  is  the  present  worth  or  cash  value,  and  the  true  discount  of  a  nt)te 
of  $9000  for  94  days  at  8  per  cent? 

Or  thus:  What  sum  loaned  or  placed  at  interest  for  94  days  at  8  per  cent, 
will  produce  $184.15  interest,  and  amount  to  $9000. 

Ans.  $8815.85  present  worth.    $184.15  true  discount. 


OPERATION. 


$4500  present  worth  assumed. 
94  luterest. 

$4594  amount  94  days  hence. 


4594 


4.500 
9000 


$9000       face  of  note. 
8815.85  jireseut   worth. 


$8815.85  present  worth.        $  184.15  true  discount. 


7l'ar/)?ann(io«.  — By  inspection  and  reason,  we  see  that  as  true  discount  is  the  interest  on  tlie 
PRESENT  woirrn,  we  cannot  tlierefore  operate  on  the  $9000,  the  face  of  the  note,  and  hence  we  are 
obliged  to  assume  some  number  to  represent  present  worth;  and  to  facilitate  the  operation  we 
assume  $4500,  the  8  i)er  cent  Interest  Divisor  to  represent  tlie  present  worth.  As  explained  in  Cash 
Notes,  page  582,  we  assume  the  Interest  Divisor  Inr  tlie  reason  that  the  interest  thereon  is  alway.s 
as  many  dollars  as  there  are  days  in  the  time.  Knowing  this  we  are  saved  the  labor  of  computing 
the  interest  on  the  sum  assumed.  Accordingly  the  interest  on  tlie  $4500  for  94  day.s  at  8  per  cent  is 
$94;  which  added  to  the  $4500  gives  $4594,  as  the  amount  of  $4500  assumed  present  worth  due  94 
days  hence  at  8  per  cent. 

We  now  observe  that  as  we  have  used  the  time  and  rate  given  in  the  problem,  there  is  the 
same  relationship  between  the  $4,">94  amount  and  the  $4500  of  assumed  pretent  u-orlh,  as  there  is 
between  the  .$9000  amount  of  note  and  the  unknown  present  worth  of  the  same ;  and  hence  to  find 
the  present  worth  of  the  .$9000  amount,  we  have  but  to  jierforni  a  proportional  operation  as  shown 
by  the  second  line  statement  in  the  solution.  The  reasoning  for  this  statement  is  as  follows;  .Since 
$4,594  amount  re(|uire  $4500  present  worth,  $1  will  require  the  4594th  part,  and  $9000  will  require 
9000  times  as  much. 


45 


$100 
94 


SECOND   OPERATION 
By  assuming  $100  as  the  present  worth. 

918.80 


&2.0Sa  =  interest. 
100.        =  assumed  present  worth. 


$102.0S|  =  amount  94  days  hence. 


100 

9 

9000.00 


$8815.85  present  worth. 
9000.00  amount  of  note. 


$184.15  true  discount. 


598  SOUI.e's    I'lIII.OSOPHIC    PRACTICAL    MATHEMATICS.  * 

GENERAL  DIKECTIONS. 

1165.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  calculating  True  Discount: 

1.  Assume  as  the  present  worth,  (he  Interest  Divisor  for  the  given  rate  per 
cent  and  find  the  amount  of  the  same  for  the  yivcn  time  and  rate  per  cent. 

2.  Then  divide  the  assumed  present  worth  multiplied  by  the  given  amount,  by 
the  amount  of  the  assutncd  jnescnt  worth. 

Note. — $100  may  be  assumed  as  tbe  present -wortb,  instead  of  tbe  interest  divisor,  if  preferred. 

2.  What  is  llie  true  discount  on  $3450  for  1  year  and  8  months  at  9  per  cent, 
counting  30  days  to  llie  month,  and  not  allowing  grace?  Ads.  $450. 

Note. — This  problem  may  be  statod  as  fjHows: 

3.  What  sum  loaned  for  1  year  and  8  months,  at  9  per  cent  will  produce  $450 
interest  and  amount  to  $3450  ? 

OPERATION   INDICATED. 


4  COO 


$4000  $3450 

3450  3000  or  thus:   COO 


$4000 
450 

$3000  =  present  worth.     $  450  =   true  discount.  $3000 

Note.— 1  year  and  8  montlis  =  20  months  =  GOO  days,  and  hence  tbe  interest  on  the  assumed 
9  per  cent  interest  divisor  is  $000, 

4.  What  sum  must  be  loaned  for  2  years  at  7  jicr  cent  to  amount  to  $8550? 

Ans.  $7500. 

5.  What  is  the  jircsent  worth  of  a  claim  of  $24C8.45  due  in  123  days,  money 
■worth  li  per  cent  jier  month?  Ans.  $2325.44. 

0.  What  principal  must  be  loaned  for  183  days  to  amouut  to  $3915.00,  inter- 
est at  0  per  cent?  Ans.  $3800. 

7.  A  merchant  sold  $5000  worth  of  goods  on  4  months  credit,  and  then 
offered  the  purchaser  3  per  cent  discount  for  cash,  which  was  accepted.  What  was 
the  discount,  and  how  much  did  it  exceed  the  true  discount  at  8  i)er  cent? 

Ans.  $150  discount.     $120.87  true  discount.    $20.13  excess. 

Note.— In  transactions  of  this  kind  count  SO  days  to  the  month  and  allow  no  grace  or  dis- 
count day. 

8.  Goods  were  bought  to  the  amount  of  $4200  on  3  months'  credit  when  the 
current  rate  of  interest  was  4i  per  cent.  Dow  much  cash  could  be  jiaid  lor  (he 
goods  without  loss  to  either  party  ?  Ans.  $4153.28. 

OPEKATION    INDICATED. 

$8000 


8090 


4200 


9.  A  owes  B  $8000  due  in  6  months.  B  proposes  to  deduct  5  per  cent  from 
the  face  of  the  bill  for  cash.  A  accepts  the  proiiosition  and  borrows  the  money  at 
6  per  cent,     ilow  much  does  A  gain  or  lose!  Ans.  $172.00  A  gains. 

OPERATION   INDICATED. 

$8000 

400  —  5  per  cent  discount.     Interest  on  $7600  for  6  months  at  C  per  cent  is  $228 

$400  —  $228  =  $172  gaiu. 

$7600 


TO    FIND    RATE    PER    CENT. 


599 


10.  What  is  tbe  present  worth  of  $4391.68  due  in  1  year  and  4  months,  money 
worth  5  per  cent,  and  no  allowance  for  days  of  grace  or  discount  day? 

Ans.  $4117.20. 

11.  What  is  the  true  discount  on  $8705.25  at  9  per  cent  for  495  days? 

Ans.  $965.25. 

PROBLEMS  INVOLVING  INTEREST  OPERATIONS. 


1166.     To  find  the  Bate  per  cent  when  the  Principal,    Time,   and   Interest  or 
Proceeds  are  given : 


1.     The  interest  on  $9000  for  94  days  is  $188;  what  was  the  rate  per  cent! 

Ans.  S% 
OPERATION 
To  find  iLe  rate  per  cent. 

188.00 


OPERATION 

To  find  tbe  interest  on  $9000 

for  tbe  given  time  at  1  per 

cent  assumed. 


23.50 


300 


9000 
94 

B23.50  interest. 


8%  Ans. 


Erplanalion.— In  all  problems 
of  this  kind  having  no  rate  per 
cent  given,  we  are  obliged  to 
assume  some  rate  to  work  from, 
and  as  the  ratio  will  be  the  same, 
whatever  rate  per  cent  we  assume, 
therefore,  to  facilitate  tbe  work, 
we  assume  1  per  cent,  then,  in 
order  to  obtain  a  sum  of  interest 
that  bears  the  same  relationship 
to  the  1  per  cent  assumed,  that 
the  $188  bears  to  tbe  required 
rate  per  cent,  we  calculate  tbe 
interest  on  tbe  principal  for  tbe  given  time  at  1  per  cent.  Accordingly  we  find,  in  this  problem, 
$L'3.50  interest.  We  then  reason  as  follows:  Since  1  per  cent  gives  $23.50  interest,  by  transposition 
$23.50  required  1  per  cent;  and  since  $23.50  interest  required  1  per  cent,  1  cent  interest  Will  require 
the  2350tii  part  and  18800c.  interest  will  require  18800  times  as  much. 


GENERAL   DIRECTIONS. 


1167.     From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions lor  finding  the  rate  jier  cent : 

J.     Assume  1  per  cent  and  find  the  interest  on  the  principal  for  the  given  time 
at  the  1  per  cent. 

a.     Then  divide  the  given  interest  hy  the  1  i^er  cent  interest  on  the  principal  for 
the  given  time. 

fJoTE.  — Wben  the  proceeds  are  given,  first  subtract  the  same  from  tbe  principal  or  face  of 
note,  and  thus  find  tbe  interest  or  discount 

PROBLEMS. 

2.  The  interest  on  $1000  for  288  days  was  $48.     What  was  the  rate  per  cent  ? 

Ans.  6%. 

3.  A  merchant  loaned  $1580  for  120  days  and  received  for  its  use  $52.6G§; 
at  what  rate  iier  cent  did  he  loan  it?  Ans.  10%. 

4.  A  note  for  $524.80  was  discounted  for  47  days  and  $518.6336  proceeds 
received.     At  what  rate  per  cent  was  it  discounted  ?  Ans.  9%. 

5.  A  received  $3679.20  for  his  note  of  $4200  which  had  ISO  days  to  run.    At 
what  rate  per  cent  was  it  discounted  ?  Ans.  24%. 


6oo 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


C.     A  note  for  $7543.80  was  discounted  for  34  days  and  $7401.300  proceeds 
received.     At  wbat  rate  per  cent  was  it  discounted?  Ans.  20%. 

7.  A  merchant  invested  in  goods  $14000  and  commenced  tlie  merchandising 
business.  He  continued  8  months,  and  then  from  his  books  lie  found  that  his  gains 
on  sales  were  $5215,  and  his  store  expenses  of  all  kinds,  $2705.  What  was  the  rate 
per  cent  per  annum  gain;  what  was  his  gain  per  cent  on  investment;  and  how 
much  more  did  he  gain  by  his  business  than  if  he'  had  loaned  his  money  at  10  per 
cent  interest  and  received  $125  per  month  for  his  services? 
Ans.  l~\l%  gain  on  investment. 
2Gis%  Kaiii  per  annum. 

$570. GG§  gain  on  his  business  in  excess  of  what  lie  would 
have  gained  by  loaning  Ins  iiioiicy  at  10  per 
cent  and  serving  others  at  $125  per  month. 

OPERATION    INDICATED. 

$2510 
100 


Gain  on  sales, 
Expense, 


^5215 
2705 


14000 


Net  gain,  ....        $2510 

(1711%  X  12)  —  8  =  26H|%  gain  per  annum. 
Interest  on  $14000,  8  mos.  at  10%  =  $933^+  salary  8  mos.  at 
=  $1933J.    Net  gam  of  business  $2510  —  $1933J  =  $57G§. 


17if%  S^'"  o"  investment. 
25  =  $1000 


TO  FIND  THE  PRINCIPAL  WHEN  THE   PATE  PER  CENT,  TIME,  AND 
INTEREST  OR  AMOUNT  OR  PROCEEDS  ARE  GIVEN. 


1168.     1.     W^hat  principal  loaned  for  94  days  at  8  per  cent  will  produce  $18S 
interest?  Aus.  $9000. 

FIRST   OPERATION. 


94 


$4,500  assumed  principal. 

188 


45 


OPERATION 

To  find  tbo  interest  on  the 
assumed  principal. 

$ 

100  assumed  principal. 
94 


$9000  principal,  Ans. 

SECOND  OPERATION. 
OPERATION 

To  find  the  principal. 


18.80 


Explanation. — In  all  prob- 
lems of  this  kind,  we  as- 
sume the  Interest  Divisor, 
or  $100  as  principal,  and 
then  calculate  the  interest 
thereon  for  the  given  rate 
and  time.  This  is  done  in 
order  to  produce  an  interest 
which  bears  the  same  re- 
lationship to  the  assumed 
])rineipal  that  the  given  in- 
terest bears  to  the  required 
principal  that  produced  it. 
In  the  first  operation  we  assumed  the  8  per  cent  Interest  Divisor,  $4500,  as  the  principal ;  and 
seeing  that  the  interest  on  $4500  lor  the  given  time  and  rate  per  cent  is  S94,  we  reason  thus:  Since 
$4500  principal  gives  $04  interest,  conversely  fOl  interest  required  $4500  principal.    And  since  $94 


$2.08|  interest. 


100 

9 

188.00 

$9000  Ans. 


*  INTEREST   TO    FIND   THE    PRINCIPAL.  6ol 

interest  requirert  $4500  principal  $1  interest  -will  raqnire  the  94th  part,  and  $188  interest  -will  require 
188  tinii'S  as  much. 

Ill  the  second  solution  ■we  found  the  interest  on  the  assumed  principal  to  he  $2.08§.  Having 
this  interest.  Tve  make  the  proportional  statement,  reasoiiiiifr  as  follows :  Since  $100  principal  gives 
$2.08§  interest,  conversely  $2.08f  interest  required  $100  principal.  And  since  Hf^c.  (|2.08|  reduced) 
interest  require  $100  principal,  |c.  interest  will  require  the  1880th  part,  and  p  or  a  ■whole  cent  will 
requires  times  as  much;  and  since  Ic.  intei'est  requires  the  result  of  the  statement  now  made, 
18800c.  (fl88  reduced  to  cents)  will  require  18800  times  as  much. 

Note. — We  much  prefer  the  first  operation,  for  the  reason  that  the  interest  on  the  Interest 
Divisor,  for  the  time  and  at  the  rate  jier  cent,  will  always  he  the  same  as  the  nuuiber  of  days.  And 
heuce,  this  heing  known,  we  save  making  one  interest  calculation  in  the  operation. 

2.  "What  principal  loaned  for  121  clays  at  10  per  cent  will  amount  to  $7442  ? 

Ans.  $7200. 

OPERATION. 

$                                                    ■  Explanation. — When  the  amount  is 

3600  r=  assumed  principal.  given,    instead   of  the   interest,   then 

7442  =  amount.  add  the  interest  on  the  assumed  princi- 

pal  to  the  assumed  principal  and  then 

$7200,  Ans.  make  the  proportional  statement. 

3.  "Wliat  principal  discounted  for  369  days  at  8  per  cent  will   give   $1836 
proceeds?  Ans.  $2000. 

OPERATION. 

Explanation. — When  the   proceeds 
4.-)n0  =  assumed  principal.  "^^  given   instead  of  the  interest,  then 

.    „ ,  .       1  ,  subtract  the  interest  on  tlie  assumed 

li>,jo  —  proceeds.  princi])al  from  tlie  assumed  principal 

and  then  make  the  proportional  state- 


Amount  =  3721 


Proceeds  =  4131 


$2000,  Ans.  ment. 

GENERAL   DIRECTIONS. 

1169.  From  tlie  foregoing  elucidations,  we  derive  the  following  general'  direc- 
tions for  finding  tbe  principal : 

1.  Assume  $100  or  the  Interest  Divisor  as  2»'i>teipal,  and  calculate  thereon  the 
interest  for  the  given  time  and  rate  per  cent. 

2.  Then  divide  the  assumed  principal  multiplied  by  the  given  interest,  by  the 
interest  on  the  asstimed  principal. 

Note. — If  tbe  amount  is  given  instead  of  tbe  interest,  then  add  the  interest  on  tbe  assimicd 
principal  to  tbe  assumed  principal,  and  divide  the  sum  into  tbe  product  of  tbe  assumed  pruiciiial 
and  tbe  given  amount. 

If  tbe  proceeds  are  given  instead  of  the  interest,  tben  subtract  tbe  interest  on  tbe  assumed 
principal  from  tbe  assumed  principal,  and  divide  tbe  ditierence  into  tbe  product  of  tbe  assumed 
principal  and  tbe  given  proceeds. 

PROBLEMS. 

4.  A  banker  loaned  a  sum  of  money  for  309  days  at  8  jier  cent  and  received 
$164  interest.     What  was  tbe  sum  loaned  ?  Ans.  $2000. 

5.  A  merchant  loaned  a  certain  sum  for  1  year  and  3  months  at  5  per  cent, 
and  received  $750  interest.     What  was  the  principal  loaned  ?  Ans.  $12000. 

6.  What  principal  loaned  for  90  days  at  6  per  cent  will  amount  to  $2030  ? 

Ans.  $2000. 


6o2  soule's  thilosophic  practical  mathematics.  * 

7.  Discounted  a  note  for  183  days  at  5  per  cent  and  received  $1037.30  pro- 
ceeds.    Wliat  was  the  face  of  the  note?  Ans.  $1080. 

8.  What  principal  loaned  for  G4  days  at  8  i)er  cent  Mill  amount  to  $1309.20? 

Ans.  $1350. 

9.  A  merchant  discounted  a  note  for  04  days  at  5  per  cent  and  received 
$178.40  proceeds.    What  was  the  face  of  the  note  ?  Ans.  $180, 

10.  A  father  desires  to  i)lace  bis  son  in  college  for  a  four  years'  course  of 
study.  The  yearly  expense  for  board,  tuition,  etc.,  is  $400,  and  for  bis  clothes, 
etc.,  he  allows  him  $150  per  year.  He  now  wishes  to  know  what  sum  of  money  be 
must  ])lace  at  interest  at  5  per  cent,  so  that  the  interest  will  just  pay  the  yearly 
e.\j)ense  of  his  son?  Ans.  $11000. 

11.  What  priucipalloaued  for  124  days  at  7  per  cent  will  produce  $130.20 
interest?  Ans.  $5400. 

OPERATION 

30000  =  1%  int.  divisor  as  assumed    l)rin.  Explanation.— S'lnce  there  is   no 

808  00         130.20  '^  1"""    '^^"t    Interest    Divisor,    we 

assume   as   jinncipal   36000,    the    1 
■;;j7  ]>er  cent  Interest  Divisor,  ami  then 

$l)400,    Ans.  iuulti|ily  tlie  interest  at  I  ])er  cent, 

which  is  etinal  in  dollars  to  the 
linniber  of  clays,  l)y  7,  the  rate  per  cent,  and  thus  ])rodueo  $868.00  interest.  The  solution  statement 
le  then  made  as  in  the  jirecediug  problems. 

12.  What  principal  loaned  for  72  days  at  11  per  cent  will  produce  $33 
interest?  An.s.  $1500. 

TO  FIND  THE  TIME,  WHEN  THE  PRINCIPAL,  RATE   PER   CENT,    AND 

INTEREST,  OR  WHEN  THE  PRINCIPAL  AND  THE  AMOUNT  OR 

PROCEEDS  AND  THE  RATE  PER  CENT  ARE  GIVEN. 

PROBLEMS. 

1170.     1.    A  note  for  $9000  was  discounted  at  8  per  cent,  and  the  interest 
amounted  to  $188.    For  liow  many  days  was  it  discounted  ?  Aus.  94  days. 

1  year  or  360  days  assumed. 
OPERATION  OPERATION 

To  find  the  interest  on  the  $9000  at  8  per  cent  To  find  the  number  of  days, 

for  1  year  assumed. 


$90.00      principal. 

8%  interest.  720 


$720.00     interest  for  300  days. 


ds. 
300 

188 

94  days,  Ans 


Explanation.— \n  all  problems  of  this  kind,  we  first  assume  360  days,  or  1  year.  Then,  in 
order  to  obtain  a  sura  of  interest  that  bears  the  same  relationship  to  the  360  days  of  assumed  time 
that  the  $188  given  interest  bears  to  the  recjuired  time,  we  calculate  the  interest  ou  the  i)rincipal 
loaned  for  the  assumed  time  at  the  given  per  cent.  Accordingly,  we  produce,  in  this  problem,  $720 
interest.  We  then  reason  as  follows:  Since  1  year's  time,  with  the  given  principal  and  rate,  give 
$720  interest,  by  transposition  $720  interest  recjuired  1  year's  or  360  days'  time;  and  since  $720 
interest  required  360  days'  time,  $1  interest  will  require  the  720th  part,  and  $188  interest  will  require 
188  times  as  many,  which  is  94  days. 

Note.— In  case  the  interest  is  computed  at  365  days  to  the  year,  then  365  would  be  the  days 
assumed. 


*  INTEREST   TO    FIND    THE    TIME,  603 

2.    A  iTiercbaiit  borrowed  $1850  at  8  per  cent,  aud  paid  $197.33J  for  its  use. 
How  Jong  did  lie  have  tlie  money  ?  Ans.  1  year  and  i  mouths. 


$18.50 

8% 


148 
$148.00     interest.  3 


OPERATION    INDICATED. 

1  year  or  300  days  assumed. 

Ti 

1 

592                                      or, 

DS. 
3G0 
148     592 
3 

1^  yrs.  =  1  yr.  4  nios. 

480 

480  ds.  =  1  yr.  4  mos. 


GENERAL   DIRECTIONS. 


1171.  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  finding  the  time: 

1.  Assume  3G0  days,  1  year,  and  find  the  interest  on  the  given  principal  at  '  i 
given  rate  per  cent  and  for  the  assumed  time. 

2.  Then  dii'ide  the  given  interest  viultiplied  by  the  assumed  time,  by  the  r  .■*  ' 
on  the  principal  for  the  rate  ])er  cent  and  assumed  time. 

Note  1. — When  tbe  amount  is  given,  first  subtract  tlie  interest  from  the  same  to  find  tlio 
principal. 

NciTE  2. — When  the  proceeds  are  given,  first  add  the  interest  to  the  proceeds  to  find  the 
principal. 

Note  3. — In  case  the  interest  or  discount  has  been  computed  at  365  days  to  the  year,  then 
assume  365  days. 

PROBLEMS. 

3.  A  note  for  $0000  was  discounted  at  5  per  cent  and  the  interest  or  discount 
was  $78.33.^.     For  how  many  days  was  it  discounted  ?  Ans.  94  days. 

4.  A  merchant  borrowed  $2500  at  4^  per  cent,  and  paid  $38.75  for  its  use. 
now  locig  did  he  have  the  money  ?  Ans.  124  days. 

5.  Loaned  a  sum  of  money  at  8  per  cent  until  it  amounted  to  $401.37.  The 
interest  was  $0.37.     How  long  was  it  loaned  "?  Ans.  63  days. 

0.  A  note  was  discounted  at  10  per  cent  and  $800  ])roceeds  received.  The 
discount  was  $200.     For  what  time  was  tlie  note  discounted  ? 

Ans.  720  days,  or  2  years. 

7.  A  broker  loaned  $50000  at  12  per  cent,  and  received  $550  interest  for  it. 
How  long  did  he  lend  it?  Ans.  33  days, 

8.  How  long  will  it  take  for  $2000  to  amount  to  $3000  at  10  per  cent  interest? 

Ans.  5  years, 

9.  A  note  dated  November  1,  1895,  was  discounted  December  23,  1895,  at  6 
percent.  The  face  of  the  note  was  $0231.50  and  the  proceeds  $0187.8795.  Grace 
and  discount  day  were  counted.  For  how  many  days  was  it  discounted  and  how 
long  did  it  run  after  being  discounted  ? 

Ans.  42  days  discounted. 

41  days  to  run  after  discounted  not  including  discount  day, 

OPERATION    INDICATED. 

$6231.50  DS. 

300 
373.89 


373.8900 


43.6205 
42  days. 


6o4 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


10.  A  note  lor  $9730  bearing  8  per  cent  interest  dated  November  24,  1895, 
payable  in  one  year,  was  discounted  at  J2  per  cent,  tlie  proceeds  being  $9883.9934. 
Wbat  was  the  time  and  tbe  date  of  discount,  allowing  grace  and  discount  day  1 

Ans.  180  days  time.     June  1,  1890  date  discounted. 


OPERATION    INDICATED. 


45 


9730. 
363 


$  784,8866+  interest. 
9730. OU 

$10514.8866+  uiaturity  value. 
98S3.[I'J34      iiroceeds. 

$    630.89:12      discount. 


1  year,  or  360  days  assumed. 
$10j14.8866+ 
1-2% 

$  1261.786392  interest  for  1  year. 

ds. 
1261.7863  I  360  =  I  year. 
630.8932 


I  180  =  days  the  note  vras  discounted. 

Note. —36.5  days  +  3  days  of  grace  and  1  discount  day  =  369  days  liad  the  note  hcen  dis. 
counted  November  24th,  369  —  180  =  189  that  the  note  was  discounted  alter  November  24th,  which 
18  June  1,  1896.     See  Time  Table,  page  298. 


1172. 


TABLE 


SHOWING  :n  how  many  years  a  given  principal  will  double  itself. 


AT  COMPOUND   INTEREST. 

COMPOUND  INTEREST 

ON  1  DOLLAR  FOR  100  TEARS. 

At  Simple 

1 

KATE. 

Interest. 

^ompoiiDtled 
i  early. 

Compounded 
Half- 
Yearly. 

Compounded 
Quarterly. 

Amount. 

Tears. 

Per  Cent. 

Accumulation. 

ox 

•  0 

YEARS. 

YEARS. 

YEARS. 

YEARS. 

$1 

100 

1 

$                     2.75 

1 

100. 

69.666 

69.487 

69.400 

100 

2 

7.25 

li 

66.66 

4  6., 556 

46.382 

46.298 

100 

2i 

11.75 

2 

50.00 

35.004 

34.830 

34.743 

100 

3 

19.25 

2* 
3 

40.00 

28.071 

27.899 

27.812 

100 

3i 

31.25 

33.33 

23.450 

23.278 

23.191 

100 

4 

50.50 

3i 
4 

28.57 

20.150 

19.977 

19.890 

100 

4J- 

81.25 

25.00 

17.673 

17.502 

17.415 

100 

5 

131.50 

4i 

22  22 

15,748 

15.576 

15.490 

100 

6 

340.00 

5' 

20,00 

14.207 

14.036 

13.946 

100 

7 

868.00 

51- 

18.18 

12.946 

12.775 

12.686 

100 

8 

2.203.00 

6 

16.67 

11.896 

11.725 

11.639 

100 

9 

5,543.00 

6i 

15.38 

11.007 

10.83G 

10.750 

100 

10 

13,809.00 

7" 

1-4.20 

10.245 

10,075 

9,989- 

100 

12 

84,675.00 

n 

8 

Si 

9 

13  33 

9.585 

9.914 

9.328 

100 

15 

1,174.405.00 

12.50 

9.006 

8.837 

8.751 

100 

18 

15,145,007.00 

11.76 

8  497 

8.346 

8.241 

100 

24 

2,551,799,404.09 

11.11 

8.043 

7.874 

7.788             1 

100 

9i 

10..50 

7.638 

7.468 

7.383 

10 

10.00 

7.273 

7.121 

7.026 

12 

8.34 

6.110 

1 

DIFFERENCE  BETWEEN  BANK  DISCOUNT  AND  TRUE  DISCOUNT, 
1173.     1.     What  are  the  proceeds  or  cash  value  of  $8000  for  Gi  days  at  10  pet 
«ent,  by  the  baukers'  system  of  discouut,  and  what  is  the  present  worth  or  cash 
value  by  the  true  discount  system  of  work  ? 

Ans.  $7857.78,  proceeds  by  bankers'  system. 
$7860.26,  present  work  by  true  discount. 


BANK    AND    TRUE    DISCOUNT    COMPARED. 


6o5 


Banking  System. 
8S000 
36       64 


OPERATIONS. 

True  Discount  System. 
$3C(i() 
3664       8000 


$  142.2li  bank  (liscouut. 
8000. 


$7860.26  present  worth. 
8000.00 


$  139.74  tine  discount. 


$7857.78  proceeds. 

2.  What  is  the  bankers'  discount  and  the  true  discount  of  $15000  for  75  days 
at  1  per  cent  per  month,  and  what  is  the  diflerence  ? 

Aus.  $375.00,  bank  discount.     $365.85,  true  discount.    $9.15,  difference. 

3.  What  is  the  difference  between  the  iiitei'est  on  $5000  for  3  years  4  months 
and  20  days  (1220  days)  at  6  jicr  cent,  and  the  true  discount  for  the  same  time  and 
rate?  Ans.  $171.79. 

4.  I  hokl  a  note  of  $10000,  due  in  10  years  or  3600  days  without  grace.  IIo^* 
much  will  I  receive  for  it  if  I  discount  it  at  10  per  cent  bank  discount,  and  hoW 
much  if  1  discount  it  at  the  same  rate  yer  cent  by  the  true  discount  system  ? 

Ans.     Nothing  by  bank  discount.    $5000  by  true  discount. 


TO  FIND  THE  KATE  TEH   CENT   OF   INTEllEST   PAID,    OR   THE   RATE 

PER  CENT  TRUE  DISCOUNT,  WHEN  NOTES,  ETC.,  ARE 

DISCOUNTED  BY  BANK  CUSTOM. 


1174.  1.  A  banker  discounted  two  notes  of  $500  each,  one  for  34  days  at  G 
per  cent,  and  the  other  lor  240  days  at  6  per  cent.  What  was  the  rate  i)er  cent  true 
discount  charged  on  each  note  ?  Ans.  65530/.,%  on  the  34  days'  note. 

6i  fc  ou  the  240  days'  note. 


60 


500  face  of  note. 
34 

$2.83J  interest. 


$497,165  proceeds. 


00 


5.00     face  of  note. 
•^iO  4 


$20.00  interest. 


$480.      proceeds. 


OPERATION   FOR  34   DAYS. 


3 

300 


14.91.50 
34 


.4695|^/  int.    on    pro- 
ceeds at  1%. 


OPERATION    FOR   240  DAYS. 


?  m 


$ 
m  4 

■^i0  SO 

$3.20    iut.  atl%on 
proceeds. 


25.3555 
3 


8^.^ 


54 
8.5000 


% 
1 

^0.00  50 


H% 


Ans. 


6o6 


SOULE  S    rillLOSOPHIC    PRACTICAL,    MATHEMATICS. 


ExfUnaiion.—ln  answering  these  questions  we  first  find,  as  shown  hy  the  first  statement,  the 
procoeils  of  each  note  according  to  the  bankers'  s.ystiiui  of  discounting.  Then,  considering  the 
proceeds  as  jirincipal,  we  have  the  (piestion  contained  in  Problem  No.  1,  on  page  509,  which  is  the 
pnncliial,  tiuiij  and  interest  given,  to  find  the  rate  per  cent.  To  answer  this  (iiiestiou  we  first  find, 
as  showu  in  the  second  statement,  the  interest  on  the  proceeds  of  the  notes  now  used  as  principals, 
for  the  respective  time  at  1  per  cent,  aud  then  make  the  proportional  statement  to  find  the  true 
discount  with  this  interest,  the  1  per  cent,  and  the  interest  obtained  by  discounting  the  notes 
Recording  to  bank  custom.  The  reasoning  for  the  last  two  statements  of  each  question  was  fully 
explained  on  page  599,  aud  hence  we  now  omit  it. 

Note  1— By  the  result  of  this  work  we  see  that  the  rate  per  cent  of  l>auk  discount  is  larger 
than  the  same  rate  per  cent  of  true  discount,  and  what  is  still  more  important,  we  see  that  this 
increased  difference  of  rate  increases  in  proportion  to  the  time  for  which  the  note  has  to  run.  This 
"being  the  case  on  all  notes  of  the  same  face  aud  at  the  same  rate,  it  seems  natural  to  conclude  th.at 
Tiaiikers  would  prefer  long  time  paper.  But  the  advantages  of  short  credits  are  generally  con- 
sidered to  bo  more  than  equivalent  for  any  excess  that  might  be  gained  by  extending  the  time. 
The  reason  for  this  increase  of  rate  per  cent  on  long  time  notes  or  transactions  is  that,  on  them  the 
(lisuount  becomes  larger,  and  hence  the  proceeds  or  jirincipal  to  which  it  is  referred  becomes  sHin/Zer. 

Note  2. — Instead  of  working  on  the  face  of  the  notes  in  the  above  questions,  we  could  as 
well  have  assumed  |I100  or  $1000,  or  better  still  the  Interest  Divisor.  By  working  from  assumed 
figures  the  operation  is  often  simplified,  and  in  questions  where  no  principal  or  face  of  note  is 
given  we  are  obliged  to  work  from  assumed  numbers 

2.  Wbat  is  the  rate  of  interest,  or  rate  i»er  cent  true  discount  on  ft  63  day 
note  at  8  per  cent  and  10  per  cent?  Ans.  8^/3%  true  discount,  at  8%. 

10/9"3%  true  discount,  at  10%. 

3.  What  is  the  rate  of  interest,  or  rate  per  cent  true  discount  on  a  360  day 
note  at  5,  6,  7,  8,  9, 10  aud  12,  per  cent  ? 

Ans.  5,%,  61?,  7^^,  8i«,  9l\,  11^,  13/,,  per  cent. 


TO  FIND  THE  KATE  PER  CENT  OF  TRUE  DISCOUNT  CORRESPONDING 
TO  A  GIVEN  RATE  OF  INTEREST. 


1175.    1.    If  I  wish  to  realize  10  per  cent  interest  on  94  day  notes,  at  what 
rate  per  cent  is  the  true  discount?  Ans.  9||2i9^g. 

FIRST    OPERATION. 


$3600  assumed  present  worth  or  proceeds. 
94  interest  on  same  for  94  days. 

$3694  principal  or  sum  loaued. 


I  $36.94 
360  I    94 

I  $  9.64JS  interest  at  1%. 


Int. 

9.64ai5 


/a 

1 

94.00 


interest. 


•^BbSU^   Ai     AUS, 


Eif\analion.—  \\\  this  solution  we  first  assume  S3600  the  Interest  Divisor  at  10  per  cent  as  the 
present  worth  or  proceeds,  and  then  add  thereto  the  interest  for  94  days  at  10  per  cent.  This 
operation  gives  l{>3694  which  is  the  principal  at  interest  for  94  days  at  10  per  cent  which  produced 
$94  interest.  The  next  step  is  to  compute  the  interest  on  the  $3694  for  94  days  at  1  per  cent,  which 
is  $9.64J'i',.  Having  this  interest,  and  the  interest  on  $3600  for  94  days  at  10  ]ier  cent,  we  then  make 
the  third  statement,  reasoning  thus:  Since  if9.643n  interest  require  1  per  cent,  $1  will  require  the 
$9.64^i;th  part,  and  $94.00  interest  will  require  94.00  times  as  much. 

SECOND   OPERATION 


36 


I  100 
94 


I  $2.6U 
$100. 


By  assumui] 
present  worth  assumed. 

,  interest, 
present  worth. 


:  $100  jircscnt  worth  or  proceeds. 

% 
9  I  92350 


360    94 


26.79,Vi,c.  interest. 


43.4045 
9 


4) 
1 

162 
23.5000 

^IWllX    Ans. 


$102. 61i  principal. 


*  INTEREST.  607 

2.    What  is  the  corresponding  rate  of  true  discount  for  a  63  day  note  at  8 
percent?  Ans.  I'slifo- 

FIRST  OPERATION. 
$4500  =  assumed  present  worth  or  jiroceeds.  $  % 

63  =  iuterest  oil  same  at  8%  for  63  days.  I  45.63  1  1 

360  I  63  7.98IJ  |  63.00 

$4563  =  principal. 


I  $7.98ii  interest.  |  7JSf  %  Ane. 

SECOND   OPERATION 


45 


By  assuming  .$100  as  tlie  present  worth  or  proceeds. 

$  $  % 

100     P.  W.  I  1.01.40 

63  360    63  .3549 


$1.40  interest.  .1774i  interest. 

100.      P.  W. 


1 

2 

1.4000 


^H'l  %  tfe  discount. 


$101.40  principal. 
3.    What  is  the  true  discount  to  correspond  with  93  day  notes  at  12  jier  cent, 
and  with  SCO  day  notes  at  15  per  cent  and  at  20  per  cent? 

Ans.  11 1^3-1%  on  93  day  note  at  12%. 
13/3%  on  360  day  note  at  15%. 
16§%  on  360  day  note  at  20%. 

ANNUAL,  SEMI-ANNUAL  AND  QUARTERLY  INTEREST. 

1176.    Annual,  Senii-Annnal  and  Quarterly  Interest  is  the  regular  simple 

interest  on  notes  or  debts,  and  the  interest  on  tiie  unpaid  simple  iuterest  from  the 

time  it  becomes  due  until  paid. 

Annual,  Sc7ni- Animal  and  Qnarierhj  Interest  is  due  after  it  lias  been  accruing 

for  the  <(Wie  s^^rtw^iVfi  for  the  interest  jiaymeuts,  except  the  interest  on  the  interest 

■which  is  not  due  till  final  payment. 

Note. — In  some  States,  a  neglect  to  collect  the  annual  interest  on  notes  drawn,  with  interest 
payable  annually,  is  considered  as  a  waiver  of  the  contract  to  pay  annual  interest. 

It  is  a  principle  of  law  that,  in  the  absence  of  any  agreement,  all  money  or 
claims  bearing  interest,  or  that  is  due  and  not  paid,  shall  draw  simple  interest  from 
the  time  that  they  become  due  until  paid.  Hence,  to  entitle  a  party  to  annnal,  semi- 
annual or  quarterly  interest,  a  special  agreement  to  that  effect  is  necessary.  In 
case  of  notes,  this  agreement  is  inserted  and  becomes  one  of  the  obligations  of  the 
instrument.    Thus,  the  following  note  contains  the  annual  interest  contract : 

$6000.  St.  Louis,  September  l.st,  1895. 

Four  years  after  date,  for  value  received,  we  promise  to  pay  to  the  order  of 
D.  G.  Uessee,  Sis  Thousand  Dollars  with  iuterest  at  8  per  cent,  payable  annually. 

Montgomery  &  Potts. 

In  this  note  the  words  in  italics  make  it  an  annual  interest  note,  by  specifying 
when  the  interest  shall  be  paid  ;  and  hence,  if  the  interest  is  not  paid  at  the  end  of 
each  year,  it  becomes  by  virtue  of  the  terms  of  the  instrument,  and  in  accordance 
with  law  and  equity,  a  new  debt,  and  will  draw  simple  interest  until  paid. 


6o8 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


The  following  work  will  sbow  tbe  operation  of  annual  interest : 
1.     In  the  foregoing  note  nothing  being  paid  until  maturity,  what  will  be  the 
amount  theu  duel  Ans.  $8150.40. 

OPERATION 
Face  of  note, 

(Interest  ou  note  for  1  year  at  8%, 
Interest  on  note  for  4  years  at  S%, 
Interest  on  $i480  for  3  years  at  S%, 
1  nterest  on  $480  for  2  years  at  8%, 
Interest  on  $480  for  1  year  at  8%, 


SnS.L'O 
7().80 
38.40— 


$6000.00 


1920.00 


230.40 


Total  amount  due  at  maturity, 


TJrpJaMa/io?!.— By  tlie  conditions 
of  the  note  we  see  that  tbe  inter- 
est was  payable  at  the  end  of 
eacli  year,  but  as  it  was  not  paid 
until  tbe  maturity  of  tlie  note, 
we  tlicrefore  lirst  liud  tbe  simple 
interest  ou  tbe  face  of  tbe  note 
for  tbe  4  years  at  8  per  cent.  We 
tben  observe  tbat  the  tirst  vear's 
S8150.40  interest,  $480,  was  due  anil  not 
]iaiil  for  3  years;  tbat  the  second 
year's  interest,  l!;480,  was  due  and  not  paid  for  2  years;  and  that  the  third  year's  interest,  $480,  was 
due  and  not  ]>aid  for  1  year  ;  and  hence,  by  the  conditions  of  the  contract,  and  in  conformity  to 
law  and  ecjuity,  we  find  tbe  interest  on  the  three  sums  of  annual  interest  for  the  respective  time 
tbat  tbey  were  due  and  not  paid.  The  last  year's  interest  being  paid  when  due,  there  is  no  interest 
to  charge  thereon.  We  now  add  tbe  total  auuiial  interest  and  the  simple  interest  on  tbe  annual 
interest  for  tbe  time  that  it  was  due  and  not  paid,  to  the  face  of  tbe  note,  and  thus  obtain  the 
amount  duo  at  the  maturity  of  the  note. 

A  Quarlcrhj  Interest  Note, 

$4373.  Chicago,  August  10,  1895. 

Two  years  after  date,  for  value  received,  I  promise  to  pay  to  the  order  of 
QuiN  &  EiLL  Four  Thousand  Three  Hundred  Seventy-two  Dollars,  with  10  per 
cent  interest  itayable  quarterly. 

J.  L.  Caillouet. 


2.     If  nothing  is  paid  ou  the  above  note  until  maturity,  what  amount  will 
then  be  due  ?  Aus.  $5322.91. 

operation 


Face  of  note. 
Interest  ou  same 
(The  interest  ou 
Interest  ou  Ist  q 
Interest  on  2d  q 
Interest  on  3d  q 
Interest  on  4th  q 
Interest  on  5th  q 
Interest  on  6th  q 
Interest  on  7tb  q 
Interest  on  8th  q 
Amount  of 


for  2  years  at  10  per  cent, 
face  of  the  note  for  1  quarter  is  $109.30.) 
uarter's  interest,  $109.30,  for  21  months, 
iiarter's  interest,  109.30,  for  18  mouths, 
uarter's  interest,  109.30,  for  15  months, 
uarter's  iuterest,  109.30,  for  12  mouths, 
uarter's  iuterest,  109.30,  for  9  months, 
uarter's  interest,  109.30,  for  6  months, 
uarter's  iuterest,  109.30,  for  3  mouths, 
uarter's  interest,  109.30,  I'or  0  months, 
interest  on  quarterly  iuterest, 


4372.00 
874.40 


S19.13 

16.40 

13.06 

10.93 

8.20 

5.46 

2.73 


Total  amount  due  at  maturity. 


76.51 
$5322.91 


IjoTP,. — Instead  of  finding  the  interest  on  each  quarter's  interest  separately,  we  could  have 
added  tbe  months  of  time  together,  making  84,  and  then  found  tbe  interest  on  1  quarter's  iuterest 
lor  the  whole  time,  84  months  or  7  years. 

A  Semi- Annual  Interest  Due  Bill. 

$GC0.  Natchez,  October  15,  1895. 

Due  A.  &  M.  Moses,  for  value  received.  Six  Hundred  and  Sixty  Dollars,  with 

interest  at  6  per  cent  payable  semi-annually. 

Fowler  &  Davis. 


INTEREST. 


609 


3.    Nothing  being  paid  on  tlie  above  due  bill  until  tLe  day  of  settlement,  two 
years  4  months  and  20  days  after  date,  what  was  then  due  ?  Ans.  $760.01. 

OPERATION. 

Face  of  due  bill, $660.00 

Interest  on  same  for  2  years,  4  moutbs  and  20  days  at  6  per  cent, 94.60 

(The  interest  on  the  face  of  due  bill  for  6  months  is  $19.80.) 
Interest  on  1st  semi-annual  interest,  $19.80,  for  22  months  20  days. 
Interest  on  2d    semi-aunnal  interest,    19.80,  for  16  months  20  days. 
Interest  on  3d    semi-annual  interest,    19.80,  for  10  months  20  days. 
Interest  on  4th  semi-annual  interest,    19.80,  for   4  months  20  days. 

Interest   on   $19.80    semi-annual    interest    for    54  months  20  days  at  6  per  cent  =  -  5.4t 

Total  amount  due  on  settlement,  ---.......        $760.01 


COMPOUND    INTEREST. 


1177.  Compound  interest  is  interest  in  which  the  principal  is  increased  at  the 
expiration  of  each  period  of  interest  payment  by  the  interest  on  the  principal  for 
such  period.  The  amount  thus  produced  forming  a  new  principal  on  which  tho 
interest  is  calculated  till  the  next  yeriod  of  interest  payment,  when  the  new  princi. 
pal  is  increased  by  the  addition  of  the  last  interest,  and  so  on  from  one  period  of 
interest  payment  to  another,  until  final  settlement  is  made. 

The  periods  of  time  for  the  interest  payments  maj-  be  annually,  semi-annually 
quarterly,  monthly,  weekly,  or  daily. 

There  is  no  law  forbidding  the  charging  and  receiving  of  compound  interest, 
provided  the  persou  to  whom  the  money  was  loaned  is  willing  to  pay  it,  but  it  cannot 
be  legally  collected  unless  there  had  been  a  previous  agreement  to  that  effect. 

1.    M'hat  is  the  compound  interest  of  $9000  for  4  years  at  8  per  cent? 

Ans.  $3244.40, 

FIRST   OPfiRATION.  SECOND   OPERATION. 

Principal,  .        -        -        .        .  $9000.00     Principal  for  the  1st  year, 

Interest  on  principal  for  1  year  at  8%, 


New  principal  for  the  2d  year. 
Interest  on  same  for  1  year  at  8%, 

New  principal  for  the  3d  year. 
Interest  on  same  for  1  year  at  S%, 

New  principal  for  the  4th  year, 
Interest  on  same  for  1  year  at  8%, 

The  amount  for  4  years  is        -        - 
First  principal  deducted, 


$9000.00 
720.00 

$9720.00 
777.60 

$10497.60 
839.80 

$11337.40 
907.00 

$12244.40 
9000.00 


Principal  for  the  2d  year. 


Principal  for  the  3d  year. 


Principal  for  the  4th  year. 


$9000.00 
1.08 

$9720.00 
1.08 

$10497.60 
1.08 

$11337.4080 
1.08 


Amount  at  the  end  of  the  4th  year,  $12244.400640 
First  principal  deducted,  -  9000.00 


Compound  interest  for  4  years,       -  $3244.40    Compound  interest,     -        -        -  $3244.40 

Explanation. — The  operation  of  this  problem  is  so  clear  that  an  explanation  is  almost  unnec- 
essary. We  first  find  the  interest  on  the  principal  for  1  year,  and  add  it  to  the  same,  which  pro- 
duces a  new  principal  for  the  second  year's  interest;  and  in  lilie  manner  we  continue  till  the  end 
of  the  4  years;  then  we  deduct  the  first  principal  from  the  amount  and  in  the  remainder  have  the 
compound  interest  for  4  years  at  8  per  cent. 

In  the  2d  operation,  we  first  assume  fl  and  then  find  the  amount  of  the  same  for  1  year  at  8 


6io 


SOULE  3    PIIILOSOI'UIC    TKACTICAL    MATHEMATICS. 


per  cent,  -nliich  is  $1.08.  Then,  liaviiig  tlie  anionnt  of  $1,  we  find  the  .imoniit  of  $9000  liy  ninUi- 
plying  the  .same  by  the  $9000,  and  thus  we  continue  to  iMuItijiIy  tlie  -SI. 08  by  the  successive  amounts 
produced  until  the  close  of  the  time.  Then  to  lind  the  interest  we  deduct  the  first  principal  from 
the  last  amount. 

Had  we  raised  the  ratio  $1.08  to  the  fourth  power  and  then  multiplied  bv  the   $9000,    we 
would  have  produced  the  same  result.     Thus,  $1.0«x4'1.08x$1.08x$l.U8x$yOOO  =  *i2244.40064000. 

2.    What  is  the  coinpouud  interest  of  $500  for  1  year  and  C  mouths  at  6  per 
cent  payable  quarterly  ?  Ans,  $46.72. 

.  OPERATION. 


Principal, $1500.00 

Interest  ou  same  for  3  months  at  6"^,  7.50 

New  principal  for  2d  quarter,       -        -  $507. .50 

Interest  ou  same  for  2d  quarter,           -  7.61 

New  principal  for  3d  quarter,       -         -  $515.11 

Interest  ou  same  for  3d  quarter,           -  7.73 

New  principal  for  4th  quarter,             -  $522.84 

Interest  on  same  for  4th  quarter,         -  7.84 

New  principal  for  5th  quarter,     -        -  $530.68 


Principal  for  5th  quarter  hroiii/lit  for'd,  $530. 6S 

Interest  on  same  tor  5th  quarter,         -  7.96- 

New  principal  for  6th  quarter,     -         -  $538.64 

Interest  ou  same  for  6th  quarter,          -  8. OS 

The  amount  for  1  year  and  6  months  is  $546.7^ 

First  iirincipal  deducted,      -        .         -  500.00 

Compound  int.  for  1  year  and  6  months,  $46.72 


3.     What  is  the  compound  interest  of  $1000  for  2  years  3  months  and  5  days 
at  10  per  cent,  payable  semiannually  ?  Ans.  $247.58. 


OPERATION. 


Princiiial, $1000.00 

Int.  ou  the  same  for  6  months  at  10%,  50.00 

New  principal  for  the  2d  6  months,  $10.50.00 
Interest  on  same  for  6  months  at  10%,  52.50 

New  principal  for  the  3d  6  months,  $1102.50 
Interest  on  same  lor  6  months  at  10%,  55.12^ 

New  principal  for  the  4th  6  months,  $1157. 62J 


Prill,  for  4tli  6  months  Irovght  for'd.  $1157.62* 

Interest  ou  same  for  6  months  at  10%,  57.884 

New  principal  for  3  mos.  and  5  days,       $1215  50J 
Int.  ou  same  for  3  mos.  and  5  ds.  at  10%,       32.08 

Amount  for  2  years  3  months  and  5  ds.    $1247.58 
First  i)rincipal  deducted,  -         -  1000.00 

Compound  int.  for  2  yrs.  3  mos.  and  5  ds.    $247.58 


4.     What  will  $1450  amount  to  at  compound  interest  for  1  year  2  months  and 
12  days  at  6  per  cent  compounded  monthly  1  Ans.  $1557.97. 

In  practice,  the  finding  of  compound  interest  is  rendered  more  expeditious  by 
the  use  of  the  following  Compound  Interest  Table  : 

Note. — To  find  the  sum  of  the  amounts  of  compound  interest  in  the  t.ablcs  from  1  year  to 
any  other  year  inclusive,  without  makiug  the  additions,  take  the  compound  interest  for  the  next 
year  greater  than  the  highest  given,  at  the  same  per  cent,  and  subtract  from  this  the  amount  of  1 
year  at  the  head  of  the  table  in  the  same  column  ;  then  remove  the  decimal  point  two  places  to  the 
right,  and  divide  by  the  rate  per  cent;  considered  as  an  integer,  the  quotient  will  equal  the  sum 
of  all  the  amounts  in  the  column  from  1  year  to  the  given  year  inclusive. 


INTEREST. 


6ii 


1178. 


A  IVI  O  U  N  T 


OF  SI,  Jt!l,  1  Fl!.,  AT  COMrOlND  INTEKF-ST  IX  ANV  NVMllEl!  OF  YEAES,  NOT  EXCEEDING  FIFTY-FIVl 


Yrs. 

1  per  cent. 

IJ  per  cent. 

2  per  cent. 

2 J  per  cent. 

3  per  cent. 

3i  per  cent. 

4  per  cent. 

1 

1.0100 

1.01,50 

1.0200  0(YIO 

1.02.50  0000 

1.0300  0000 

1.0350  0000 

1.0400  0000 

9 

1.0201 

1.0302 

1.0404  0000 

1.05(16  2.500 

1.0609  OCCO 

1.0712  2500 

1.081()  0(100 

3 

1 .0303 

1.0457 

1.0612  08(0 

1.0768  9062 

1.0927  2700 

1.1087  1787 

1.1248  6400 

4 

1.0406 

1.0614 

1.0824  3216 

1.1038  1289 

1.12,55  0881 

1.1475  2300 

1.1  (!98  58,56 

5 

1.0510 

1.0773 

1.1040  8080 

1.1314  0821 

1.1592  7407 

1.1876  8631 

1.2166  5290 

6 

1.0615 

1.0934 

1.1261  6242 

1.1.596  9342 

1.1940  5230 

1.2292  5.533 

1.2653  1902 

7 

1.0721 

1.1098 

1.1486  8567 

1.1886  8.575 

1.2298  7387 

1.2722  7926 

1.31.59  3178 

8 

1.0829 

1.1265 

1.1716  5938 

1.2184  0290 

1.26(i7  7008 

1.3168  0904 

1.3685  6905 

9 

1.0937 

1.14.34 

1.19,50  9257 

1.2488  6297 

1.3047  7318 

1.3628  9735 

1.4233  1181 

10 

1.1016 

1.1605 

1.2189  9442 

1.2800  8454 

1.3439  1638 

1.4105  9876 

1.4802  4428 

11 

1.1157 

1.1779 

1.24,33  7431 

1.3120  8666 

1.3842  3387 

1.4.599  6972 

1..5394  5406 

12 

1.1268 

1.19,56 

1.2682  4179 

1.3448  88,82 

1.4257  8089 

1.5110  6866 

1.6010  3222 

13 

1.1381 

1.2136 

1  2936  061  i3 

1  3785  1104 

1.4685  3371 

1  5639  5606 

1.66.50  7351 

14 

1.1195 

1.2318 

1.3194  7876 

1.4129  7382 

1.5135  8972 

1.61,^6  9452 

1.7316  7645 

15 

1.1610 

1.2502 

1.3458  6834 

1.4482  9817 

1.5579  6742 

1.6753  4883 

1.8009  4351 

16 

1.1720 

1.2690 

1.3727  8570 

1.4845  0562 

1.6047  0644 

1.7339  8601 

1.8729  8125 

17 

1.1843 

1.2880 

1.4U02  4142 

1.5216  1826 

1  6528  4763 

1.7946  7555 

1.9479  0050 

18 

1.1961 

1.3073 

1.4282  4625 

1.5,596  5872 

1.7024  3306 

1.8574  8920 

2.0258  l(;5-2 

I'J 

1.2081 

1.3270 

1.4.568  1117 

1..5986  5019 

1.7535  0005 

1  9225  0132 

2.1068  4918 

20 

1.2202 

1.3469 

1.4859  4740 

1.6386  1644 

1.8061  1123 

1.9897  8886 

2.1911  2314 

21 

1.2.324 

1.3671 

1.51.56  6634 

1.6795  8185 

1.8602  94.57 

2.0594  3147 

2.2787  6807 

22 

1.2117 

1.3876 

1.5459  7967 

1.7215  714(1 

1.9161  0341 

2.1315  1158 

2,3699  1879 

23 

1.2.572 

1.4084 

1.5768  9926 

1.7646  1(168 

1.9735  8(i51 

2.2061  1448 

2,4647  15.55 

24 

1.2i;97 

1.4295 

1.6084  3725 

1.8087  2595 

2.0327  9411 

2.2833  2849 

2.5633  0417 

25 

1.2824 

1.4,509 

1.6406  0599 

1.8539  4410 

2.0937  7793 

2.8632  4498 

2.6658  3633 

26 

1.2953 

1.4727 

1.6734  1811 

1.9002  9270 

2.1.565  9127 

2.4459  5856 

2.7724  6979 

27 

1.3082 

1.4948 

1.7068  8648 

1.9478  0002 

2.2212  8901 

2.5315  6711 

2.8833  6858 

28 

1.3213 

1.5172 

1.7410  2421 

1.9964  9502 

2.2879  27(18 

2.6201  7196 

2.9987  0332 

21t 

1.3345 

1.5400 

1.7758  4469 

2.0464  07:!9 

2.3.565  6.551 

2.7118  7798 

3.1186  5M5 

30 

1.3478 

1.5631 

1.8113  6158 

2.0975  6758 

2.4272  6247 

2.8067  9370 

3.2433  9751 

31 

1  3613 

1.5865 

1.8475  8882 

2.1500  0077 

2.5000  8035 

2.9050  3148 

3.3731  3341 

32 

1.3719 

1.6103 

1.8845  4059 

2.2037  5694 

2.57,50  8276 

3,0067  0759 

3.. 5080  5875 

S3 

1.3887 

1.6345 

1.9222  3140 

2  2588  5086 

2.6523  3524 

3.1119  4235 

3.6483  8110 

31 

1.4026 

1.6590 

1.9606  7603 

2.31,53  2213 

2.7319  0530 

3.2208  6033 

3.7943  1634 

35 

1.4166 

1.6839 

1 .9998  8955 

2.3732  0519 

2.8138  6245 

3.3335  9045 

3.9400  8899 

36 

1.4308 

1.7091 

2.0398  8734 

2  4325  3532 

2.8982  7833 

3.4.502  6611 

4.1039  3255 

37 

1.4451 

1.7348 

2.0806  8509 

2.4933  4870 

2.9852  26(18 

3.5710  2.543 

4.21.180  8986 

38 

1.4595 

1.7608 

2.1222  9879 

2.5.556  8242 

3.0747  8348 

3  6960  1132 

4.4388  1345 

39 

1.4741 

1.7872 

2.1647  4477 

2.6195  7448 

3.1670  2698 

3.82.53  7171 

4.6163  6.599 

40 

1.4889 

1.8140 

2.2080  3966 

2.6850  6384 

3.2620  3779 

3.9592  5972 

4.8010  2063 

41 

1.5038 

1.8412 

2.2522  0046 

2.7521  9043 

3.3.598  9893 

4.0978  3381 

4.9930  6145 

42 

1.5188 

1.8688 

2.2972  4447 

2.8209  9520 

3.4606  9589 

4.2412  5799 

5.1927  8391 

43 

1.5340 

1.8969 

2  3431  8936 

2.8915  2(108 

3.5645  1677 

4.3,X97  0202 

5.4004  9527 

44 

1.5493 

1.9253 

2.3900  5314 

2.9638  0.808 

3.6714  5227 

4.5433  4160 

5.6165  1.508 

45 

1.5648 

1.9542 

2.4378  5421 

3.0379  0328 

3.7815  9584 

4.7023  5855 

5.8411  7568 

46 

1.5805 

1.9835 

2.4866  1129 

3.1138  5086 

3.89.50  4372 

4.8669  4110 

0.0748  2271 

47 

1..5963 

2.0133 

2.5363  4351 

3.1916  9713 

4.0118  9503 

5.0372  8404 

6.3178  1562 

48 

1.6122 

2.0435 

2.5870  7039 

3.2714  8956 

4.1322  5188 

5.2135  8898 

6.5705  2824 

49 

1.6283 

2.0741 

2.6388  1179 

3  3532  7680 

4.2.562  1934 

5.3960  6-1.59 

6.8333  4937 

50 

1.6446 

2.1052 

2.6915  8803 

3.4371  0872 

4.3839  0602 

5.5849  2686 

7.1066  8335 

51 

2  7454  1979 

3  Tt^?,0  3644 

4.5154  2320 
4.6508  8590 
4.7904  1247 
4.9341  2485 
5.0821  48.59 

5.7803  99.30 
5.9827  1327 
6.1921  0824 
6.4088  3202 
6.6331  4114 

7.3909  5068 
7.6865  8871 
7.9940  5226 
8.3138  1435 
8.6463  6692 

52 

2  8003  2819 

3  111  1 1    1''35 

53 

2  8563  3475 

3  7013  9016 

54 

2  91,34  6144 

3  7939  2491 

55 

2.9717  3067 

3.8887  7303 

Subtract  f  1  from  (be  amount  iu  this  table  to  find  the  interest. 


6l2 


1179. 


rillLOSOPHlC    PRACTICAL    MATHEMATICS. 


AMOUNT 


OF  $1,  £.  1,  Oil  1  Fit.,  AT  COMl'Ol'Nr)  IKTKHEST  IN  ANY  NUMBKU  OF  YF.AllS,  NOT  KXCEFDIXG  FIFTV-FIVK.     i 


Yr.s. 

4  A  per  ccut. 

5  per  cent. 

G  per  cent. 

7  per  cent. 

8  per  cent. 

9  per  cent. 

10  per  cent. 

1 

1.0450  0000 

1.0500  000 

1.0600  000 

1.0700  000 

1.0800  000 

1.0900  000 

1.1000  000 

2 

1.0920  2500 

1.1025  000 

1.1236  000 

1.1449  000 

1.1664  000 

1.1881  000 

1.2100  000 

3 

1.1411  6612 

1.1576  250 

1.1910  160 

1.2250  430 

1.2597  120 

1.2950  290 

1.3310  000 

4 

1.1925  1860 

1.21.55  063 

1.2624  770 

1.3107  960 

1.3604  8896 

1.4115  816 

1.4641  000 

5 

1.246x  8194 

1.2702  816 

1.3382  256 

1.4025  517 

1.4693  281 

1.5386  240 

1.6105  100 

6 

1.3022  6012 

1.3400  956 

1.4185  191 

1.5007  304 

1.5868  743 

1.6771  001 

1.7715  610 

7 

1.3608  6183 

1.4071  004 

1.5036  303 

1.6057  815 

1.7138  243 

1.8280  391 

1.9487  171 

8 

1.422i  0061 

1.4774  554 

1.5938  481 

1.7181  862 

1.8509  302 

1.9925  626 

2.1435  888 

9 

1.4860  9514 

1.5513  282 

1.6894  790 

1.8384  592 

1.9990  046 

2.1718  933 

2.3579  477 

10 

1.5529  6942 

1.6288  946 

1.7908  477 

1.9671  514 

2.1589  250 

2.3673  637 

2.5937  4i5 

11 

1.6228  5305 

1.7103  394 

1.8982  986 

2.1048  520 

2.3316  390 

2.5804  264 

2.8531  167 

12 

1.6958  8143 

1.7958  563 

2.0121  965 

2.2521  916 

2.5181  701 

2.8126  648 

3.1384  284 

13 

1.7721  9610 

1.8856  491 

2.1329  283 

2.4098  450 

2.7196  237 

3.0658  046 

3.4522  712 

14 

1.8519  4492 

1.9799  316 

2.2609  040 

2.5785  342 

2.9371  936 

3.3417  270 

3.7974  983 

15 

1.9352  8244 

2.0789  282 

2.3965  582 

2.7590  315 

3.1721  691 

3.6424  825 

4.1772  482 

16 

2.0223  7015 

2.1828  746 

2.5403  517 

2.9521  638 

3.42.59  426 

3.9703  059 

4.5949  730 

17 

2.1133  7681 

2.2920  183 

2.6927  728 

3.1588  152 

3.7000  181 

4.3276  334 

5.0544  703 

18 

2.2084  7877 

2.4066  192 

2.8.543  392 

3.3799  323 

3.9960  195 

4.7171  204 

5.5.599  173 

19 

2.3078  6031 

2.5269  502 

3.0255  995 

3.6165  275 

4.31.57  Oil 

5.1416  613 

6.1159  390 

20 

2.4117  1402 

2.6532  977 

3.2071  355 

3.8696  845 

4.6609  571 

5.6044  108 

6.7275  000 

21 

2.5202  4116 

2.7859  626 

3.3995  636 

4.1405  624 

5.0338  337 

6.1088  077 

7.4002  499 

22 

2.6336  5201 

2.9252  607 

3.6035  374 

4.4304  017 

5.4365  404 

6.6586  004 

81402  749 

23 

2.7521  6635 

3.0715  238 

3.8197  497 

4.7405  299 

5.8714  637 

7.2578  745 

8.9543  024 

24 

2.8760  1383 

3.2250  999 

4.0489  346 

5.0723  670 

6.3411  807 

7.9110  832 

9.8497  327 

25 

3.0054  3416 

3.3863  540 

4.2918  707 

5.4274  326 

6.8484  752 

8.6230  807 

10.8347  059 

26 

3.1406  7901 

3.5556  727 

4.5493  830 

5.80T3  529 

7.3963  532 

9.3991  579 

11.9181  765 

27 

3.2820  0956 

3.7334  563 

4.8223  459 

6.2138  676 

7.9880  615 

10.2450  821 

13.1099  942 

28 

3.4296  9999 

3.9201  291 

5.1116  867 

6.6488  384 

8.6271  064 

11.1671  395 

14.4209  936 

29 

3.5840  3649 

4.1161  356 

5.4183  879 

7.1142  571 

9.3172  749 

12.1721  821 

15.8630  9:;o 

30 

3.7453  1813 

4.3219  424 

5.7434  912 

7.6122  550 

10.0626  569 

13.2676  785 

17.4494  023 

31 

3.9138  5745 

4.5380  395 

6.0881  006 

8.1451  129 

10.8676  694 

14.4617  695 

19.1943  425 

32 

4.0899  8104 

4.7649  415 

6.4533  867 

8.7152  708 

11.7370  830 

15.7633  288 

21.1137  7I1.S 

33 

4.2740  3018 

5.0031  885 

6.8405  899 

9.3253  398 

12.6760  496 

17.1820  284 

23.2251  514 

34 

4.4663  61.54 

5.2533  480 

7.2510  253 

9.9781  185 

13.6901  336 

18.7284  109 

25.5476  6!t;i 

So 

4.6673  4781 

5.5160  154 

7.6860  868 

10.6765  815 

14.7853  443 

20.4139  679 

28.1024  369 

36 

4.8773  7846 

5.7918  161 

8.1472  520 

11.4239  422 

15.9681  718 

22.2512  2,50 

30.9126  805 

37 

5.0968  6049 

6.0814  069 

8.0360  871 

12.2236  181 

17.24.56  256 

24!2538  353 

34.0039  486 

38 

5.3262  1921 

6.3854  773 

9.1.542  524 

13.0792  714 

18.6252  756 

26.4366  805 

37.4043  434 

39 

5  5658  9908 

6.7047  512 

9.7035  075 

13.9948  204 

20.1152  977 

28.8159  817 

41.1447  7T.X 

40 

5.8163  6454 

7.0399  887 

10.2857  179 

14.9744  578 

21.7245  215 

31.4094  200 

45.2592  556 

41 

6.0781  0094 

7.3919  882 

10.9028  610 

16.0226  699 

23.4624  832 

34  2362  679 

49.7851  811 

42 

6.3516  1548 

7.7615  876 

11.5570  327 

17.1442  568 

25.3394  819 

37.3175  320 

54.7636  992 

43 

6.6374  3818 

8.1496  669 

12.2504  546 

18.3443  548 

27.3666  404 

40.6761  098 

60.2400  692 

44 

6.9361  2290 

8.5.^71  503 

12.98.54  819 

19.6284  596 

29.5.-09  717 

44.3369  597 

66.2640  761 

4. J 

7.2482  4843 

8.9850  078 

13.7646  108 

21.0024  518 

31.9204  494 

48.3272  801 

72.8904  837 

4G 

7.5744  1961 

9.4342  582 

14.,5904  875 

22.4726  234 

34.4740  853 

52.6767  419 

80.1795  321 

47 

7.9152  6849 

9.9059  711 

15.4659  167 

24.0457  070 

37.2320  122 

57.4176  486 

88.1974  8.^.3 

48 

8.2714  5557 

10.4012  697 

16.3938  717 

25.7289  065 

40.2105  731 

62.5852  370 

97.0172  33S 

49 

8.6436  7107 

10.9213  331 

17.3775  040 

27.5299  300 

43.4274  190 

68.2179  083 

106.7189  572 

50 

9.0326  3627 

11.4673  998 

18.4201  543 

29.4570  251 

46.9016  125 

74.3575  201 

117.3908  529 

51 

9.4391  0490 

12.0407  698 

19.5253  635 

31.5190  168 

50.6.537  415 

81.0496  969 

129.1299  382 

52 

9.8638  6463 

12.6428  083 

20.6968  853 

33.7253  480 

54.7060  408 

88.3441  696 

142.0429  320 

53 

10.3077  3853 

13.2749  487 

21.9386  985 

36.0861  224 

.59.0825  241 

96.2951  449 

1.56.2472  252 

54 

10.7715  8677 

13.9368  961 

23.2550  204 

38.6121  509 

63.8091  260 

104.9617  079 

171.8719  477 

55 

11.2563  0817 

14.63,^6  309 

24.6.503  216 

41.31.50  015 

68.9138  561 

114.4082  616 

189.0591  42.-. 

Subtract  $1  fiom  tke  amount  iu  this  table  to  find  the  interest. 


*  COMPOUND    INTEREST.  '  613 

TO  COMPUTE  COMrOUND  INTEEEST  BY  TUE  USE  OF  THE  COMPOUND 

INTEREST   TABLE. 

1180.     To  find  the  amount  or  interest  ulien  the  principal,  time  and  rate  are 
given : 

1.     What  is  the  compound  interest  on  $9000  I'or  4  years  at  8  per  cent? 

Aus.  $3244.40. 

OPERATION.  Explanation.— To  find 

the  eompouiul  iuterest  by 

$1.3G04S9        =  compound  amount  of  $1  tor  4  years.  1!:^;^  tbll^.'^Crilni 

StOOO  =  principal.  in  the  table  the  comimuul 

iimount  of  $1  for  the  time 


$12244.401000  =  compound  amount  of  $9000  for  4  years.  :"i.<l  rate,  th.n  nmltiply 
qoOO  —  iirincimi  ^^^^  amount  by  the  prin- 

^"^'"-  —  ptincipai.  cipal,  and  from  the  prod- 

iictsubtracttheprincipal; 

$   3244.40  =  compoand  interest.  the  remainder  will  bo  the 

compound  iuterest. 

KoTE  1. — If  the  interest  is  roniponnded  semi-aiiTinally,  then  find  the  compound  amouut  on  §1 
fur  twi.-e  tbs  imiiiber  of  years  and  for  |  the  rate  per  cent. 

Tbti,'<.  the  compound  interest  on  $1  for  8  years  at  6  per  cent  payable  semi-annuaHy,  is  the  same 
as  the  compound  interest  on  $1  for  16  years  at  3  per  cent,  paijahle  annually. 

Note  2. — If  the  interest  is  compounded  quarterly,  then  find  the  compound  amount  on  $1  for 
4  times  the  number  of  years  and  for  i  the  rate  per  cent. 

Thus,  the  compound  interest  on  $1  for  6  years  at  8  per  cent,  payable  quarterly,  is  the  same  as 
the  comjiound  interest  on  $1  for  2-t  years  at  2  per  cent,  imijable  annually. 

Note  3. — When  compounded  naontbly,  consider  each  mouth  a  year  and  divide  the  rate  jter 
cent  by  12. 

Note  4. — When  tlie  amount  is  reqnircd  for  a  number  of  years  not  contained  in  the  table, 
compute  the  same  by  multiplying  together  the  amonnts  of  any  two  number  of  years  whose  sum 
equals  the  number  of  years  reijuired.  Thus,  if  the  amount  of  86  years  at  4  per  cent  is  required, 
multiply  together  the  compound  amount  of  46  and  40  years,  or  36  and  50  years:  For  example,  the 
amount  for  46  years  is  6.(i748l'+  and  for  40  years  it  is  4.80102+  and  the  product  is  29. 16ti3323-J- 
■which  is  the  amount  of  $1  for  86  years  at  4  per  cent. 

Note  5. — If  partial  payments  are  made  on  notes  or  debts  bearing  compound  interest,  the 
compound  amount  of  the  principal  must  be  first  found,  and  the  sum  of  the  comiiouud  amounts  of 
the  payments  subtracted  from  it. 

I  2.     "What  is  the  compound  interest  on  $800  for  6  years,  4  months  at  4J  per 

cent?  Alls,  $257.43. 

OPERATION. 

■      $1.30220012      =  compound  amount  of  $1  for  6  years  at  4J  per  cent. 
SdO  =  iirincipal. 


$1041.80809000  =  compound  amount  of  $800  for  6  years. 

15.G275  =  interest  on  $1041.81  for  4  months,  at  4J  per  cent. 

$1057.43  =  compound  amount  on  $800  for  6  years,  4  months  at  4J  per  cent. 

800.  =  principal. 


$257.43  =  compound  interest  on  $800  for  C  years,  4  months  at  4J  per  cent. 

Note — AVhen  the  time  includes  months  or  days,  the  interest  for  such  number  of  months 
or  days  is  computed  on  the  compound  amouut  for  the  even  years,  and  added  thereto,  as  shown 
above. 


6 14  ■      soule's  philosophic  practical  mathematics.  *■ 

3.     What  is  tlie  coinpoiiiKl  interest  ou  $5000  for  12  years,  1  moiitli,  12  clays  at 
10  |)cr  cent  eoiiipouiided  quarterly  V  Ans.  $10548.2!)  eompouiid  ainouiit. 

$11548.29  coiiipoiiiul  interest. 
operation. 

$3.27148956        =  coinpoiind  amount  of  $1  for  48  years  at  2J  per  cent. 
5000  =  ])rineipal. 


$10357.44780000        =  compound  amount  of  $5000  for  12  years  at  10  per  cent. 

190.84  r=  interest  on  compound  amount  of  $10357.45  at  lo^o  i'ov  42  ds., 

■ or  for  108  ds.  at  2^%. 

$10548.29  =  compound  amount. 
5000.00  =  principal  deducted. 


$11548.29  =  compound  interest. 

4.     What  is  the  compound  interest  ou  $3000  for  20  years,  3  montbs,  16  days 
at  3  per  cent,  payable  seuiiannually?  •  Ans.  $2490.07. 

TO  FIND  THE  PRINCIPAL  WEEN  THE  COMPOUND  AMOUNT  Oil  INTER- 
EST, THE  TIME,  AND  THE  RATE  ARE  GIVEN. 

1181.  1.     Wliat  principal  at  8  per  cent  compound  interest   ■will   amount   to 
$12244.401,  in  4  years?  Aus.  $9000. 

OPERATION. 

$12244.401  -  $1.300489  =  $9000. 

lixplnnntion. — The  componnd  amount  of  $1  for  4  years  at  8  per  cent  is  S1.3C0489  and  lience, 
$12244.401  IS  the  couipounil  amount  of  as  many  dollars  as  it  is  ecjual  to  $1.360489,  which  is  $9000. 

Therefore,  in  all  problems  of  this  kind  to  find  the  ])rincipal,  we  divide  the  amount  or  interest 
by  the  amount  or  interest  of  |1  i'or  the  given  time  and  rate. 

2.     What  principal  at  4  per  cent  compound   interest   will   produce   $10000 
interest  in  9  years  ?  Ans.  $23623.25. 

OPERATION    indicated. 

$10000  -^  .42331181. 

Note— The  $.42331181  is  the  compound  interest  on  |ll  for  9  ye.ars  at  4  per  cent.  See  table 
acd  subtract  $1  from  compound  amount. 

TO  FIND   THE   TIME,   WHEN   THE   PRINCIPAL,   THE   RATE   AND   THE 
COMPOUND  AMOUNT  OR  INTEREST  IS  GIVEN. 

1182.  1.     In  what  time  will  $9000  amount  to  $12244.401  at  8  per  cent! 

Ans.  4  years. 

OPERATION. 

$12244.401 -H  $9000  =  $1.300489  =  compound  amount  of  $1  which,  iu  the  table  cor- 
responds to  4  years. 

Explanation. — Since  $12244.401  is  the  componud  amount  of  $9000  at  8  per  cent  in  a  certain 


*  COMPOUND    INTEREST.  6l5 

time,  tbe  amount  of  $1  -nill  l>e  tlie  OOOOtli  part,  -nlueli  is  $1.3G04S9  and   which  we  linil  in  the  table 
under  8  i)er  cent  to  correspond  to  4  years. 

Tlierefore,  in  all  ])rol)lems  of  tins  kind  to  find  the  time,  we  divide  the  amonnt  by  the  jriven 
principal  and  in  the  ijnotient  we  have  the  amoniit  of  $1  for  the  tmie.  Then  we  lind  this  amount  iu 
the  table  nnder  the  gi\'en  rate,  and  Ojiposite  thereto  is  the  requiretl  time. 

2.     Ill  wliiit  time  will  $'2000  amonnt  to  $45S4.03GG  at  5  per  cent  compound 
interest?  Aus.  17  years. 

TO  FIND  THE  HATE  WHEN  TDE  PRINCIPAL,  TEE  COMPOUND  AMOUNT 
OK  INTEREST,  AND  THE  TIME  ARE  GIVEN. 

1183.     1.     At  what  rate  will  $9000  amount  to  $12244.401  in  4  years? 

Aus.  Sfo- 

OPERATION. 

$12244.401  ^  $9000  =  $1.300489,  winch  ni  the  table  corresponds  to  8  per  cent. 

Explanation.— Smce  S9000  amounts  to  $1224  1. 101  in  4  years  at  the  reijuired  rate,  $1  for  the  same 
time  and  rate  will  amonnt  to  the  9UU0th  part,  which  is,  Sl.oGU48'J,  and  which,  in  the  table  o]iposito 
4  years  corresponds  to  8  per  cent. 

Therefore,  in  all  jirohlems  of  this  kind  to  find  the  rate,  we  divide  the  amount  by  the  giveu 
principal  and  the  quotient  will  be  the  amount  of  $1  for  the  given  time  at  the  re(|uired  rate. 

Then  wo  refer  to  the  talde  and  tind  this  amount  ojiposite  the  given  time,  and  the  rate  at  the 
top  of  the  column  is  the  required  rate. 

2.     At  ^vhat  rate  will  $8000  amount  to  $22072.252  in  15  years  ?       Aiis.  7%. 

NOTK. — See  the  subject  of  Annuities  for  further  work  involving  the  jirinciples  of  simple  and 
compound  interest  investments. 

SIMPLE,  ANNUAL,  AND  COMPOUND  INTEREST  COMPARED. 

1184.     What  is  the  xinqAv,  the  annual,  and  the  coniponnd  interest  on  $5000,  for 
4  years  at  8  per  cent?  Ans.  $1600,  simple  interest. 

$1792,  annual  iuteix'st. 
$1802.45,  compound  interest. 


OPERATION 
FOK  SIMPLK  INTEREST. 

$50.00 
_«% 

$400.00 

4   yrs. 

$1600.00  Ans. 


OPERATION   I'OR  ANNUAL  INTEREST. 

Interest  on  the  $5000  principal  for  4  years  at  8%,  -        -        $1000 

(The  annual  interest  is  $400  I'er  year;. 
Interest  ou  $400  I'or  3  years, 
Interest  on    400  for  2  years, 
Interest  on    400  for  1  year,  = 

the  int.  on  $400  for  6  years  at  8  per  cent,      -  ...  jg2 

$1792 


OPERATION  KOU  CO.MPOUND   INTEREST   BY  THE   COMPOUND   INTEREST  TABLE. 

$1.36048896        =  compound  amonnt  of  $1  for  4  years  at  8  jier  cent. 
5000  =  principal. 

$6802.4480000  =r  compound  amount  of  $5000  for  4  years  at  8  per  cent. 

5000.  =  principal  deducted. 

12.45  =  compound  interest  on  $5000  for  4  years  at  8  per  cent. 


6i6  soule's  philosophic  practical  mathematics.  * 

By  the  foregoing  -work  we  see  that  the  difference  between  simple,  annual,  and 
compound  interest  in  their  effect,  depends  upon  the  time  when  interest  money,  if 
due  and  not  paid,  begins  to  draw  interest.  As  remarked  in  annual  interest,  a  debt 
begins  to  draw  interest  when  due;  but  the  time  when  a  debt  of  interest  becomes 
due,  is  a  question  in  some  cases  governed  by  business  custom,  and  in  other  cases  it 
is  a  question  to  be  decided  by  the  parties  interested.  In  bank  discount,  the  interest 
is  payable  in  advance. 

In  simple  interest,  it  is  not  considered  due  until  the  iinal  payment  of  the 
principal. 

In  annual  interest,  it  is  due  after  it  has  been  accruing  for  the  time  specified 
for  interest  payments,  except  the  interest  on  interest  which  is  not  due  till  final  pay- 
ment. 

Compound  interest  supposes  all  interest,  whether  upon  principal  or  interest, 
to  be  due  at  the  end  of  such  equal  successive  intervals  of  time  as  may  be  agreed 
upon. 

When  the  interest  is  considered  due  the  instant  it  has  accrued  and  all  inter- 
est is  made  to  draw  interest,  it  is  termed  instantaneous  compound  interest. 

PARTIAL  PAYMENTS,  OR  PAYMENTS  BY  INSTALLMENTS. 

1185.  Partial  payments,  or  payments  by  installments  as  the  terms  imply,  are 
the  payments  made  at  different  times,  of  promissory  notes,  acceptances,  bonds, 
personal  accounts,  or  other  obligations. 

At  the  time  these  partial  payments  are  made,  the  creditor  should  and  usually 
does  specify  in  writing,  on  the  back  of  the  note  or  other  instrument,  the  sum  paid 
and  the  date  of  payment,  and  then  signs  his  name. 

It  is  a  principle  of  law  that  a  note  does  not  draw  interest  until  it  is  due, 
unless  the  words  "  with  interest"  are  contained  therein. 

There  are  several  systems  of  computing  interest  upon  notes,  bonds,  personal 
accounts,  and  other  obligations  upon  which  partial  payments  have  been  made;  and 
as  the  different  systems  generally  give  diflerent  results,  according  as  the  time 
intervening  between  the  payments  is  more  or  less  than  a  year,  or  the  interest  more 
or  less  than  the  payment,  it  is  therefore  impossible  in  advance  of  knowing  the  exact 
facts  of  each  case,  to  say  which  system  would  be  strictly  just  to  all  parties. 

Among  these  various  systems,  the  five  following  are  mostly  used  :  The  United 
States  System;  the  Mej-chnnts^  System;  the  Vermont  System;  the  New  Hampshire 
System  ;  and  the  Connecticut  System. 

These  five  systems  we  will  define,  and  then  illustrate  by  examples  the  United 
States  and  the  Merchants'  Systems,  which  are  in  general  use  outside  of  the  bound- 
aries of  the  States  of  Vermont,  Connecticut  and  New  Hampshire. 

THE  UNITED  STATES  SYSTEM. 

I 

1186.  The  following  is  the  decision  of  Chancellor  Kent  for  computing  interest 
where  partial  payments  have  been  made,  and  having  been  adopted  by  the  Sui)reme 


*  PARTIAL    PAYMENTS.  blj 

Court  of  the  United  States,  as  also  by  several  otlier  States,  is  Leiice  termed  tbe 
United  States  System. 

I. — The  rule  for  casting  interest  ■nLcn  partial  payments  have  been  made,  is  to  apply  the  pay- 
ment, iu  the  first  place,  to  the  discharge  of  the  interest  due. 

II. — If  the  payment  exceeds  the  interest,  the  surplus  goes  towards  discharging  the  principal, 
and  the  subsequent  interest  is  to  be  computed  on  the  balance  of  principal  remaining  due. 

III. — If  the  payment  be  less  than  the  interest,  tbe  surplus  of  interest  must  not  be  taken  to 
augment  the  principal ;  but  interest  continues  on  the  former  principal  until  tbe  period  when  the 
payments  taken  together  exceed  the  interest  due,  and  then  the  surplus  is  to  be  applied  towards 
.   discharging  the  jirincipai,  and  interest  is  to  be  computed  on  the  balance  as  aforesaid. 

[See  Chancellor  Kent,  Johnson's  Chancery  Rep.,  Vol.  1,  p.  17], 

The  practical  result  of  this  system  is  the  loss  of  the  use  of  money  paid  by  the 
person  owing  the  debt  when  the  same  is  less  than  the  interest,  from  the  time  paid 
until  the  sum  of  the  payments  exceeds  the  interest,  and  also  the  compounding  of 
the  money  ia  favor  of  the  creditor  as  often  as  the  amount  paid  exceeds  the  interest 
then  due,  and  hence  two  reasons  for  the  debtor  to  defer  payment  as  long  as  possible. 
Thus,  presuming  a  debtor  owes  a  note  for  $10000  bearing  6  per  cent  interest,  and 
that  he  pays  thereon  $50  per  month,  at  the  close  of  the  year  he  still  owes  the 
$10000.  Had  he  loaned  the  $50  monthly  payments  at  6  per  cent  he  would  have  had 
at  the  end  of  the  year  861 6.50  to  pay  on  his  note  and  the  accrued  interest,  $10000) 
leaving  a  debt  of  $9983.50  instead  of  $10000. 

THE  MERCHANTS'  SYSTEM. 

1187.  The  Merchants'  System  consists  in  computing  the  interest  ou  the  prin- 
cipal, or  original  debt,  from  the  time  it  became  due  until  the  close  of  the  civil,  or  of 
the  fiscal  year,  and  adding  it  to  the  principal ;  and  then  computing  the  interest  also 
upon  the  payments  made,  if  any,  during  the  year,  from  the  time  they  were  made  to 
the  end  of  the  civil,  or  fiscal  year,  and  adding  the  same  to  the  payments  made ;  then 
deducting  the  sum  of  the  payments  and  interest  from  the  amount  of  jirincijial  and 
interest.  The  remainder  will  be  a  new  principal.  In  this  manner  tlie  operation  is 
continued  from  year  to  year,  or  from  settlement  to  settlement,  if  made  oftener  than 
once  a  year,  until  the  final  settlement. 

The  periods  of  time  for  which  the  interest  is  computed  and  the  balance 
brought  down  are  termed  resis. 

Merchants  frequently  close  their  accounts  every  six  months,  and  bankers  every 
quarter,  in  which  case  the  directions  given  above  for  yearly  settlements  would  be 
made  semi-annually  and  quarterly.  The  effect  of  this  system  is  the  same  as  the 
account  current  and  interest  account  with  the  creditors,  made  as  often  as  the 
accounts  are  balanced,  and  the  practical  result  is  the  compounding  of  money  for 
both  parties  as  often  as  settlements  are  made. 

The  Merchants'  System  is  the  only  system  that  is  applicable  to  book  accounts 
as  well  as  notes,  bonds,  etc.,  and  it  is,  we  think,  practically  nearer  true  justice  to 
both  parties,  debtor  and  creditor,  than  the  United  States  System. 

1188.  VERMONT  SYSTEM  IS 

I.   "  JThen pnyments  are  made  on  notes,  hills,  or  similar  oiligations,  uhetlicr  payable  on  demand  or 


6i8 


souLE  s  rini.osopiiic   tractical  mathematics. 


at  a  specified  lime,  uitli  interest,  siuh  jini/T^enls  shall  he  applied:  First,  to  liquidate  the  interest 
that  has  arrriud  at  the  time  (if  such  pai/mcvis,  and  SECONDLY,  TO  THE  KXTINQIUSIIMENT  OF  THE 
ritlNUII'AL. 

II.  "  The  annval  interests  that  shall  remain  unpaid  on  notes,  bills,  or  similar  ohli(jaiions,  whether 
pai/ahle  on  demand  or  at  a  sjieci/ied  time,  'tcith  interest  annualli/,'  shall  be  SUIUECT  to  simple  INTEREST 
from  the  time  they  become  due  to  the  time  of  fnal  settlement. 

III.  "  If  payments  hare  not  been  made  in  any  year,  rechoning  from  the  time  such  annual  interest 
began  to  accrue,  the  amount  of  such  payments  at  the  end  of  the  year,  tcith  interest  thereon  from  the  time 
of  payment,  shall  he  applied:  First,  to  liquidate  the  simple  interest  tliat  has  accrued  from  the 
i!Ni>AiD  ANNUAL  INTERESTS;  SECONDLY,  (o  LiQUi DATE  (Ae  ANNUAL  INTERESTS  that  have  become  due; 
Thirdly,  to  the  extinguishment  of  the  niiNCiPAL." 

THE  NEW  HAMPSHIRE  SYSTEM. 

1189.  The  New  Hampshire  System  is  essentially  the  same  as  the  preceding 
•when  partial  payments  are  made  on  notes  "with  interest  annually."  But  "where 
])ayiiients  are  made  esjnessly  on  account  of  interest  accruing  but  not  then  due, 
they  are  applied  when  the  interest  falls  due,  without  interest  ou  such  payments." 

THE  CONNECTICUT  SYSTEM. 


1190.     The  Connecticut  System,  according  to  the  Supreme  Court  of  Connecti- 
cut, is  as  follows : 

Compute  the  interest  on  the  principal  to  the  time  of  the  first  payment ;  if  that 
be  one  year  or  more  from  the  time  the  interest  commenced,  add  it  to  the  principal, 
and  deduct  the  payment  from  the  sum  total.  If  they  be  after  payments  made, 
compute  the  interest  on  the  balance  due  to  the  next  payment,  and  then  deduct  the 
payment  as  above;  and  in  like  manner  from  one  payment  to  another,  till  all  the 
payments  are  absorbed,  jnovidcd  the  time  between  one  payment  and  another  be  one 
year  or  more. 

If  any  payments  be  made  before  one  year's  interest  has  accrued  then  compute 
the  interest. on  the  principal  sum  due  on  the  obligation  for  1  year,  add  it  to  the 
principal,  and  compute  the  interest  on  the  sum  paid  from  the  time  it  was  paid  up  to 
the  end  of  the  year;  add  it  to  the  sum  paid,  and  deduct  that  sum  from  the  principal 
and  interest  added  as  above. 

If  any  payments  be  made  of  a  less  sum  than  the  interest  arising  at  the  time 
of  such  payments,  no  interest  is  to  be  computed,  but  only  on  the  principal  sum  for 
any  period. 

Note — If  a  ye.ar  extends  beyond  tlie  time  of  settlement,  find  the  amonnt  of  the  remaining 
jirincipnl  to  the  time  of  seltlement;    find  also  tlie  amount  of  the  jiayiiieiit  or  payments,  if  any, 
from  the  time  they  were  paid  to  the  time  of  settlement,  and  subtract  their  sum  Irom  the  amount  J 
of  the  principal. 


*  PARTIAL    PAYMENTS.  619 

1191.      PEOBLEMS  WORKED  BY  TDE  UNITED  STATES  SYSTEM. 

1. 
$0000.  New  Orleans,  February  3d,  1S94. 

On  demand,  for  value  received,  we  promise  to  pay  to  the  order  of  Cummings 
&  Cenas,  Nine  Tliousand  Dollars  with  interest  at  six  (C)  per  cent. 

MoYSE  &  Wilson. 

Tlie  following  payments  were  made  and  indorsed  on  tlie  foregoing  note: 


September  lOtli,  1894,     -     $1500.00 
December  IStli,  1S94,     -        100.00 


Marcli  oth,  1895, 
July   20lb,  1895, 


$3953.75 
2000.00 


Counting  30  days  to  tbe  montb,  wliat  was  the  amount  due  Ai)ril  1,  189C? 

Aus.  $2177.43. 

OPERATION. 


Fob. 

Sept. 


Dec. 


1895 
March 


July 


1896 
Ai)ril 


18 


20 


To  face  of  note  ............ 

"  juterest  on  same  from  Feb.  3  to  Scjjt.  10,  217  days  at  6  ]>er  cent 

"  amount  due  ..-.. 

By  cash,  (1st  payment),  ......-.- 

"  balance  due  

"  interest  on  balance  due  from  Sep.  10  to  Dec.  18,  98  days  at  G'X       -        $127.82 
By  cash,  (2d  payment), "    -         -  100.00 

"  balance  of  interest      ...        - J27.82 

"  int.  on  hal.  due  ($7825.50)  from  Dec.  18  to  March  :,,  77  days  at  6%^          100.43 
which  added  to  balance  of  interest  due  makes  -         -  

"  amount  due     ---.-...-.-■- 

By  cash,  (3d  payment),  ......... 

"  balance  due  - 

"  interest  on  balance  due  from  March  5  to  July  20,  135  days  at  6% 

"  amount  due     -..----..-... 

By  cash,  (4th  payment),  ......... 

"  balance  due     .-.---....-.. 

"  interest  on  halance  due  from  July  20  to  April  1,  251  days  at  6% 

"  amount  due  April  1,  1896       .......... 


9000 
325 


9325 
1500 


7825 


128 


7953 
3953 


1000 
90 


4090 
2000 

2090 

87 


2177 


00 
50 

50 
00 

50 


25 

75 

75 

00 
00 

00 
00 

00 

43 

43 


Explanalion. — In  the  operation  of  this  prohlem  wo  found  that  the  interest  due  on  the  18th  of 
December  was  greater  than  the  payment  then  made;  hence,  according  to  the  principles  of  this 
system  of  Tvork,  we  did  not  add  the  same  to  the  principal.  Had  we  done  so  the  result  wonld  have 
been  compound  interest  on  the  excess  of  interest,  $27.82,  to  the  disadvantage  of  tbe  debtor,  and  in 
a,  sense  that  the  law  of  this  system  does  not  allow.  We  accordingly  deducted  the  payment  from  the 
interest,  and  reserved  the  excess  until  the  next  payment,  to  which  excess  wo  then  added  the  interest 
on  balance  due  from  December  18  to  March  5,  1895,  and  thus  obtain  $128.25  interest,  which  beiug 
less  than  the  jiayuieut  then  made,  we  added  to  the  balance  due  September  10,  1894. 

Instead  of  deducting  the  payment  made  from  the  interest,  we  could  have  set  the  whole 
interest  aside  until  the  sum  of  the  payments  exceeded  the  sum  of  the  interests,  and  then  proceeded 
in  the  regular  manner  by  adding  the  sum  of  the  interest  to  the  amount  due,  and  then  deducting 
the  sum  of  the  payments  therefrom.  But  we  think  that  the  method  shown  in  the  operation  is  tbe 
clearest,  and  hence  prefer  it. 


620 


SOULE  S    rillLOSOPHlC    PRACTICAL    MATHEMATICS. 


$5200.  Marshall,  Texas,  July  28,  1894. 

Due,  from  date  lieicof,  to  WoMACK  &  Williams,  or  order.  Five  Thousand 
Two  Hundred  Dollars,  for  value  received,  -with  interest  at  10  per  cent. 

Patterson  &  Turman. 

The  following  payments  were  made  and  indorsed  on  the  above  due  bill : 


October  1st,  1894,  -  $1200 
December  8th,  1894,  -  1000 
June  25th,  1895,      -         100 


August         1st,  1895,     -     $1500 
IJovember  11th,  1895,     -        900 


A  final  settlement  was  made  December  15th,  1895.    Counting  actual  time, 
what  was  then  due  ?  Ans.  $938.30. 


OPERATION. 


1894 
Jnlv 
Oct. 


Dec. 


1895 

JlltIO 


Alls 


iVov. 


Dec. 


11 


15 


To  amount  iliie      -- 

"  interest  ou  same  from  July  28  to  Oct.  1st,  65  days  at  10  per  cent 

"  amount  due  .......... 

By  cash,  (1st  payment),  --..... 


balance  due     - 

interest  ou  same  from  Oct.  1  to  Dec.  8,  68  days  at  10  i>er  cent    - 

amount  due     --..-.--... 

By  cash,  (2d  payment),      -- 


balance  due 


int.  on  same  from  Dec,  8,  '94,  to  June  25,  '95,  1S9  days  at  10,"^ 
By  cash,  (3d  payment),  ...-..- 


balance  of  interest  due       • 

int.  on  bal.  due  ($3171.22)  from  June  25  to  Aug.  1,  37  days  at  lOJ'^ 


■which  added  to  balance  of  interest  makes 


amount  due 

By  cash,  (4th  payment), 


balance  due     --.----.-.. 

interest  on  same  from  Aug.  1  to  Nov.  11,  102  days  at  10  per  cent 

amount  due     -.-.--.-... 

By  cash,  (5th  payment),  ..--... 


balance  due  ---.-..... 

interest  on  same  from  Nov.  11  to  Dec.  15,  34  days  at  10  per  cent 

amount  due  on  6ual  settlement  December  15,  1895, 


$175.30 
100.00 


$75.30 
32.59 


$    I  c. 

5200  00 

93  89 

.529389 
1200'00 

4093  89 

77,33 

417122 

1000  00 

L 

317l'2:i 


107 


3279 
1500 


1779 
50 


1829 
900 


929 

8 


938 


89 


11 

00 


00 

52 
78 

30 


Expl<l7laiion.—^B  the  operation  of  this  problem,  we  have  counted  actual   time,    which    w© 
regard  as  the  better  .way  at  all  times. 


PARTIAL    PAYMENTS. 


621 


1192.        PEOBLEMS  WORKED  BY  THE  MERCHANTS'  SYSTEM. 

1. 
$9000.  New  Orleans,  February  3,  ]S94. 

On  demand,  for  value  received,  Me  ])rouiise  to  ])ay  to  tlie  order  of  Cummings 
«S:  Cenas,  Nine  Thousand  Dollars  with  interest  at  6  i)er  cent. 

]\IOYSE  &  Wilson. 

The  follovring  payments  were  made  and  indorsed  on  the  foregoing  note: 

September  10th,  1894,    $1500.00  March    5th,  189.5,       -     $3953.75 

December  18th,  1894,         HIO.OO  July     20th,  1895,       -       2000.00 

Counting  30  days  to  the  month,  and  making  an  annual  settlement  on  the  first 
of  Januarj',  1895,  what  was  the  amount  due  April  1st,  1890?  Aus.  $2165.40. 

OPERATION. 


1894 

Fell. 

3 

1«95 

Jan. 

1 

18!U 

'  ' 

Sept. 

10 

1895 

J.TU. 

1 

1894 

Dec. 

18 

1895 

Jan. 

1 

1896 
Jan. 


1895 
Mareh 

1896 
Jan. 

1895 
Julv 

1896 
Jan. 


April 


To  face  of  note     ...---.--- 
"  interest  on  same  from  Feb.  3  to  Jan  1,  328  days  at  G  per  cent 
"  amount  due  to  date        ..-..--- 
By  casb,  (Ist  payment),         ...--. 


"   interest  on  same  from  September  10  to  January  1,   111 
days  at  6  jier  cent       - 

By  casb,  (2d  payment),  .-..--. 


"   interest  on  same  from  December  18  to  January  1,   13 
days  at  6  per  cent 


$1500.00 

27.75 
100.00 


"  total  credits  to  date      ...-..--- 
To  balance  due  this  date      ..-...---.- 

"  interest  on  same  from  Jan.  1  to  Jan.  1,  360  days  at  6  per  cent 

"  amount  dne  to  date        .-.-------- 

By  casb,  (3d  payment), $3953.75 

"    interest  on  same  from  Marcb  5  to  January  1,  296  days 

at  6  per  cent       --------  195.05 

"   casb,  (4tb  payment), 2000.00 

"   interest  on  same  from  July  20  to  January  1,   IGl  days 

at  6  per  cent       --. 53.67 


9000 
492 


9492 


c. 

00 

00 

00 


total  credits  to  date 


To  balance  dne  tbis  date  ...... 

"  interest  on  same  from  Jan.  1  to  April  1,  90  dajs  at  6";^ 


amount  due  on  final  settlement 


1627  97 
03 


78C4 
471 


8335 


6202 


47 


40 


2133 
32J0O 

216540 


Explanation. — In  tbe  solution  of  tbis  problem,  as  sbown  by  tbe  operation,  we  made  tbe  first 
partial  settlement  at  tbe  close  of  tbe  civil  year,  notwitbstanding  a  year  bad  not  elajised  from  tbe 
date  of  tbe  obligaticni.  We  did  tbis  because  it  is  strictly  in  conl'ormity  witb  tbe  mercbant's 
system  of  closini;;  bis  accounts  at  tbe  end  of  each  civil  or  bis  fiscal  year,  and  because  we  believe 
tbat  it  is  tbe  only  correct  way  of  applying  tbe  mercbants'  system  of  partial  payments.  Mercbants 
generally  not  only  desire,  but  by  tbe  nmnicipal,  State  and  general  government  laws  are  reiiuired  to 
know  tbeir  gains  and  losses,  resources  and  liabilities  on  tbe  first  of  January  of  each  year.  With- 
out this  information,  correct  returns  for  the  assessment  of  various  taxes  could  not  be  rendered,  nor 


622 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


would  their  books  show  a  correct  condition  of  Uicir  business.  Had  we,  as  some  authors  direct, 
computed  the  interest  from  Fel>rnary  3,  1894,  to  February  3,  1S95,  and  then  from  February  3,  1895, 
to  February  3,  1S96,  and  then  from  February  3,  IKHi,  to  April  1,  1S96,  the  iiual  result  would  have 
l)een  slightly  ditlerent,  and  when  the  books  were  closed  on  tlie  first  of  January  of  each  year,  the 
trno  gains  and  resources  of  the  business  would  not  have  been  shown  by  the  books. 

In  counting  the  time  we  used  30  days  to  the  month  in  order  to  compare  the  result  of  the 
problem  with  the  result  of  the  hrst  and  same  problem  worked  by  the  United  States  System,  on 
page  619. 

By  the  United  States  System  the  amount  due  on  final  settlement  is  -  $2177.43 
And  by  the  Merchants'  System  as  above, 2165.40 


Excess  produced  by  the  United  States  System,      -        -        -  $12.03 

2. 
$5200.  Marshall,  Texas,  July  28,  1894. 

Due  from  date  hereof  to  Womack  &  Williams,  or  order.  Five  Thousand 
Two  Hundred  Dollars  for  value  received,  with  interest  at  10  per  cent. 

Patterson  «&  Tubman. 
The  following  ijaymeuts  were  made  and  indorsed  on  the  above  due  bill: 

October        1st,  1894,      -     $1200  August  1st,  1895,     -     $1500 

December    Sth,  1894,      -       1(»00  November  11th,  1895,     -         900 


June 


25th,  1895, 


100 


A  final  settlement  was  made  December  15,  1895.  Counting  actual  time,  and 
making  a  ])artial  settlement  at  the  close  of  the  civil  year  1894,  what  amount  was 
due  on  final  settlement?  Ans.  $927.77. 

OPERATION. 


1894 
Jiilv 

1895 
Jan. 


1894 
Oct. 


Dec. 


1895 
Jan. 


Dec. 


Juno 
Dec. 

Aug. 
Dec. 

Nov. 
Dec. 


15 


To  face  of  due  bill        ---....., 
"  interest  on  same  from  July  28  to  Jan.  1,  157  days  at  10% 
"  amount  due  this  date  .-.,.. 


By  cash,  (1st  payment), $1200.00 

"   interest  on  same  from  October  1  to  .launary  1,  92  days 

at  10  per  cent -         -         -  30.ti7 

"   c.ish,  (2d  payment), 1000  00 

"   interest  on  same  from  December  8  to  January   1,   24 

days  at  10  per  cent 667 


"   total  credits  to  date 


To  balance  due  to  date         .         -         . 

"  interest  on  same  from  January  1  to  December  15,  348  days  at  10% 


amount  due  this  date  -        - 

By  cash,  (3d  payment), 

"   interest  on  same  from  June  25  to   December   15,    173 

days  at  10  per  cent     ------- 

"   cash,  (4th  payment ),         -        - 

"   interest  on  same  from  August  1  to  December  15,  13G 

days  at  10  per  cent     ------- 

"    cash,  (5th  payment), -         - 

"  interest  on  same  from  November  11  to  December  15,  34 

days  at  10  per  cent     -.--.-. 


5200 

226 

.-.42G 


c. 
OO 

78 

8 


1100.00  i 

4.81 
1500.00 

56.67 
900.UO 

8.50 


2237j34 

:il89,44 
308  31 

3497  75 


"    total  credits  to  date 
To  amount  due  on  final  settlement 


2569  <)S 
927  77 


ICxpfanation. — By  comparing  the  result  of  this  problem  with  the  result  of  the  second  and 
same  problem  worked  by  the  United  States  System,  on  page  620,  we  have  the  following  figures: 


*  PARTIAL    PAYMENTS.  623 

Amount  due  on  final  settlement  by  tbe  United  States  System,  -         -        $938.30 

Amount  due  as  above  by  the  jMercbauts' System,  ....  927.77 

Excess  produced  by  the  United  States  System,        -        -        -  $10.o3 

Where  the  payments  are  small,  or  the  interval  of  time  between  them  is  more 
than  a  j'ear,  the  Merchants'  System  would  produce  an  excess  over  the  United  States 
System. 

1193,  MISCELLANEOUS    TROBLEMS 

INVOLVING    THE    UNITED    STATES   AND    MERCUANTs"   SYSTEMS    OF    M'ORIC. 
1. 

S3740.  ViCKSBUKG,  May  1,  1894. 

Two  years  after  date,  we  promise  to  pay  to  the  order  of  Stewart  &  Br  adston, 

Three  Thou.sand  Seven  ilundred  Forty  dollars,  with  interest  at  eight  (8)  per  cent, 

value  received. 

Bowles  &  Stewart. 

The  following  payments  were  made  and  indorsed  on  the  foregoing  note: 

September  16th,  1894,    -       $500  October  28tli,  189."),    -     -     $1200 

February      1st,  1895,    -  75  April         1st,  189G,    -     -         300 

May  4th,  1895,    -       1500 

Counting  actual  number  of  days,  and  working  by  the  United  States  System, 
what  was  the  amount  due  at  the  maturity  of  the  note,  JMay  4th,  1S9G  ? 

Ans.  $500.68. 

o 

$2000.  Little  Bock,  Ark.,  January  12,  1894. 

Due  Scull  &  JIcRey,  Two  Thousand  Dollars,  -with  interest  at  8  ])er  cent. 

Foster  &  Sullivan. 

The  following  payments  were  made  and  indorsed  on  the  foregoing  note: 

November    1st,  1894,       -     $C00  .Alay  9th,  1895,    -     -     $400 

January     18th,  1895,       -       750  December  1st,  1895,    -     -       200 

Counting  actual  time,  and  working  by  the  Merchants'  System,  what  was  due 
on  final  settlement,  April  10,  1S9G?  Ans.  $249.84. 

3. 
$6840.50. 

On  tbe  18tb  of  October,  1894,  a  judgment  was  rendcied  by  one  of  the  District 
Courts  of  New  Orleans  for  Six  Thousand  Eight  Hundred  Forty  and  fj'^  Dollars. 
(The  legal  rate  of  interest  in  Louisiana  is  5  per  cent). 

The  following  payments  were  made  thereon  : 

January    1st,  1895,   -     -     $1000  November  25th,  1895,     -     $1000 

May         14th,  1895,   -     -       1000  March         10th,  1890,     -       30U0 

A  final  settlement  was  made  April  21,  189C.  Counting  actual  time  and 
working  by  the  United  States  System,  what  was  then  due  ?  Ans.  $1225.24. 


624 


SOULE  S    nilLOSOPHIC     PRACTICAL    MATHEMATICS. 


4. 
$20000.  Waco,  Texas,  January  1,  1894. 

On  demand,  for  value  received,  vre  promise  to  pay  to  the  order  of  Hightower 
&  Rose,  Twenty  Thousand  Dollar.s,  with  10  jier  cent  interest. 

Polly,  Yinyard  &  Co. 
Semi-annual  payments  of  $3000  each  were  made  on  the  above  note  from  date 
to  January  1,  1897,  at  which  time  a  final  settlement  was  made.    Counting  actual 
time,  what  was  then  due  by  the  United  States  System,  and  also  by  the  Merchants' 
System  of  partial  payments'?  Aus.  $9460.19,  by  the  United  States  System. 

$9329.23,  by  the  Merchants'  System. 


1194. 


PARTIAL    PAYMENTS 


ON    NOTES    DRAWN    FOR    A     SPECIFIED     TIME     AND     BEARING     INTEREST     AFTER 

MATURITY. 

$7000.  Summit,  Miss.,  June  1,  1894. 

One  year  after  date,  for  value  received,  we  promise  to  pay  to  the  order  of 
Curtis  &  Newman,  Seven  Thousand  Dollars,  with  8  per  cent  interest  after 
maturity  until  paid.  Tyler  &  Mills. 

The  following  payments  were  made  and  indorsed  on  this  note: 

December    1st,  1894, $2000 

April  loth,  1895, 2500 

October      17th,  1895, 1500 

A  final  settlement  was  made  December  20, 1895.  Counting  actual  time,  what 
was  then  due?  Ans.  $972.48. 

operation. 


1894 
June 
Dec. 

1805 
April 


June 

Oct. 
it 

Dec. 


Face  of  note  dne  June  1/4,  1895  .----.... 

Uy  cash,  (1st  payment), $2000.00 

"   interest  on  same  from  Decemljer  1  fo  Jnne  4,  1895,  185 

days  at  8  per  cent 82.22 

"   cast,  ('2d  payment), 2500.00 

"   interest  on  same  from  April  10  to  June  4,  1895,  55  days 

at  8  per  cent      .-.--.-.  30.56 

"   total  amount  of  credits  before  the  maturity  of  the 

note  June  4,  1895    --------- 


To  balance  due  at  the  maturity  of  the  note       ------ 

"  interest  on  same  from  June  4,  1895,  to  Oct.  17,  1895,  135  days  at  8% 


amount  dne  this  date 

By  cash,  (3d  payment), 


"  biilance  due  .-.-...- 

"  interest  on  same  from  Oct.  17,  to  Dec.  20,  64  days  at  8% 

"  amount  due  on  final  settlement  .        .        -        - 


f    ic. 
7000,00 


4612,78 


238722 
7162 


2458  84 
1500|00 

958'84 

13J64  a 

972I48 


Explanation. — Wliere  partial  payments  are  made  on  notes  of  this  character  before  they 
mature,  we  must  lirst  find  the  balance  due  at  maturity.  To  do  this  we  deduct  from  the  face  of  the 
note  the  payments  made  belore  due,  and  the  interest  on  such  payments  from  the  date  of  payment 
till  the  maturity  of  the  note.  Having  thus  produced  the  balance  due  on  the  note  or  obligation  at 
the  time  of  maturity,  we  then  proceed  with  it,  and  the  remaining  payments  if  any,  the  same  as  in 
the  preceding  work"  of  partial  payments,  either  by  the  United  States  or  Merchants'  System.  In 
the  operation  of  the  above  problem,  we  worked  by  the  United  States  System,  which  is  the  system 
in  general  use  for  transactions  of  this  kind. 


PARTIAL    PAYMENTS. 


625 


1195. 


PARTIAL   PAYMENTS 


ON   NOTES,    ETC.,    DKAWN    FOR    A  SPECIFIED   TIME,   AND    WHICII    BEAR    ANNUAL, 
SEMI-ANNUAL    OR   QUARTERLY   INTEREST. 

$0000.  New  Orleans,  March  17,  1889. 

Three  years  after  date,  for  value  received,  I  ])roniise  to  pay  to  the  order  of 
Francis  &  Baker,  Niue  Thousand  Dollars  with  eight  (8)  per  cent  interest,  i^ay- 
able  annually.  E.  W.  Jones. 

Tlie  following  payments  were  made  and  indorsed  on  this  note: 
November    1st,  1889,     -     $1000  May         15th,  1891,   -     -     $2000 

July  10th,  1890,     -       1500  January    4th,  1892,   -     -       1800 

Counting  30  days  to  the  month,  what  was  due  June  17, 1892,  at  M'hich  time  a 
final  settlement  was  madel  Aus.  $4552.09,  by  1st  operation. 

$4559.63,  by  2d  operation. 

FIRST    OPERATION. 


18S9 

1890 
Jlarch 

1889 
J>ov. 

1890 
Jilarch 


1891 
March 


1890 
July 

1891 
March 


1892 
March 


1891 
May 

1892 
Jau. 


March 

it 
.Jnne 


17  To  face  of  note  

"  interest  on  same  from  March  IT,  1889,  to  March  17,  1890,  1  year  at  8% 


amount  due  this  date 


By  cash,  (1st  payment), $1000.00 

"   interest  on  same  from  Noveniher  1,  1889,  to  March  17, 

1890,  4  months  16  days  at  8  per  cent         -        -         -  30.22 

"   total  credits  to  date  ...----        


9000 
720 


9720 


c. 

00 

00 

00 


1030 


To  biilance  due  this  date  ....----- 

"  interest  on  same  from  Jlarch  17,  1890,  to  March  17,  1891,  1  year  at  8^ 
"  amount  due  this  date     ....-.-.-- 


By  cash,  (2d  payment), $1500.00 

''   interest  on  same  from  July  10,  1890,  to  March  17,  1891, 

8  months  7  days  at  8  jier  cent            .        -         -         -  82.33 

"   total  credits  to  date  ....---        

"  halance  due  this  date  .....-.--. 

"  interest  on  same  from  March  17,  1891,  to  March  17,  1892,  1  year  at  8%     - 

"  amount  due  this  date     .-.....---• 


By  cash,  (3d  payment),  ...--.- 
"   interest  on  same  from  May  15,  1891,  to  March  17,  1892, 

10  months  2  d.ays  at  8  per  cent  .... 

"  cash,  (4th  payment),  -..---- 
"   interest  on  same  from  January  4,  1892,  to  March  17, 

1892,  2  months  13  days  at  8;^  .... 

"  total  credits  to  date  .-...-. 


$2000.00 


134.22 
1800.00 

29.20 


To  halance  due  this  date     ...-.------ 

"  interest  on  same  from  March  17,  1892,  to  June  17,  1892,  3  months  at  8%  - 


halance  due  on  final  settlement 


8689 
695 


J384 


1582 


i802 

624 

8426 


22 
78 
18 
96 

33 

63 
1 

84 


3963  42 

4463,42 
89,27 

4552  69 


£x7)?ajmf!on.^Transactions  of  this  character  have  not  occurred  with  sufficient  frequency  to 
•establish  any  real  custom  pertaining  to  the  method  of  their  solution;  and  hence,  iu  the  absence 
■«f  both  custom  and  law,  and  in  consequence  of  there  being  so  many  different  systems  of  partial 


^26 


SOULE  S    rillLOSOPHIC    TRACTICAL    MATHEMATICS. 


payments,  accountants  and  authors  are  divided  in  their  opinions  as  to  -nbich  of  several  niethoil* 
should  lio  used  in  cases  of  this  kind.  For  this  reason  we  present  two  solutions,  to  the  first  of 
which  ve  give  preference.  'J'ho  contract  of  aiinu.nl  interest  specifies  the  )ienod  of  time  that  the 
same  shall  he  paid,  and  therefore  it  seems  reasonable  that  the  settlements  tor  the  jiartial  payments- 
should  1)0  made  at  the  same  time ;  and  iu  accordance  with  this  view,  we  have  made  them  at  the 
close  of  each  year  from  the  date  of  the  note.  JJad  the  iiilercst  been  2>aijable  ecmi-anuuuUij  or  quar- 
terly, we  would  hare  made  the  jx^rlial  puijinenta  accordingly. 

As  shown  in  tlie  o])eratioii,  ■we  first  find  the  interest  on  the  face  of  the  note 
for  one  year  and  add  it  to  the  same,  and  then  from  this  amount  we  deduct  the  ])ay- 
jnent  made  during  the  year,  inci'eased  by  the  interest  on  the  payment  from  tlie  date 
that  it  was  uiade  until  the  end  of  the  year;  and  thus  M-econtinueuntil  thematurity 
of  tlie  note,  and  then  for  the  remainder  of  the  time  up  to  the  final  settlement,  we 
proceed  iu  the  same  manner  as  in  the  Merchants'  System  of  partial  payments.  In 
transactions  of  this  nature,  wliere  there  are  payments  made  subsequent  to  the 
maturity  of  the  note,  they  would  be  treated  the  same  as  in  the  Merchants'  System. 

In  fact  the  only  diflerence  between  the  operations  of  this  work  and  the 
Merchants'  System  is  the  difl'erent  time  or  rests  at  which  the  partial  payment 
settlements  are  made. 

By  this  system  of  work  both  parties  receive  yearly  compound  interest. 

SECOND   OPERATION. 


1889 
March 

1893 
June 

17 
17 

tt 

a 

ti 

'< 

ti 

" 

ti 

(C 

1880 
Nov. 

1 

1890 
July 

10 

1891 
May 

15 

1892 
Jan. 

i 

June 

17 

It 

f  I 

To  face  of  note 


int.  on  same  from  March  17,  '89,  to  Juue  17,  '92,  3  yrs.  and  3  mos.  at  8% 

amount  of  note  at  time  of  settlement -        - 

interest  on  S720  (the  1st  annual  interest)  from  March  17,  1890,  to 

June  17,  1892,  2  years  and  3  months  at  8",;  -         .         .         .  $129.60 

interest  on  $720  (the  3d  annual  interest)  from  March  17,  1891,  to 

Juue  17,  1892,  1  year  and  3  months  at  8%  ....  72. 00 

amount  of  interest  on  annual  interest  ....... 


total  amount  due  this  date 


By  cash,  (1st  payment), $1000.00 

"   interest  on  same  from  Nov.  1,  1889,  to  June  17,  1892,  2 

years  7  mouths  and  16  days  at  8;'o             -        -        -  210.22 

"   cash,  (2d  payment), 1500.00 

"   interest  on  same  from  July  10,  1890,  to  June  17,   1892, 

1  year  11  mouths  and  7  days  at  8%            -         -         -  232.33 

"   cash,  (3d  payment), 2000.00 

"   interest  on  same  from  May  15,  1891.  to  June  17,   1892, 

1  year  1  mouth  and  2  days  at  8%      ....  174.22 

"  cash,  (4th  payment), 1800.00 

"   iutercst  on  same  from  January   4,  1892,    to   June   17, 

1892,  5  mouths  and  13  days  at  8%             -        -        -  65.20 


"   total  credits  to  date 
To  balance  due  on  final  settlement 


9000  00 
2340  00 
00 


11340 


201 


11541 


6981 


60 
60 


97 


4559,63 

Explnnntion.—'in  this  solution  -we  first  find  the  interest  on  the  face  of  the  note  from  the  day 
of  date  to  the  day  of  final  settlement,  and  then  the  interest  on  the  annual  interest  from  the  time  it 

became  due  nntirthe  final  settlement.     Then,  fr the  total  of  these  three  amounts  we  deduct  tlie 

sum  of  the  payments,  increased  by  the  interest  on  each  from  the  time  paid  uutil  the  day  of  settle- 
ment.    The  remainder  is  what  is  then  due, 

j^oTE.— In  the  forcRoinf;  operations  of  interest  and  partial  payments,  -where  the  actual  time 
•was  counted,  29  days  were  reckoned  for  February  in  all  the  leap  years  that  occurred. 


*  EQUATION    OF    ACCOUNTS.  627 

EQUATION  OE  AVEEAGE  OF   PAYMENTS  AND  ACCOUNTS. 

1196.     Equatiou,  or  Average  of  Payments  and  Accounts,  is  the  process  of 
finding  tlie  nieau  or  average  time  lor  Ibe  itayuieiit,  111  one  amount,  of  several  sums 
■  of  monej',  notes  or  debts  due  at  difl'erent  dates;  or  of  the  balance  of  accounts  tbat 
contain  botli  debit  and  credit  items. 

The  Equated  Time  is  the  date  at  whicb  tlie  various  sums  due  at  different 
times  may  be  paid  in  one  amount  without  injustice  to  either  i^arty. 

The  Adjustment  of  Interest  is  the  real  object  of  equating  or  averaging 
payments  and  accounts. 

The  principles  upon  -which  the  process  of  the  work  is  based  are  that  the 
interest  of  any  sum  paid  any  length  of  time  before  it  is  due  is  equitably  balanced  by 
the  interest  of  an  equal  su7ti  paid  an  equal  time  after  it  is  dne.  Hence  in  all  equa- 
tions or  averages,  the  interest  on  the  sums  j)aid,  before  due  by  equation,  must  equal 
the  interest  on  the  sums  paid  after  they  are  due  by  equation. 

The  Assumed  Due  Date,  or  Focal  Date  is  the  date  at  which  we  assume  all 
items  of  debit  or  credit  or  both  to  be  due,  and  from  which  we  count  the  debit  or 
credit  of  time  in  the  equation. 

The  Debit  or  Credit  of  Time  is  the  time  an  item  has  to  run  from  the  assumed 
due  date  until  it  is  due. 

The  Earliest  Date  in  an  account  or  of  the  items  to  be  equated,  is  used  in 
this  work  as  the  assumed  due  date  or  focal  date. 

Note. — Authors  and  accountants  differ  in  regard  to  the  date  to  assume  as  the  due  or  focal 
date.  Some  assume  the  earliest  date,  some  the  latest  date,  some  the  first  of  the  month  in  which 
the  earliest  d;ite  occurs  and  some  assume  a  date  prior  or  subsequent  to  the  earliest  date.  The 
result  is  the  same  whatever  the  date  assumed. 

To  work  from  the  earliest  date  each  item  of  debit,  except  the  first,  is  subject 
to  banlc  discount ;  and  to  work  from  the  latest  date,  each  item,  except  the  latest,  is 
entitled  to  interest. 

The  subject  is  divided  into  simple  and  compound  equation  or  average. 

Simple  Equation,  or  Average,  is  the  process  of  finding  the  average  time  of 
payment  for  debts  or  accounts,  which  contain  debit  or  credit  items  only. 

Compound  Equation,  or  Average,  is  the  process  of  finding  the  equated 
average  time  of  payment  for  the  balance  of  accounts,  which  contain  both  debit  and 
credit  items. 

Equations  with  Special  Conditions.  For  equations  involving  special  con- 
ditions, see  pages  643  to  652  and  100"J  to  1013  of  this  book. 

There  are  numerous  methods  given  by  various  authors  for  the  equation  and 
averaging  of  accounts;  but  we  believe  that  the  introduction  of  all  of  them  would 
confuse  and  bewilder  the  student  rather  thau  elucidate  and  make  clear  the 
operations  and  principles  of  tlie  work. 

The  methods  we  present  are  universal  in  their  application,  have  received 
the  approval  of  practical  and  scientific  accountants,  and  are  short,  concise  and 
require  iu  their  use  the  exercise  of  reason  instead  of  the  memorizing  of  rules. 


628  SOULe's    I'lIILOSOPHIC    I'RACTICAL    MATHEMATICS.  * 

PROBLEMS 

TO  FIND  THE  EQUIVALENT  OR  AVERAGE  TIME  FOR   THE  PAYMENT 

OF  SEVERAL  SUMS  OF  MONEY  DUE  AT  DIFFERENT  TIMES, 

OR  WITH  DIFFERENT  TERMS  OF  CREDIT. 

1197.  1.  January  1,  1895,  Z  purchased  from  A  goods  amounting  to  $3000, 
Mil ich  he  agrees  to  pay  as  follows:  $800  in  three  months;  $1200  in  eight  months, 
and  the  remainder,  $1000,  in  twelve  months.  What  is  the  equated  or  average  time 
for  the  payment  of  the  whole  amount,  so  that  neither  party  will  gain  or  lose  by  the 
use  of  money?  Ans.  8  months. 

OPERATION   BY   THE    PRODUCT   SYSTEM. 


i 


Z's 

Z's                                       Z's 

• 

Debit  of  Dolls. 

Credit  of  Mouths.        Credit  of  Months  ou 

1895. 

January  1, 

$800 

X 

3              =                 2400 

1, 

1200 

X 

8              =                 9C00 

"          1, 

1000 

X 

12              =               12000 

Total  debit  of  dollars  $3000        Total  credit  of  months  on  $1,     24000 

24000  -^  3000  =  8  months. 

Ans.  8  months  equated  time. 

Explanation.— By  the  terms  of  the  contract  we  see  that  Z  is  debited  -n-ith  S800,  $1200  and  ' 
$1000  in  moneij,  and  credited  on  the  respective  amounts  with  three,  eight  and  twelve  months  in 
time,  from  January  1,  which  is  the  due  date  of  all  the  items;  and  since  he  is  debited  with  various 
Bums  of  money  and  credited  with  various  periods  of  time,  it  is  clear,  by  the  exercise  of  our  reason, 
that,  in  order  to  solve  the  question,  we  must  first  find  Z's  total  credit  of  time  on  $1,  from  the  due 
date.  To  Cud  his  total  credit  of  time  on  $1,  we  reason  thus:  Three  months'  credit  on  $800  is  equal 
to  800  times  three  mouths'  credit  ou  $1,  which  is  2400  months  ;  then  eight  months'  credit  on  $1200 
is  equal  to  1200  times  eight  months'  credit  ou  $1 ;  and  then  twelve  mouths'  credit  on  $1000  is  equal 
to  1000  times  12  mouths'  credit  on  $1;  which  added  gives  24000  months'  credit  on  $1.  Having  this 
total  credit  of  time  ou  $1,  we  then  find  the  total  amount  of  his  dibit  in  dollars  by  adding  the 
various  sums  together.  We  thou  reason  thus:  Since  Z  has  a  credit  of  24000  months' time  ou  $1,  on 
$3000  he  will  have  the  3000th  ]>art,  which  is  eight  months.  Here  it  may  be  asked,  why  Z  should 
not  have  3U00  times  as  much  credit  on  $3000  as  on  $1,  instead  of  the  3000th  part  ?  To  this  question 
we  reply  that  Z's  credit  on  $1,  for  24000  months,  is  but  an  equivalent  for  the  various  terms  of 
credit  due  him  by  the  conditions  of  the  problem:  and  hence,  if  he  is  to  receive  a  credit  ou  the 
whole  amount,  $3000,  that  he  owes,  instead  of  ou  Ijl,  the  credit  allowed  to  produce  an  equivalent 
to  the  24000  months  on  $1,  or  tlio  various  terms  of  credit,  according  to  the  first  conditions,  must 
necessarily  be  in  reciprocal  jiroportiou  to  the  amount  to  which  it  is  to  apply. 

To  illustrate  the  correctness  of  the  foregoing  operation,  by  the  principles  of 
interest,  we  present  the  following  figures: 

By  the  first  conditions  of  the  i)robleni,  Z  was  to  pay  $800  in  three  months, 
$1200  in  eight  months  and  $1000  in  twelve  months  ;  but  as  Z  did  not  pay  the  $800 
until  the  equated  time,  eight  months,  he  therefore  gained  and  A  lout  the  interest  of 
the  same  for  five  mouths,  which  at,  say,  8  per  cent,  is  $20.07. 

The  $1200  Z  was  to  have  paid  in  eight  months,  and  as  the  equated  time  is 
eight  months,  there  is  ueither  gain  uor  loss  to  either  Z  or  A  on  the  same,  by  reason 
of  the  equation. 


*  EQUATION    OF    ACCOUNTS.  629 

The  $1000  Z  was  to  have  paid  in  twelve  months;  but  by  reason  of  the  equa- 
tion, he  paid  it  in  eight  months;  and  hence  he  lost  and  A  gained  the  interest  of  the 
same  for  four  months,  which,  at  the  same  rate  as  above,  8  per  cent,  is  $20.67.  By 
this  we  see  that  A's  gains  and  Z's  losses,  or  Z's  gains  and  A's  losses  are  equal,  and 
that  tiie  interest  on  the  sums  ixdd  before  they  are  due  by  equation,  is  equal  to  the 
interest  on  the  sums  paid  after  they  are  due  by  equation.  This  will  always  be  the  case, 
no  matter  how  many  sums  are  paid  before  or  after  the  equated  time. 

To  elucidate  the  work  of  equation  further,  we  present  the  following  operation : 

12%  =    $24  interest. 
12%  =       96 
12%  =     120         " 

$3000  $30  )  $240  interest. 


1895. 

Mouths  Cr. 

January  1, 

$800 

3    at 

1200 

8    at 

1000 

12    at 

8    months. 

Expta^iation.—^'By  tliis  metliod  ve  see  tb.it  if  Z  Lad  paid  tlie  various  debits  of  dollars,  at  tlio 
time  of  |>m'cbase,  witliout  waiting;  for  the  expiration  of  the  various  teruis  of  credit,  he  would  have 
been  eutitled,  as  shown  iu  tlie  operation,  at  I'i  per  cent  interest,  to  $'2iO  interest  deduction  ou'  the 
$3000.  Hut  as  bo  does  uot  wish  to  pay  the  whole  auuiuut  iu  advance  of  the  whole  credit,  but  at  a 
period  of  time,  so  that  the  interest  ou  tbo  sums  pai<l  before  due  will  equal  the  interest  ou  the  sums 
paid  after  tbcy  are  due ;  it  is,  therefore,  by  the  exercise  of  our  reason,  plain  tbut  the  time  it  vill 
require  the  |3000,  at  12  per  cent,  to  produce  $240,  will  be  the  equated  time,  or  the  time  to  pay  the 
whole  amount,  so  that  Z's  gains  and  A's  losses,  or  Z's  losses  and  A's  gains,  by  interest,  shall  be 
equal.  To  fiud  this  time,  we  compute  the  interest  ou  the  $;.'iOOO,  for  one  month  at  1"2  percent,  which 
gives  us  $30.  We  then  reason  thus  :  If  $30  interest  reciuircs  one  month's  time,  $1  will  require  the 
thirtieth  part,  and  $240  will  require  240  tinu's  as  many  mouths,  which,  worked,  gives  eight  months, 
the  equated  time.  Had  the  terms  of  credit  been  days  iustoad  of  mouths,  we  would  theu  have 
found  the  interest  on  the  whole  sum  for  one  day. 

The  equated  time  will  be  the  same  regardless  of  the  rate  of  interest  used.  We 
used  12  per  cent  because,  in  the  most  of  cases,  it  is  easier,  and  requires  less  time  to 
calculate  interest  at  12  per  cent  than  at  any  other  rate. 

By  this  method  of  work,  interest  tables  can  be  used  to  advantage  by  those 
who  desire.  The  interest  method,  however,  we  do  not  regard  with  much  favor.  Tlie 
product  method,  or  system,  is  better  and  more  easily  comprehended  by  learners. 

Note. — The  accuracy  of  the  result  of  the  foregoing  methods  of  -work  has  been  questioned  by 
several  good  authors,  and  much  warm  discussion  has  been  indulged  in  on  the  subject;  but  as  the 
arguments  advanced  against  the  correctness  of  these  methods  are  alike  subtle  and  fallacious,  we 
shall,  therefore,  be  very  brief  iu  considering  them. 

The  following  example  and  argument  is  generally  used  to  elucidate  the  asserted  incorrectness 
of  the  foregoing  methods  of  fiudiug  the  eciuated  time: 

"  If  a  man  owes  me  $200,  $100  of  which  is  payable  now,  and  $100  payable  in  two  years,  the 
equated  time  is  not  one  year.  For  in  deferring  the  payment  of  the  first  $100  one  year,  he  ought  to 
pay  the  $100,  plus  the  interest,  which,  at  6  per  cent,  is  $106 ;  but  for  the  $100  which  he  pays  one 
year  before  it  is  due,  he  ought  to  pay  the  present  worth  of  $100,  which,  at  6  per  cent,  is  $94.33|i  ; 
aud  $106  +  94.335^  =  $200.33f ^ ;  whereas,  by  methods  in  general  use,  he  only  pays  $200." 

Ey  this  argument  it  is  contended  that  I  lose  $6,  the  interest  on  $100,  for  2  years  only,  and 
gain  oniy  $5.66,  the  true  discount  on  $100  for  1  year,  making  a  diB'ereuce  of  nearly  34  cents. 

Rut  the  argument  is  fallacious,  and  the  error  is  in  supposing  that  I  lose  $6,  when  really  I 
only  lose  the  present  worth  of  $6,  for  the  $6  interest  is  not  due  nutil  the  close  of  two  years,  and 
heuce  I  should  receive  $100,  plus  the  present  worth  of  $6,  due  in  one  year,  which  is  $5.86+  and 
which  added  to  the  $100  due  the  first  year  =  105.66+  that  I  should  receive ;  but  as  I  only  receive 
$100,  I  therefore  actually  lose  $5.66+ which  is  just  equal  to  the  gain  on  the  $100  received  for  the 
present  worth  ($94.33|;  j  of  the  $100  for  one  year  before  it  was  duo. 


6x0 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  A  owes  Z  $200,  payable  now;  $500  payable  in  4  months;  $1000  payable 
in  6  mouths;  and  $14:00  payable  in  9  mouths.  What  is  the  equated  time  lor  paying 
all  1  Aus.  6  months  and  19  days. 


OPERATION. 


Del)it  of  Pollars. 

$2pp 

5pp 

lOp0 

14p0 


X 
X 
X 
X 


Credit  of  Months. 

0 
4 
6 
9 


Credit  of  Months  on  $1. 

0 
20 
60 

126 


Total  debit  of  dollars  $31 


Total  credit  of  months  on  $1 


206 


206  -^  31  =  6|o  months  =  G  mouths  19^  days. 


Explanation. — We  here  see  that  A  owes  -varioHs  sums  of  money,  and  has  due  hira  various 
tsrms  of  credit  of  time,  and  as  shown  in  the  first  prolileni,  wi  first  find  a  total  credit  of  time  on  $1. 
The  reasoning  for  the  work  is  the  same  as  given  in  the  first  problem,  hence  omitted.  Before 
making  the  multiplications,  we  cancel  each  sum  of  dollars  by  100,  and  thus  save  figures.  When 
the  fraction  of  a  day  in  the  answer,  or  efjuated  time,  is  more  than  i,  )t  is  customary  to  couut  1 
■whole  day  ;  aud,  when  less  thau  i  to  omit  it. 

3.  A  merchant  owes  4  notes,  dated  January  1,  1895.  The  first  is  for  $500, 
payable  in  30  days;  the  second  is  for  $800,  payable  in  40  days;  the  third  is  for 
$1200,  payable  in  90  days ;  and  the  fourth  is  for  $2000,  payable  in  120  days.  What 
is  the  equated  time  for  the  payment  of  all  ? 

Aus.  88  days,  which,  added  to  Jan.  1,  1805,  the  date  of  the  notes, 
gives  March  30,  1895,  as  the  date  to  pay  all  at  one  time. 


Dr.  of  Dolls. 

Or 

of  Days. 

Cr. 

of  days  on  ijil 

$500 

X 

30 

= 

15000 

800 

X 

40 

= 

32000 

1200 

X 

90 

= 

108000 

2000 

X 

120 

= 

240000 

$4500  total  debit  of  dollars.  395000  total  credit  of  days  on  $1. 

395000  ^  4500  =  87^  days. 
4.     A  loaned  Z  $600  for  9  months,  how  long  ought  Z  to  loan  A  $800  to  equal 
the  favor?  Aus.  6  months,  22  days. 

OPERATION. 


Dr   of  Dolls.       Cr.  of  Months.      Months  Cr.  on  $1.  800 

$600         X         9         =  5400 

5.    A  bought  of  Z  the  following  invoices : 

January    2,  -        -        -        ■ 

J,  -        -        -        - 

"         15,  - 

"         21  -         -         -         • 

"         30,'  .... 

What  is  the  equated  time  for  paying  all  ? 


5400 

G;}       months  =  6  inos.  22J  ds.  *  ' 


$400 

210 

350 

800 
1100 

Ans.    January  20. 


EQUATION    OF    ACCOUNTS.  63 1 


FIRST   OPERATION   WORKING  FROM   THE   EARLIEST   DATE. 

Cr.  of  days  011  $11. 

1470 

4550 

15200 

30S00 


Dr.  of  Dolls. 

Cr.  of  ilays 

January    2,    $400 

X 

0 

"           9,      210 

X 

7 

'•          15,      350 

X 

13 

"         21,      800 

X 

19 

"         30,    1100 

X 

28 

$2860  )  52020  (  18  dnys. 

Explanalion. — In  this  problem  tbe  various  debits  of  dollars  Mere  credited  at  difl'erent  dates, 
and  as  tliere  was  no  credit  of  time  allowed,  each  item  of  debit  is  therefore  due  at  the  time  of  pur- 
chase. 

In  this  operation  we  assume  the  earliest  date  January  2,  as  the  due  date  or  focal  date  and 
tbeu  proceed  to  liiul  the  time  that  each  purchase  or  debit  item  is  due  after  tbe  assumed  due  dale. 
We  reason  thus:  The  $400  beiiij;  due  tbe  day  of  purchase,  it  has  0  days  credit  of  time.  Tbe  $210 
purchase  on  tbe  9th  is  really  due  that  day  ;  but  presuming  it  dine  on  the  earliest  date,  January  2,  Z 
therefore  owes  A  iime  from  January  2  to  January  9,  which  is  7  days'  credit  ou  the  same.  The  |i350 
hoiigbt  ou  the  15th  is  really  due  that  day  ;  but  presumiug  it  due  on  the  earliest  date,  January  2,  Z 
owes  A /i»ie  from  Jauiiary  2  to  January  15,  =  13  days'  credit  on  tbe  same.  The  same  reasoning 
applied  to  tbe  $800  and  §1100,  purchases  of  the  21st  and  30th,  gives  respectively  19  and  28  days' credit 
due  to  A.  Having  thus  made  or  presumed  all  the  purchases  due  January  2,  we  proceed  to  tiud  tbe 
credit  that  Z  owes  on  SI,  and  tlien  on  the  whole  amount  in  tbe  same  manner  as  explained  lu  the 
first  problem  of  ef|iiations.  Tbe  result  of  the  e(|uation  is  18  days'  credit  duo  A  on  tbe  whole 
amount,  $2860,  which,  for  the  reason  that  we  make  all  tbe  debit  items  due  January  2,  must  be 
ad<led  to  tbat  date  in  order  to  give  A  tbe  credit  due  to  bim;  18  days  added  to  January  2  gives 
January  20  as  the  equated  time  for  paying  the  whole  amount,  S28G0. 

To  more  fully  elucidate  tbe  Trork  of  equation,  we  present  tlie  following: 

SECOND   OPERATION   WORKING   FROM   THE   LATEST   DATE. 

Dr.  of  days  ou$I. 
11200 

4410 
5250 
7200 

$2oGp  )  2SO60  (9||| days, practically  10  days. 

Explanation.— 'By  working  from  the  latest  date,  January  30,  we  see  that  as  each  debit  item 
*was  due  the  day  that  it  was  purchased,  that  the  $400  bought  January  2  was  therefore  due  28  days 
prior  to  January  30;  and  hence  A  owes  the  time  as  vvell  aa  the  money.    The  $210  bought  on  the 
prior  to  January  30;  and  hence  A  owes  the  tune  as  well  as  the  money.     The  $210  bought  on  the 
t)th  was  due  21  days;  and  again  A  owes  Z  the  time  and  money  ;  tbe $350  aud  $800,  purchased  respec- 
tively January  15  and  21,  are  due  respectively  15  and  9  days,  for  which  A  owes  Z  tbe  time;  tbe 
$1100  bought  on  tbe  30th  is  due  that  day,  aud  hence  A  owes  no  time  thereon.     Having  now  the 
days'  time  that  A  owes  Z  on  the  various  items  of  dollars  to  lind  his  debit  of  days  on  $1,  and  then 
on  the  whole  amount  that  be  owes,  we  proceed  to  multiply  aud  divide,  as  shown  in  the  operation, 
and  for  the  reasons  given  in  tbe  first  jiroblem  ou  equations. 

Tbe  result  of  tbe  work  is  10  day.s'  debit  of  time  tbat  A  owes  Z  on  tbe  wbole 
amount  of  $2800 ;  and  to  give  Z  tbe  credit  or  advantage  of  tbis  time,  it  is  clear  tbat 
tbe  10  days  must  be  counted  backward  from  January  30,  tbe  day  tbat  all  of  tbe  debit 
items  of  dollars  were  due;  and  by  deducting  tbe  10  days,  from  January  30,  we 
obtain  January  tbe  20tb  as  the  equated  time  for  tbe  i)aymeut  of  tbe  wbole  amount. 


Dr 

of  Dolls. 

Dr. 

of  Days 

January    2, 

$400 

X 

28 

9, 

210 

X 

21 

»       15, 

350 

X 

15 

"       21, 

800 

X 

9 

"       30, 

1100 

X 

0 

632 


SOULE  S    rHILOSOPHIC    I'RACTICAL    MATHEMATICS. 


G.     What  is  the  equated  time  for  the  payment  of  the  following  account? 

Ans.     October  5,  1895. 

S.  J.   &   M.   iM.   KOIILMAN. 

To  Levy  &  IIiller,  Dr. 


181)5. 

May 
June 
Aii};nst 
September 


To  inercbandise  at  2  months 
"  cash  .         .         .         - 

"  merchandise  at  3  mouths 
"  merchandise  at  3  months 


541 
210 
800 
525 

$2078 


85 
30 
00 
90 

05 


SOLUTIONS. 
FIRST  OPERATION,    ASSUMING   THE   EARLIEST   DATE,    MAY    1,    To   WORK   FROM. 


Dr.  of  Dolla. 

Contract  Cr 

.  Ds. 

Cr. 

Davs 

assumin^r  each 

Cr.  of 

1' 

urchase  due  May  1. 

Ds.  on  $1. 

May  1, 

$541.85 

61 

+ 

00 

= 

61 

33063 

June  28, 

210.30 

00 

+ 

58 

=:r 

58 

12180 

August  16, 

800.00 

92 

+ 

107 

— 

199 

159200 

September 

20, 

525.90 

91 

+ 

142 

= 

233 

122558 

$2078.05 


327000 


327000  (hiys  credit  on   $1  -^  2078  =  157  days  equated  time  for  the  payment 
of  the  whole  amount,  alter  May  1,  1S95,  which  is  October  5,  1895. 

^^^P''"><''>o»-—ln  this  problem,  like  the  preceding,  the  yarious  debits  of  dollars  were  created 
at  different  periods  of  time ;  but  unlike  the  preceding  problem,  all  the  items,  except  the  cash  have  a 
credit  "t  time  allowed  the  ]Hirchaseis.  The  principles  involved  in  the  preceding  problems  are 
alike  applicable  to  the  solution  of  this  and  all  other  questions  in  equations. 

In  order  to  find  the  total  credit  of  time  due  the  purchasers  on  $1,  we  first  find  the  credit  that 
each  debit  of  dollars  is  entitled  to  by  the  terms  of  the  contract  from  the  assumed  due  date  Mav 
1,  in  the  operation.  The  debit  of  |541.85,  of  May  1,  has,  by  the  conditions  of  the  purchase,  2 
mouths  (61  days)  credit.  -^  ' 

The  debit  of  |210.30.  Juno  28,  has  no  contract  credit  of  time ;  but  because  we  presume  it  due 
May  1,  It  has  from  May  1  to  June  28,  58  days  credit  from  that  presumed  time. 

...  -■      T?"  •'•^^i*  "f  ^*^^'  August  16,  has  3  months,  93  days  contr.-jct  credit ;  and  because  we  assumed 
it  due  May  1,  it  has,  Irom  Jlay  1  to  August  16,  107  days  additional  credit,  making  199  days  credit. 

The  debit  of  $525.90,  of  September  20,  has  3  months,  91  days  contract  credit;  and  because 
■we  presume  or  assume  it  due  May  1,  it  has,  from  Jlay  1  to  September  20,  142  davs  additional  credit 
making  233  days  credit  from  May  1.  Having  thus  the  credit  of  time  for  each  item  of  debit  of 
dollars,  we  proceed  to  find  the  equated  time,  the  same  as  in  the  preceding  problems  with  this 
''^".^^^IZ":  I"  ™»ltiply"igtli«  first  and  last  terms  of  days  and  dollars,  we  used  resnect'ive Iv  *542 
by  rif  i -'^'adlf ^.V/o'  tr'  •'";-"r%*5-fl«5  a."l  $525.90 ;  and  in  th'e  second  termCe'm^il^ipl^l 
by  $210,  instead  of  $210.30;  and  in  the  division,  we  omitted  the  cents  in  the  divisor. 

on  operations  of 
by  $1 ;  a-nd  when- 
■  ,  ,     ,.^     •      P'"'af"'K  thus  with  the  cents,  time  and  labor 
my  perceptible  difference  in  the  result. 

The  result  of  the  operation  is  157  days  credit  due  the  purchasers  on  the  whole 
amount,  $2078.05,  affer  May  1,  the  assumed  diite  that  each  debit  of  dollars  was 
due.    This  added  to  May  1,  gives  October  5,  1S95,  as  the  equated  time 


_  In  practice  the  cents  need  not  and  by  accountants,  are  not  used  in  enuatin 
iplyiug  and  dividing.  Whenever  they  excee.l  fifty  the  amouut  is  increased  bv 
they  are  less  than  fifty,  they  are  omitted.     Hy  operating  thus  with  the  cent.s 


mnltipl 
ever 

are  saved  without  etl'cctiu; 


*  EQUATION    OF    ACCOUNTS.  633 

SECOND  OPERATION,   ASSUMING   TUE   LATEST   DATE,    SEPTEMBER  20,    1S95,  AS  THE 

DUE   OR   FOCAL   DATE. 

Dr.  of    Coutract 
U0II9.    Cr.  ol"  Us. 

May  1, 
June  28, 
Aug.  IG, 
Sept.  20, 


Dr.  of 

Coutract 

Dr. 

Cr. 

Dr.  of  Ds. 

C 

r.  of  Ds. 

U0II9. 

Cr.  of  Us. 

ol'  l)s. 

of  Us. 

on  $1. 

on  $1. 

$541.85 

Gl 

— 

142 

^^ 

81 

43902 

210.30 

00 



84 

= 

84 

17640 

800.00 

92 

— 

35 

57 

45000 

525.90 

91 

— 

0 

91 

47SCG 

$2078.05                                                                                         G1542  934GG 

Debit  of  days  on  $1 G1542 


Excess  of  credit  of  days  on  $1  is 31924 

31924  —  $2078  =  lo/oVs  days,  practically  15  days. 

Explanation. — In  this  oper.atiou -we  assume  the  Litest  d.ite  in  the  account  to  -work  from  and 
reason  thus:  The  $541.8.5  purchaseil  May  1,  uot  being  settled  for  until  September  20,  the  purchasers, 
therefore,  owe  time  ou  the  same  from  May  1  to  .September  20,  -nhich  is  142  days ;  this  we  set  in  the 
column  of  debit  days,  and  deduct  therefrom  the  61  days  credit  allowed  by  the  terms  of  the  contract, 
whicli  leaves  an  excess  of  81  debit  days. 

Then  the  $210.30  of  June  28,  not  being  settled  for  until  the  assumed  date,  September  20,  the 
purchasers  owe  time  on  the  same  from  J  uue  28  to  Sejitember  20,  -which  is  84  days;  this  -we  set  in 
the  column  of  debit  diiys,  and  having  no  contract  credit  to  deduct  therefrom,  it  remains  a  net 
debit  of  days.  Then  tlie  $800  debit  of  August  16  not  being  settled  for  until  September  2<>,  the 
purchasers  owe  time  on  the  same  from  the  date  of  purchase  to  the  assumed  date  of  settlement, 
which  is  35  days;  this  we  also  set  In  the  column  of  debit  days,  and  deducting  the  same  from  the 
92  days  credit  allowed  by  the  contract  of  purchase,  \ee  have  an  excess  of  57  credit  days,  which  we 
extend  to  the  column  of  credit  days.  Then  the  $525.90  debit  of  September  20  being  by  assumption 
settled  on  that  day,  there  is  no  debit  of  time,  and  hence  the  whole  credit  of  d.iys,  91,  allowed  by  the 
terms  of  the  agreement,  is  extended  to  the  column  of  credit  days.  Having  now  found  the  varioas 
debits  and  credits  of  days  on  the  different  debits  of  doll.ars,  wo  next  find,  by  multiplication,  the 
debit  and  credit  of  days  on  §1,  and  then  the  difference,  which  is  31924  days  credit  on  $1 ;  this  we 
divide  by  the  wliole  amount  of  debit  dollars,  and  thus  obtain  15  days,  the  equated  time  whicli,  for 
reasons  previously  given,  is  counted  forward  from  the  assumed  date,  and  gives  October  5,  1895,  as 
the  final  result. 

7.     A  owes  Z  $2000  due  January  1,  1895 ;  May  1,  1894,  Le  pays  Z  $800 ;  when 
is  tlie  balance  due?  Ans.     June  12,  1895, 

OPERATION. 

$2000 

800  245  days  credit  on  $800  (from  .May  1,  1894,  tu  January  1, 1895,)  =  19GO0O 

days  credit  on  $1. 

$1200 

12J30  )  196000 


163^      days  credit  on  $1200  after  January  1, 1895. 
Explanation. — 'WTien  the  debtor  makes  partial  payments  on  the  amount  that  he  owes  before 
it  is  due,  it  is  clear  that  he  is  entitled  to  interest  on  the  money  paid,  or  to  a  credit  on  the  balance 
after  it  becomes  due,  which  is  equal  to  the  credit  on  the  amount  paid  before  it  is  due. 

8.  A  is  owing  Z  $2000,  which  is  due  by  contract  July  4,  1894.  On  the  14th 
of  February,  1894,  A  wishes  to  pay  Z  such  a  sum  as  will  Justly  exttud  the  payment 
of  the  balance  until  October  7,  1894.  What  amount  must  he  pay,  and  what  amount 
■will  remain  unpaid  ?  Ans.  $808.51,  to  be  paid. 

$1191.49,  to  remain  unpaid. 

From  Feb.  14,  1894,  to  July  4,  1894,  are  140  days. 
From  July    4,  1894,  to  Oct.  7,  1894,  are    95     " 

235     " 


634 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Here  we  see  that  tbe  time  from  the  date  of  payment  (Feb.  14)  to  maturity  of 
tbe  wbole  sum  (July  4)  is  greater  tliaii  tlie  time  from  the  maturity  to  tbe  date  of 
tbe  extension,  Oct.  7  ;  and  as  tlie  credit  of  days  on  $1  is  in  direct  proportion  to  tbe 
amount  of  dollars  multiplied  by  tbe  credit  of  time  thereon,  it  is  clear  that  we  here 
require  a  less  sum  to  be  i)aid  before  due  than  to  remain  unpaid  after  it  is  due,  to 
])r()duce  equivalent  credits;  hence,  tbe  amount  to  be  paid  before  maturity  and  also 
the  atnount  to  remain  uni)aid  after  maturity  ■will  be  in  reciprocal  proportion  to  tbe 
two  xieriods  of  time,  140  and  95  days. 

Tbe  following  statements  produce  tbe  desired  reciprocal  results: 


OPERATION 
To  Dud  the  amount  to  be  paid  before  maturity. 


235 


2000 
95 


$808.51  Ans. 


OPERATION 
To  find  tbe  amount  to  be  paid  after  maturity. 


235 


2000 
140 


$1191.49  Aus. 


i 


To  show  tbe  whole  work  of  reciprocal  proportion  involved  in  tbe  foregoing 
statement,  we  i^reseiit  the  following: 

1     -f.     140     = 
1     -       95     = 


1  4  0 


_ J5  _ 
13  3  0  0 

_14  D_ 
13300 


_S3a_ 
13300 


By  this  we  see  that  tbe  reciprocal  of  140  is  y 
added  give  tI.too-     With  these  reciprocal  proportional  numbers,  we  make  the  fol- 
lowing solution  statements: 


OPERATION 
To  find  tbe  amount  to  be  paid  before  maturity. 


235 
13300 


2000 

13300 

95 

$808.51  Ans. 


OPERATION 
To  find  tbe  amount  to  be  paid  after  maturity. 


235 
13300 


2000 

13300 

140 

$1191.49  Aus. 


To  show  that  the  credit  of  time  on  $808.51  for  140  days,  is  equivalent  to  tbe 
credit  of  time  on  $1191.49  lor  95  days,  we  present  tbe  following: 

Or.  of  Ds.  Cr.  of  Ds.  on  $1. 

$80S.ai  X  140  =  113191  + 

1191.49  X  95  =  113191  + 


$2000.00 
Note. — To  tbose  accustomed  to  worlj  on  the  interest  system,  we  could  say  that,  in  the  main, 
the  same  process  of  reasoning  as  given  above,  would  be  used.     Where  we  use  the  term  credit  of 
days,  the  interest  method  of  reasoning  would  require  the  term  credit  of  interest. 

9.     Sold  merchaudise  to  Lacy  &  KoGERS,  as  follows: 
1895.  1895. 

Feb.        1,      $500.00  at  CO  days.  ]\Iay  21,    $200.00  at  00  days. 

March  18,      1019.08  "  90     "  July  28,      400.00  "  60     " 

M;iy        6,         704.25  "  30     " 

AVhen  does  tbe  account  mature  by  equation  1  Ans.     June  13,  1895. 


EQUATION    OF    ACCOUNTS. 


635 


OPERATION   BY 

FIRST 

METHOD. 

1895. 

Dr.  of  Dolls. 

Cr.  ol 

Ds. 

Feb.        1, 

$500.00 

GO 

+ 

0 

=               GO 

March  IS, 

1019.CS 

90 

+ 

45 

=         135 

May       G, 

704.25 

30 

+ 

94 

=          124 

4i                 01 
.-  1, 

200.00 

00 

+ 

109 

=          109 

July     28, 

400.00 

CO 

+ 

177 

=         237 

Or.  of  Ds.  on  $1. 
30000 
137700 
87296 
21S00 
94800 


$2823.93 


371596 


371596  days  credit  on  $1  -  $2824  =  132  days  credit  on  $2823.93,  wbich 
added  to  February  1,  the  assumed  maturity  of  each  item  of  debit,  gives  June  13  as 
the  equated  maturity  of  the  account. 

10.  Suppose  in  the  above  i)robleni  that  the  purchasers,  Lacy  &  Rogers, 
Lad  desired  on  August  1,  to  settle  the  account  by  a  "  cash  "  note,  payable  November 
1/4,  1S95,  allowing  interest  at  8  jier  cent,  for  what  face  must  the  note  be  drawn  ? 

Ans.  $2916.91. 

OPERATION    INDICATED. 

Interest  on  $2823.93  from  June  13  to  August  1,  =  49  days  at  8  per  cent  = 
$30.75.  $2823.93  +  $30.75  =  $2854.68,  due  Augu.st  1,  and  for  which  a  cash  note  at 
8  jier  cent  is  given  payable  November  1/4,  =  95  days,  +  discount  day,  =  96  days. 
4500  X  $2854.68  -^  (4500  —  96)  =  $2916.91,  face  of  cash  note. 

C.  B.  LoTT  &  Co. 

To  CAniLL  &  Leidenheimer  J)>: 


1896. 

Jan. 

20 

Feb. 

15 

M.irrh 

9 

April 

1 

To  nieroli;in(lise  at  4  nionllis 

,,   3       ,.  .         . 

••   4       ••  -         - 

"  acceptance      "   2       "         (with  grace), 


1500 
840 
200 
500 


3041 


00 
75 
28 
50 

53 


11.  May  1,  LoTT  &  Co.  settle  liy  note  drawn  for  such  a  time  as  will  mature 
at  the  equated  date  of  the  account.     For  how  many  days  will  the  note  be  drawn  ? 

Ans.  23  days  without  grace,  or  20  days  with  grace. 

NoTK  1. — In  tbe  operation,  count  actual  days  and  allow  for  leap  years. 

Note  2. — The  equated  date  is  125  days  after  January  20  =  May  25,  less  1  day  for  leap  year 
=  May  24th. 

12.  Suppose,  in  the  above  problem,  that  Lott  &  Co.  had  settled  the  account 
May  1,  by  paying  cash,  what  amount  would  they  have  paid,  money  worth  8  per  cent?  ■ 

Ans.  $3025.98. 
OPERATION   INDICATED. 

Since,  as  shown  in  the  above  problem,  the  equated  date  for  the  payment  of 
the  account  was  May  24,  therefore  if  the  same  is  paid  on  May  1,  tbe  amount  is  to 
be  discounted  for  23  days. 


Thus 


45 


$3041.53 
23 

$15.5455+  discount. 


$3041.53  amount. 
15.55  discount. 


$3025.98  net  proceeds. 


636 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


liEMARKS. — It  will  be  observed  that  in  all  of  our  work,  where  dates  were 
given,  we  have  counted  the  actual  days  intervening  between  two  dates.  This  we 
believe  to  be  the  only  correct  systein. 

Many  accoiiiitants  however,  to  save  a  little  time,  count  30  days  to  the  month, 
by  whi(di  the  final  result  is  often  changed. 

To  find  the  time  between  dates  and  also  in  extending  the  equated  time  forward 
or  counting  it  backward  from  the  date  assumed,  whether  tlie  earliest  or  latest,  a 
Time  Table  is  often  used  to  advantage.     See  Time  Table  on  page  298. 


COMPOUND  EQUATIONS,  OR   AVERAGE  OF  ACCOUNTS. 

1198.  When  the  equated  time  is  counted  forward  and  when  it  is  counted 
backward. 

In  all  comjwnnd  cquatians  0/  payments  or  average  of  accounts,  when  the  balance 
of  dollars  and  the  balance  of  dai/s  are  on  the  same  side  of  the  account  (the  debit  or  the 
credit)  the  equated  time  is  counted  forward.  When  the  balance  of  dollars  and  the 
balance  of  days  are  on  the  opposite  sides  of  the  account,  the  equated  time  is  counted 
backward. 

An  account  with  the  earliest  date  on  the  debit  side  and  the  final  balance  of 
dollars  and  balance  of  days  on  the  debit  side,  equated  time  is  counted  forward  irhen  the 
earliest  dale  is  assumed,  and  backward  when  the  latest  date  is  assutned  as  the  maturity 
of  the  dollar  items. 

1.     What  is  the  equated  time  for  the  payment  of  the  following  account  1 

Ans.     July  17,  1895. 
Dr.  Patton,  Swift  &  Co.  Cr. 


1895. 
Mar. 
April 
Jiiue 


To  mdse.,  at  60  days 

"         "       "  'JO     " 
"         "       "  30     " 


1895. 

$800 

00 

April 

10 

2000 

00 

May 

28 

1200 

00 

Aug. 

9 

400 

00 

$4400 

00 

By  cash,     - 

tt       it 

"    balance, 


$500 

1000 

800 

2100 


$4400 


00 
00 
00 
00 

00 


FIRST    OPERATION 
Assuiiiins  the  earliest  date  in  the  account  to  work  from. 


Dr  of 

Contrart 

Cr.  of 

Cr.  of  U.S. 

1895 

Cr.  of 

Dr.  of 

Dr.  of 

Dolls. 

Cr   ol  Us. 

Us. 

oil  -f  1. 

UoUs. 

Us. 

Ds.  ou$l. 

$800 

60   + 

0 

=     00 

48000 

April  10, 

$500 

3G 

18000 

2000 

90  + 

48 

=  1,38 

27(;()00 

May 

28, 

1000 

84 

840(10 

1200 

30  + 

88 

=  118 

141000 

Aug. 

9, 

800 

157 

125G0O 

400 

00 

107 

42800 

$2300 

$4400 

508400 

227600 

2300 

227000 

1895. 

Mar.     5, 

AjHil  22, 

June     1, 

"      20, 


Bal,  of  $        2100  Bal.  of  days  280800 

280800  -^  2100  =  134  days  credit  due  Patton,  Swift  &  Co.  on  the  balance 


EQUATION    OF    ACCOUNTS. 


637 


$2100,  and  wliicb,  for  reasons  given  in  the  first  problem,  is  counted  forward  from 
tbe  assumed  maturity  of  all  tbe  debit  and  also  credit  items  of  dollars,  and  gives 
July  17, 1S95,  as  tbe  equated  time  for  settling  tbe  balance  of  tbe  account. 

Explanation. — In  tliis  problom,  Patton  Swift  &  Co.  have  both  a  debit  and  credit  of  dollars, 
and  also  a  debit  and  credit  of  days,  but  by  our  method  of  work,  the  solution  is  not  in  the  least 
complicated  by  reason  of  the  credit  of  dollars  and  debit  of  days.  The  operation  is  necessarily 
longer,  but  the  two,  and  only  two  principles  that  we  use  the  (debit  of  dollars  and  credit  of  dai/s)  are 
as  api>licable  to  questions  of  this  character  as  to  those  containing  a  debit  of  dollars  and  credit  of 
days  only.  In  the  main,  the  operation  of  this  jirolilem  is  tbe  same  as  in  the  preceding  exauiples. 
We  lirst  assume  the  earliest  date,  March  5,  as  tbe  due  dale  of  all  tbe  items  of  debit  and  credit  of 
dollars  on  both  sides  of  the  account,  and  then  found,  as  is  fully  explained  in  the  preceding  work, 
and  adjusted  the  proper  credit  and  debit  of  dai)3  on  each  item  of  debit  and  credit  of  dollars,  accord- 
ing to  the  conditions  of  tbe  contract  and  the  assninption  of  March  5,  as  the  maturity  of  all  the 
dollar  items  of  the  account.  We  thus  produce  a  net  credit  of  280800  days  on  $1,  due  to  Patton, 
Swift  &,  Co.,  ■which,  as  shown  m  the  operation,  is  equal  to  131  days  ou  thebalauce,  $i2100  that  they 
owe. 

SECOND   OPERATION 
Assuming  the  latest  date  in  the  account  to  work  from. 


1895. 

Pr  of 
Uolls. 

Cr  of 
Us 

Dr 
JJ. 

or 

- 

Dr    of 
Da  ou$L 

1895. 

Cr.  of 
Dolls. 

Cr.  of 
Us. 

Cr.  of 
Da.  on  |I. 

Mar.     5, 

$  800 

60     - 

-     157 

= 

97 

77600 

Apr.  10, 

$  500 

121 

60500 

Ai.r.  22, 

2000 

ilO    - 

-    109 

z=r 

19 

38000 

May  28, 

1000 

73 

73000 

June    1. 

1200 

30    - 

-      69 

■=: 

39 

46800 

Aug.    9, 

800 

0 

"    20, 

400 
$1400 

00    - 

-      50 

— 

50 

20000 

182400 

$2300 

133500 

2300 

133500 

Bal.  of  $ 


2100 


Bal.  of  days    48900 


48900  —  2100  =  23  days  tbat  Patton,  Swift  &  Co.  owe,  August  9,  on  tbe 
balance  $2100,  and  to  give  tbe  parties  to  wbom  tbis  23  days  credit  is  due,  tbe 
advantage  of  it  in  tbe  settlement  of  tbis  account,  it  is  clear  by  tbe  exercise  of  our 
reason  tbat  we  must  count  backward  from  August  9,  which  gives  us  July  17,  1895, 
as  tbe  equated  time. 

Having  in  the  last  problem  and  in  preceding  work  on  equations  illustrated  the 
principles  and  operation  of  work  by  assuming  either  tbe  earliest  or  latest  date  as 
tbe  maturity  of  all  tbe  debit  and  credit  items  of  dollars,  and  preferring  tbe  method 
of  assuming  tbe  earliest  date,  to  save  space,  we  shall  therefore  give  but  one  solution 
of  subsequent  problems,  and  that  by  the  earliest  date  method. 

An  account  icith  the  earliest  date  on  the  credit  side,  the  balance  of  dollars  and 
days  on  the  debit  side,  equated  time  counted  foricard  from  the  earliest  date. 

2.    What  is  the  equated  date  of  the  balance  of  tbe  following  account  ? 

Ans.     Jan.  G,  1S9G. 


Br. 

Weil,  Honor  &  Fox. 

Cr 

1895. 

1895. 

Oct. 

4 

To  Mdse.,  at  2  months 

$2,500 

00 

Sept. 

8 

By  Cash, 

$  800 

00 

Nov. 

1 

"  Cash, 

3000 

00 

Oct. 

20 

&4                tfc 

1500 

00 

Dec, 

18 

"  Acpt.  at  1  month 

1000 

00 

i( 

2r, 

U                U 

200 

00 

Nov. 

14 

U                1( 

700 

1 

00 

638 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


1895. 

Oct.     4, 
Nov.     1, 
Dec.  18, 

Dr.  of           Cr.  of 
Dolls.             Ds. 

12500             61             26         = 
8000               0 
1000            34          101        = 

87 

54 

135 

119M, 

Or.  of 

Ds.  on  $1. 

217500 
162000 
135000 

514500 
119300 

1895. 

Sept.     8, 

Oct.     20, 

"      25, 

Nov.    14, 

Cr.  of 
Dolls, 

$  800 

l.-i00 

200 

700 

Dr.  of 
Da. 

0 
42 
47 
67 

Dr.  of 
Ds.  on  .$1. 

63000 

9400 

46900 

16500 
3200 

$3200 

119300 

$3300  l)a].  of  dolls. 

395200  -^  3300  = 

395200  bal.  of  days. 

jiractically  120  days 

Explanalion. — Thi.s  120  days  credit  of  time  on  the  balance  of  dollars,  is  due  to  Weil,  Honor 
&  Fox  ;  hence  the  time  of  payment  is  extended  120  days  after  the  due  date. 

An  accomit  tcith  the  earliest  date  on  the  debit  side,  the  balance  of  dollars  and 
days  on  the  credit  side,  equated  time  counted  forward  from  the  earliest  date. 
3.    When  is  the  balance  of  the  following  account  due  ? 

Ans.    Feb.  21, 1895. 

Br.  Woods  &  Winn.  Cr. 


1895. 
Jan. 
Feb. 
Ajjiil 


21  To  Cash, 
1    "   Mdse., 


at  2  mouths, 


1895. 

$1420 

00 

Feb. 

1 

1800 

()(» 

Mar. 

10 

950 

40 

By  Cash, 

a         u 


$5000 
7000 


00     II 

00 


1895. 

Jan.  21, 
Feb.  1, 
Apr.      7, 


OPERATION. 
Assuming  the  earliest  date  as  the  maturity  of  the  debit  and  credit  items. 


Dr.  of 
DoUs. 

$1420.00 

1800.00 

9.50.40 

$4170.40 


Cr.  of 
Ds. 

0 

0 

61 


11 
137 


Cr.  of 

Ds.  on  $1. 


19800 
]3()15() 


149950 


1895, 

Feb. 
JIar. 


1, 
10, 


Cr.  of 
Dolls. 

$5000.00 
7000.00 


$12000.00 
4170.40 


Dr.  of 

Ds. 

11   : 
48   : 


Dr.  of 
Ds.  on  %\. 

55000 
336000 


391000 
149950 


241050  days 


Hal.  of  .lollars,     $7829.60 

7830  ($7829.60)  gives  31  days. 


Bal.  ds.,     241050 


Explanalion. — This  31  days  time  Woods  &  Winn  owe  to  the  parties  with  whom  they  have 
this  account,  on  the  balance  $7829.60,  and  as  this  balance  of  money  is  due  to  them,  it  is  clear  that, 
in  order  to  settle  the  31  days  time,  the  paynuMit  of  the  balance  must  be  extended  31  days  after 
January  21,  the  assumed  maturity  of  all  the  dollar  items  of  the  account,  which  gives  February  21. 

An  account  with  the  earliest  date  on  the  debit  side,  the  balance  of  dollars  on  the 
debit  side  and  balance  of  days  on  the  credit  side,  equated  time  counted  bacTcicard  from 
the  earliest  date. 

4.    What  is  the  equated  time  for  the  payment  of  the  balance  of  the  following 


account? 


Ans.     Aug.  5,  1894. 


EQaATION    OF    ACCOUNTS. 


639 


Dr. 


Spencer  &  Folsom. 


Vr. 


'     1895. 
1   July 
1    Aug. 
1    Sept. 
Oct. 

1 

10 
25 

6 

To  Mdse.,  at  60  ds 

u         u       u  60   " 
"         "       "  30   " 
((         u       t.i  10    ■' 

• 

$1200 
1300 
1100 
2000 

00 
00 
00 

(to 

1895. 
July 

Aug. 

u 

Oct. 
Dec. 

')0 

10 

10 

G 

1 

By  CasL, 

''   Acpt.  at  30  ds. 

"   Cash, 

"   Note  at  90  ds. 

$  500 

GOO 

500 

1500 

2000 

00 
00 
00 
00 
00 

OPERATION. 

AVorkiug  from  tbe  earliest  date. 

' 

1895. 

Dr.  of     Cr.  of 

Dolls.         1)8. 

Cr.  of 
Ds.  on  $1. 

1 SQCL                 Cr.  of         Dr.  of 

Dr.  of 

Us.  OD  $1. 

July     1 
Aug.  10 
Sept.  25 
Oct.      6 

$1200        60      0    = 
1300        60    40    = 
1100        30    «6     = 
2000        10    97     = 

60 
100 
116 
107 

72000 
130000 
127600 
214000 

July   22, 
Aug.   10, 
"     10, 
Oct.      6, 
Dec.      1, 

%  DOO 
600 

500    33      40     = 
1500 
2000     93     153     = 

21 
40 
73 

97 

246 

10500 

24000 

36500 

145500 

492000 

$5600 
5100 

543G00 

$5100 

708500 
543600 

$500  balance  of  dollars. 


Balance  of  days,     164900 


164900  days  divided  by  500  =  330  days. 

Explanation. — By  the  operation  of  this  problem,  v;e  find  that  Spencer  &  Folsom  owe  both 
a  balance  of  dollars  and  a  balance  of  d.ays,  and  hence,  by  the  exercise  of  our  reason,  we  see  that 
in  order  to  settle  or  pay  in  kind  the  330  days,  the  balance  of  dollars  must  be  m.ade  due  330  days 
prior  to  July  1,  the  assumed  due  or  focal  date  of  all  the  debit  and  credit  items  of  dollars.  330 days 
counted  backward  from  July  1  gives  August  5,  1894,  as  the  equated  time  for  paying  the  balance  of 
dollars. 

By  the  foregoing  problems  and  illustrations,  we  see,  for  reasons  given  in  the  solutions,  that, 
when  the  lialance  of  dollars  and  balance  of  dai/s  occur  on  the  same  side  of  the  account,  whether  the  debit 
or  the  credit,  the  equated  time  is  counted  forward  from  the  earliest  date ;  and  that,  when  the  balance 
of  dollars  and  hiilance  of  dai/s  occur  on  opposite  sides  of  the  account,  the  equated  time  is  counted 
backward  from  the  earliest  d.ate. 

An  account  with  the  earliest  date  on  credit  side  and  balance  of  dollars  on  credit 
side,  and  balance  of  days  on  debit  side,  equated  time  counted  backward  from  the  earliest 
date. 


Br. 


What  is  the  average  date  of  the  following  account  f 


Ans. 


Lacy  &  Eobinson. 


April  10,  1805. 

Cr. 


1895. 

1895. 

July 

6 

To  Mdse.,  90  days, 

$2100 

00 

June 

14 

By  Cash, 

$5441 

00 

Aug. 

30 

"         "      90     " 

500 

00 

July 

28 

U            ii 

3000 

00 

Sept. 

18 

a         a       60     " 

200 

00 

Aug. 

1 

a        u 

300 

00 

Oct. 

1 

"         "       30     " 

900 

00 

640 


SOULE  S    PHILOSOPHIC    TRACTICAL    MATHEMATICS. 


OPERATION. 


1895. 

July  6, 
Auk.  30, 
.Sept.  18, 
Oct.      1, 


Dr.  of 
DoUs. 

12100 
500 
200 
900 

$3700 


Cr.  of 

1)3. 

90 
90 
60 
30 


22 

77 

96 

109 


112 
167 
156 
139 


Cr.  or 
Ds.  ou  $1. 

235200 
83500 
31200 

125100 


475000 
146400 


1895. 

June  14, 
July  28, 
Aug.  1, 


Cr.  of 
BoUs. 

$5441 

3000 

300 


$8741 
3700 


Dr.  of 
Ds. 

00 
44 
48 


Dr.  of 
Ds.  ou  $1. 


132000 

14400 


146400 


Bal.  of  days,    328600 
32SC00  days  -^  5041  =  65  days. 


15041  bal.  of  dolls. 


Explanation.— By  the  operation  of  this  problem,  we  find  that  Lacy  &  Robinson,  have  dne 
to  them  both  a  balance  of  days  and  a  balance  of  dollars,  and  heuce,  in  order  to  give  them  credit 
for  the  65  days  due  to  them,  we  must  pay  the  balance  of  money  65  days  before  the  assumed  date. 
June  14,  and  consequently  the  time  is  here  counted  backward  from  June  14. 


C.    A  owes  B  $1G00,  due  in  6  months. 
the  equated  date  for  paying  the  balance  f 


A  pays 


in  2  months.    What  is 
Ans.  10  months. 


FIRST  OPERATION. 


$1600  X  6  months  =  9600  =  months  credit  dne  A  on  $1. 
800  X  2  months  =  1600  =  mouths  debit  due  by  A  on  $1. 


Dr.  of  D0II9. 
$1600 
800 


800 
8000 


Cr.  of  M08. 
X         6 


8000  =  months  credit  balance  due  A  on  fil 
=  10  =  months  credit  due  A  on  $800. 

SECOND   OPERATION. 


$  800 


Cr.  of  M03.  on  $1. 
9000 
1600 

8000 


Cr.  of  DoUa. 
$800 


Dr.  of  Mos. 
X  2 


Dr.  of  Mos.  on  $1| 

1600 


Explanation. — In  the  operations  of  the  above  problem,  we  first  observe  that  while  A  owes  B 
SlfiOO  in  money,  B  owes  A  6  months  time  to  pay  the  money.  And  6  rriouths  credit  on  $1600  is  equal 
to  9600  months  credit  on  $1.  Then,  a.s  A  paid  .$800  in  2  months  B  owes  A  for  the  $800,  and  A  owes 
B  fur  the  2  months  credit  on  $800,  which  equals  1600  months  credit  on  $1.  The  balauce  of  account 
then  stands,  A  debit  $800  and  credit  8000  mouths  ou  $1.  8000  —  800  =  10  mouths  credit  ou  the 
balance  $800. 

7.  A  owes  B  a  note  dated  March  10, 1895,  due  in  183  days  for  $1200.  June 
14,  1895,  A  paid  on  account  $300.  What  is  the  equated  time  for  paying  the  bahmce, 
counting  the  actual  days  in  each  month  ?  Ans.    Oct.  8, 1895. 

FIRST    OPERATION. 

$1200  X  183  =  219600  =  days  credit  dne  A  on  SI. 
300  X    96  =    28800  =  days  debit  that  A  owes  B  on  $1. 

$  900  190800  =  balance  of  days  credit  due  A  on  $1. 

190800  -7-  900  =  212  days  credit  due  A  on  $900.  And  212  days  after  Marcll 
10,  1895,  =  Oct.  8,  1895.     See  Time  Table. 

Explanation. — From  March  10  to  June  14,  is  96  days.  See  above  problem  for  further  explana- 
tion. 


EQUATION    OF    ACCOUNTS. 


641 


SECOND   OPERATION. 


Mar.  10, 


Dr.  of  Dolls.    Ci 
$1200 
300 


.  of  Ds. 

183 


Cr.  of  Ds.  on$l. 

219600 

28800 


June  14, 


Cr.  of  Dolls. 
$300 


Dr.  of  Ds. 
96 


Dr.  of  Da.  on  $U 

28800 


$  900  190800 

190800  -~  900  =  212  days  credit  on  $900  after  Marcli  10,  1895,  =  Oct.  8,  1895. 

8.  Jlay  14, 1895,  A  owes  B  $800  due  in  4  months  and  $500  due  in  3  montbs. 
A  paid  ou  account  $300  in  1  mouth  and  $100  in  3  months.  When  is  the  balance 
due  by  equation?  Ans.  4  months  17  days. 

OPERATION    INDICATED. 


Dr.  of  Dolls. 

Cr. 

of  Mos.  on  $1. 

Cr.  of  D0U3. 

Dr.  of  Mos.  on  $1 

$  800 
500 

X 
X 

4 
3 

= 

3200 
1500 

$300 
100 

X 
X 

1 

3 

= 

300 
300 

$1300 
400 

4700 
600 

$400 

600 

$  900 

4100 

45  mos. 

=  4  11103.  16§  ds. 

9.  ]\ray  14,  1895,  A  owes  B  $800  due  in  4  months,  and  $500  due  in  3  months. 
A  paid  on  account  $700  iu  4  months  and  $500  in  6  months.  When  is  tlie  bahince 
due !  Ans.    June  14,  1894. 

OPERATION    INDICATED. 


$  800 
500 

$1300 
1200 


X 
X 


3200 
1500 

4700 


$700 
500 

$1200 


X 
X 


2800 
3000 

5800 
4700 

1100 


$  100 

1100  4-  $100  =  11  months,  i.  e.  the  $100  balance  was  due  11  months  before 
the  date  of  May  14,  1895,  which  is  June  14, 1894.  In  other  words  A  owes  B  interest 
on  $100  for  11  months.    Hence  the  time  is  counted  backward. 

10.  May  14,  1895,  A  owes  B  $800  due  in  4  months  and  $500  due  in  3  months, 
A  paid  $900  in  4  months  and  $000  in  5  months.  What  is  the  equated  date  for  the 
settlement  of  the  balance  due  A  ?  Ans.    Feb.  29,  1896. 

OPERATION    INDICATED. 


500 


X 
X 


$1300 


3200 
1500 

4700 


$900 
600 

$1500 
1300 


X 
X 


3600 
3000 

6600 
4700 


1900 


I  200  1900 

$200  —  9|  months,  after  May  14,  1895,  =  Feb.  29,  1896. 


Explanation. — The  result  of  the  operation  shows  that  B  owes  A  $200  and  that  A  owes  B  9i 
months  credit  on  the  same,  from  May  14,  1895.  Hence  while  B  owes  A  the  $200  he  need  not  pay  it 
until  9J  months  after  May  14,  1895.  It  must  not  he  forgotten  that  all  items  of  debit  and  credit 
«re  assumed  to  be  due  ou  the  earliest  date. 


642  SOULe's    nilLO-SOPHlC    PRACTICAL    MATHEMATICS.  * 

AN  ACCOUNT  EQUATED  AND  SETTLED  BY  NOTE. 


1190.     1. 
days  each : 


The  following  bills  were  sold  to  Tyler  &  Mills  on  a  credit  of  90 


January  4,  1895, 
February  19,  1895, 
April  1,  1895, 


$824.15 

371.08 

1410.00 


May  21,  1895, 
Juue    7,  1895, 


The  following  payments  were  made  on  the  above  purchases: 


March  1,  1895, 
April    1,  1895, 


$000  cash. 
350     " 


May  10,  1895, 
Juue    8,  1895, 


$718.15 
2120.80 


$800  cash. 
1000     " 


What  is  the  equated  time  for  paying  the  balance  of  the  account? 

Ans.     Oct.  4,  1895. 

OPERATION. 


1895. 

Dr.  of 
Dolls. 

Cr.  of 

Ds. 

Cr.of 
Ds.  ou  $1. 

1895. 

Cr.of 
Dolls. 

Dr.  of 

Ds. 

Dr.  of 

Ds.  ou  $1. 

Jan.      4, 
Feb.    19, 
April    1, 
May    21, 
Juue    7, 

$824.15 
371  68 

1410.00 
718.15 

2120.80 

90 
90 
90 
90 
90 

0 

46 

87 

137 

154 

^ 

90 
136 
177 

227 
244 

74160 

50592 

249570 

162986 

517524 

Mar.     1, 
April    1, 
May    10, 
Juue     8, 

$600 
350 
800 

1000 

56 

87 

126 

155 

33600 
304.50 

lonsoo 
155000 

$5444.78 
2750.00 

1054832 
319850 

$2750 

319850 

$2694.78   bal.  of  dolls. 


734982  bal.  of  days. 


734982  days  -^  2095  =  273  days,  which  counted  forward  fron;  January  4, 
1895,  gives  October  4,  1895,  as  the  equated  time. 

2.  Suppose  in  the  above  problem  that  Tyler  &  Mills  had,  on  the  8th  of 
June,  given  their  note  due  at  the  equated  date  for  the  balance  due,  lor  how  many 
days  would  the  note  have  been  drawn  ? 

Ans.  118  days  without  grace,  or  115  days  with  grace. 

Note. — The  equated  date  being  October  4,  1895,  we  simply  Cud  tbe  iutcrveuuig  tuucbctweeQ 
June  8  and  October  4,  wbieh  i.s  118  days. 

3.  Suppose  again,  in  problem  No.  1,  that  Tyler  &  Mills  had  settled  June 
8,  the  balance  due  in  cash,  allowing  8  ])er  cent  for  the  use  of  money  and  not  count- 
ing discount  day,  what  sum  would  be  required  to  ]>ay  the  balance? 

Ans.  $2024.12;  which  is  called  the  cash  balance. 

OPERATION. 

Explanation. — Tbe  equated  maturity  of  tbe  balance  being  Oct.  4, 
1895,  it  is  clear  that  if  Tyler  &  Mills  pay  it  Juue  8,  tbey  abould 

be  allowed  tbe  interest  ou  tbe  same  for  tbe  intervening  time,  118 

$70.00  days,  which  is,  as  shown  by  the  opposite  statement,  $70.66. 


45 


2094.78 

lis 


*  EQUATION    OF    ACCOUNTS.  643 

4.     May  10,  1S95,  A  owes  Z  83000.  due  in  3  niontbs.     June  4,  1895,  A  paid  Z 
$2000  on  account,  wlien  is  tlie  balance  due?  Ans.     Dec.  22,  1S95. 


OPERATION 
1895.  Ds.  Ds.  on  $1. 

May  10,  $3000  92  =  270000 
2000         50000 


)  22Cpf)p 


226  ds. 


1895.  Ds.  Ds,  on  $1. 

June  4,       $2000     25     =     50000 


EQUATION  OF  ACCOUNT  SALES. 

1200.  An  Account  of  Sales  is  a  statement  of  tbe  articles  and  price  of 
merchandise  or  other  jiroperty  sold,  the  charges  incurred  iu  efl'ecting  the  sales,  and 
the  net  proceeds  due  tbe  party  or  parties  for  whose  account  the  goods  were  sold,  and 
to  whom  the  account  sales  is  rendered. 

Tbe  net  proceeds  are  due  as  cash  at  tbe  equated  date  of  tbe  different  sales 
and  charges. 

When  equating  account  sales,  it  is  usual  to  consider  the  freight  and  drayage 
due  on  the  receipt  of  the  goods.  Tbe  commission  is  often  considered  not  due  until 
the  last  day  of  sale,  but  some  accountants  equate  tbe  time  for  the  charge  of 
commission.  In  large  sales,  or  where  there  is  a  long  interval  of  time  between  sales, 
tbe  commission  should  be  equated.  The  other  charges  are  sometimes  considered 
due  on  tbe  last  day  of  sale  and  sometimes  on  the  day  that  each  item  was  paid,  or 
tbe  average  date. 

In  case  of  account  sales  for  merchandise  sold  on  joint  account,  tbe  interest 
j  of  the  consignee  is  sometimes  considered  due  at  the  date  of  shipment  and  sometimes 
i  at  tbe  date  the  goods  were  received.  In  justice  to  both  parties,  and  in  order  to 
preserve  a  uniformity  in  the  accounts  of  each  party  interested  iu  tbe  goods,  this 
i  matter  should  be  agreed  upon  at  tbe  time  the  parties  contract  for  tbe  joint  account 
'  operation.  Tbe  consignee's  gain  or  loss  iu  joint  account  transactions  is  generally, 
'  and  should  always  be  made  due,  the  same  as  the  commission,  at  the  equated  date 
'  of  tbe  sales. 

I  NoTK,  — It  is  the  cnstom  of  cotton  factors  in  New  Orleans,  to  consider  tbe  proceeds  of  cotton 

'  sales  due  7  days  after  tbe  sale  was  made.  Tbis  7  days  use  of  the  proceeds  is  considered  a  just 
]  eoinpensation  V(ir  tlio  interest  on  tbo  mouey  advanced"  for  cbarges,  and  saves  tbe  time  and  labor 
I    of  equating  tbe  account  sales. 

i  rionr  and  produce  dealers  consider  tbe  proceeds  of  goods  sold  on  commission  due  ten  days 

I   after  tbe  sale  was  made.     Tbis  allowance  is  made  iu  place  of  er|uating  tbe  account  sales,  and  is 
considered  just  to  both  i^arties. 


644 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


PROBLEMS. 

1.     When  are  the  net  proceeds  of  the  following  consignment  due? 

Ans.     Nov.  4,  1895. 
Ko.  1820. 

Sales  by  W.  H.  Morgan  &  Co.  of  1200  bbls.  flour,  received  Oct.  11,  1895,  per 
steamer  Susie  Silver,  for  account  of  Schreiber  &  Eicke,  Belleville,  III. 


Oct. 


Oct. 


Nov. 


11 


Sold    100  bbls.  Sucker  City,  cash,  ®  $8 

-  -     ■•  "  "  "  to  .$75 

"  "      20  lis.,  ®  .$7i 

"  "      cash,  ®  $8i- 

"      10  ds.,  ®  .$7i 

"  "      15  ds.,  ®  $7i 


200 
284 
8 
200 
408 

1200 


-CHARGES. 


Freight,  40  eta.  per  bbl.,  $480 ;  Charges  per  B.  L.,  $240,      - 
29|Liibor,  $12;  Tarpaulins,  $6  ;  Inspection,  $24,         .... 
"IDrayage,  $49;  Storage,  $54  ;  Insurance,  Fire  $21.84— River  $67.50, 

"'skidding,  $1.60;  Cooperage 

2  Commission  ou  sales  at  2i  per  cent,        ..-..- 


Net  proceeds  to  your  credit  due  by  equation  Nov.  4,  1895, 


720  00 

4200 

19234 

160 

228 1 28 


800 
1575 
2130 
66 
1500 
3060 

$9131 


1184 


22 


$7946178 


E.  4-  0.  E. 


New  Orleans,  November  1,  1895. 


WM.  H. 

MORGAN 

&.  CO., 

per  F, 

Horn. 

OPERATION. 

1895. 

Ds. 

Ds. 

1895. 

Da. 

Ds. 

Ds. 

Oct.   11, 

"    29, 

Nov.    2, 

$720.00 
235.94 
228.28 

$1184.22 

0 
18 
22 

4248 
5016 

9264 

Oct.  12, 
"   15, 
"   15, 
"   21, 

"   26, 
"   29, 

$800.00 
1575.00 
2130.00 
66.00 
1500.00 
3060.00 

$9131.00 
1184.22 

0 

0 

20 

0 

10 

15 

4 

15 

18 

1 

4 

=    24 

10 

=    25 

=    33 

800  1 

6300 

51120 

660 

37500 

100980 

197360 
9264 

I 


$7946.78  188096 

1SS096  4-  7947  =  24  days,  which  counted  forwaid   from   October   11,   gives 
November  4,  1895,  as  the  equated  date. 

Explanation. — In  this  equation  we  considered  the  freight  and  drayage  due  on  the  receipt  of 
the  goods,  the  commission  on  the  average  date  of  the  sales,  and  the  other  charges  on  the  last  day 
of  sale. 

OPERATION 
To  find  the  equated  date  of  the  sales  only,  at  which  date  the  commission  is  due. 


Da. 

Da. 

Oct.  12, 

$  800 

0 

"     15, 

1575 

3 

= 

4725 

"     15, 

2130 

20 

+ 

3     =    23 

^^ 

48990 

"    21, 

66 

9 

rr: 

594 

"    26, 

1500 

10 

+ 

14     =     24 

= 

36000 

"    29, 

3060 

15 

+ 

17     =     32 

= 

97920 

$9131  188229 

188229  -4-  $9131  =  21  days. 
This  equation  gives  us  21  days,  -which  counted  forward  from  October  12, 
gives  November  2,  as  the  equated  date  of  the  sales. 


EQUATION    OF    ACCOUNTS. 


645 


EQUATE  THE  FOLLOWING  ACCOUNT  SALES  ON  JOINT  ACCOUNT. 

2.  Sales  of  1000  bbls.  potatoes,  received  per  steamer  E.  E.  Lee,  Nov.  2,  and 
sold  ou  tbe  joint  account  of  the  shippers,  Spencer  &  Frederick,  aud  ourselves, 
each  J,  by  IMoses,  Lehman  &  Co. 


1895 
Nov. 


Dec. 


1  Nov. 


Sold    300  bbls.  Pe.nch  Blows,        casb,     ®  $3 


450 
60 
50 

140 

1000 


Nesbauuock, 

Pink  Eyes, 
Peach  Blows, 


20  ds.,  @  $3.20 
casb,     <a>  $3^ 
10  ds.,  ®  $3i 
15  ds.,  ®  $3i 


CUARGES. 


Freight,  40  cts.  per  bbl., 
Drayage,        .         -         -         - 
Iiisurauce,      -        -        .        . 
Storage,  $51.25;  labor,  $8.50, 
Commission  ®  2J  jier  cent. 


Net  proceeds, 
Cost  of  iuvoicc. 


Net  gain, 


Our  \  net  gain,      -        .         .         . 
Sjieiicer  &.  Frederick's  \  net  gain, 
"  "  i  invoice. 


"  "  net  proceeds,  ...... 

(exclusive  of  i  invoice  previously  credited,)  due  bv  equatioii  Dec. 
19,  1895,  -        -        -         -     '  -         -         -         -     ■    - 


$900 

1440 

195 

175 

490 


I 


400 
50 
21 
59 
80 


$294 
1000 


$3200 


00 


611 


$2588 
2000 


$294 


1294 


25 

75 
00 

75 


37 


38 


E.  4-  O.  E. 


New  Orleans,  December  5,  1895. 


MOSES,  LEHMAN  &  CO., 

Per  Louis  Levy. 


OPERATION. 


!     1895. 

i  Nov.  2, 

'■  Dec.  1, 

'  Nov.  24, 

:      "  24 

I      "  2,' 


D3. 


Ds. 


Fr't  &  Dray. 
Ins.,  Sto.,  &c. 
Commission 
Our  i  gain 
"     i  iuv. 


$450.00 

81.25 

80.00 

294.37 

1000.00 

$1905.62 


10577 


1895. 


0 

Nov.  3, 

$300.00 

29 

2349 

"   9, 

1440.00 

22 

1760 

"  15, 

195.00 

22 

6468 

"  28, 

175.00 

0 

Dec.  4, 

490.00 

D». 


20  7 

0  13 

10  26 

15  32 


Ds. 
1 

27 
13 
36 
47 


$3200.00 
1905.62 


61068  4- 


$1294.38 
L294  =  47  days. 


Ds. 

900 

S8880 

2535 

6300 

23030 

71645 

10577 

61068 


Explanaiion. — In  this  operation  we  have  considered  the  freight  and  drayage  one  the  day  that 
the  produce  was  received.  Tbe  other  charges,  except  commission,  we  considered  due  December 
1,  as  the  greater  portion  of  them  must  by  custom  have  been  paid  then,  if  not  paid  -when  they  were 
respectively  incurred.  Strict  justice  would  require  the  equated  date  of  charges  in  all  account 
sales,  but  in  practice  it  is  not  always  done.  The  commission  and  gain  are  due  on  the  average  date 
of  the  sales,  and  the  i  invoice  the  day  that  the  produce  was  received,  or  the  date  of  purchase 
according  to  agreement.  We  have  here  considered  the  J  invoice  due  the  date  received,  aa  the 
account  sales  does  not  give  the  date  of  imrchase. 


646 


SOULE  S    PIIILOSOPIIIC    PRACTICAL    MATHEMATICS. 

OPERATION 
To  equate  the  sales  only,  at  which  date  the  commission  and  gain  are  due. 


LSI  15. 

1)8. 

1)8. 

D8. 

Nov.  3, 

$900 

0 

9, 

1410 

20 

+ 

6 

=  2(i 

= 

37440 

"   15, 

195 

0 

+ 

12 

=  12 

= 

2340 

"   28, 

175 

U) 

+ 

25 

--^  35 

;= 

6125 

Dec.  4, 

490 

15 

+ 

31 

=  46 

= 

22540 

$3200  •  68445 

68445  -4-  $3200  =  21  days. 

Note. — In  the  equation  of  account  sales  for  joint  account  transactions,  in  which  losses  are    S 
sustained,  they  are  entered  on  the  credit  side  of  the  account  in  the  o|)eratiou  of  equation. 

ACCOUNT  SALES. 


1895 
Jan. 


Jan. 


Sold  for  cash,  500  libls.,  Ex.  Flour  at  $9. 
Sold  at  10  days,  100  bbls.,  Flour  at  $8.50, 


CHARGES. 


Susdry  charjjes, 

Coniniission  on  sales  at  2J  per  cent, 

Net  proceeds,         .... 


$4500 
850 


$290 
133 


What  is  the  equated  date  of  the  proceeds  in  the  above  account  sales 

Ans.     Jan.  G, 

OPERATION. 


00 


25 


Jan.  2,  Charges, 
"    8,  Commis.sion, 


$291    0     =        0 
134    6     =    804 


$425 


804 


Jan.  3,  Sales, 
"    S,       " 


$4500       1 
850     16 

$5350 
425 


17296  -H  $4925  =  4  days,  after  Jan.  2,  =  Jan.  6/95. 


$4925 


$5350 

424 
$4925 

1895. 


4.. WO 
13,6u0 

18,100 
804 

17,296 


SALES  OF  MDSE.  ON  JOINT  ACCOUNT. 

F.ilin. 


1895 
Jan. 


COST   AND   CHAKGES. 

Received,  per  Steamer  Golden  Rule,  from  W.  B.  Henry,  to  be  sold 
for  his  and  onr  joint  account,  each  i  : 

400  bbls.  Mess  Pork  at  $20, $8000 

Com.  Sales  Dr.  to  W.  13.  Henry,  our  i         -         -         - 
To  Charges: 


Cash  for  Freight, 

"     for  Drayage,       .         -        - 
Insurance,  Fire,  ... 

"  River, 

Inspection,  $12 ;  Cooperage,  $3, 

Com.  S.  Dr. — Our  J  and  charges, 


$120; 

24  ( 

20 

180 

15 


To  Coininission  on  Sales  at  2\  per  cent, 
"   W.  B.  Henry's  i  Inv. 
"  do  I  Net  Gain, 


$4000 
232.38 


W.  B.  Henry's  Net  Proceeds, 

Due  by  Equation  February  7/95: 
To  P.  and  Loss,  our  J  Net  Gain, 

Com.  Sales  Dr.  to  close  account. 


11 
C.  1 

7 
11 
11 

11 
12 
11 


4000 


359 


43.59 
226 

4232 
232 


4691 


00 


37 


00 


4000 

359 

226 

4232 
23: 

9050 


00 

00 

25 

38 
37 

00 


SALES    ON    JOINT    ACCOUNT. 

SALES  OF  THE  ABOVE  JMDSE. 


647 


Folio. 


1K95 
Jan. 


\V.  J.  Vizaril, 

100  bills.  Mess  Pork  (damaged)  ®  : 

S.  J.  I'.lessnis;, 

100  bills.  Mess  Pork  ®  $26, 

Geo.  W.  Wcingart,  Jr., 

200  bbls.  Pork  ®  ^24. 75, 


Cash. 

C.  1 

Note  15  ds. 

. 

2(11 

10  ds. 

"         ' 

8/11 

When  are  W.  B.  Henry's  net  proceeds  due? 

OPERATION. 


i50o;oo 

2600  00 

4950!0O 

9050I0O 
Ans.     Feb.  7,  1895. 


JaD. 


4,  i  luv.  &  Cbgs 
11,  Com    iV;  Gaiu, 


$1359 
4.yj 

§4818 


0     = 

7     = 


0 
3213 


3213 


Jan. 


6,  Sales, 
y,     " 
11,     •• 


$1,500  2 

2600        23 
4950         17 

$9050 
4818 


143737-$4232=34  days,  after  Jan.  4,  which  is  Feb.  7/95.       $4232 


SALES  ON  JOINT  ACCOUNT. 


Folio. 


3,000 
69,800 
84,150 

146,950 
3,213 

143,737 


1895 
Jau. 


20 


20 


COST    A.NO    CUAl'.GES. 

Received,  per  Steamer  J.  M.  Wbite,  from  I).  Hughes,  to  be  sold  on 
tbe  joint  account  of  liimself,  J.  W.  Moses  and  ourselves  each  i  : 
.500  bills.  XX  Flour  ®  $7.50  =  $3750. 

To  wbicb  we  have  added  from  store  : 
100  bbls.  Molasses  ®  $18  =  S1800Cr.  Mdse.,      .        -        -        - 
D.  Hughes,  Cr.  for  our  i  of  bis  Inv.,  .        -         .         -         - 

Com.  Sales,  Dr.  for  our  J  of  our  Iiiv.,  .         -         .         -         - 

D.  Hughes,  Dr.,  for  J  of  our  luv.,  $600.  .        -        .        - 

J.  W.  Moses,  Dr.,  for  J  of  our  luv.,  $600.  -        .        -        - 

Charges : 


Casli  for  Freight,  $80;  Drayage,  $30, 
Labor,  .52;  Inspection,  $10, 
Insurance,  Fire,  .         .         .         - 

do  River,  .         .        .         . 


Com.  Sales,  Dr.  for  our  \  cost  and  charges. 


Commission  on  sales  ®  2^  ]ier  cent,  .         .         .         -        , 

Cash  refunded  to  W.  J.  Vizard —  Amount  allowed  on  damaged 
Molasses,  .___.-.--- 

Charges  : 


2S  Storage,  $10;   Cooperage,  j 
D.  Hughes,  i  of  his  Inv., 
"         j  of  our  Inv., 


29 


Less  his  J  Net  Loss,         .        ,        - 
D.  Hugbes'  Net  Proceeds,  Cr. 

Due  by  Equation  Feb.  6/95: 
J.  W.  Moses,  I  of  D.  Hughes'  Inv., 
do  j  of  our  Inv., 


Less  his  J  Net  Loss, 


$1250.00 
600.00 

$1850.00 
620.95 


$1250.00 
600.00 

$1850.00 
620.95 


J.  W.  Moses,  Net  Proceeds,  Cr. 
Due  by  equation  Feb.  6/95. 
Com.  Sales,  Dr.,  to  close  account. 


1 
12 

V 

12 
12 


C.  1 


11 

11 
C.  2 


12 


1250  00 
600,00 


11000 
12  00 
18JO0 
75  00 


2005  00 


105,55 


14  00 


1229  05 


1229 


2577 


05 


65 


1250  00 
600  00 


21500 


105  55 
200  00 


14  00 


1229  05 


1229 


05 


4842,65 


64S 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


SALES  OF  THE  ABOVE  MDSE. 


Folio. 


1895 
Jan. 


29 


29 


I 


W.  .1.  Vizard, 

500  bbls.  XX  Flour  ®  $5,     - 

100  bbls.  Molasses,  3826  gallons,  ®  45c., 

By  Profit  aud  Loss  for  our  i  Not  Loss, 


Cash, 


C.  2 

12/11 


2500 
1721 


4221 
620 


4842 


70 
95 

65 


When  are  the  net  proceeds  of  D.  Hughes  and  J.  W.  Moses,  due  by  equation 
in  the  above  accouut?  Aus.    Feb.  6/95. 


OPERATION. 


Jan.  20,  i  Inv.  &  Chgs.,        $2065    0    =  0 

"   29,  Com.  &  Chgs.,  320    9    =        2880 


$2385 


2880 


Jan.  29,  Sales, 
"   29,  J  Loss, 


$4222    9     = 
621    9    = 


$4843 
2385 


37,998 
5,589 

43,587 
2,880 


$2458  )40,707(17ds. 

40707  -^  2458  =  17  days,  after  Jan.  20/95,  =  Feb.  G/95. 


EQUATION  OF  RENT  NOTES. 


1201.  1.  January  1,  1895,  A.  leases  of  Z.  a  store  for  1  year,  at  $500  per 
mouth,  amounting  to  $0000,  on  the  following  conditions:  The  rent  to  be  paid 
monthly  in  advance,  as  follows:  The  first  month's  rent  is  to  be  paid  in  cash,  and 
monthly  maturity  rent  notes  given  for  the  remaining  11  months  rent. 

After  the  lease  is  signed,  A.  proposes  to  give  one  note  for  the  whole  rent 
payable  at  such  a  time  that  neither  party  will  gain  nor  lose  by  the  change  of  terms; 
to  this  proposition  Z.  agrees.    What  is  the  time  for  the  payment  of  the  new  note? 


FIRST    OPERATION. — BY    MONTHS     WITHOUT 
LOWING  FOR  GRACE. 
Mos.    Mo3.  Or.  on  $1 


$500 

0 

0 

500 

1 

500 

500 

2 

1000 

500 

3 

1500 

500 

4 

2000 

500 

5 

2500 

500 

6 

3000 

500 

7 

3500 

500 

8 

4000 

500 

9 

4500 

500 

10 

5000 

500 

11 

5500 

$6000 

)  33000 

51      months. 


Ans.     June  16,  1895. 

WITHOUT     AL- 

SECOND  OPERATION 

—BY  ACTUAL  DAYS   AND    Ali- 

B. 

LOWING 

FOR  3  I>A^ 

S  OE  GRACE  ON  EACH  NOTE. 

Ds. 

Ds.  Cr.  oa  $1. 

C. 

$500 

0 

0 

1 

500 

33 

16500 

2 

500 

61 

30500 

3 

500 

92 

46000 

4 

500 

122 

61000 

5 

500 

153 

76500 

6 

500 

183 

91,500 

7 

500 

214 

107000 

S 

500 

245 

122500 

9 

500 

275 

137500 

10 

500 

306 

153000 

11 

500 

336 

168000 

$6000 

) 

1010000 

June  16,  '95. 

168J  days,    June  18,  '95. 

EQUATION  OF  RENT  NOTES. 


649 


2.  Suppose,  in  the  above  problem  the  lessee  had  desired  to  give,  and  tlie 
lessor  bad  consented  to  receive,  two  notes  of  equal  amount,  payable  in  ])eriods  of 
time  in  the  ratio  of  1  and  4,  wbat  would  have  been  the  time  of  each  note,  allowing 
three  days  of  grace  on  each  of  the  11  notes? 

OPERATION. 

As  sbown  in  tbe  second  operation  above,  A.,  the  lessee,  has  a  credit  of  1010000  days  on  St>  on 

1         the  presumption  tliat  the  $€000  are  due  Januarii  1,  1S95.     Now,  by  tbe  terms  of  tbe  problem,   we  must 

I         find  tbe  respective  credits  (m  tbe  two  new  notes  so  that  tbe  total  credit  on  SI  shall  be  equivalent  to 

t         the  1010000  days  ou  the  11  notes  which  be   tirst  proposed  to  give.     This   we  do  by  the  following 

statements: 


Ist  note,  $3000  x  1  =    3000 
2d   note,     3000  X  4  =  12000 

15000 


15000 


1  dav. 
1010600 


15000 


67J  ds.  =  2  mos.  7i  ds. 
PROOF, 
let  note  S3000  x    67J  ds.  =    202000  ds.  credit  on  $1. 
2d  note  $3000  x  269i  ds.  =    808000  ds.  credit  ou  $1. 


4  ds. 

1010000 

2694  ds.  =  8  mos.  29J  ds. 


1010000  total  ds.  credit  on  $1. 
The  aggregate  days  credit  on  $1  is  tbe  same  as  shown  above  on  the  eleven  notes. 
See  pages  651  and  652  for  full  discussion  of  this  kind  of  work. 


EQUATION  OF  ACCOUNTS  AND  NOTES  BEARING  INTEREST. 

1202.  1.  A.  owes  Z.  $500  due  March  8,  1895,  and  bearing  interest  at  8  per 
cent;  $1700,  due  May  17,  1S95,  and  bearing  interest  at  C  ])er  cent;  and  $350,  due 
June  1,  1805,  and  bearing  interest  at  8  per  cent.     "What  is  the  equated  time  for  the 


payment  of  the  whole  debtl 


Ans.    May  3,  1895. 


OPERATION. 

1895 

Da. 

Mar.    8, 
May  17, 
June    1, 

$  500 
1700 
350 

8%        =        $  4000 
6"^         =           10200 
8»o'        =            2800 

0 

70 
85 

714000 

238000 

$17000  )  952000 

952000  days  -^  17000  =  5G  days. 

2.  A.  holds  Z's  note,  due  August  10,  1895,  for  $5000,  bearing  8  per  cent 
interest  from  maturity,  and  Z.  holds  A's  acceptance,  due  October  24,  1895,  for 
$14000,  bearing  interest  at  5  per  cent  from  maturity.  "What  is  the  equated  time 
for  paying  the  balance?  Ans.    Feb.  1,  189G. 

OPERATION. 


1895. 
Aug.  10,        $5000 


8%    =    $40000 


Ds. 

Da. 

1895. 

Da. 

Ds. 

0 

0000 

Oct.  24, 

$14000  5%  = 

$70000 
40000 

$30000 

75 

=     5250000 

)  5250000 
175  ds. 

65o 


SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS. 


5250000  days  -^  30000  =  175  days,  which  counted  forward  from  August  10, 
1895,  gives  February  1,  1S9G,  as  the  equated  date  for  the  payment  of  the  bahxuce. 

By  this  equation,  we  see  that  the  $5000  due  August  10, 1895,  was  paid  175 
days  after  it  was  due;  and  that  the  $14000,  due  October  24, 1895,  M-as  paid  100  days 
after  it  was  due. 


Interest  on  $  5000  at  8%  for  175  days  =  $194.44 
"         "      14000   "  5%   '<    100     "       =     194.44 


3.     A.  holds  notes  and  acceptances  against  Z.  as  follows  ; 


His  note  due  Jan.     28,  '95,  for  $2000,  bearing  S%  interest. 
"       "       "     Feb.      4,  '95,    "     4200,        "        G%        " 
"     acpt.  "     Mar.    10,  '95,    "      7000,        "        6%        " 


"    note    "        "       31,  '95,    "     2500, 
Z.  holds  the  following  claims  against  A. : 


r% 


His  note  due  Feb.     14,  '95,  for  $2400,  bearing  5%  interest. 

"   acpt.    "    Mar.    21,  '95,    "      3000,         "       8%         " 
■    "  note     "  April     9,   '95,    "     5000,  without  interest. 

What  is  the  equated  time  for  ^laying  tlie  balance  of  tlio  above  account,  allow- 
ing 8  per  cent  interest  on  the  balance?  Ans.     March  5,  1895. 


1895. 

utrtua,^ 

1895. 

Jan. 

28, 

$2000 

8%' 

= 

$16000  0 

=: 

Fel).  14, 

$2400 

oV 

:= 

$12000 

17 

— 

204000 

Feb. 

4, 

4200 

ev 

:= 

25200  7 

=: 

176400 

Mar.  21, 

3000 

«"o 

= 

240110 

52 

= 

1248000 

Mar 

10, 
31, 

7000 
2500 

$15700 

7% 

= 

42000  41 
17500  62 

$100700 

= 

1722000 
1085000 

2983400 

April  9, 

5000 

o^- 

00000 

71 

0000000 

$10400 

$36000 

1452000 

10400 

1452000 

$5300  ®  S^'  =  $42400 


)  1531400 


1531400  -4-  42400  =  3G-/ft-  days. 


Explanation. — By  the  operation  we  obtain  practically  36  days,  wbicli  counted  forward  from 
January  '28,  gives  March  5, 1895,|as  the  equated  date  for  the  payment  of  the  balance,  allowing  8  per 
cent  interest  on  same. 

Note. — The  operation  also  shows  a  debit  of  1.531400  days  on  $1  at  1  per  cent,  and  as  the 
product  of  the  lialance  of  .$,5300  multiplied  by  the  8  pef  cent  and  the  36i'i^£  days  just  equals  1531400 
days,  it  is  clear  that  the  equated  time  is  March  5,  1895. 


EQUATION    OF    NOTES, 


65i 


EQUATION  OP  NOTES  IMATUEING  AT  DIFFERENT  DATES,  AND    THE 

SUBSTITUTION  OF  OTHERS  OF  DIFFERENT  AMOUNTS  AND 

MATURING  AT  EQUAL  INTERVALS  OF  TIME. 

1203.  1.  A.  owes  Z.  three  uotes  bearing  date,  and  drawn  for  the  amount  and 
time  respectively,  as  follows:  August  10,  1S05,  $2000  payable  90  days  after  date; 
September  12,  1S95,  $1200  payable  G  months  after  date;  and  November  25, 1895, 
$4800  payable  1  year  after  date. 

A.  wishes  to  take  up  or  redeem  these  three  notes  with  four  others  of  equal 
amounts  and  maturing  at  equal  intervals  of  time,  so  that  there  will  be  no  loss  or 
gain  to  either  party.     What  is  the  time  of  each  new  note  ? 


OPERATION. 

1895. 

Dr.  of 
DoUs. 

Contrart 
Cr.  of  Da. 

Total  Cr. 
ul'  Da. 

Cr.  of  Ds. 

oii$l. 

Aug.   10, 
Sept.  12, 
Nov.    25, 

$2000 
1200 
4800 

90          + 
182          + 
366         + 

0 
33 

107 

= 

90 
215 

473 

180000 

258000 

2270400 

4  ;  lj8000 

2708400 

$2000 

eacli  new  note. 

Having  now  found  the  number  of  days  credit  due  A.  on  the  three  old  notes, 
from  August  10,  1895,  the  day  that  we  assume  them  all  due,  we  have  next  to  find 
the  respective  number  of  days  credit  on  each  of  the  4  new  notes  in  propoi'tion  to  1, 
2,  3,  and  4,  as  required  by  the  conditions  of  the  problem,  so  that  the  total  credit  of 
the  same  shall  be  equivalent  to  the  2708400  days  on  the  3  old  notes.  We  therefore, 
in  order  to  produce  the  necessary  proportional  numbers  to  make  a  solution  state- 
ment, i^roceed  as  follows : 

1)9.  Cr.  aasnnied.        Da.  Cr.  on  $1. 


1st   New  Note 

S2000 

1 

2000 

2d      " 

2000 

2 

4000 

3(1      " 

200U 

3 

6000 

4tli     "         " 

2000 

4 

8000 

20000 

By  this  work  we  obtain  20000  days  credit  on  the  4  new  notes,  and  as  the 
assumed  credit  of  1,  2,  3,  and  4  days  gave  this  20000  days,  by  transposition,  we  see 
that  the  20000  days  credit  required  the  1,  2,  3,  and  4  days  time,  and  as  we  wish  to 
distribute  270S400  days  credit  on  $1,  on  the  time  of  the  4  new  notes,  the  following 
proportional  statements  will  give  the  correct  time  of  each  note. 

statement  for  the  lat  Xote.        Statement  for  the  2d  Note.         Statement  for  tlie  3d  Kote.        Statement  for  the  4tU  Note. 


B. 

I  1 

20000  2708400 


20000  2708400 


20000 


idOsflJuO  J 


20000 


406/oWa- 


D.S. 
4 

2708400 


652 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


The  fractious  of  a  day  in  tlie  result  cannot  be  avoided,  and  to  use  tbem  as 
equitably  as  the  conditions  of  tlie  problem  -will  admit  of,  we  omit  the  fractions  in 
each  result  and  increase  the  whole  numbers  of  the  2d  and  4th  results,  1  day  each. 

The  reasoning  for  the  first  statement  is  as  follows:  If  20000  days  credit  on 
$1  require  1  day's  time,  or  a  note  of  $2000  payable  1  day  after  sight,  1  day's  credit 
will  require  the  20000th  part,  and  270S400  days  credit  will  require  2708400  times  as 
much.  For  each  of  the  other  statements  the  reasoning  is  the  same,  except  that  we 
have  respectively,  2,  3,  and  4  days  assumed  time  on  the  2d,  3d,  and  4th  notes. 

Instead  of  making  these  4  statements,  we  could,  in  this  problem,  have  multi- 
plied the  result  of  the  first  by  the  assumed  numbers,  2,  3,  and  4,  and  thus  have 
obtained  the  result.  This  method  of  work  is  applicable  to  all  questions  of  this 
character,  regardless  of  the  proportion  of  time  that  the  new  notes  may  be  desired 
to  run. 

To  elucidate  still  further  that  the  credit  on  the  new  notes  is  equivalent  to  the 
credit  on  the  old  notes,  we  present  the  following: 


D3.  Or.  OD  $1 

1st  New  Note 

$2000 

135  days 

270000 

2d   "   " 

2000 

271   " 

542000 

3d   "   " 

2000 

406   " 

812000 

4tli  "    " 

2000 

542   " 

1084000 

Total  credit  on  $1, 

"         "         "   |1,  of  old  notes, 

Excess  of  credit  on  old  notes, 


2708000 
2708400 


400 


Note. — The  small  excess  of  credit  on  the  old  notes  results  from  the  fact  that  ■n-e  did  not  and 
cannot  use  the  exact  time  .as  shown  in  the  four  stiitements  above.  In  the  first  and  third  notes  the 
small  fraction  of  time  was  omitted.  While  in  the  second  and  fourth  notes  the  fractions  of  time 
being  each  more  than  one-half  day,  they  were  each  counted  as  one  day.  The  small  excess  thus 
produced  is  only  equal  to  one-fifth  of  a  day  on  one  of  the  new  notes,  which  is  too  small  a  credit  to 
he  used 


V 


\©©/ 


/ 


m 


.ccounts  Current  and  Interest  Accounts,  and  Cash  Balances. 


1204.  "WTien  it  is  desired  to  settle  an  account  by  cash,  or  to  find  the  balance 
that  is  due  at  any  specified  date,  for  the  purpose  of  closing  the  books  or  determining 
the  full  resources  and  liabilities  of  the  firm,  one  of  two  methods,  the  interest  or 
iheproduct,  is  generally  used. 

THE  INTEEEST  METHOD. 

1205.  The  first  and  most  usual  method  is  to  find  the  interest  on  each  item  of 
debit  and  credit  of  dollars  from  the  time  it  became  due  up  to  the  date  of  settle- 
ment, or  balancing  of  the  accounts,  and  then  charging  or  crediting  the  debtor  with 
the  balance  of  interest,  according  as  it  is  against  him  or  in  his  favor, 

THE  PRODUCT  METHOD. 

1206.  The  second,  and  next  most  usual  method,  is  to  find  the  balance  of 
interest  dne  the  debtor  or  creditoi',  by  first  finding  the  balance  of  the  product  of 
the  debit  and  credit  of  days  on  $1,  the  same  as  in  equation  of  accounts  when  work- 
ing from  the  latest  date,  and  then  dividing  the  same  by  the  interest  divisor,  and  thns 
obtaining  the  interest,  which  is  charged  or  credited  to  the  debtor  as  in  the  first 
method. 

In  our  problems  and  operations,  we  will  illustrate  both  of  these  methods  and 
also  a  new  method  by  which  the  operation  is  shortened  and  the  accountant  facili- 
tated in  the  discharge  of  his  duties. 

THE  EQUATION  METHOD. 

1207.  This  is  a  new  method  which  we  name  the  Equation  Method,  for  the 
reason  that  we  assume  all  of  the  dollar  items  due  on  the  earliest  date  in  the  account, 
and  proceed  with  all  the  debit  and  credit  items  of  dollars,  the  same  as  we  worked 
equations  when  we  assumed  the  earliest  date  to  work  from. 

PROBLEMS. 

1.  What  is  the  cash  balance  of  the  following  account,  December  31,  1S95, 
interest  at  8  per  cent  ?  Ans.  $1531.42. 

Dr.  Eench  &  WuKZLOW.  Cr. 


1895 

Feb. 

4 

Mar. 

L'8 

May 

3 

Sept. 

14 

To  Mdse.  ®  90  days, 

Cash,     .     .     -     - 

Mdse.  ®  30  days, 

"       ®  30  days, 


1895 

$  500 

00 

Mar. 

10 

1200 

00 

April 

25 

700 

00 

June 

"i 

1500 

00 

Oct. 

19 

By  Cash, 

"   Acpt.  ®  60  days,    -     -     - 

"   Cash, 

"  Proceeds  of  flour  due  by 
equation  October  22, 


I  300  00 


500 
400 

1250 


00 
00 

00 


(653) 


654 


SOULE  S    PHILOSOPHIC    PRACTIC.\L    MATHEMATICS. 


FIRST  OPEKATIOX  BY  THE  INTEREST  METHOD. 

Kenxii  &  WiRZLOW  in  Account  Current  and  Interest  Account  at  8'i  to  December  31,  1895.  with 

MlTCUEU.  &  Braxdxek. 
Dr.  Cr. 


ToMdse.®90ds. 
•  Cash, 

"Mdse.'ffSOds. 
"     "       '•  30  •' 

"  Bal.  of  int. 

■Wlien 
doe. 

.! 

lot 

1 
1 

Amt. 

1895. 
Mar. 
Apr. 
June 
Oct. 

Dec. 

10 

25 

7 

19 

31 

THien 
due. 

LI ,..  i 

Amt. 

1?95. 
Feb. 
Mar. 
Jlay 
Sept 

Dec. 

4 

28 

3 

14 

31 

Mav 
Mar. 
June 
Oct. 

5 

28 

14 

240 
278 
212 

78 

1 
i 

26  67 
74  13 
3298 
2600 

159'^ 

-550000 

120000 

700  Oti 

150000 

398142 

By  Cash. 
•  Acpt.®60d8. 

"  Cash. 

"  Proceeds  of 
flonr.    due    by 
equatiou  Oct." 22 

Mar. 
June 

Oct. 

10296 
27187 

22  70 

19  73 

20  78 
1840 

i 

1945 

$300'00 
50000 
40000 

125000 

"" 

By  hal.    due  to 
date,                  ' 

78  36 

153142 
398142 

FxpJanation. — lu  this  solution  we  first  find,  after  maturius  the  debit  and  credit  items  of  dol- 
lars, the  number  of  days  that  each  debit  and  credit  item  of  dollars  has  been  due  prior  to  December 
SI.  1895.  and  then  compute  the  interest  on  the  same  for  the  time  at  8  per  cent.  We  then  take  the 
difference  between  the  debit  and  credit  of  interest,  and  as  the  excess  or  balance  is  against  Rench  & 
Wurzlow.  we  chartre  it  to  them,  aud  then  to  find  the  cash  balance  we  take  the  diflerence  between 
the  debit  aud  credit  amounts  of  the  account. 


SECOXD  OPERATION"  BY  THE  PRODUCT  METHOD. 
Eexch  &  AVuRZLOW  in  Account  Current  and  Interest  Account  at  8^^  to  December  31,  1895,  with 

MlTCHEU.   &   BEAXDSER. 

Dr.  Cr. 

DK.       DR.  CE.       CR. 


doe      [Ds.lProdct 


1895.' 
Feb.    4 
M.ir  28 
Mav ,  3 
Sept  14 


I 
ToMdse.^gCds.  May  :  5240  1200O1    b(HnX< 
Cash.  M.ir.  28278  333aX>  120000 

Mdse.ff  oOds.  Jnne    2212  148400 
••      «  30  ••  ,Oct.  14  78  1170a>  15000e|Oct 


Dec.  31  Xo.  ds.  int.  on  §1, 


31 


To  Bal.  of  days 
interest  on  $1 
which  divided 
by  the  8  per 
cent  int.  div- 
isor 4500  gives 
§81.42  int. 


719000 
352600 


386400 


1895.1  I     I 

Mar.  lOBy  Cash,  Mar.  10 

Apr.  25  "  Acpt.®60ds.  June  27 
OOOOlJuue;  7  "   Cash. 

19  "  Proceeds  of  ' 
flonr.  due  by  ' 
equation  Oct.  22  Oct.  i22 


81}4 


398142 


296 
187 
207 


Dec.  31  Xo.  ds.  int.  on§l. 
.31  £y   Halaace    due 
to  date, 


70 


I 


I 


88800 
93500 
82800 


30000 
cOOOO 
40000 


87500  125000 


S52600 


153142 


398142 


Erplanaiion. — In  this  solntion.  we  first  mature  all  of  the  debit  and  credit  items  of  dollars, 
and  then  find  the  number  of  days  that  each  item  of  dollars  on  the  debit  and  credit  sides  of  the 
account  has  been  due  prior  to  the  day  of  settlement,  December  31.  Having  these  niimber  of  days, 
we  proceed  thus :  Taking  the  first  transaction  on  the  debit  side,  we  see  that  the  J500  was  due  May 
5,  ^40  davs  before  December  31.  and  hence  Reuch  &  Wurzlow  owe  interest  on  the  same  for  240  days, 
or  ^n  $1  ior  500  times  240  davs.  which  is  120000  days.     In  this  manner  we  continue  with  all  of  the 


ACCOUNTS    CURREXT    AND    INTEREST    ACCOUNTS. 


655 


debit  Hems,  and  produce  a  debit  of  719000  days  interest  on  SI  against  Kench  &  Wnrzlo-n-.  AVe  then 
pass  to  the  crtilit  side,  and  in  tlie  lirst  transaction  we  see  that  Keiich  i  AVurzlow  iiaid  §300,  211(3 
days  before  December  31,  and  hence  they  have  interest  due  them  thereon  for  296  days,  or  on  SI  for 
SlXI  times  296  days,  -n-hich  is  88800  days.  In  this  manner  we  contiuue  with  all  of  the  credit  items 
and  ijnnluce  a  credit  of  352600  days  on  $1  iu  favor  of  Kench  tt  AVurzlow.  We  then  take  the  dift'er- 
ence  between  the  debit  and  credit  of  days  interest  on  $1.  and  tind  the  balauce  to  be  366400  days 
interest  that  Eench  and  Wurzlow  owe  on  §1.  This  366400  we  divide  by  the  8  per  cent  interest 
divisor,  and  obtain  S81.42  interest,  which  is  charged  against  Kench  j;;  Wuxzlow,  and  the  account 
balanced  as  in  the  first  solution. 

THIRD  OPEKATION  BY  THE  EQUATION  METHOD, 
it  WinzLOW  iu  Account  Current  and  Interest  Account  at  S^o  to  December  31,  1895,  with 

illTCHELL    it    BliAXDXER. 


I       RE>-cn 
Dr. 


CR. 


DR. 


Cr. 

DR.  CB. 


1895. 
Feb. 
Mar. 
May 

Sept 


ToMdse.®90ds.+     0 

"  Cash. 


I  I' 

Ds.  Prod'ct      Amt. 

I  I 


90   45000  '  500'00 
;2    62400  12tX)|00 


Mdse.®30ds.  4-  88  11.S    82600  i  700  00 
•'      ®30ds.+222  252  378000„1500  00 


|No.  of  ds.  Cr.  on  $1, 
Ko.  of  ds.  Dr.  on  $1, 

Bal.  of  Dr.  ds.  on  §1, 
which  -r  the    8    per 
cent  interest  divisor 
4500    gives    balauce 
of  iuterest 


1 


568000^  3900 
934400 


366400 


81 


00 


42 


3981142 


1895. 
Mar. 
Apr. 
June 
Oct. 


Dec. 


Dec. 


10  By  Cash, 
25   "  Acpt.  ®  60  ds.  4-80 
7   "  Cash, 
19 1  "   Cash    proceeds  of 
flour    due    bv    equa- 
tion Oct.  22,  " 
$3900 
2450 

3l[By  $1450  X  330 

No.  of  ds.  Dr.  on  §1. 
31  III)  Balance  due  to  date, 


Ds.  Prod'ct 


34 
143 
123 


260 


10200 
71500 
492001 


400 
325000  1250 


Amt. 


soo'oo 

500100 
00 


455900 
478500, 


934400 


2450 


1531 


3981  42 


OO 
00 

42 


Explanation. — In  this  solution,  we  first  assume  the  earliest  date  as  the  maturity  of  all  the 
debit  and  credit  items  of  dollars,  and  then  allow  an  equivaleut  credit  of  days  ou  the  purchases, 
and  charge  the  Corresponding  deliit  of  days  on  the  payments,  in  tlio  same  manner  as  we  worked 
equation  of  accounts,  when  we  assumed  the  earliest  date  as  the  maturity  of  all  the  dollar  debit  and 
credit  items.  By  this  work  we  obtain  a  credit  of  568000  days  on  $1,  in  favor  of  Kench  iV-  AVnrzlow, 
and  a  debit  of  455900  days  ou  81  against  Kench  &  Wurzlow.  It  should  be  borne  in  mind  that  the 
debit  and  credit  of  days  are  shown  ou  the  sides  of  the  account,  reverse  to  the  debit  and  credit  of 
dollars;  i.  e.,  the  debit  of  days  on  the  credit  side  of  dollars,  and  rice  rersa. 

These  debits  and  credits  of  days  on  SI,  represent  the  contract  credit  of  days  on  the  various 
transactious,  including  tlie  debit  and  credit  of  days  obtained  b.v  assuming  all  the  dollar  items  due 
ou  the  earliest  date,  February  4,  and  hence,  iu  eflect,  justly  make  both  the  debit  and  credit  of  all 
the  dollar  items  due  on  the  earliest  date,  February  4;  and  as  we  settle  or  close  the  account  Decem- 
ber 31,  it  is  clear  that  Eench  ct  Wurzlow  owe  interest  on  S3900,  from  February  4,  to  December  31, 
330  days,  which  is  equal  to  1287000  days  on  SI;  and  that  they  have  due  to  them  by  Mitchell  & 
Branduer,  interest  on  the  money  paid,  $2450,  from  Februiiry  4,  to  December  31,  330  days,  which  is 
equal  to  808500  days  on  $1.  and  when  deducted  from  the  1287000  days  that  they  owe,  leaves  a  bal- 
ance of  478500  days  that  they  still  owe  on  SI. 

It  is  obvious  that  instead  of  talcing  the  difference  between  the  debit  and  credit  of  days,  we 
conld  hare  added  the  resjiective  numbers  to  the  debit  and  credit  of  days  first  found.  By  this  elu- 
cidation, we  see  clearly  how  the  balance  of  days  is  obtained. 

But  in  practice,  and  in  the  operation  of  this  account,  to  save  time  and  labor,  instead  of  mul- 
tiplying both  the  total  debit  and  credit  of  dollars  bv  the  days  iuteivening  between  the  earliest  date 
and  the  date  of  settlement,  we  first  find  the  difl'ereuce  or  balance  between  the  debit  and  credit  of 
dollars,  and  then  multiply  the  same  by  the  intervening  time,  and  thus  produce  tlie  correct  balance 
of  debit  or  credit  of  days.  In  this  account  the  diti'erence  is,  $3900  —  S2450  =  S1450,  which  multi- 
plied by  330  days,  the  intervening  time,  gives  the  478500  days  debit  on  $1,  which  we  add  to  the  debit 
of  days  obtained  by  the  first  work,  and  thus  obtain  a  total  debit  of  934400  days  interest  on  $1. 
Having  now  the  total  days  interest  on  SI  that  Eench  &  Wurzlow  owe  to  Mitchell  &  Brandner,  and 
that  Jlitchell  &  Brandner  owe  to  Kench  &  AVnrzlow,  we  take  the  difference  and  obtain  a  balance 
of  366400  days  interest  on  $1  that  Kench  &  Wurzlow  owe,  which  we  divide  by  the  8V  interest 
divisor.  4500.  and  obtain  $81.42.  This  interest  we  charge  to  Eench  &,  Wurzlow,  and  balance  the 
account  as  iu  the  preceding  solutions. 


656 


SOULE  S    PHILOSOPHIC    TRACTICAL    MATHEMATICS. 


THE  EQUATION  METHOD 

presents  advantages  -n-liich  no  other  method  can.  It  is  shorter  than  most  other 
methods,  and  identical  with  e(iuatioii  of  payments  when  the  earliest  date  is  assumed 
as  the  Jiiatnrity  date  of  tlie  dollar  items. 

i;y  the  Interest  and  I'rodiiet  JNIethods,  the  interest  or  prodnct  on  any  item  of 
dollars  cannot  be  found  until  the  day  that  the  account  is  settled  or  closed,  or  the 
date  for  closing  determined,  and  by  reason  of  this,  accountants  are  pressed  with 
work  at  stated  ])eri()ds  of  rendering  accounts  or  of  closing  books,  even  when  "mak- 
ing out"  iiidividtial  accounts.  But  by  the  Equation  Method  the  i)roducts  maybe 
coniptited  on  all  of  the  debit  and  credit  items  of  dollars  attlie  pleasure  of  the  book- 
keeper, and  when  the  account  is  settled  or  closed,  one  subtraction,  one  nuiltiplica- 
tion,  and  one  division  operation  is  performed,  and  the  balance  of  interest  is  obtained. 

By  this  method,  should  it  be  desired  to  equate  the  account  after  the  intei-est 
products  are  found,  one  subtraction  and  one  division  operation  accomplishes  the 
object  as  shown  by  the  following  oi)eratiou : 


No.  of  days  Cr.  on  $1  is  508000 
No.  of  days  I)r.  ou  $1  ia  455900 


Debit, 
Credit, 


$3900 
2450 


Credit  balance,       112100  Debit  balance,     $1450 

112100  —  14.50  =  77i\  days,  which  added  to  Feb.  4,  gives  April  22,  tlie  equated  date..  Inter- 
est on  $1450  from  April  22,  to  Dec.  31,  =  253  days,  is  $81.52,  Tliis  gives  10  cents  excess  of  interest 
over  tbe  former  metliods  and  results,  from  tlie  fact  tliat  vre  did  not  use  tlio  yi|th  of  a  day  in  tlie 
equated  date,  and  hence  the  253  days  used  in  comiiutiug  the  interest  were  too  many  by  jjthof  a  day 

2.     What  is  the  cash  balance  of  the  following  account,  September  1,  1895, 
interest  at  8  per  cent 'i  Ans.     $1-4075.37. 

Br.  EoBERSON  &  Brice.  Gr. 


1895. 

— 

1895. 



Mar. 

21 

ToMdse.,        -         -         -        - 

1500 

00 

.Ian. 

24 

By  Cash, 

... 

5000 

00 

May 

ir, 

"   Cash,          .         .         -         - 

2100 

00 

Mar. 

s 

"       " 

. 

7500 

00 

June 

4 

It      It 

500 

00 

Apr. 

28 

it             K 

. 

1200 

00 

.July 

.s 

"   Mdse.  ®  4  months 

3200 

00 

June 

1 

it             tt 

- 

1400 

00 

It 

n 

"   Cash,          -         -        -         - 

400 

00 

July 

10 

"    Arpt. 

®  90  days. 

8000 

00 

Aug. 

20 

"   Mdse.  ®  2  months, 

1800 

00 

FIRST  OPERATION  BY  THE  INTEREST  METHOD. 
ROBERSON  &  Brice  in  Account  Current  and  Interest  Account  at  S/'o  to  September  1,  1895,  with 

Sl.MON    &.   EllKKN. 

Br.  Cr. 


When 

due. 

Us. 

lut. 

Amt. 

1895. 

due. 

1 

Us. 

Int. 

Amt. 

1895. 

1 

t 

1 

Mar 

21 

To  Mdse., 

Mar. 

21 

164 

54 

67 

150000 

Jan. 

24 

By  Cash, 

Jan. 

24 

220 

244 

44 

500000 

May 

16 

"  Cash, 

May 

16108 

50 

40 

210000 

Mar. 

8 

tt       tt 

Mar. 

8 

177 

295 

00 

7500100 

June 

4 

((      (f 

June 

4 

89 

9 

89 

500|0f 

Apr. 

28 

tt      It 

Apr. 

28 

12fa 

33 

60 

120000 

July 

3 

"  Mdse.,®  4mos. 

Nov. 

3 

fi.f 

44 

SO 

320000 

June 

1 

tt      It 

June 

1 

92 

28 

62 

1400l0O 

11 

"  Cash, 

July 

11 

52 

4 

62 

400^00 

Julv 

10 

"  Acpt.  ®  90 

Oct. 

11 

40 

?1 

11 

8000|00 

Aug. 

20 

"  Mdse.,®  2mo8. 

Oct. 

20 

^9 

19 

GO 

180000 

days. 

— 

— 

Dr.  of  red 

interest  $64.40 
Cr.  of  red 

interest    71.11 
"  Bal.    of 

6 

71 

Total  credit 
of  interest. 

Total  debit  of 
interest. 

By  bal.  of  cre- 
dit of  int. 

601 
126 

66 
29 

475 

37 

126 

29 

Sep. 

1 

To  Balance  due  this 
date, 

14075 

37 

23575 

37 

23575 

37 

ACCOUNTS    CURRENT   AND    INTEREST   ACCOUNTS. 


65/ 


Explanation. — la  this  account,  we  have  what  is  called  " red  interest,"  by  which  is  meant 
interest  which  appears  by  reason  of  the  method  on  either  the  debit  or  credit  side  of  the  account 
when  injustice  it  belongs  and  will  be  carried  to  the  opposite  side.  The  transactions  of  July  3  and 
August  20,  on  the  debit  side,  and  of  July  10,  on  the  credit  side  of  the  account,  did  not  mature 
until  after  the  date  of  settlement,  and  hence  it  is  clear  that  the  interest  on  each,  from  the  day  of 
settlement  until  the  maturity  of  these  transactions,  belongs  to  the  reverse  side  of  the  account,  and 
to  facilitate  the  addition  of  the  column  without  including  the  same,  and  also  to  facilitate  transfer- 
ring it  to  the  side  to  which  it  belongs,  it  is  generally  entered  in  redink.  In  this  account,  not  being 
able  to  print  with  red  ink,  we  have  indicated  the  numbers  that  the  accountant  would  have  made 
red,  by  printing  them  in  italic  figures. 

Instead  of  thus  indicating  by  red,  the  interest  that  appears  on  one  side  when  it  belongs  to 
the  other,  some  accountants  have  a  debit  and  credit  column  for  interest  on  both  the  debit  and  credit 
sides  of  the  account. 


SECOND  OPERATION  BY  THE  PRODUCT  METHOD. 


EoBERSON  &  Brice  in  Account  Current  and  Interest  Account  at  8^q   to  September  1,  1895,  with 

Simon  &,  EHRE>f. 


Br. 


DR. 


CR. 


Cr. 

CR. 


1895. 
Mar. 
May 
June 
July 


Aug. 


Sep. 


To  Mdse., 
"  Cash, 

"  Mdse.  -S    4 

mouths, 
"  Cash, 
"  Mdse.  'S>    2 

months, 
"  Balance  of 

prod,  in  red, 


To  Balance  due 
this  dale, 


■When 
due. 


Mar. 
May 
June 

Nov. 
July 

Oct. 


Da. 


Prod'ct 


164 

108 
89 

GS 


49 


246000 

226800 

44500 

soieoo 

20800 


30200 


568300 


i5oo;oo 

210001 
50000 

320000 
40000 

I 
180000 


1407538 
23575  38 


1895. 
Jan. 
Mar. 
Apr. 
June 
July 


By  Cash, 


'  Acpt.  -3)  90 
days. 

No.  of  ds.   Cr 

on  SI, 
No.  of  ds.  Dr 

on  $1, 

Bal.  of  ds.  Cr. 
on  §1, 
which  divid 
ed  by  the  8% 
int.  div.  4500 


Jan. 
Mar. 
Apr. 
June 

Oct. 


"WTie„    ,„ 
due.     1"3. 


24220 

8177 
28  12(5 


Prod'ct. 


1100000 

1327500 

151200 

128800 

330000 


2707500 
568300 


2139200 


Amt. 


5000 
7500 
120000 
140000 


00 
00 


800000 


47538 


23575 


38 


Explanation. — By  the  Product  Method  of  work,  we  have  "red  products"  instead  of  "  red  inter- 
est," which  are  treated  in  every  respect  the  same  as  "red  interest,"  and  for  the  same  reasons  aa 
given  in  the  solution  of  this  problem  by  the  Interest  method.  By  reason  of  the  fractions  of  a  cent 
interest  in  the  operation  by  the  Interest  Method,  we  obtain  1  cent  more  interest  by  this  Method, 
than  by  the  Interest  Method. 


658 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


THIRD  OPERATION  BY  THE  EQUATION  JIETHOD, 

EOBERSON  &  limcE  in  Account  Current  and  Interest  Account  at  8%  to  September  1,  1895,  with 

Simon  &  Eiiuen. 
Jh:  Cr. 

CR.  VR.  DK.  CR. 


1895. 

Miir. 

21 

May 

16 

June 

4 

July 

3 

iC 

11 

Aug. 

20 

Sept. 

1 

11 

1 

It 

1 

To  Mdse., 

"   Cash, 
it       it 

"  Mdse. 'S)  4  months 

(123  days), 
"   Ca.sh, 
' '   Mdse.  'a'  2  months 

(61  days), 


By  $13600  mult,   by 

220, 
No.  of  ds.  Cr.  on  $1, 


To  Bal.  due  Ihia  elate, 


Ds. 

Product. 

Amt. 

1805. 
.I;ni. 
Mill-. 
Apr. 

24 

8 

28 

56 
112 
131 

84000 

235200 

65500 

1500 

2100 

500 

00 
0(1 
00 

283 
168 

905600 
67200 

3200 
400 

00 

00 

June 
July 

1 

10 

269 

484200 

1800 

00 
00 

Sept. 

1 

1841700 
2992000 

9500 

4833700 

14075 

38 

23575 

38 

By  Cash, 


'   Acpt.  &  90  ds.  (93 
plus  167), 


By  Bal.  of  days  Cr. 
on  $1, 
which  divided  by 
the   8%  int.    div. 
4500  gives 


Ds. 


0 

43 

94 

128 

260 


Product. 


322500 
112800 
179200 

!080000 

2694500 

2139200 


4833700 


A  int. 


5000  00 
7500J0O 
1200  00 
1400|00 


8000 


23100 


OO 


OO 


475  38 


23575  38 


Explanation. — By  the  Equation  Method,  vre  have  no  red  interest  or  red  products.     The  reason 
ing  for  tlie  work  is  the  same  as   given   iu  the  tirst  problem  by  the  Et|uatiou  Method,  and  henco 
is   omitted. 

3.     What  is  tlie  cash  balance  of  the  following  account,  October  24,    1895, 
interest  at 6  percent?  Ans.     $1704.35 

Dr.  A.  DoEKR.  Cr. 


1895. 

) 

Jan. 

1 

To  Mdse.  ®  60  days,  - 

$500 

00 

Mar. 

10 

"   "   (no  credit),  - 

800 

00 

Apr. 

1 

"  Ca,sh,    -    .    .    . 

1.50 

00 

Sept. 

20 

"  Mdse.  ®  60  days,   - 

200 

00 

OPERATION  BY  THE  EQUATION  METHOD. 
Dr.        A.  DoEKR  in  Account  Current  and  Int.  Acct.  at  6,%'  to  Oct.  24,  1895,  with  A.  Rivier.       Cr, 

CR.  DR.  DR.       CR. 


1895. 

Jan. 

1 

Mar. 

10 

Apr. 

1 

Sept. 

20 

Oct. 

24 

it 

24 

To  Mdse.  ®  60  ds. 
"       •'      (uo  credit), 
"  Cash, 
"  Mdse.  ®  60  ds. 

No.  of  ds.  Cr.  on  $1, 
To  bal.  (326100)  of  ds. 
Dr.  on  $\.  which-l-the 
6,"„  iut.  divisor  gives 


Ds. 

Prod'ct 

Amt. 

1895. 

60 

30000 

500 

00 

Oct. 

•24 

68 

54400 

800 

00 

90 

13500 

150 

00 

ti 

24 

322 

64400 

200 
1650 

00 
00 

162300 

54 

35 

1704 

35 

By  $1650  X  296,  No.  of 

ds.  Cr.  on  $1, 
By  Bal,  due  this  date, 


Ds. 


Prod'ct 


488400 


Amt. 


1704 


1704  35 


35 


ACCOUNTS    CURRENT   AND    INTEREST    ACCOUNTS. 


659 


4.     TTliat  is  tlie  cash  balance  of  the  following  accouut,  December  31, 1895, 
interest  at  10  per  cent  ?  Ans.     $5549.72. 

Br.  C.  Enderlin.  Cr. 


1895. 

Oct. 

8 

Nov. 

(i 

Dec. 

10 

Bv  Cash, 

Dft.  <S'  30  (l.iys, 
Acpt.  fai  60  days 


$3000 

500 

2000 


00 
00 
00 


I 


OPERATION  BY  THE  PRODUCT  METHOD. 

C.  Endeklin  in  Account  Current  and  Jut.  Acct.  at  10%  to  Dec.  31,  1895,  with  Hagan  &  Henderson. 

Br.  Cr. 

DR.        DR.  CE.       CE. 


1895. 
Dec. 


31 


To  Hal.  due  this 
date, 


"Wlien 
due. 


Ds. 


Prod'ct 


5549 


1895. 
Oct. 
72  Nov. 

Dec. 


554972 


8  By  Cash, 

6  "   Dft.®  30  ds. 

10  "  Acpt.®60ds. 

No.  of  ds.  Cr.  on 

II. 
No.  of  ds.  Dr.  on 

$1  (red), 

Bal.   of  ds.   Cr. 

on  $1, 
which  div.  by 
the    10"^    int. 
div.  3600  gives 


Oct. 
Dee. 
1896. 
Feh. 


When 
due. 


Da. 


11 


Prod'ct 


252000 
11000 

S4000 


263000 
84000 


179000 


Amt. 


300000 
50000 

200000 


49 
5549 


72 
72 


AN  ACCOUNT  WITH  SPECIAL  CONDITIONS  OP  DISCOUNT  FOR  CASH 
PAYMENTS  FEOjM  WHICH  AN  ACCOUNT  CURRENT  AND  INTER- 
EST ACCOUNT  AND  AN  EQUATION  OF  ACCTS.  ARE  MADE. 


1208. 

ditions: 


J.  McDonald  bought  of  John  Gunn,  on  the  following  dates  and  con- 
Jan.  3,  1895.  $237.65  on  90  days,  or  3?;^  discount  for  cash  in  5  days. 
"  17,  •■  123.50  on  60  "  3%  "  "  "  " 
"  24,  "  324.00  on  60  "  2%  "  "  "  '• 
"  31,  "  150,00  ou  30  "  2%  "  "  "  •' 
Feb.  10,  "  200.00  on  120  "  ^«i  "  "  "  " 
"  1.5,  "  100.00  on  120  "  3%  "  "  "  " 
"24,      "           75.85  net             " 

J.  McDonald  paid  cash  on  account  of  said  purchases  as  follows : 
January  10,  1895,  $150.00.  January  21,  1895,  $100.00  April  5,  1895,  $300.00. 

Required:    1.  What  is  the  balance  due  July  1,  1895,  at  6  jier  cent  interest? 


66o 


SOULE'S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  If  equated,  wliat  would  be  tlie  equated  date  for  paying  the  balance  of  the 
account  i 

IvEMARKS.  It  will  be  observed  that  tliere  was  a  conditional  discount  privi- 
loge  on  each  bill  purchased.  But  since  the  purcliaser  did  not  avail  him.self  of  said 
])rivilego  by  paying  cash  within  tlie  specified  time,  tlie  privilege  became  inoperative 
and  cannot  be  considered  in  the  solution  of  the  problem. 

Special  or  adjunct  conditions  and  terms  of  contracts,  are  not  binding  upon 
either  party  thereto,  unless  strictly  conformed  to  by  both  parties. 

OPERATION  BY  THE  PRODUCT  SYSTEM  TO  FIND  BALANCE  DUE  JULY  1,  1895. 

J.  McDonald  iu   Accouut  Curreut  and  Interest  Account  wit.li  John  Gunn,  interest  at  G^^  to 

July  1,  1895. 

Dr.  ■  Cr. 

DR.  DR.  CR.         CR. 


When 

Wbcn 

due. 

Us. 

Prod'ct 

Amt. 

1895. 

due. 

Ds. 

Prod'ct 

Amt. 

1895. 

Jan. 

3 

90  ds.  or  3%  for 

J.an. 

10 

Cash, 

172 

25800 

1.50 

00 

cash  in  5  ds. 

Apr. 

3 

89 

21182 

237 

65 

a 

21 

(*          _        _ 

Ifil 

16100 

100 

00 

tt 

17 

60  ds.  or  3%  for 
cash  in  5  ds. 

Mar. 

18 

105 

12915 

122 

50 

Apr. 

5 

"          . 

87 

26100 

300 

00 

It 

24 

60  ds.  or  2"^  for 

No.  ds.  Cr.  on  |1, 

68000 

cash  iu  5  ds. 

ti 

25 

98 

31752 

324 

00 

.Inlv 

1 

By  Balance  due, 

665 

S4 

ti 

31 

30  ds.  or  2%  for 
cash  in  5  ds. 

It 

2 

121 

18150 

150 

00 

Feb. 

10 

120  ds.  or  i%  for 
cash  in  5  ds. 

June 

10 

21 

4200 

200 

00 

ti 

15 

120  ds.  or  3%  for 
cash  iu  5  ds. 

ct 

15 

16 

1600 

100 

on 

t( 

24 

Net, 

No.  ds.  Dr.  on$l. 

To  Bal.  ds.   Dr. 

on$l, 
■vvhieh  div.  by 
6"„'    int.     div. 
6000  gives 

Feb. 

24 

127 

9652 

99451 
S8000 

31451 

75 

5 



1215 

85 

24 
24 

1215 

24 

OPERATION  TO  EQUATE  THE  ACCOUNT. 


1895.                   Ds.  Cr. 

Jan.    3,  $237.65    90 

17,    122..50    60  +  14=    74 
24,    324.00    60  +  21  =    81 
31,    150.00    30  +  28=    58 

Feb.  10,    200.00  120  +  38  =  1.58 
15,    100.00  120  +  43  =  163 
24,      75.85    net                 52 

Ds.  Cr.  on  $1. 

-  -     21420 

-  -       9102 
.     -     26244 

-  -       8700 

-  -     31600 

-  -     16300 
-       3952 

1895. 
Jan.    10, 

21, 
Apr.      5, 

$1.50.00 
100.00 
300.00 

1550.00 

$1210.00 
550  00 

117318 
30450 

Ds.  Dr.  Ds.  Dr.  on  $1 

7 1050 

18 1800 

92 27600 

30450 


$660.00  86868  H-  660  =  131?f  ds.  practically,  132  ds.,  after  Jan.  3, 

1895,  —  May  15,  1895,  equated  date. 

NoTR. — Interest  on  $660,  May  15,  to  July  1,  47  days,  =  S5.17.  Interest  shown  in  the  Account 
Current  ami  Interest  Account  is  $>5.24.  The  dilference  results  from  the  fraction  of  a  day  lost  in  the 
equated  date. 


ACCOUNTS    CURRENT    AND    INTEREST    ACCOUNTS. 


66l 


In  practice,  instead  of  arranging  the  credit  of  the  account  to  the  right  of  the 
debit  side,  it  is  frequently  placed  below  it,  in  oi'der  that  it  may  be  made  ou  paper 
of  a  convenient  width  to  copy  in  the  accouut  current  book. 

The  following  account  current  illustrates  this  form  : 


E.  Kicker  in  Account  Current  ami  Iiit.  Acct.  at  Sj!^  to  October  1,  1895,  with  T.  IT.  Erode  &  Co. 

DEBITS. 


I 


1895. 
Mar. 
April 
June 
Oct. 


1895. 
July 

Sept. 


Oct. 


To  Mdse.  ®  30  days, 
"  Cash,  -        .        . 

"  Mdse.  ®  60  (lays, 
"  Red  interest  from  credit. 

Total  debit  of  interest, 
"      credit  of  interest, 

To  Balance  of  debit  interest, 


CREDITS. 


By  Cash,     ... 
"   Acpt.  <3>  30  days, 

By  Balance  due  this  dale, 


Wbeiidue. 

D.s. 

Int. 

Amt. 

April 
Aug. 

24 
10 

in 

160 

174 

46 

64 

46 

20 

4 

135 
30 

00 
40 
44 
67 

51 
67 

1800 
1200 
2000 

104 
5104 

00 
00 
00 

84 

84 

When  due.    Ds.        Int 


.July 
Oct. 


92 

7 


30  67 
4  67 


30 


67 


Amt. 


1500 
3000 


604 


5104 


00 
00 


84 

84 


ACCOUNTS  CURRENT  AND  INTEREST  ACCOUNTS  IN  ENGLISH  MONET. 

'-i209.     5.    What  is  the  cash  balance  of  the  following  account,  July  1,  1895, 
inteiest  at  6  per  cent  ?  Ans.    £462  5s.  l|d. 


Dr. 


E.  G.  MusaROVE,  Jr. 


Cr. 


£ 

s. 

d. 

£ 

s. 

d. 

1895. 

— 

— 

— 

1895. 

— 

— 

Jan. 

15 

To  Cash, 

500 

8 

6 

Feb. 

10 

By  Cash,     ... 

1100 

00 

8 

April 

8 

"   "       ... 

1000 

14 

3 

xMar. 

5 

"   "       ... 

800 

10 

4 

June 

18 

"  Ex.  ®  60  days, 

1000 

00 

0 

May 

21 

"   " 

1050 

6 

11 

662 


SOULE's    rHILOSOPHlC    PRACTICAL    MATHEMATICS. 


OPEEATION  BY  THE  INTEREST  METHOD. 
R.  G.  MUSGROVK,  Jit.,  ill  Account  Current  and  Int.  Acct.  at  G"o  to  July  1,  1893,  with  A.  CAr.ON. 

Br.  Cr. 


1895. 
Jan. 

Apr. 
June 


July 


ro  Cash, 

it  tl 

"  Kx.     at 

60  da. 


To  BaJ.  (hie 
thin  date, 


When 
[due. 


J.nn. 
Apr. 

Aug. 


l.T  167 
84 


30 


50 


Int. 


18'6J 
02i 


£ 

500 
1000 

1000 


00 


47S 


7,9 


298o'  1 


18!)o. 
Vvh. 
Miir. 
May 

July 


By  Cash, 


Ani'tof  Cr. 

int. 
Red  Int. 

Am'lofJiit. 

l|l?al.   of  Cr. 
int. 


When 
due. 


Feb. 
Mar. 
May 


Int. 


10141  I'DlTOO 

")  lis  l.")14  10 

21  41    7   3  6i 


£ 
1100 

800 
1050 


29 


2980 


U 


Note. — See  page  507  lor  a  similar  account. 


Dec.  1,  1895,  A.  and  B.  agreed  to  enter  into  a  joint  speculation,  each  J  in  tbe 
purchase  of  200  shares  of  factory  stock.  A.  is  to  advance  the  margins  and  to  settle 
with  the  broker  for  the  purchase  of  the  stock,  and  to  render  an  account  to  B. 

Accordingly,  Dec.  5,  A.  advanced  to  his  broker,  H.  Lee,  $2000,  and  Dec.  20 
he  advanced  $1000  more. 

Dec.  5,  the  broker  purchased  200  shares  of  factory  stock  at  93^,  $18750.  Com- 
mission for  purchasing  at  ^  per  cent,  $25. 

Dec.  22,  the  broker  collected  a  li  per  cent  dividend  $300,  and  March  23, 1894, 
he  collected  a  second  IJ  per  cent  dividend  $300. 

May  G,  the  broker  rendered  his  account  to  A.  which  A.  settled  on  the  8th  of 
May,  and  the  stock  was  delivered  to  him.  June  26,  A.  received  a  $300  dividend  on 
the  stock. 

June  30,  189G,  A.  rendered  his  account  to  B, 

Required:  1.  The  broker's  account  showing  the  balance  due  him  May  6th 
and  8th.    2.  The  account  of  A.  showing  balance  due  him  by  B.  June  30, 1894. 


SALES    ON    JOINT  .ACCOUNT. 


66' 


OPERATION. 

Two  Hundred  Shares  Stock  at  6  per  cent  per  annnin  to  May  6th,  1895. 


Date. 

Amt. 

Ds. 

1894, 
Bee. 

5 
5 

6 

8 

To  Cash,  200  shares  ) 

®  931 
"   Com.  ®  i%           ) 
Less  Or.  lut. 

I5aL  Dr.  lut. 

18775 
393 

00 
18 

152 

19168 

18 

18 
19 

1895. 
May 

To  Balance  due, 

lut.  ou  same  for 
2  ds.  ®  6%, 

Amount  due, 

15568 

5 

May 

15573 

37 

lut. 

Date. 

Amt. 

Ds. 

Int. 

1894. 

Dec. 

5 

Cash, 

2000 

00 

1,-2 

■■^O 

67 

475 

63 

" 

2(1 

" 

1(1110 

(Id 

137' 

22 

S3 

it 

22 

Dividend  1^^^, 

30(.i 

00 

135 

6 

75 

82 

45 

1895. 

Mar. 

23 

Total  Cr.  Int. 

300 

3600 

00 

00 

44 

2 

82 

20 
45 

21(11/ 

6' 

By  Balance  due, 

19168 

IS 
18 

ACCOUXT  OF  A.  WITH  A.  AXD  B. 

A.  and  B.  each  J  in  account  tvith  A.  from  Dec.  5,  1894,  to  June  30,  1893,  at  (J  per  cent 

per  annum. 

Date.  Amt.  Ds.        Int.  Date.  Amt,      Ds,      Int. 


1894. 
Dec. 


1895. 
May 


Jnne 


30 


Cash — margin     to 

broker, 
Cash— margin     to 

broker. 

Cash — paid  liroker 
as  per  his  account 

Total  Dr.  Int. 
Less  Cr.  Int. 

Bal.  Dr.  Int. 


To  Balance, 

r.EilAEKS. 

A'sJ  -        -        $9255,80 

Li's  i"  (B.  owes  A. 
Jiiue  30)         -       9255.87 


2000  00 
00 


1000 
15573 

238 


18811 


18511 


37 


36 


207 
192 

53 


69 
32 

137 

238 


1895,1 
June 


$1»511.73 

Wlien  B,  pays  A.  $9255.87 
then  A.  &  B,  will  each 
own  A  of  200  shares  of 
stock  as  per  account  with 
broker. 

Note, — In  this  account  we  change   the   locations  of  the   columns   for   amount,  days   and 
interest,  and  omit  the  '■  when  due"  column. 

This  is  in  accordance  with  the  form  in  general  use  by  Stock  Brokers. 
See  "Stocks  and  Bonds"  in  this  work  for  further  accounts  with  brokers. 


By  Cash,  Div.  1*%, 
"   Balance, 


300 
ISoll 


00 


18811  73 


20 


664 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


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lankers'  Interest  on  Daily  Balances. 


1310.  Ill  the  foregoing  Interest  Accounts  Current,  we  have  illustrated  the 
different  methods  of  finding  the  cash  balance  of  accounts  at  stated  periods,  or  at 
such  irregular  times  as  a  settlement  may  be  desired. 

It  is  a  general  custom  with  most  baukers  to  balance  accounts  with  their 
correspondents  every  month,  or  every  3,  6,  or  12  mouths,  allowing  interest  on  all 
credits,  and  charging  interest  on  all  debits,  and  bringing  the  balance  of  interest  to 
the  debit  or  credit  side  of  the  account,  as  the  case  may  be.  The  balance  of  interest 
thus  entered  and  brought  down  to  the  new  account,  will  subsequently  draw  interest 
the  same  as  any  other  item  in  the  account.  The  effect  of  thus  closing  and  bringing 
down  the  interest  balances  is  the  compounding  of  money  as  often  as  the  accounts 
are  thus  closed,  as  we  fully  explained  in  the  Merchants'  System  of  Partial  Pay- 
ments. 

But  in  addition  to  the  accounts  current  and  interest  accounts  with  their 
correspondents,  some  bankers  allow  interest  to  their  depositors  on  the  daily  bal- 
ances of  their  respective  accounts,  and  to  elucidate  the  method  of  computing 
interest  on  such  balances,  as  well  as  the  interest  on  the  daily  balances  of  accounts 
with  correspondents,  we  present  the  following 

PROBLEMS: 

1.  What  is  the  balance  due  February  1,  1895,  on  the  following  account,  at  5 
per  cent  interest  on  daily  balances,  counting  365  days  to  the  year? 


Br. 


Albert  Lee  &  Edward  E.  Soul6. 
Interest  at  5fo- 


Or. 


1895. 

• 

1895. 

Jan. 

1 

To  Checks, 

300 

00 

Jan. 

1 

By  Cash, 

1200  00 

2 

ti          it              _    _ 

900 

00 

0 

"   "       -    . 

800  00 

3 

ti    ~     ii              ,    _ 

400 

00 

4 

1020,00 

6 

a            tt                  ^     _ 

1100 

00 

5 

750  00 

7 

il              n                     _      _ 

800 

00 

fi 

500  00 

9 

"     "       -    - 

500 

00 

8 

900  00 

14 

"     "       -    - 

600 

0(1 

9 

200  00 

1(1 

it            ii 

:i-.o 

00 

11 

1000  00 

21 

ti            It 

2000 

00 

15 

400  00 

25 

tt            tt 

100 

00 

17 

650  00 

28 

tt            tt 

210 

00 

18 

150  00 

30 

It            tt                  _     _ 

75 

00 

22 

"   '<       .    - 

340  00 

Feb. 

1 

"  Balance, 

2902 

73 

24 

27 

"   "       -    - 

220  00 
2100  00 

73 

Feb. 

1 

Interest  to  date, 

7 

73 

10237 

10237 

73 

(665) 


666 


SOULE  S    rillLOSOPIIIC    PRACTICAL    MATHEMATICS. 


FIRST  OPERATION  BY  THE  USUAL  METHOD. 

1211.  The  above  account  presents  the  figures  as  tbey  stand  on  the  Individual 
Ledger  of  tlie  bank,  and  from  tliis  ledger  or  the  pass  book  the  daily  balances 
are  taken  and  arranged  on  a  separate  sheet  of  paper,  in  a  statement  form  as  fol- 
lows : 

Equity  National  Bank,  N.  O.,  in  account  witli  A.  L.  &  E.  E.  Soule,  interest  at 

5  per  cent  to  February  1,  1895. 


DR. 

CE. 

1 

900  00 

2 

800;00 

3 

400|00 

4 

1420  00 

5 

2170  00 

6 

1570  OO; 

7 

770  00, 

8 

167000 

9 

1370,00 

10 

137000 

11 

237000 

12 

2370:00 

13 

2370,00 

14 

1770,00, 

15 

2170'00 

16 

1830  00 

17 

2470'00 

18 

2620'00| 

19 

2620  00 

20 

2620'00' 

21 

620  OU, 

22 

960'00 

23 

960,00; 

24 

1180  00, 

25 

1080  00 

26 

1080  00 

27 

3180  00 

28 

297000 

29 

2970  00 

30 

2895  00 

31 

2895  00 

Total  daily  hal. 

56430100 

Explttnnlinn. — In    finding  the   daily  balance,  we 
observe  tliat.  nu  tbo  1st  ol'  January,  A.  L.   &.  E.   E. 

Soule  deposited - 

and  drew  out  -------- 

leaving  a  balance  of 

The  second  day  they  deposited      -         -         -        . 

Tvbiph  added  to  the  balance  of  tbe  first  day  gives 
of  this  amount  they  drew  out         -         -        -        - 


leaving  a  balance  of        -        -        -        -'       - 
On  the  3d,  they  made  no  deposit  but  drew  out 


leaving  on  deposit 

On  the  4th,  they  deposited 

this  producing  a  balance  of 


$1200 
300 

$  900 
800 

$1700 
900 

$800 
400 

$400 
1020 

$1420 


In  this  manner  we  continue  tlirougli  the 
month  and  find  a  total  daily  balance  of  $50430.00 
for  1  day  against  the  bauk  and  in  their  favor, 
which  divided  by  7300,  the  5%  interest  divisor 
for  305  days  to  the  year,  gives  $7.73  interest  due 
to  the  depositors. 

Had  the  rate  of  interest  been  4  per  cent, 
we  would  have  first  found  the  interest  at  5  j>er 
cent,  and  then  reduced  it  \  of  itself.  Had  it  been 
3  per  cent,  we  would  have  found  the  interest  at  5  jier  cent,  and  then  reduced  it  f  of 
itself.  Had  it  been  6  per  cent,  or  7  i^er  cent,  we  would  liave  found  it  at  5  per  cent, 
and  then  increased  it  \  for  6  per  cent  and  |  for  7  per  cent. 

Or  to  find  the  interest  at  any  rate  i>er  cent,  we  multiply  the  balance  of  the 
daily  balances  by  the  rate  and  divide  by  3G500  when  365  days  are  counted  a  year^ 
and  by  36000  when  360  days  are  counted  a  year. 


ledger   accounts. 

second  operation.— (ledger  account). 
Albert  Lee  &  Edward  E.  Soiile. 


667 


■ 

Daily  Balances. 

lloinhly  Interest. 

1895. 

F. 

Di: 

F. 

Cr. 

Dr.   1 

Cr. 

^           1 

Jan. 

1 

To  Checks, 

300 

00, 

Ky  Cash  Dep. 

1200 

00 

000  00 

1 

900 

no 

■> 

..        .. 

900 

111 

..      ..        .. 

800 

10 

800  00 

1 

. 

800 

on 

(4 

3 

.t       *> 

400 

on 

400  00 

1 

400 

on 

11 

4 

>«      It        It 

1020 

00 

1420,00 

1 

, 

1420 

nn 

1( 

5 

It      1.        11 

750 

no 

2170;00 

1 

2170 

on 

It 

(i 

.(        11 

1100 

on 

II      11        ti 

500 

no 

1570  00 

1 

«      ' 

1570100 

t* 

7 

tt       ti 

800 

on 

77000 

1 

77000 

•  * 

8 

II      II        11 

900 

[)0 

1070  00 

1 

1070  00 

*« 

9 

4*       «» 

500 

0(J 

II      11        11 

200 

00 

1370,00 

2 

274000 

«t 

n 

1*      II        11 

1000 

1)0 

2370  00 

3 

7110  00 

41 

14 

4«              t« 

600 

on 

1770  00 

1 

1770(10 

4( 

IS 

11      11        11 

400 

00 

2170,00 

1 

2170,00 

•  « 

Ifi 

11              M 

350 

0(1 

1820,00 

1 

1820,00 

** 

17 

11      11        11 

650 

00 

2470  00 

1 

2470(10 

•  t 

18 

11      II        11 

150 

00 

2020  00 

3 

1 

7800,00 

t( 

?1 

11              11 

2000 

no 

f)20  00 

1 

020,00 

>« 

'>•> 

11      11        11 

340 

(Id 

000  00 

2 

,      ^ 

1920,00 

•  4 

'>4 

11      11        II 

220 

00 

1180  00 

1 

1 

118000 

(4 

?.■) 

11              11 

100 

no 

108(1  00 

2 

210000 

14 

"7 

11      11        11 

2100 

00 

3180  00 

1 

3180,00 

•  4 

'>8 

•  1              11 

210 

on 

2970  00 

2 

' 

5940  00 

Feb. 

30 
1 

"  Balance, 

2902 

00 
73 

•'  iDterest, 

^ 

73 

2895  00 

2 

579000 

div.  br 

7300 

10237 

5043000 

1 

~ 

7 

73 

10237 

Explanation. — We  here  present  a  ne-n-  form  of  Individual  Ledger  in  ^hicli  we  keep  the 
depositors'  accouut.s,  and  perform  the  operation  to  find  the  interest  on  the  daily  balances.  Where 
there  are  a  large  nnmher  of  accounts  on  -vvhith  interest  is  allowed  on  daily  balances,  this  form  of 
Ledger  may  be  used  to  advantage. 

2.  "What  is  the  balance  of  the  followiug  account  on  March  1,  1896,  making 
monthly  settlements  and  allowing  interest  on  the  debit  of  the  daily  balances  at  8  per 
cent,  and  on  the  credit  of  the  daily  balances  at  G  per  cent,  and  connting  3C5  days 
to  the  year?  Aus.    $17372.18. 

l^EW  York  City  National  Bank. — Interest  8%  on  debits  and  6%  on  credits. 


.g         ~ 
189G. 

- 

Dr. 

- 

• 

Cr. 

30 

Daily  Balances. 

Ds. 

Tot.  Daily  Bal. 

Monthly  Int. 

Jan. 

4 

To  Cash, 

1500  00 

By  Casli, 

2000' 

500  0( 

6 

1 

3000,00 

10  '•    •■ 

5(100(10 

1 

4500  K 

1 

5 

22500 

10 

1 

«* 

15  "      " 

1000  00 

It      II 

900000 

350000'   6 

2100000 

44 

21 

II      II 

sonnoo 

B-wonn  7 

4550000 

4t 

28  '•       '• 

7500  00 

II      II 

11400,00 

10400  0 

',  4 

41  GOO  00 

Felt. 

1  "  Balance, 

10413  33 

*'  Interest, 

13  33 

— 

22500 

.10 

111100  no 

4 

93 

18 

IfB 

25413  33 

J541333 

== — 

— 

1 



— 

T*«I> 

1 

By  Ilalauce. 

1041333 

104133 

3    4 

41653  32 

5  To  Cash. 

40000' 

"  Cash, 

1600,00 

11613  3 

i\  3 

348119  99 

•' 

f 

II      .1 

1800  00 

2100  00 

11913  3 

V   5 

59506  05 

'I 

13 

•1      11 

6000  00 

200,00 

61133 

'    + 

24453 

12 

'. 

17 

11      II 

1    1000  00 

8000  00 

131133 

1    6 

78679 

98 

>> 

23 

1 

380000 

16913  3 

3|   C 

101470 

98 

41 

29 

.1      11 

1700  00 

2100,00 

17313  3 

3    1 

17313 

IB 

Mar. 

1 

'*  Balance, 

17372|18 

"  Interest 

58  85 

-^..— 

— 



— 

2827218 

28272 

18 

357980 

5V 

58 

Mb 

— 

- 

— 

1 

ExpJatialion. — In  this  account  we  have  interest  on  the  debit  of  daily  balances  at  8  per  cent, 
and  on  the  credit  of  daily  balances  at  6  per  cent,  and  hence  to  find  the  interest  on  the  monthly- 
footings  of  the  dally  balances,  we  multiply  the  same  by  the  respective  rates  per  cent,  8  and  6,  and 
divide  by  the  one  per  cent,  interest  divisor  36500. 


668 


SOULES   PHILOSOPHIC   PRACTICAL   MATHEMATICS. 


S.    Wliat  is  tLe  balance  of  the  following  account  June  1,  1895,  allowing 
interest  on  daily  balances  at  6  per  cent,  and  counting  3G0  days  to  the  year? 

Ans.    $7083.75. 

Third  National  Bank,  Baltimore. 


Daily  Balances. 

Tot.  Daily  Bal. 

Us. 

1 

395. 

14 

Dr. 

1 

395. 

30 

Cr. 

Dr. 

Cr. 

Dr. 

Cr. 

V 

1 

To  Cash, 

5000  00 

1 

ly 

1 

By  Casb, 

400000 

1 

1 

ti      tt 

3000  00 

1 

1 

2500  00 

6300 

00 

350000 

1 

3500 

1)0 

1 

11           41 

2000  00 

1000000 

2 

it       t 

3000  00 

f. 

(t           It 

1000  00 

2 

tt       t 

201)0  00 

•>. 

tl           it 

4000  00 

2 

it       t 

1200  00 

"/. 

it           ii 

2500  00 

7500  00 

2 

it       t 

2100  00 
500  00 

2200  00 

8 

17000 

JO 

ro 

it           it 

2000  00 

2 

It       t 

8800 

00 

10 

ii           ii 

1500  00 

' 

12 

it        1 

3000  00 

7700  00 

2 

15400 

00 

10 

ti           ii 

2000  00 

5500  00 

12 

11       t 

1000  00 

Vfi 

ii           ii 

400000 

12 

it       i 

800  00 

4800 

M 

2900 

00 

13 

37700 

JO 

V5 

ii           it 

2500  00 

25 

it       i 

5000  00 

?5 

it           it 

1000  00 

25 

it       t 

3000  00 

V5 

ti           it 

500  00 

25 

ti       i 

2000  00 

VS 

ii           ii 

1500  00 

25 

11       t 

1000  00 

11000 

00 

WOO 

00 

5 

34500 

00 

vn 

11           Ii 

2000  00 

3f 

tl       t 

1400O  00 

14000 

00 

4100 

00 

1 

4100 

OO 

?ri 

11           it 

3500,00 

1500000! 

31 

ti       i 

3000  00 

;io 

it           ii 

2000  00 

1 

31 

tl       1 

9000 

00 

12000 

00 

7100 

00 

1 

7100 

00 

:io 

ai 

1 

"   ;; 

1000  00 

900000 

16  25 

3000  00 
9000  00 

June 

"  Interest, 

108700 

00 

1 

"  ijala»ce, 

7083  75 

00 

11200 

OJ 
00 

11200 

0(). 

57100  00 

67100 

97500 

Explanation. — In  this  account  we  present  another  style  of  ruling  for  the  General  Ledger. 
Where  each  debit  and  credit  item  in  posted  separately,  this  form  of  ledger  may  be  used  to  advan- 
tan-e.  The  interest  being  the  same  on  both  sides  of  the  account,  the  difference  of  the  total  daily 
balances  is  here  divided  by  6000,  the  6  per  cent,  interest  divisor  for  360  days  to  the  year. 


V@(t)  (g  fe)  (i>©)X 


^^^•X^^^©"^ 


lavings  Banks  and  Savings  Bank  Accounts. 


1212.  Savings  Banks  are  Financial  Institutions  for  the  receipt  on  deposit  of 
small  Mums  of  jnouey  lor  which  a  moderate  rate  of  interest  is  allowed  ou  stated 
conditions. 

A  savings  bank  furnishes  each  depositor  with  a  pass  book,  in  which  are 

recorded  all  sums  deposited  and  all  sums  withdravrn.    This  book  contains  the 

conditions  ou  which  dejiosits  are  received  and  interest  is  allowed  and  credited.    It 

also  contains  the  interest  credited  to  the  depositor  at  the  end  of  each  interest  term. 

Such  interest  is  entered  in  red  ink. 

Note. — It  is  not  the  custom  of  savings  banks  to  calculate  interest  on  a  fraction  of  a  dollar, 
but  in  liudiug  the  amount  at  the  end  of  each  interest  term,  fractious  of  dollars  are  included. 

1213.  The  Deposits  in  Savings  Banks  are  usually  paid  on  demand,  though 
the  charter  of  the  bank  and  the  conditions  of  deposit  give  the  institution  the  right 
to  require  60  or  90  days  notice. 

1214.  The  Interest  Term  is  the  period  of  time  at  the  close  of  which  interest 
is  computed  and  credited  as  a  deposit.     It  is  thus  compounded,  if  not  withdrawn. 

The  Interest  Term  varies  with  diflereut  savings  banks;  with  some  it  is  1 
month,  with  others  3  and  with  others  C.  Interest  is  usually  declared  and  credited 
June  30  and  December  31.  In  some  savings  banks,  deposits  commence  to  draw 
interest  on  the  first  of  each  quarter,  January  1,  April  ],  July  1,  and  Oct.  1.  la 
others  the  money  must  be  on  deposit  at  least  three  mouths  before  it  will  be  entitled 
to  receive  any  interest. 

Some  of  the  Savings  Banks  of  New  Orleans  give  interest  on  net  deposits  that 
have  been  in  the  bank  3  months ;  others  require  dei)osits  to  have  been  in  bank  4 
months  before  giving  interest  thereon. 


PROBLEMS. 


1.    Allowing  interest  at  3  per  cent  per  annum,  and  declaring  interest  semi- 
annually,  what  was  due  on  the  following  account,  January  1,  1S9G  ? 


Note. — Deposits  to  he  iu  bank  3  months,  before  they  bear  interest. 
Dr.  Soule  College  Savings  Bank  in  account  with  C.  McGiugln. 


(669) 


Cr. 


1895. 

— 

1895. 

Jan, 

1 

To  Balance,       -        -        - 

421 

65 

Feb. 

10 

By  Check,          -        .        - 

90 

30 

n 

22 

"  Cash, 

50 

00 

June 

16 

"         **               ... 

65 

15 

Mar. 

Ifi 

it      ti                 .        _        _ 

115 

50 

Nov. 

1 

tt        ti               ... 

38 

80 

May 

10 

a        li                    _          -          _ 

80 

25 

Aug. 

14 

li        ti                    ,          .          . 

100 

00 

Oct. 

1 

tl        it                    ... 

75 

00 

670 


soule's  philosophic  practical  mathematics. 


FIRST   operation. 

NoTK. — In  this  operation,  ve  luive  coiinte<l  actual  time  and  given  the  depositor  credit  of  lo. 
for  every  fraction  of  ic.  or  over  every  alternate  time. 


Date. 


Dt>po.sit.s. 


1895. 

Jan. 

ITo  Balance, 

It 

22  "   Dep. 

Mar. 

16 

a       11 

May 

10 

tt        a 

June 

30 

"  Inttresl, 
"  Jialance, 

July 

1 

To  Balance, 

Aug. 

14 

"  Dep. 

Oct. 

1 

it        a 

Dec. 

31 

"  Interest, 
"  Jialance, 

1896. 

Jan. 

1 

To  Balance, 

421 

50 

115 


8025 

675 


674 


15 

51870 

10000 

75;00 

91 


702 


87 


Checks. 


90 


30 


65 

SIS 


674 


38 


6G4 


15 


80 


07 


702 


87 


Int.  on. 


38100 
115U0 


,Jan.     1 
Mar.  'l6 


15 


00' 


518,00 
10000 


May 


July 
Aug. 


10 


June 


To,  in- 
clusive. 


Dec. 


Actual  !  _  ^ 
days.    I  Interest. 


181 
107 

52 


184 
140 


07 


Tot.  Int. 


75 


17 


664|07 
Note. — The  depositor  is  credited  1  cent  for  each  i  cent  interest  every  alternate  time. 

SECOND    OPERATION. 

Operation  reckoning  in  months  of  30  days  and  counting  odd  days  to  the  nearest  convenient 
fraction  of  a  niontli. 


Date. 


Deposits. 


1895. 

Jan. 

1 

To  Balance, 

" 

22 

"  Dep. 

Mar. 

16 

li      it 

May 

10 

«      II 

June 

30 

"  Interest, 
"  Balance, 

July 

liTo  Balance, 

Aug. 

14 

"  Dep. 

Oct. 

1 

Dec. 

31 

"  Interest, 
"  Balance, 

1898. 

Jan. 

1 

To  Balance, 

421 

50 

115 


8025 

0,72 


674112 


51867 

10000 

75,00 

8\96 


702  63 


Checks. 


9030 


65 


15 


S1S\67 
67412 


3880 
663,83 


702163 


Int.  on. 


38100, 
11500 

15,00 


Jan.     1 
Mar.  16 


From 


To 


May 


51800  July 
10000  Aug. 


10 


June  30; 
30 

30 


Dec.  31 

"    31 


Moa. 


3i 
If 


6 


Interest. 


572 
100 


06 


Tot.  Int 


72 


89S 


663183' 

Explanation. — Since  the  balance  $421.65  was  not  in  for  3  months  before  a  check  was  drawn 
•which  would  decrease  it,  it  is  clear  that  interest  will  not  be  calculated  on  that  amount.  The  clieck 
of  February  10th  is  to  be  deducted  tirst  from  the  last  deposit  preceding  February  10th,  (deposit  of 
January  22d,  |I50),  and  then  the  excess  of  cheek  over  this  deposit  from  the  bahmce  of  January  1st, 
1895 ;  $471.65  sum  of  deposits  of  January  1st  and  22(1,  less  $90  leaves  $381,  on  which  we  calculate 
Interest  for  6  months  at  3"o  lier  annum,  which  is  .f5.72.  Next  calculate  interest  on  $115  for  3*  mouths 
*t  3%  per  annum,  which  is  $1.00.  Deposit  of  May  10th  not  being  in  3  months,  no  interest  is 
credited  on  same  at  Juno  period.     Total  interest  to  be  credited  Jviue  30th,  1895,  is  $6,72. 

Note. — Balance  the  book  and  bring  balance  down* 


SAVINGS    BANKS    AND    SAVINGS    BANK    ACCOUNTS, 


671 


To  calculate  interest  to  he  credited  December  31st,  1S95. 

Calculate  interest  on  balance  $518  for  6  months  at  Z^a  per  ann>im  =  $7.77  and  add  to  this  sum 
the  interest  on  $15  (the  excess  of  deposit  of  May  10th  over  check  of  June  16th),  for  IJ  months  =  6 
cents.  The  same  result  Tvould  be  obtained  by  calculatinj;  interest  on  the  balance  $518.67  minus  the 
$15  excess  of  deposit  of  May  10th  over  check  of  June  10th,  for  6  months  =  $7.55,  and  adding  to  this 
sum  the  interest  on  the  $15  for  a  period  of  7j  months  =  28  cents.  Worked  by  either  method  the 
interest  amounts  to  $7.83.  Then  calculate  interest  on  $100  for  4i  months  at  Z%  per  annum,  =  $1.13. 
Total  amount  interest  December  31st,  1895,  =  $8.96.    Balance  of  account  January  1st,  1896,  =  ^663.55. 

To  find  interest  when  hook  is  halanced  hut  once  a  year,  say  on  January  1st, 

First  find  interest  on  $381  (difference  between  the  sum  of  the  deposits  of  January  1st  and 
22d,  and  the  check  of  February  10th)  for  6  months  =  $5.72.  Next  get  interest  on  $115  for  3i  months 
=  11.00.     Total  amount  of  interest  to  be  credited  June  30th,  1895,  =  $6.72. 

To  find  interest  due  on  December  31st,  1S95. 

First  calculate  interest  on  $381  for  6  months  =  $5.72;  next  calculate  interest  on  $115  for  6 
months  =  $1.72;  next  calculate  interest  on  $15  (dili'erence  between  deposit  of  May  10th  and  check 
of  June  16),  for  7f  months  =  29  cents;  next  calculate  interest  on  $6  for  6  months  =  9  cents;  next 
calculate  interest  on  $100  for  4i  months  =  $1.12.  Total  interest  to  be  credited  on  December  31st, 
1895,  =  $8.94.     Balance  of  account  January  1st,  1896,  =  ■f-GGS.Sl. 

Note  1. — On  general  principles,  interest  would  not  be  allowed  on  balance  of  deposit  0/ 
October  1st,  ($75  —  38.80).    According  to  circumstances,  interest  would  or  would  not  be  allowed. 

Note  2. — The  second  operation  above  is  the  method  generally  adopted  by  Savings  Banks. 
It  is  labor  saving  and  on  the  general  run  of  savings  accounts,  there  is  very  little  difference  iu  the 

final  result. 

Note  3. — Fractional  jiarts  of  a  dollar  are  not  considered  in  calculating  interest. 

Note  4.  In  subtracting  checlcs  from  deposits  there  is  no  settled  custom.  Some  banks  sub- 
tract checks  from  1st  deposit,  other  Itanks  subtract  checks  from  the  deposit  immediately  preceding 
date  of  check  and  again,  other  banks  count  back  3  or  4  months  (according  to  the  time  the  deposits 
must  be  on  hand  before  drawing  interest)  and  subtract  the  check  from  the  deposit  nearest  to  that 
date. 


2.     How  miicli  was  clue  on  tlie  following  account  January  1,  1S95,  interest  at 
3  per  cent  per  annum,  declared  semi-annually,  and  allowed  after  3  niontlis? 

Ans.  $707.37, 


r>r. 


SoulS  College  Savings  Banlz  in  account  icith  Katie  Childress. 


Cr. 


1894. 

1895. 

Dec. 

3 

To  Cash, 

200 

00 

Jan. 

18 

By  Check, 

50 

00 

1895. 

May 

4 

* '         *  * 

70 

00 

Jan. 

15 

a      it                 _        ■        . 

110 

80 

Feb. 

20 

it       ti                 .        ,        . 

50 

40 

Ajir. 

28 

ti       ti                 .        _        . 

150 

00 

June 

10 

"      "                 ... 

300 

00 

672 


SOULE  S    rillLOSOPHIC    rUACTICAL    MATHEMATICS. 
OPERATION. 


Date 

. 

1894. 

Dec. 

1895. 

3  To  Deposit, 

Jan. 

15  " 

Feb. 

201  "          " 

Apr. 
June 

28 
10 

It              n 

li           it 

(1 

30 

"  Interest, 
"  Balance, 

July 

1 

To  Bahiuce, 

Apr. 
June 

28 
10 

"  Deposit, 
11         11 

Dec. 

31 

"  Interest, 
"  Balance, 

1896. 

Jan. 

1 

1 

To  Balance, 

Deposits. 


$20000 

11080 

50:40 

15000 

30000 

4\S2 


J81l)05 


ijiGOli'Ua 


11\35 


$70737 


Checks. 


50  00 
7000 

69602 


lut.  on. 


$81002 


70757 


570737 


$20000 

6000 
5000 


From 


Dec. 

.Jan. 
Feb. 


31600  July 

80  00  Apr. 

30000  Juno 


To 


June  30 


Dec.  31 
"  31 
"    31 


M03. 


6i-a 
4i 


6 

8,V 
6i 


Interest. 


474 
161 
500 


Tot.  Int. 


482 


11 


35 


1707:37 


NOTK. — This  operation  for  calculating  interest  due  on  Dec.  31,  1895,  applies  to  2d  part  of 
following  explanation  tor  liniluig  interest  due  on  December  31,  1895. 
lu  this  account  no  interest  would  be  credited  January  1,  1895. 

To  find  interest  due  on  June  SO,  1S95. 

Find  interest  on  $200  for  6i^  months  at  3  per  cent  per  .annum  =  $3.45.  Then  calculate  inter- 
est on  excess  of  deposit  of  January  15,  over  check  of  January  18,  =  $60  for  5i  months  at  3  per  cent 
per  annum  =  83  cents.  Then  calculate  interest  on  $.50  deposit  of  Feb.  20,  for  4^  months  at  3  per 
cent  per  annum  =  54  cents.  Total  interest  to  be  credited  on  Juue  30,  is  $4.82,  Balance  of  account 
ou  July  1,  1895,  is  $696.02. 

To  find  interest  due  Dec.  SI,  1S95. 

1.  Ciilcul.ate  interest  on  balance  duo  for  6  months  =  $10.44  and  add  to  it  the  interest  on 
excess  of  deposit  of  April  28,  over  check  of  May  4,  $80  for  21^  mouths  =  41  cents,  also  add  interest 
ou  $300  for  I  months  =  50  cents  =  total  interest  $11.35. 

2,  The  same  result  would  bo  obtained  by  getting  Interest  on  $696  —  $380  =  $316  for  6  months 
=  $4.74  and  adding  to  thi.s  interest  the  interest  on  $80  for  8i-'i!-  months,  =  $1.61  jilus  the  interest 
on  $300  for  6|  months  =  5.00  =  total  interest  credited  Dec.  31,  1895,  =  $11.35.  Tlie  reason  that 
we  subtract  the  $380  and  not  $4.50  from  $696,  is  because  the  check  of  May  4,  for  $70  cancels  that 
much  of  the  $150  deposit  of  April  28.     Balance  due  Jau.  1,  1896,  $707.37. 

SECOND  OPERATION'. 

Where  the  hooh  is  balanced  only  on  Jan.  1,  of  each,  year. 

Since  the  deposit  of  Dec.  3,  was  not  in  for  three  months  previous  to  the  crediting  up  of  inter, 
est,  none  will  be  credited  ou  the  account  on  Dec.  31,  1894. 

To  find  interest  due  June  30,  1895,  proceed  as  follows:  First  calculate  interest  on  $200  for 
6rcs  months  =  $3.45.  Next  calculate  interest  on  $60  (the  difference  between  the  deposit  of  the  15th 
January  and  the  check  of  the  18th),  for  5^^  months  =  83  cents.  Next  calculate  interest  ou  $50  for 
i\  mouths  =  54  cents.     Total  interest  to  be  credited  Juue  30,  1895,  =  $4.82. 

To  find  interest  to  be  credited  Dec.  31,  1895,  first  calculate  interest  on  $200  for  6  months  ^ 
$3.00;  next  calculate  interest  ou  $60,  difference  between  check  of  Jan.  18,  and  deposit  of  Jan.  15, 
for  6  months  =  90  cents;  next  calculate  interest  on  $50  for  6  months  =  75  cents;  next  calculate 
interest  on  $80,  (difference  between  deposit  of  April  28  and  check  of  May  4),  for  8/5^  months  = 
$1.61;  next  calculate  interest  on  .$300  for  6|-  months  ^  $5.00;  next  calcuLate  interest  on  $4  for  6 
mouths  =  6  cents.     Total  amount  of  interest  to  be  credited  Dec.  31,  1895,  =  $11.32. 

The  difference,  3  cents  between  the  two  statements,  arises  from  the  fact  that  the  odd  cents 
■which  were  considered  when  the  account  is  balanced  ou  July  1,  1895,  amounts  to  $2  and  the  inter- 
est on  the  same  for  6  mouths  is  3  cents. 


SAVINGS    BANKS    AND    SAVINGS    BANK    ACCOUNTS. 


673 


THIRD    OPERATION. 

Operation  ivhen  the  cltecls  draini  are  cancelled  into  the  first  deposit  regardless  of  the  fact  that  other  deposits 
have  been  made  between  the  time  of  the  first  one  and  the  first  check  drawn. 

To  calcul.ato  the  interest  due  June  30,  1893,  first  calculate  the  interest  due  on  $150  (difference 
between  lirst  deposit  and  first  check)  from  Dec.  3,  to  May  4,  =  5  months  =  §1.88;  next  calculate 
interest  on  §80,  (difference  between  balance  of  $150  and  check  of  May  4),  to  July  1,  =  lj|  months 
=  37  cents  ;  next  calculate  interest  on  $110  for  5|-  months  =  $1.51 ;  next  calculate"  interest  on  $50  for 
4i  mouths  =  54  cents.     Total  amount  of  interest  to  bo  credited  on  June  30,  18'J5,  =  §4.30. 

To  find  interest  to  be  credited  Dec.  31,  1S95,  without  book  being  balanced  on  Juhj  1,  1S95. 

First  calculate  interest  on  §4  for  6  months  =  6  cents  ;  next  calculate  interest  on  $80  (differ, 
ence  between  deposit  of  Dec.  3,  cid  sum  of  checks  of  Jan.  18  and  May  4),  for  6  months  =  §1.20 
next  calculate  interest  on  $110  for  S  month.s  =  .$1.65;  next  calcnlate  interest  on  $50  for  6  months  = 
75  cents;  nest  calcnlate  interest  on  $150  for  S-fV  months  =  $3.02;  next  calculate  interest  on  §300 
for  6f  months  =  |5.00.  Total  amount  of  interest  to  be  credited  Dec.  31,  1895,  =  §11.68.  Balance 
<lue  Jan.  1,  1896,  =  §707.18. 

FOURTH    OPERATION. 

To  find  interest  due  Dec.  31,  1895,  if  book  had  been  balanced  on  July  1,  1895.  Balance  at 
that  date  by  above  third  method  would  have  been  $695.50. 

Interest  on  balance  for  6  months  =  §10.43.  Interest  on  $150  for  2,\  months  =  77c.  Interest 
on  $300  for  |  months  =  50  cents.  Interest  to  be  credited  Dec.  31,  1895,  =  $11.70.  Balance  due 
Jan.  1,  1896,  =  $707.20. 

The  difference  between  the  two  amounts  due  is  the  result  of  the  odd  cents  adding  $1.50  and 
drawing,  interest  for  6  months  =  2  cents. 

3.     Allowing  interest  at  3  per  cent  per  annum  and  declaring  interest  semi- 
annually what  was  due  on  the  following  account,  Jan.  1,  1S96. 
Note. — Deposits  to  be  in  bank  3  months  before  they  bear  interest. 

Dr.  Soule  College  Savings  BanJc  in  account  ivith  F.  E.  Gillespie.  Cr. 


1895. 

War. 

16 

Apr. 

10 

June 

3 

tiept. 

15 

To  Cash, 


500 

00 

200 

00 

130 

00 

240 

00 

OPERATION. 


Date. 


Deposits. 


1895. 
Mar.  116 
Apr.  10 
June    3 
"     30 


July 

Apr. 

Jnne 

Sep. 

Dee. 


1896. 
Jan. 


To  Deposit,    - 

it  a 

I    tt  II 

"  Interest, 


1  To  Amt.  brought  down, 
10  "  Deposit,    - 

3'  "         "  .        . 

15  "         "  .        . 

31  "  Interest,     - 


To  Amt.  brought  down, 


I 

§.50000 

20000 

15000 

4'39 


§834  39 


$854 


39 


24000 

16,5S 


§1110,97 


$111097 


Checks.       Int.  on, 


00 


Mar. 


From 


504  00  July 
200  OO  Apr. 
150,00  June 
240  OO:  Sep. 


To 


16  June  30 


Dec. 


Mos. 


3i 


6 
3i 


Interest. 


4  39 


756 
433 
259 
210 


Tot.  Int. 


4  39 


16 


58 


674 


SOULEs  I'liiLosoinirc  ikacticai^  mathematics. 


EXPLANATION. 

To  find  the  inhrcut  to  In-  vntlitcd  June  30,  1S95. 

First  calculate  interest  on  $500  for  3i  months  at  3  jier  cent  per  annnm  =  $4.39.  The  deposits 
of  April  10  and  June  3,  do  not  draw  interest  since  they  were  not  on  dejiosit  for  3  months.  Interest 
to  be  credited  on  Juno  30,  =  |4.3y. 

To  find  inliri'st  to  br  credited  Dec.  31,  1S95. 

First  calculate  interest  on  total  anioiint  deposited  and  interest  =  $854.38  for  6  months  at  3 
per  cent  ])er  annum  =  $12.81.  To  this  amount  add  interest  on  $200  for  2j  months  =  $1.33  and  also 
add  interest  on  $150  for  i'^  months  =  34  cents.  Then  add  interest  on  $240  for  3i  mouths  ^  2.10. 
Total  interest  to  be  credited  on  Dec.  31,  =  16.58. 

The  same  result  can  be  obtained  by  getting  the  interest  on  $854  —  $350  =  $504  for  6  months  = 
$7.56.  Then  add  to  this  amount  interest  on  $200  for  8|  months  =  $4.33,  and  interest  on  $150  for 
6iTr  months  =  $2.59.  Then  add  interest  on  $240  for  34  months  =  $2.10.  Total  interest  to  be  credited 
Deo.  31,  =  $16.58.     Amount  due  Jan.  1,  1896,  =  $1110.97. 


! 


loan  and  Investment  Companies. 


1215.     To  find  the  interest  antl  tbe  rate  per  cent  of  interest  realized  by  a  Loan 
and  Investment  Co.  on  certain  conditions. 

1.  A  Loan  and  Investment  Co.  loans  $3000  for  5  years,  at  8  per  cent  per 
annnm.  The  interest  ($11.'0U)  is  added  in  advance  and  tlie  whole  amount  $4200  is 
made  payable  in  60  notes  of  $70  each  payable  monthly,  Avithout  interest.  Allowing 
that  the  company  may  re-invest  the  monthly  collections  of  the  notes  at  8  per  cent 
simple  interest,  what  wonkl  bo  the  total  interest  received  by  the  Company,  and  what 
woiild  be  the  rate  per  cent  per  annum  'i 

Ans.  $2026  total  interest;  13 ff  rate  per  cent  per  annum. 

OPERATION. 

Interest  on  S3000  for  5  years  at  8  per  cent  is, $1200.00 

"  "  *7j  each  monthly  payment  at  8  per  cent  for  1770  months  which  is  the  sum 

of  ti- 1  arithmetical  series  from  1  to  59  months,  for  the  60  notes,  the  last  note's 
interest  not  bein<r  invested  is,  ...........  826.00 


Total  interest  received  by  the  Company  for  5  years,  .......        |2026.00 

$2026  -I-  5  =  $405.20  =  average  interest  for  1  year. 
($405.20  X  100)  -H  $3000.00  =  1338  per  cent  rate  per  annum. 
Note. — The  sum  of  the  arithmetical  series  is  found  by  adding  the  first  and  last  terms  of  pay- 
ment and  multiplying  the  sum  by  one-half  the  number  by  terms  or  notes.     Thus,  0  +  59  =  59  X  30 
^  1770  or,  by  multiplying  half  the  sum  of  the  extremes  by  the  number  of  terms,  thus:  (0  +  59) 
-T-  2  =  29i  X  CO  =  1770. 

2.  Suppose  in  problem  1,  above,  that  the  monthly  note  and  interest  collec- 
tions were  re-invested  at  simple  interest  as  therein  stated,  what  would  have  been 
the  respective  interest  gain  for  each  year? 

OPERATION. 
Interest  on  $3000  for  1  year  at  8%,  $240.00 

"       "  $70at8%.  66mos.  Csumof  seriesO  to  11),  ($30.80)  30.80  $270.80  1st  year. 

"       "  $3000  for  1  year  at  8  V  240.00 

"       "  $70  at  S^,  210  mos.  (.sum  of  series  12  to  23)  (98.00)  98.00  338.00  2d  year. 

"       "  $3000  for  1  year  at  8%,  240.00 

"       "  $70  at  8%,  354  mos.  (sum  of  series  24  to  35)  (165.20)  165.20  405.20  3d  year. 

"       "  $3000  for  1  year  at  8,"o,  240.00 

"       "  $70  at  8"^,  498  mos.  (sum  of  series  36  to  47)  (232.40)  232.40  472.40  4th  year. 

"       "  $3000  for  1  year  at  8%,  240.00 

"       "  $70  at  8%,  642  mos.  (sum  of  series  48  to  59)  (299.60)  299.60  539.60  5th  year. 

Total  interest  for  5  years,  as  in  the  first  problem,  (.$826.00)  $2026.00 

3.  Suppose  in  problem  1,  above,  that  the  monthly  collection  of  interest  on 
the  notes  and  the  interest  on  such  interest  were  used,  at  the  close  of  each  year,  as 
a  part  of  the  dividend  fund,  what  Avould  be  the  amount  of  interest  earned  yearly? 


OPERATION  FOR  THE   FIRST  YEAR. 

Interest  on  $3000,  1  year  at  8  per  cent,       $240.00 
Interest  on  $'iO  for  66  months  at  8  per 
cent,  -which  is  the  sum  of  the  arith- 
metical series  from  11  to  0,  is  30.80 

Total  interest  revenue,  first  year,  $270.80 


Note. — The  first  year  there  were  but  11  invest- 
ments made  during  the  12  months,  as  the  first 
note  was  not  collected  till  the  close  of  the  first 
month. 


operation  for  second  year. 

Interest  on  $3000,  1  year  at  8  per  cent,        $240.00 
Interest  on  $70  for  78  months  at  8  per 
cent,  which  is  the  sum  of  the  arith- 
metical series  from  12  to  1,  is  36.40 


Total  interest  revenue,  second  year,  $276.40 

Note.— The  $70  collected  on  the  first  of  the 
first  month  of  the  second  year  was  invested,  thus 
making  12  investments.  The  operation  of  the 
three  subsequent  years  will  be  the  same  as  for 
the  second  year. 

Remarks.— Seepages  590 to 596,  for  the  Gain  and  Gain  Per  Cent  per  Annum,  for  certain  kinds 
of  Loans  and  Investments. 

(675) 


Itoclcs  and  Bonds. 


^"xj^ 


Note. — See  Soul^'s  New  Pciciir-ci  and   rractii'ii   of  Accounts  for  200  pages  of  Corporation 
Accounting. 

121G.     Stocks  are  the  transferable  sliares  of  corporations-stock  companies- 
specifyiiig-  tlie  aniouiit  of  capital  owned  tlioreiu  by  the  stock  or  sliarelioldeis. 

1217.  The  Stock  or  the  Capital  Stock  of  a  Stock  Company  is  the  capital 
contributed  by  stockholders  and  invested  in  the  business. 

1218.  A  Corporation  or  Stock  Company  is  an  Association  of  men  created 

and  authorized  by  law  to  act  as  a  single  i)erson,  under  a  common  name.  The  mem- 
bers of  a  corporation  succeed  each  other,  so  that  the  Association  continues  always 
the  same,  notwithstanding  the  change  of  the  individuals  which  compose  it.  Cor- 
porations are  in  law  artificial  jiersons. 

The  capital  of  corporations  is  divided  into  shares  of  equal  value,  which  are 
transferable  and  may  be  bought  and  sold  like  any  other  property. 

The  management  of  corporations  is  generally  nested  in  a  Board  of  Directors 
who  are  named  in  the  Charter,  or  elected  by  the  stockholders,  and  who  select  the 
officers  to  conduct  the  business  of  the  Company  or  Corporation. 

1219.  Certificates  of  Stock  are  instruments  issued  by  a  corporation,  certi- 
fying under  tlie  signatures  of  the  i>roper  oflicers  that  the  holder  is  the  owner  of  a 
specified  number  of  shares  of  its  capital  stock.  Thus,  the  following  is  a  general 
form  of  a 

certificate  of  stock: 


^^^T£  OF   LOU/SM/,^ 


\Shares. 


SOULE  COLLEGE  BHNK. 

^\^n  \^  lo  4"*'fe'  That 

the  Proprietor  of Shares  of  One  Hundred  Dollars 

each,  ill  the  Capital  Stock  of  SOULE  COLLEGE  BANK,  which  is 
transferable  on  the  hooks  of  the  Bank  hi/  the  said  Stockholder, 

or Attorney  or  Agent,   on   the  surrender   of   this 

Certificate. 


Witness,  tli(>  Seal  of  the  Bank  attested  by  the  Si^atiires 
of  the  Tresideut  and  Cashier  tliereof. 

New  Orleans, 189 


President. 


(676) 


*  STOCKS    AND    BONDS.  677 

1220.  The  Par  or  Nominal  Yalue  of  Stock  is  tlie  sum  named  in  tbe  instru- 
meut  or  certiticiite  of  stock. 

Wbeu  stock  sells  for  more  than  tlie  original  or  par  value,  it  is  said  to  be  above 
par  or  at  a  premium.     When  for  less,  it  is  below  par  or  at  a  discount. 

1221.  The  Market  Value  of  stocks  or  bonds  is  what  they  will  sell  for  in  open 
market.  In  the  case  of  stocks,  this  value  depends  upon  the  dividends  they  pay 
compared  with  the  current  rate  of  interest  in  the  money  market;  and  in  the  case 
of  bonds,  the  market  value  depends  upon  the  degree  of  confidence  entertained  by 
capitalists  of  their  being  paid  at  maturitj',  and  the  rate  of  interest  they  bear 
compared  with  the  current  rate  of  interest  in  the  money  market. 

The  success  of  incorporated  companies  has  much  to  do  with  the  market 
value,  and  the  sobriety,  industry,  fidelity,  and  capacity  of  the  officers  and  directors 
of  companies  have  almost  all  to  do  with  the  success  of  the  company. 

1222.  The  Intrinsic  or  Liquidation  Talue  of  Stock  is  the  sum  that  each 
share  would  be  worth  in  case  the  corporation  went  into  liquidation,  or  it  is  the 
quotient  of  the  net  resources  after  all  liabilities  are  x>aid,  by  the  number  of  shares. 

1223.  Preference  or  Preferred  Stock  is  stock  issued  by  some  corporations 
which  is  entitled  to  an  extra  amount  of  interest  or  to  a  specified  per  cent  dividend 
out  of  the  profits,  before  the  common  stock  dividend  is  declared. 

The  circumstances  which  give  rise  to  preferred  stock  are  varied  and  are 
generally  based  upon  equity.  A  corporation  may  be  in  great  distress  for  money,  and 
to  relieve  the  company,  certain  stockholders  or  other  parties  may  advance  the 
required  money  for  which  preferred  stock  is  issued  to  them.  Again,  in  the  re-organ- 
ization of  railroads  or  banks,  preferred  stocks  may  be  issued  to  save  the  sacrifice 
of  the  company's  property,  or  to  obtain  the  means  to  carry  on  the  company's  busi- 
ness. 

1224.  Guaranteed  Stock  consists  of  shares  on  which  a  certain  amount  of 
money  or  interest  is  due,  payment  of  which  is  guaranteed,  thus  rendering  the  shares 
especially  valuable. 

1225.  Forfeited  Stock  is  sixbscribed  stock  on  which  installments  may  or  may 
not  have  been  paid,  and  on  which  the  subscriber  fails  to  pay  installments  when 
called  for.  Such  stock  is  forfeited  to  the  company  according  to  the  terms  of  the 
Charter. 

1226.  Debenture  Stock  is  a  writing  acknowledging  a  debt.    2.   A  deed  of 

mortgage  given  by  railway  companies  for  borrowed   money.     3.    The  bonds   and 

securities  issued  for  bt)rrowed  money,  by  towns  and  corporations. 

NoTE.^This  term  is  in  general  use  ■witli  English  Fiuauciers,  but  it  is  rarely  usetl  l)y  Aiucriean 
Bankers  or  Security  dealers. 

1227.  An  Installment  is  a  certain  per  centum  of  the  capital  stock  which 
subscribers  are  required  to  pay  at  a  specified  time. 

1228.  An  Assessment  is  a  per  centum  tax  levied  and  collected  from  the 
stockholders  to  make  up  losses  or  deficiencies. 

1229.  Gross  Earnings  are  the  total  receipts  of  the  company. 


678  soule's  riiiLosopiiu:  practical  mathematics.  * 

1230.  Net  Earuings  or  Net  Profits  are  what  remains  after  all  expenses  ai'e 
deducted. 

1231.  A  Reserve  Fund,  or  Contingent  Fund,  or  Eedemption  Fund  is  a 

certain  portion  of  the  yearly  i)rolits  of  the  tonipany,  reserved  or  set  aside  for  the 
purpose  of  creating  a  fund  to  meet  unforeseen  emergencies  or  accidents. 

DIVIDENDS. 

1232.  A  Dividend  is  a  pro  rata  division  of  the  profits  or  a  portion  thereof, 
among  the  stockholders  of  a  corporation. 

1233.  A  Cash  Dividend  is  a  certain  amount  of  the  profit  of  a  company  dis- 
tributed in  cash  among  the  stockholders  annually,  semi-annually,  or  quarterly. 

Note. — Divi(len(Ts,  -when  declared,  are  generally  a  certain  per  cent  on  tbe  par  value  of  tlie 
shares,  but  sometimes  they  are  a  certain  per  ceut  ou  the  paid-up  stock. 

1234.  An  Installment  Dividend  is  a  certain  amount  of  the  profits  of  a  com- 
pany apportioned  to  the  stockholders  and  applied  to  the  payment  of  their  unpaid 
.subscriptions.  An  installment  dividend  is  in  effect  the  i)aying  of  a  cash  dividend 
aud  the  collection  of  an  installment  of  equal  amount. 

1235.  A  Stock  Dividend  is  a  certain  amount  of  the  profits  of  a  company 
apportioned  to  the  stockholders  and  retained  by  the  company,  for  which  new  paid-up 
•stock  is  issued  to  the  stockholders.  Or,  in  the  absence  of  a  profit,  it  may  be  addi- 
tional stock,  to  be  jjaid  out  of  the  capital  or  the  future  profits  of  the  company,  in 
which  latter  case,  it  is  equivalent  to  the  operation  of  "  watering"  stock. 

1236.  A  Forced  Cash  Dividend,  or  a  Capital  Stock  Dividend,  or  a  Fictitious 
Dividend  is,  in  the  absence  of  a  profit  or  reserve  fund,  a  certain  per  cent  of  the 
capital  of  the  company  distributed  among  the  stockholders. 

The  method  usually  adopted  is  to  place  an  inflated  value  on  the  real  estate, 
plant,  bonds,  personal  accounts,  or  other  resources  of  the  corporation,  or  by  charg- 
ing exjiense  and  repairs  to  the  property  accounts,  by  which  means  false  i)rofits  are 
shown,  and  then  Forced  or  Fictitious  dividends  arc  declared.  The  object  of  a  forced 
or  fictitious  dividend  is  deception,  and  the  result  dishonest.  For  by  it,  the  public 
are  led  to  believe  that  the  company  is  prosperous  when  it  is  not,  and  hence  pay 
more  for  the  stock  than  it  is  worth. 

Note. — In  this  connection  it  must  not  be  forftotten  that  sometimes,  from  unusual  causes, 
companies  do  not  make  a  yearly  or  semi-annual  profit,  and  then  they  declare  a  dividend  from  a 
portion  of  the  previous  gains  of  the  company,  which  have  been  set  aside  under  the  head  of  resehve 
FUND  for  that  iHirpose.  A  dividend  of  this  cliaracter  is  not  to  bo  classed  as  a  forced  or  fictitious 
dividend.  In  the  early  history  of  railroads  and  some  other  companies,  forced  or  lictitious  dividends 
■were  of  frequent  occurreuce. 

1237.  Watered  Stock  is  that  which  has  been  increased  above  the  authorized 
capital  by  the  issuance  and  distribution  among  the  stockholders  of  new  stock,  for 
■which  no  payment  is  or  will  be  made. 

When  these  "watered  shares  are  made  transferable,  the  act  of  watering  is 
violative  of  every  principle  of  ethics,  and  should  secure  for  the  parties  thereto  the 
judgment  of  a  Criminal  Court. 


*  BONDS.  679 

When  the  tvateked  shares  are  not  made  transferable,  then  the  transaction 
is  one  of  deception,  the  motive  for  -which  will  appear  jilain  when  we  consider  that 
the  charters  of  some  companies  rcqnire  that  their  dividends  shall  not  exceed  a  cer- 
tain per  cent  on  the  Capital  Stock,  and  that  whenever  the  net  profits  of  the  company- 
are  sufficient  to  produce  a  dividend  in  excess  of  the  specified  per  cent,  that  the 
comi)any  shall  reduce  its  rates  of  tariff.  Hence,  to  avoid  this  reduction  of  its  rates, 
new  stock  is  issued  and  distributed  pro  rata  to  the  stockholders,  by  which  means 
the  cajiital  stock  is  increased  and  the  rate  per  cent  dividend  decreased  and  kept 
within  the  prescribed  limits,  and  the  company  is  thus  enabled  to  continue  its  high 
charges  and  heavy  profits. 

1238.  Stock  is  often  watered  in  the  re-organization  of  a  railroad  or  the  con- 
solidation of  different  railroads,  or  the  consolidation  of  other  coriJorations,  when 
the  property  or  plants  are  estimated  at  an  increased  valuation.  If  the  proi:)erty  has 
increased  in  value  from  any  cause,  then  such  watering  might  be  honest,  though  it  is 
not  the  legal  way  to  increase  the  capital  stock  of  a  corporation. 

Note. — Tbe  waterini;  of  stock  Imsbcen  indulged  in  liy  some  of  the  leading  railway  companies 
of  tbe  country  and  by  other  corporations,  to  an  alarming  extent.  Watered  stock  amount  does  not 
represent  any  real  investment,  but  only  increased  earnings,  and  hence,  with  the  reduction  of  freight 
and  passenger  transportation  charges,  which  the  comjietitiou  of  new  roads  now  building  and  to  be 
l)uilt  will  certainly  effect  all  watered  stock,  will  eventually  prove  a  loss  to  tbe  holders.  For,  when 
the  incomes  of  the  companies  are  reduced  to  their  normal  level  by  tbe  means  indicated,  the  entire 
issue  of  stock  will  in  like  manner  be  reduced  to  the  original  value  of  the  first  issue. 

AVe  can,  in  a  measure,  forgive,  if  we  cannot  justify,  the  vendor  of  watered  whiskey  or  rum, 
on  the  moral  plea  of  bis  having  jierformed  a  jdiilauthropic  act  by  selling  a  less  poisonous  article 
when  watered  than  when  pure,  but  no  justiticatiou  whatever  can  be  offered  in  extenuation  of  the 
sin  and  crime  committed  by  watering  stock. 

If  the  capital  of  the  company  is  not, commensurate  with  the  workings  and  demands  of  the 
company,  the  capital  stock  can  be  increased  in  a  legitimate  and  honest  manner 

1239.  Clandestine  Stock  is  first,  a  new  edition  of  stock  issued  without 
public  notice  and  iilaced  on  the  market  to  raise  money  to  cover  losses  or  exiienses 
of  which  the  public  have  no  knowledge. 

Note. — Clandestine  Stock  has  also  been  issued  under  the  following  circumstances : 

The  market  value  of  stock  is  very  high,  and  the  company  issues  and  distributes  pro  rnia,  at 
par,  to  the  stockholders  of  the  company,  a  certain  amount  of  new  stock.  This  new  stock  is  placed 
on  the  market,  and  the  full  market  value  of  the  original  stock  is  realized  before  the  fact  is  known 
that  it  is  Clandestine. 

BONDS. 

1240.  Bonds  are  written  or  printed  obligations  of  Individuals,  Corporations, 
Cities,  States  or  nations,  to  pay  a  specified  sum  of  money  at  a  certain  time.  Bonds 
generally  bear  a  fixed  rate  of  interest  payable,  annually,  semi-annually  or  quarterij', 
and  when  issued  by  individuals  or  corporations  for  loans  are  secured  by  mortgages 
upon  their  property. 

The  security  of  the  Bonds  of  States  or  nations,  usually  have  no  other  security 
than  the  confidence  of  the  people.  Bonds  are  issued  in  denominations  varying 
from  ten  dollars  to  fifty  thousand  dollars. 

1241.  Coupon  Bonds  are  those  having  interest  tickets  or  coupons   attached, 


68o  soule's  ruiLOsopHic  practical  mathematics.  * 

which  specify  tlie  interest  due  each  year,  half  year,  or  quarter,  and  uliich  are  cut 
off  as  they  are  i)aid,  and  heUl  as  receipts. 

124r2o  Eegistered  Bouds  are  bonds  that  are  payable  to  the  order  of  the  holder 
or  owner,  and  cousequently  cauDot  be  negotiated  without  assignment  dulj'  acknow- 
ledged. 

Bonds  are  known  as  First  Mortgage,  Second  Mortgage,  Debenture,  etc., 
Consols,  Income,  Sinking  Fund,  Adjustment,  etc.,  according  to  their  i)ri(irity  of 
lieu,  the  class  of  i)roperty  upon  which  they  are  secured  or  other  characteristics. 

Bonds  are  also  named  from  the  rate  of  interest  they  bear,  or  from  the  dates 
at  which  they  are  payable,  or  from  both.  Thus :  Louisiana  4's,  United  States  4's  of 
1907,  New  Orleans  City  I'remium  Bonds,  United  States  3's. 

■     12-1:3.     A  Mortgage  is  a  conditional  conveyance  or  sale  of  real   or  personal 
property  as  security  for  the  jiayment  of  a  debt  or  for  the  performance  of  a  duty. 

The  written  instrument  is  termed  the  mortgage  and  is  proi^erty  or  a  resource 
to  the  holder. 

1244.  Consols  (from  consolidated),  are  bonds  issued  in  redemption  of  two  or 
more  classes  of  bonds  which  are  consolidated  into  the  new  bond,  and  generally  at  a 
reduced  rate  of  interest. 

ENGLISH  CONSOLS. 

1245.  English  Consols  are  3  per  cent  English  bonds.  The  term  is  abbrevia- 
ted from  the  words  consolidated  annuities. 

Note  1. — Tlicso  bonds  were  issued  1>y  TJrtue  of  .tii  Act  of  Parliament  to  redeem  several 
separate  issues  of  government  bouds  bearing  various  rates  of  interest,  and  with  tlie  condition  tliat 
while  the  interest  was  regularly  jniid  tbe  government  could  not  be  required  to  ])ay  the  jirinci]ial. 
They  are  licnco  iiractically  perpetual  annuities.  In  the  act  creating  them,  they  are  termed  consoli- 
dated annul  lies 

The  quotations  of  these  3  per  cent  bonds  or  "  Consols"  indicate  ordinarily  the  state  of  the 
money  market,  as  they  form  a  large  part  of  the  Knglish  jiublie  debt. 

Note  2. — In  England  tlie  term  stock  is  limited  to  the  Government  Stocks  and  Annuities; 
and  the  term  siiauk  is  used  for  the  Capital  Stock  of  Corjioratious. 

1246.  French  Kentes  are  the  Government  Stocks,  which  bear  various  rates 
of  interest. 

Note. — In  France  the  term  rkntes  bas  the  same  limitation  as  stocks  in  England. 

UNITED  STATES  BONDS. 

The  following  are  the  interest  bearing  bonds  of  the  United  States,  outstanding 
April  1,  189.5,  as  per  the  statement  of  the  United  States  Secretary  of  the  Treasury: 

1247.  1.  Funded  Loan  of  1891.  These  are  bonds  which  were  authorized 
by  an  Act  of  Congress  of  July  14,  1870,  and  Jan.  20,  1871,  to  refund  the  national 
debt.  They  bear  interest  at  the  rate  of  4i  per  cent,  payable  quarterly  in  gold. 
They  are  redeemable  at  the  option  of  the  United  States  1.5  years  after  Act  of 
Authorization.    Amount  outstanding  April  1, 1895,  was  $25,364,500. 


*  STOCK    EXCHANGE.  68 1 

1248.  2.  Fuuded  Loan  of  1907.  These  bonds  were  authorized  by  an  Act 
of  Congress  of  July  11,  1870,  and  Jan.  20, 1S71.  Tliey  bear  4  per  cent  gold  interest, 
payable  quarterly,  and  are  redeemable  July  1,  1907.  Aniuuut  outstanding  April 
1,  1895,  was  $555,624,850. 

1249.  3.  Refuuding  Certificates.  These  certificates  were  authorized  by  an 
Act  of  Congress  of  February  2(j,  1879.  They  bear  4  per  cent  interest,  payable 
quarterly.     $54,710  are  outstanding  April  1,  1895. 

1250.  4.  Loail  of  1904.  These  bonds  were  authorized  by  Act  of  Congress 
of  Jan.  14,  1875,  they  bear  5  per  cent  interest,  payable  quarterly,  and  are  redeemable 
February  1,  1904.  Amount  outstanding  April  1,  1895,  $100,000,000.  They  were 
jdaced  on  the  market  in  1894. 

1251.  5.  Loau  of  1925.  These  bonds  were  also  authorized  by  the  Act  of 
Congress  of  Jan.  14,  1875.  They  bear  4  per  cent  interest,  payable  quai'terly,  and  are 
redeemable  Feb.  1,  1925.     Amount  outstanding  April  1,  1895,  was  $28,807,900. 

1252.  G.  U.  S.  Pacific  R.  R.  Currency  Sixes.  These  are  30  year  G  per 
cent  currency  registered  bonds,  -which  the  government  issued  to  the  various  com- 
panies chartered  by  Congress  to  build  railways  to  aud  from  the  Pacitic  coast.  They 
were  issued  on  the  completion  of  20  miles  of  road  at  the  rate  of  $10000,  $32000,  and 
$48000  per  mile,  according  to  the  expense  and  difficulty  of  construction.  The 
interest  is  payable  Jan.  1,  and  July  1.  Amount  outstanding  April  1,  1895,  was 
$64,623,512. 

In  addition  to  the  foregoing  interest-bearing  debt  the  United  States  had, 
April  1,  1895,  the  following  indebtedness: 

1.  Bonded  debt  on  which  interest  had  ceased  since  maturity,  $1,770,250.20. 

2.  United  States  aSTotes,  National  Bank  Xotes  and  Fractional  Currency, 
$381,025,090.92. 

3.  Gold  and  Silver  Certificates,  aud  Certificates  of  Deposit,  all  of  which  is 
offset  by  au  equal  amount  in  the  Treasury,  $567,944,442.00. 

STOCK  EXCHANGE. 

1253.  Stock  Exchanges  are  Associations  of  Brokers,  generally  incorporated 
bodies,  whose  business  it  is  to  buy  and  sell  stocks,  bonds,  and  other  securities.  The 
regular  commission  for  buying  and  selling  seciirities  at  the  New  York  Exchange  is 
f  per  cent  on  par  value,  except  for  mining  stocks,  on  which  contract  rates  are 
usually  charged. 

The  New  Orleans  Stock  Exchange  is  au  Association  organized  for  the 
purpose  of  buying  and  selling  stocks,  bonds,  and  other  securities.  The  regulations 
of  the  association  are  very  stringent  and  specific,  in  order  to  prevent  unworthy 
persons  from  being  admitted,  and  to  insure  fidelity  and  capacity  in  the  performance 
of  contracts.  To  become  a  full  member  of  the  New  Orleans  Stock  Exchange,  the 
applicant  nrast  be  of  good  character,  must  be  regularly  engaged  as  exchange,  stock, 
coin,  or  bill  broker,  and  own  at  least  one  share  of  the  stock  of  the  exchange.     The 


682  soule's  riiiLosopiiic  i'Ractical  mathematics.  * 

number  of  shares  of  the  New  Orleans  Stock  Exchange  is  70,  and  the  par  of  each  is 
SilO.     Tlie  present  (1S95,)  market  vahie  per  shaie  is  about  $S00. 

The  New  Orleans  Stock  Exchange  has  an  insueance  fund  which  was 
ci'eated  by  an  assessment  of  $15  per  member.  Upon  the  death  of  any  member  of  the 
Exchange,  each  surviving  nienibcr  is  assessed  $15,  which  is  paid  into  the  insurance 
FUND,  and  within  CO  days  after  tlie  jjroof  of  the  death  of  any  member  the  amount 
of  money  tlius  realized  is  paid  to  the  legal  heirs  of  the  deceased  member. 

The  commission  for  buying  or  selling  in  the  New  Orleans  Stock  Exchange  is 
as  follows:  ^  per  cent  on  the  par  of  City  and  State  Bonds.  On  all  otlier  bonds 
selling  at  and  above  par,  i  per  cent  on  par  value. 

On  stock  selling  at  and  over  $00  per  share,  50  cents  iier  share  is  charged.  On 
stocks  selling  at  $25  and  under  $C0  per  share,  25  cents  per  share  is  charged. 

Ou  stocks  selling  below  $25  per  share,  a  contract  charge  is  made.  On  Scripts, 
Warrants,  etc.,  J  per  cent  is  charged  on  par  valne. 

lu  the  sale  of  stocks,  the  following  regulations  are  also  observed  by  the  New 
Orleans  Stock  Exchange: 

1.  Ou  all  contracts  of  time  exceeding  3  days,  one  day's  written  notice  must 
be  given  before  the  stocks  or  securities  can  be  delivered  or  demanded. 

2.  All  sales  not  "cash"  are  considered  time.  No  jmrchase  ou  sale  at  the 
option  of  the  buyer  or  seller  for  3  days,  shall  bear  interest. 

3.  All  purchases  and  sales  beyond  3  days  shall  be  "flat"  unless  otherwise 
expi'essed. 

4.  On  all  time  bargains,  the  rate  of  interest  shall  be  8  per  cent,  to  be  cal- 
culated by  days  according  to  bank  usage. 

1254.     DEFINITIONS    OF    WORDS    AND    PHRASES    USED   EY    FINAN- 
CIERS,  BROKERS   AND   BANKERS. 

A  Truat  is  a  rorpor.itinn  formpiT  l)y  large  dealers  in  a  certain  article,  Tvliose  oliject  is  to  control 
tlie  market,  thercliy  forcing  prices  above  the  level  established  by  tree  competition,  or  rednciug 
them  below  the  cost  to  produce  the  article  in  order  to  sujipress  opposition.  Trusts  sometimes 
fairly  and  honorably  reduce  prices,  by  reason  of  an  extensive  trade,  of  buying  material  and 
manufaclaring  articles  in  large  quantities  at  a  few  centers  of  trade,  etc. 

A  Siindimie  is  a  combination  of  Broters,  Bankers  or  C'ai)italists,  who  undertake  to  jilaco  large 
loans  and  transact  other  financial  business. 

Collaterals,  are  .Securities,  Bills  Receivable,  Stocks.  Bonds,  etc.,  which  are  pledged  and  placed  in 
the  possession  of  an  obligee,  in  addition  to  the  princiiial  security,  to  secure  the  payment  of 
money  or  the  performance  of  an  obligation  entered  into  by  the  obligor. 

Uypnlhecatinrf  Sfocls  and  JJonds  is  depositing  them  as  collateral  security  for  money  borrowed  or 
debts  contracted. 

A  Marq'ui  is  a  deposit  nmde  with  a.  broker  by  a  person  who  wishes  to  speculate  in  buying  or  selling 
stocks  or  bonds,  to  enable  the  broker  to  carry  the  stocks  or  bonds  and  protect  himself  against 
any  loss  that  may  occur  by  :iu  unfavorable  change  in  the  market  price. 

The  margin  varies  from  5  to  20  per  cent  of  the  value  of  the  stocks  or  bonds. 

Note. — In  case  the  market  moves  so  as  to  lessen  the  per  cent  margin  and  threaten  less  to  the 
broker  who  is  "  carrying"  the  stock  or  bonds,  the  margin  must  be  made  good,  on  the  notice  of 
the  broker,  by  the  deposit  of  an  additional  sum.  If  tliis  is  not  done  the  broker  is  justified  in 
selling  the  securities  to  protect  himself  against  the  loss  of  the  money  he  has  advanced  to  or  for 
his  principal. 

Brokers  charge  interest  on  all  sums  advanced  for  carrying  stocks  or  bonds,  allow  interest  on 
margins,  and  charge  the  regular  per  cent  brokerage  for  buying  and  for  selling. 


*  DEFINITIONS    OF    WORDS    AND    PHRASES.  683 

Difference  is  the  variation  of  the  price  for  -n-hicli  stock  is  contracted  and  the  market  rate  the  day 
of  delivery.  lu  cases  like  this  the  broker  against  w-honi  the  variation  exists  freqneutly  Jiaya 
the  diliereuce  in  money  instead  of  snpplyiug  or  receiving  the  stock. 

Banker,  the  proprietor  of  a  hank;  one  who  deals  in  money,  and  negotiates  hills  of  exchange. 

Broker,  one  ^vho  receives  and  eseciites  orders  for  other  parties.  . 

Factor  or  Commission  Mercliant,  one  who  sells  goods  on  commission. 

Jobber  or  Merchant,  cue  who  deals  in  Stocks  or  Merchandise  on  his  own  account. 

Stag  or  Outsider,  one  who  does  business  outside  of  the  exchange. 

Bear  is  au  operator  who  is  "  Short  of  Stocks.  "  or  who  wishes  to  purchase  stocks;    and  hence  one 

who  labors  to  decrease  or  depress  the  price  that  he  may  buy  or  "  till "  at  lower  rates. 
The  term  is  also  applied  to  all  parties  who  wish  to  purchase  goods.    The  name  is  derived 

from  the  nature  of  tlio  bear  to  jjull  things  down  with  his  paws. 

Bull  is  an  operator  who  is  "  long  of  stocks,  "  or  who  wishes  to  sell  stocks ;   hence  one  who  labors 

to  advance  or  inflate  the  prices. 

The  term  is  also  applied  to  all  parties  who  liave  goods  to  sell.     The  name  is  derived  from  the 

nature  or  character  of  the  bull  to  throw  things  up  with  his  horns. 
Future  Sales  or  Seller's  or  Buyer's  Option,  the  privilege   given   to   either  party   of  delivering   the 

Stock  at  any  time  within  the  blank  days  specitied. 
Cash  Sales,  to  be  delivered  and  paid  for  the  same  day  before  2  P.  M.,  if  sold  before  1  i'.  M.,  and 

before  3  P.  M.,  if  made  after  1  P.  M. 
A  Hcgular  Sale,  payable  the  next  day  before  2  r.  M. 
Selling  Short,  selling  what  you  have  not  in  the  hope  of  buying  for  less  before  called  upon  for  a 

delivery. 
Zong  of  Stocks,  buying  and  holding  the  same  in  the  hope  of  a  rise. 
Short  of  Stocks  means  that  you  have  sohl  that  which  you  do  not  possess  in  the  hope  of  buying  at 

lower  prices  and  thus  realizing  a  profit. 
A    Wash  Sale,  a  sham  transaction  between  two  brokers  for  the  purpose  of  causiug  an  advance  or 

decline  in  prices. 

Wash  sales  are  not  allowed  in  well  managed  exchanges. 
To  make  a  Turn,  sell  for  Cash  and  then  buy  it  back  on  regular  sale. 

Milking  or  Lowering,  lowering  prices  to  induce  holders  to  sell,  and  then  the  clique  step  in  and  buy. 
Salting  Down,  saving  what  you  have  made. 
Busy  Bees,  a  clique  or  several  parties  working  together  with  some  special  object  in  view  to  inlluenco 

prices. 
To  Fill  or  to  Cover,  buying  Stock  or  Merchandise  to  fill  contract  of  short  sales. 
To   Unload  or  to  Get  Out,  to  sell  Stock  or  Merchandise  that  is  burdensome,  either  at  a  gain  or  a 

loss. 
Lame  Duck,  one  who  has  lost  heavily  or  has  been  badly  crippled,  but  who  has  not  failed. 
Dead  Duck,  one  who  is  unable  to  fullill  his  contracts  and  fails.  , 

Opening  Dag,  the  first  day  that  the  books  are  opened  after  a  dividend  has  been  declared. 
Settling  Dng,  a  day  that  the  things  bought  or  sold  are  delivered  or  paid  for,  or  a  day  that  the  con- 
tract matured. 
Contango,  the  postponement  of  settling  day. 

Point,  one  of  any  unit  or  measure  of  price,  either  increase  or  decrease. 
Black  List,  a,  record  kept  of  members  of  the  exchange  or  customers  of  banks  who  violate  their 

contracts. 
A  Put  Contract,  a  contract  by  which  you  have  the  right  to  put  or  sell  to  your  broker,  One  Hundred 

Shares  of  specified  stock  at  a  fixed  i)rice,  during  thirty  days.     For  this  contract  vuu  pav  your 

Broker  $;10G.25.  '  '    ' 

A  Call  Contract,  a  contract  by  which  you  have  the  right  to  call  or  buy  from  your  Broker,    One 

Hundred  Shares  of  specitied  stock  at  a  lixed  price,  during  thirty  days.     For  this  contract  you 

pay  your  Broker  $106.25. 
A  Spread  is  a  contract  by  which  vou  have  the  right  to  put  to,  or  call  from  your  Broker,   One  Hun- 
dred Shares  of  specified  stock  at  a  fixed  price,  in  thirty  days.     For  this  contract  you  will  jiay 

your  Broker  $212.50. 

Note. — A  Straddle  is  where  a  dealer  is  "long"  of  one  option  and  "short"  of  another.  He 
has  thus  straddled  the  market.  Also  where  a  dealer  buys  for  future  delivery  is  one  market  and 
sells  in  another. 


684 


souLE  s  rniLosoPHic  practical  mathematics. 


5.  S,  10,  ,'>0,  etc.,  iiiciiii  lliiit  tho  Si/lltT  lias  tlircc,  ten,   thirty,  etc.  days  in   which  to  deliver  Stocks 

or  MercliaiiiUso  bought. 

B.  S,  10,  iO,  so.  etc,  mean  that  the  Buyer  lias  tliree,  ten,   twenty,  thirty,  etc.  days  in  -wLicli  to 

receive  Stocks  or  Merchandise  boujtiit. 
Ex.  Dividend  means  that  the  sale  is  made  with  tlie  dividend  oil'. 

C.  means  a  cash  transaction. 
Jt.  means  a  regular  sale. 

P.  F.  means  preferred  stock. 

G.  T.  D.  means  guaranteed. 

Ex,  Coup,  means  without  coupons  or  interest. 

B.  ^  Int.  means  that  the  buyer  is  to  receive  the  interest  on  the  stocks  bought. 

B.  Flat  means  that  the  buyer  is  not  to  receive  the  interest. 

6.  c,  between  calls,  inearis  that  the  sale  was  made  between  the  first  and  second  call  of  the  stock, 

there  being  two  or  more  calls  each  day. 

Corner  \ii  !>n  operation  by  several  .jobbevs  or  operatorf.— "longs" — who  form  a  clique  to  buy  and 
holil  stocks  or  luercliandiso,  thus  jiroduciug  a  scarcity  in  the  market  wliiclx  results  iu  high 
prices,  aud  a  loss  to  the  "shorts"  or  others  who  are  obliged  to  buy,  and  are  hence  cornered. 

Corners,  frequently  by  result  disastrously  to  the  cliques  or  "  longs"  who  organize  tlieni  as  well  as  to 
the  "shorts"  as  tliey  are  left  with  a  load  that  they  cannot  discharge  except  at  a  great  loss. 

Spread  Eai/Ie  is  the  operation  of  a  liroker  wlio  sells  a  given  quantity  of  stoclis  on  time,  say,  60,  and 
buys  tlie  same  quantity  at  a  lower  price,  say,  GO.  If  both  contracts  run  60  daj^  he  will  make 
the  difference  ;  but  if  he  is  comiielled  to  deliver  by  buyer  or  seller  before  the  full  time  he  may 
Buffer  loss. 

Bucket  or  Curhxtone  Brolers  are  brokers  who  are  not  members  of  the  stock  exchange  and  thongli 
some  of  them  are  men  of  honor,  many  are  lame  ducks  and  some  are  only  speculators  without 
capital;  tliey  are  not  governed  by  as  strict  rules  and  do  not  abide  by  the  strict  letter  of  the 
law  as  do  members  of  the  exchange. 

Kite  Flying  is  a  term  applied  to  a  certain  class  of  transactions  of  an  extra  hazardous  character, 
and  engaged  in  only  liy  persons  who  are  non-observing  of  the  ]irinciples  of  ethics.  The  iollow- 
ing  will  servo  to  illustrate  the  transaction  :  A.  is  without  cajiital ;  he  solicits  of  B.  the  privilege 
of  drawing  on  him  at  10,  30,  or  60  days,  with  the  assurance  that  bclbre  the  maturity  of  the 
draft  ho  will  plaie  funds  iu  his  hands  to  meet  the  same.-  B.  consents;  A.  draws,  negotiates  the 
draft,  and  vises  the  proceeds.  When  this  draft  is  about  to  mature,  ho  draws  anil  negotiates 
another  sufficiently  larger  than  the  first  to  net  the  face  of  the  first.  This  process  of  drawiuc 
and  selling,  ho  keeps  up  until  by  prosperous  business  he  clears  enough  to  cover  the  discounts 
on  his  drafts,  and  to  meet  the  last  one  ;  or,  until  his  repeated  losses  and  clandestine  dealings 
involve  him  iu  inextricable  difficulties,  aud  he  becomes  a  disgraced  aud  ruiued  mau. 


PROBLEMS  IN  STOCKS  AND  BONDS. 
1255.  To  buy  and  sell  Stocks  and  Bonds  on  our  own  account. 

1.  What  will  50  shares  of  Germaiiia  National  Bank  stock  cost  at  15  per  cent 
premium?  Ans.  $5750. 

OPERATION. 

$115  X  50  =  .$5750.     Ans. 

2.  Bouglit  20  shares  of  National  Bank  stock  at  95  per  cent.     What  did  it 
cost!  Ans.  $1900. 

3.  Sold  40  shares  of  City  Railroad  stock  at  125i.     What  did  I  receive  for  it? 

Ans.  $5020. 

4.  Bought  75  shares  of  Chattanooga  Railroad  stock  at  90  per  cent,  and  sold 
the  same  at  5  per  cent  premium.    How  much  did  I  gain  ?  Ans.  $1125. 


PROBLEMS    IN   STOCKS    AND    BONDS. 


685 


1256. 


To  invest  in  stoclcs  and  Bonds  on  our  oirn  account. 


1.     How  many  shares  of  $100  each  of  the  City  National  Bank  stock  can  I 


buy  for  $2280,  the  stock  being  14  per  cent  premium  ? 


Ans.  20  shares. 


FIRST   OPERATIOX. 

1  sliare  =:  $100  par  value. 

14  =  14%  premium. 

$114  cash  value. 


114 


100 
2280 


I  $2000  =  par  of  stock. 
wbich  divided  Ijy  $100  =  20  shares. 

SECOND  OPERATION. 
SHARE. 
1 

2280 


Aus. 


114 


20  shares.     Ans. 


Explanation. — In  this  solution,  -(vo  first  find  the 
par  amouMt  of  stock  that  ve  can  buy  -with  the 
money,  and  then  the  number  of  shares.  In  the 
operation  we  first  added  to  the  $100,  par  value 
of  1  share,  the  14  per  cent  premium,  which  gave 
us  $114  cost  value ;  we  then  placeil  the  $100  par 
A-alue  on  our  statement  line,  and  reasoned  thus: 
Since  .$114  cash  will  bnv  $100  of  stock,  $1  cash 
will  buy  the  114th  ])art',  and  $2280  will  buy  2280 
times  as  much.  The  result  of  which  is  $2000. 
It  is  then  clear  that  as  1  share  is  $100,  there  are 
as  many  shares  as  $2000  are  etjual  to  $100,  which 
is  20. 


Explanation. — Hero'we  place  1  share  on  onr 
statement  line,  and  reason  thus:  Since  $114 
cash  will  buv  1  share,  $1  will  buv  the  114th  part, 
and  .$2280  w'ill  buy  2280  times 'as  much.  The 
result  of  which  is  20  shares. 


2.  Invested  $15375  in  City  bouds,  at  3S|  per  cent  discouut.  What  amount 
did  I  buy  ?  Aus.  $25000. 

OPERATION    IXKICATED. 

$100  —  $38i  =  $6U ;  $1537.5  —  $61+  =  $25000.  Ans. 

3.  A  merchant  invested  $3675  in  Louisiana  Frear  Stone  Manufacturing 
Company  stock,  at  5  per  cent  premium,  and  sold  the  stock  at  $114i.  What  was  bis 
gain  ?  Ans.  $31».37i,  gain. 

4.  Invested  $10450  in  Louisiana  4's  at  104i.  What  is  my  annual  income 
from  the  same?  Ans.  $400. 


1257. 


To  buy  and  sell  Stocks  throuyh  a  Broker. 


1.  Bought  25  shares  of  New  Orleans  Wrecking  Company  stock,  at  95  per 
cent.    Paid  New  Orleans  Board  of  Brokers  rates  of  brokerage.     What  did  it  cost? 

Ans.  $2387.50. 

OPERATION    INDICATED. 

25  shares  at  $100  =  $2500  par ;  $2500  X  95%  =  $2375.  $2375  +  i%  ^^o^-  $12.50  =  $2387.50.  Ans. 

2.  Sold  100  shares  of  New  Orleans  5  per  cent  Water  Works  bonds  at  40  per 
cent  discouut.     Paid  J  per  cent  brokerage.    What  were  the  proceeds  ? 

Ans.  $5950. 

3.  Bought  80  shares  of  $50  each  of  the  Mechanics'  &  Traders'  Bank  stock, 
at  75  per  cent,  and  sold  the  same  at  10  per  cent  discount.  Paid  brokerage  J  per 
cent  for  buying  and  i  per  cent  for  selling.    What  was  my  gain  f  Ans.    $560. 


686 


SOULE  S    rillLOSOI'IIIC    rUACTICAL    MATHEMATICS. 


125S. 


To  invest  ill  Slads  mitl  JUhkIs  IhroKf/li  a  Brolcr. 


1.     Invested  $9056.1,*")  in  Louisiaiiii  Savings  Bank  stock,  at  a  pieminm  of  20J 
percent.     Bi-okerage  A  per  cent.     How  many  shares  did  1  buy  ?  Aus.  75. 


rlI!ST  OPERATION. 

1  share  =  $100    par  value. 

20i-  =  20^,",;  jirciiiium. 
i  =  i?,i  brokeraKc 


483 


|120f  cost  of  $100  par. 

$   PAR  VAL. 

100 

4 

9056.25 


100  )  $7500.00  amount  of  stock. 
75       shares.     Ans. 
$100,  it  is  plain  that  we  hijve  as  many  shares  as  the  $7500  is  eriual  to  $100,  which  is  75 


Explanation. — We  hero  first  find  the  cost  of 
$100  par  value  at  the  rate  of  premium  and  hro- 
kerage  given  in  the  problem.  This  we  do  by 
addini;  to  the  $100  Jiar  value  tlio  20J  ]ii'r  cent 
I)reniium  and  |  per  cent  brokerage,  which  gives 
$120^;  with  these  ratio  figures  we  make  the 
proportional  solution  statement,  as  shown  in 
the  operation,  to  tind  tlie  jiar  amount  of  stock 
hought.  The  reasoning  for  the  statement  is  as 
follows:  Since  ^^  (120J  reduced)  dollars  cash 
•will  buy  $100  par  value  of  stock,  J  of  a  dollar 
cash  will  buy  the  483d  part,  and  j  or  a  whole 
dollar  will  buy  4  times  as  much,  and  $9056.25 
cash  will  buy  9056.25  times  as  much.  The  result 
of  which  is' .17500.     Then  as  1   share  is   worth 


483 


SHARE. 

1 

4 

9056.25 


SECOND   OPERATION. 

Explanation. — In  this  solution  we  work  the 
same  as  in  the  first  to  obtain  the  cost  of  $100 
par  value  of  stock  ;  then,  instead  of  finding  the 
amount  of  stock  that  we  can  buy,  we  find  at 
once  the  number  of  shares.  This  wo  do  by 
placing  1  share,  instead  of  the  par  value  of  the 
same,    on   our    statement    line,    and    reasoning 

thus  :  Since  %^^  ($120|  reduced)  cash  will  buy  1  share,  ($i  will  buy  the  483d  part,  and  $}  will  buy 

4  times  as  many,  and  $9056i"u^o-  will  buy  9056-r(ftr  times  as  many  as  $1.     The  result  of  this  gives  us  75 

shares. 

Note. — We  present  the  two  operations  because  in  some  problems  we   wish  to  know  the 

number  of  shares,  and  in  others  we  wish  to  know  the  par  value  or  amount  of  stock. 


75       shares.     Ans. 


2.     Paid  $1162.50  for  City  of  New  Orleans  4's  at  3g  per  cent  discount.    Bro- 
kerage J  per  cent.    What  is  my  semi-annual  interest  on  tliem  ?  Ans.  $24. 


3f  %  discount. 
i%  brok.  deducted. 


34"o'  not  discount. 
$100       par  val.  of  stock. 


$96i      cost  of 


par. 


OPERATION    INDICATED. 

$   PAR. 

I    100 
96i  I    1162.50 

$1200     bonds  bot. 


$1200 
4% 

$48.00    int.  for  1  year. 

8  —  2  =  .$24  iut.  for  6  mos. 


3.  Invested  $15912  in  City  of  New  Orleans  5  per  cent  bonds  at  103.1,  bro- 
kerage J  per  cent.     What  is  the  annual  income  ?  Aus.  $705.00. 

4.  What  amount  must  I  invest  in  3J  per  cent  bonds  which  are  selling  at 
104,  brokerage  J  per  cent,  so  that  I  may  have  an  income  of  $525  per  quarter  ? 

Ans.  $62700.00. 

OPERATION    INDICATED. 

$525  X  4  =  $2100.  3 J  I  f^^^  \  =  $60000  X  4J%  =  $2700  -f  $60000  =  $62700.  Ans. 


*  ■  PROBLEMS    IN    STOCKS    AND    BONDS.  6S7 

1259.  The  capital  and  rate  per  cent  given,  to  find  the  dividend. 

1..  A  mercliant  owns  200  sliares  cif  bank  stock,  tlie  par  value  of  wliicli  is 
$50.     How  much  will  lie  receive  when  a  divideud  of  G  per  ceut  is  declared  ? 

Aus.  8000. 

OPERATION. 
$50  X  200  =  $10000  X  6,%'  =  $600  dividend. 

2.  What  is  the  amount  of  a  6  per  cent  dividend  ou  120  shares  of  New  Orleans 
Gas  Light  Company  stock  ?  Ans.  $720. 

3.  Invested  $5747.50  in  State  National  Bank  stock,  at  104.  Paid  brokerage 
for  buying  J  per  cent.  How  many  shares  did  I  buy,  and  what  is  my  annual  income 
therefrom  when  the  dividends  are  4  per  cent  semiannually  ? 

Ans.  55  shares.     $440  annual  income. 

1260.  The  capital  or  investment  and  net  r/ain  (/iven,  to  find  the  rate  per  cent  dividend 

or  per  cent  (jain. 

1.  The  capital  of  a  Bank  is  $1500000.  The  net  profits  for  six  months'  busi- 
ness are  $192814.58.  The  Board  of  Directors  resolve  to  pass  $12814.58  to  the  credit 
of  "reserve  fund,"  and  to  distribute  tlie  remainder  to  the  stockholders.  What  is 
the  rate  per  cent  of  the  semi-annual  dividend?  Aus.  12%. 

OPERATION. 

$192814.58  uet  prolits.  $ 

12814.58  passed  to  reserve  fund.  I  180000 
1500000     100 


$180000.00  profit  for  a  dividend. 


12'''      Aus. 


2.  I  subscribed  for  100  shares  of  bank  stock.  After  paying  three  install- 
ments of  25  per  cent  each,  a  dividend  is  declared.  The  capital  stock  of  the  Bank 
is  $300000;  the  net  luofits  are  $3S4G0.50.  Passing  $8400.50  to  reserve  fund,  what 
is  the  rate  per  ceut  dividend  and  the  rate  per  cent  profit  ou  my  investment? 

Ans.  10%  dividend.    13^%  gain  on  investment. 

3.  An  investor  has  $14700  in  manufacturing  stock,  and  receives  $3G7.50  per 
quarter,  dividend.     What  is  his  annual  jier  cent  gain  1  Ans.  10%. 

4.  Invested  $20000  and  gained  $1500.     What  was  the  gain  per  cent  f 

Aus.  7J%. 

OPERATION    INDICATED. 

.       $ 

I  1500 
20000  I  100 

5.  Bought  50  shares  of  stock  at  $00  per  share,  and  sold  the  same  at  12.^  per 
cent  discount.     What  per  cent  did  I  lose  ou  the  investment?  Aus.  2|%  loss. 

OPERATION    INDICATED. 

$ 
50  X  $90  =  $4500  cost.  I  125  =  loss. 

50  X  $87i  =  4375  selling  price.      4500  |  100 


688 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


1261.     The  rate  iper  cent  dividend  or  interest,  and  the  price  of  Stock  given,  to  find 

the  rate  jier  cent  gain  on  investment. 


1.     Bought  Kew  Orleans  City  C's  sit  25  per  cent  discount.    What  is  my  per 

Ans.  &%. 


cent  gain  on  investment? 

OPERATION. 


$100      ])ar  value. 
25%  discouut. 

$75      cost. 


$  GAIN. 

I  6 

75     100 


8;o'  Ans 


ExplanaUon. — As  shown  in  the  operation,  $100  par  value 
cost  lint  $75,  and  hence  the  6  per  cent  interest  is  really 
fjaiiuMl  on  the  $75  cash  investment.  In  uialiing  the  pro- 
]iiirtional  statement  to  find  the  gain  per  cent,  we  reason 
thns :  iSinoo  $75  gain  $G,  $1  will  gain  the  75th  jiart,  and 
$100,  100  times  as  much. 


2.  Bought  C  per  cent  bonds  at  5  jier  cent  premium.  What  is  my  per  cent 
gain  on  investment,  making  no  allowance  for  the  use  of  the  semi-annual  interest? 

Ans.  5f%. 

3.  If  I  buy  60  shares  of  bank  stock  at  190,  and  pay  J  per  cent  brokerage, 
and  the  bank  declares  a  semi-annual  dividend  of  15  per  cent,  what  is  my  per  cent 
gain?  Ans.  lo^yV^- 


$190        cost. 

^%  hrok.  on  par  value. 

|190J      total  cost  of  stock. 


OPERATION    INDICATED. 

V^^  semi-annual  dividend. 
2 

30%  yearly  dividend. 


381 


30 

2 

100 


4.  Which  is  the  better  investment,  S's  bought  at  a  premium  of  20  per  cent, 
or  6's  at  a  discount  of  25  per  cent?  Ans.  6's  are  1J%  better. 

5.  A  capitalist  buys  Louisiana  C's,  having  10  years  to  run,  at  a  discount  of 
15  per  cent.  What  will  be  his  rate  ])er  cent  interest  gain  if  he  holds  them  until 
maturity  and  then  collects  the  face  and  interest?  Ans.  8|4%. 

OPEP.ATION. 


$100  par  of  l)ond. 
15  ^  15"o  discount. 

$85  cost  of  bond. 


par  of  bond. 
(j%  interest. 

i6.00  interest  for  1  year. 
10  vears. 


$100  face  of  liOTid. 
GO  int.  for  10  years. 

$160  value  at  maturity. 
85  cost  of  bond. 

$75  gain  in  10  years. 


160.00  interest  for  10  years. 

We  now  have  to  find  at  what  rate  per  cent  interest  $85,  for  10  years,  will 
produce  $75  interest. 

OPEKATION. 


10       years. 

$8.50  int.  for  10  years  at  \%. 


8.50 


1 

75.00 

8|?  ."o-  Ans. 


or, 


85 
10 


75 
100 


8|4  % 


Explanation. — For  the  reasoning  of  this  last  work,  see  Interest  Problem,  on  page  599. 
C.     Bought  5  per  cent  interest  bearing  bonds  at  98  per  cent.    These  bonds 


PROBLEMS    IN    STOCKS    AND    BONDS. 


689 


have  7  years  and  6  months  to  run,  "WTiat  will  be  tlie  gain  per  cent,  the  interest 
being  paid  senii-auuually,  making  no  allowance  for  the  use  of  the  semi-annual 
interest?  Ans.  5^4%%  or  5.102+%. 

7.  In  the  above  problem  allowing  6  per  cent  simple  interest  on  the  annual 
interest  of  the  bonds,  what  would  be  the  annual  gain  jier  cent  made  on  the  invest- 
ment? Ans.  5iifo- 

OPERATION    INDICATED. 

Interest  on  $5  for  12  mos.  at  6J!^  =  30c.     $5  +  30c.  =  $5.30. 


98.00 


5.30 
100 


1262.     TJie  per  cent  that  StocJcs  bear,  and  the  per  cent  that  is  desired  to  be  gained, 

given,  to  find  the  price  at  which  Steele  must  be  bought  in  order  to 

realize  the  desired  gain  per  cent. 

1.    At  what  price  must  stock  be  bought  which  pays  6  per  cent,  in  order  that 
the  investment  will  pay  8  per  cent?  Ans.  $75  per  share  of 


FIRST  OPERATION. 

$100       assumed  price. 
8^  gain. 

$8.00  amount  of  gain. 


100 
6 


$75  Ana. 


Explanation. — Here,  Laving  no  value  to  work 
from,  we  assume  $100,  and  tlien  find  the  8  per 
cent  gain  thereon  to  be  $8.  We  then  see  that 
flOO  value  of  stock  gives  |8  gain,  and  by  trans- 
position that  $8  gain  requires  $100  value  of  stock. 
The  question  is  now  simply  this:  Since  $8  gaia 
require  $100  value,  what  value  will  $6  gaia 
require?  The  exact  reasoning  for  the  statement 
is,  since  $8  gain  require  $100  value,  $1  will 
require  the  8th  part,  and  $6  will  require  6  times 
as  much. 


2.  If  I  loan  my  money,  I  can  get  10  per  cent  interest.  At  what  discount 
must  I  buy  Louisiana  Levee  bonds,  that  bear  8  per  cent,  in  order  to  obtain  the  same 
per  cent  on  the  investment  ?  Aus.  20%  discount. 

OPERATION    INDICATED. 


10  1 


100 

8 


i-> 


100  —  80  =  20. 


3.     Interest  on  money  is  4J  per  cent. 
pays  6  per  cent  dividends  ? 


What  price  can  I  pay  for  stock  that 

Ans.  $133J. 

OPERATION    INDICATED. 


I  100 
9      2 
!6 


4i 


100 


4.    Which  is  the  best  investment  to  make  with  $52000 :  To  loan  it  at  12  per 
cent  interest  on  good  collaterals;   to  invest  it  in  stocks  at  65  per  cent  that  pay 


690 


SOULE  S    nilLOSOPIIIC    PRACTICAL    MATHEMATICS. 


semi-ammal  dividends  of  4  per  cent;  or  to  buy  merchandise  -wliicli  we  can  sell 
during  tbe  year  for  $07380,  with  an  attendant  expense  of  $9400 1 

Aus.  Stocks  by  -,^3  per  cent  over  the  12  per  cent  loan,  besides  the  use  of 
one  semi-annual  dividend  for  G  months;  and  by  |^  per  cent  over 
tbe  merchandise  investment,  not  making  any  allowance  for  the 
use  of  the  money  collected  jirior  to  the  close  of  the  year. 


OPERATION    INDICATED. 


65  I  100  I  ~  ^^i"^^-    $67380  —  S52000  =  $15380  gain  on  sales ;  less  $9400  expense.     = 
X  100)  -H-  52000  =  11^%.     12A%  -  12%  =  -,%%.     12^%  -  lli?%  =  IJ%. 


net  gaic 


1263.     The  rate  per  cent  that  Stoehs  hear,  the  cost  of  them,  and  the  gain  or  income 
desired  to  be  realized  from  the^n  given,  to  find  the  sum  to  be  invested. 

1.    What  sum  must  be  invested  in  Louisiana  6's,  at  75  per  cent,  to  realize  an 
income  of  $3000  per  year  ?  Aus.  $37500. 

OPERATION. 


$100      par  of  investment  assumed. 
65'o   rate  of  interest. 

$6.00     amount  of  income  on  $100  stock. 


100 
3000 

I  $50000     par  amount  of  investment. 
15^  cost  of  $100  stock. 


137500.00  cost  of  $50000  stock  or  amt. 
of  investment. 

Explanation. — Here  •we  assnme  $100  aa  the  par  of  the  investment,  and  find  the  6  per  cent 
Income  on  the  same  to  be  $6.  Then,  with  tlicse  ratio  figures  and  tlie  $3000  income,-  we  make  the 
proportional  solution  statement  by  placing  the  $100  par  of  investment  on  the  statement  line,  and 
reason  thus:  If  $6  income  require  $100  par  of  investment,  $1  will  require  the  6th  part,  and  $3000 
income  wiU  require  3000  times  as  much.  The  result  of  this  is  $50000,  par  of  the  investment,  ■which 
multiplied  by  the  price,  75  per  ceut,  gives  a  cash  investment  of  $37500. 

2.  How  many  shares  of  bank  stock,  of  $.50  each,  must  be  bought  at  10  per 
cent  premium  in  oider  that  an  income  of  $1500  may  be  realized,  on  the  presumption 
that  the  semi-annual  dividends  will  be  6  per  cent,  and  what  will  be  the  amount 
invested  ?  Ans.  250  shares.     $13750,  amount  invested. 


12 


OPERATION. 


100 

1500 

$12500  par  of  stock  4-  $50  =  250  shares. 


$12500  p.ar  of  stock. 

1250  10  per  cent  premium, 

$13750  amount  invested. 


3,    If  6  per  cent  bonds  are  selling  at  5  per  cent  discount,  what  amount  must 
I  invest  in  them  in  order  to  receive  a  yearly  income  of  $1200  ?  Ans,  $19000. 


*  PROBLEMS   IN   STOCKS    AND    BONDS.  09I 

1264.  The  cost  of  Stocks,  and  the  premium  or  discount  at  tchich  they  icere  bonghty 

given,  to  find  the  par  value. 

1.  BoitgLt  railroad  stock  at  70  per  cent  discount,  and  paid  $7.50  per  share. 
What  is  the  par  value  ?  Aus.  $25. 

OPEKATION.  Explanation. — Havin'i;  no  face  of  stock  to  work 

on,  we  hence  assume  |l()0  as  the  par  value  on 

$100  par  value  assumed.  which  we  calculate,  antl  then  deduct  therefrom 

70  =  70%  discount,  •  the  70  per  cent  discount,  and  thus  obtain  $80  as 

the  cost  of  $100  stock  at  TO  per  cent  discount. 

$30  cost  of  $100  par.  This  gives  us  the  necessary  ratio  figures  to  solve 

the  question;  and  in  the  line  statement  of  the 

$  solution  we  place  the  $100  par  value  on  the  line, 

I  100  and  reason  thus:    Since  $30  cost  at  70  per  cent 

30  I  7.50  discount  requires  $100  par  value,  $1  will  require 

the  30th  part,  aud  $7.50  -nill  requiie  7,50tiiue3 

I  $25.00    Ans,  as  much. 

2.  Sold  bank  stock  for  $55.50  and  gained  11  per  cent.  What  is  the  par  of 
the  stock?  Aus.  $50. 

OPERATION. 

$100  par  assumed.  $ 

11  =  11%  premium.  100 

Ill    oo.iiO 

$111  cost. ■ 

$50    Ans. 

1265.  The  time  that  Bo7ids  have  to  run  and  the  interest  they  draw,  given,  to  find  the 
price  at  ichich  they  must  he  purchased  so  that,  when  paid  at  maturity  without 

estimating  the  use  of  the  annual  interest,  the  amount  received  icill 
equal  a  certain  rate  per  cent  on  the  investment. 

1.  What  nui.st  be  paid  for  State  bonds  bearing  4  per  cent  interest  and  having 
8  years  to  run,  in  order  to  make  the  investment  equal  to  5  per  cent?  No  allowance 
to  be  made  lor  interest  on  the  interest  of  the  bonds  paid  annually,  semi-annually  or 
quarterly?  Ans.  $yJ:.i;S5+. 

OPERATION. 

$100       face  of  bond.  $100       investment.                                       $ 

4%  interest.  5%  interest.                                            I  100 

— _  140  I  132 

$4.00  interest  for  1  year.  $5.00  interest  for  1  year.                          

8  years.               *  8  years.                                                |  $94,285+  investing  price 

of  the  bonds. 

$32.00  interest  for  8  years.  $  40.00  interest  for  8  years. 

100.00  face  of  bond.  100.00  face  of  bond. 

$132.00  =  value    of    bond   at        $140.00  =  value  of  investment 
maturity.  at  5%  at  the  end 

of  8  years. 

Explanation. — As  shown  in  the  operation,  we  first  find  the  value  of  a  $100  bond  and  the  value 
of  $100  investment  each  for  the  time  at  the  respective  rates  per  cent.  Having  these  respective 
values,  wo  see  that  $100  at  4per  cent  produces  $132,  and  at  5  per  cent,  $140;  aud  as  we  wish  to  realize 
5  per  cent  on  the  investment,  we  place  the  $100  assumed  value  on  the  statement  line,  and  with 
the  ratio  numbers  produced  reason  thus:  Since  $140  value  in  8  years  at  5  jier  cent  requires  $100 
investment,  $1  will  require  the  140th  part,  and  $132  value  ■will  require  132  times  as  much,  which 
gives  $94,285+  investing  price  of  the  tjonds. 

Note. — This  method  does  not  take  into  consideration  the  interest  on  the  interest  received 
annually  on  the  bonds.  To  make  such  an  allowance  the  priqe  of  the  bonds  would  be  a  little  higher 
according  as  the  payments  of  interest  were  made  annually,  semi-annually  or  quarterly.  See  the 
problems  following,  for  such  allowance.     Also  see  problem  on  page  591. 


69: 


SOULE  S    rillLOSOPHIC    rUACTICAL    MATHEMATICS. 


1260.     The  face  of  Bonds  or  Notes,  the  time  to  run,  and  the  rate  per  cent  of  yearly, 

half-yearly  or  quarterly  interest  that  they  hear,  given,  to  find 

the  value  at  maturity. 

1.     What  is  the  value  at  iiiatuiity  of  $1000  Louisiana,  S's  due  in  5  years, 
estimating  simple  iutei'est  on  the  annual  interest  at  8  per  cent?  Aus.  $14:04. 


OPEnATION. 


Face  of  bond,        ............... 

Auiiual  interest  for  5  years  at  8  per  cent,  ......... 

Interest  on  $80  (the  yearly  interest  on  the  $1000)  for,  4  +  3  +  2  +  1  =  10  years  at  8%, 


Value  at  maturity, 


$1000 

400 

64 

$1464 


2.  What  is  the  value  at  maturity  of  a  $10000  bond  due  in  10  years  with  5 
per  cent  semi-auuual  interest,  allowing-  simple  interest  on  the  semi-annual  interest 
at  5  per  cent?  Ans.  $16187.50. 


OPERATION. 
To  find  the  terms  and  years. 
1 
19 

20  X  9i  =  190  —  2  —  95  years. 


OPERATION. 
Face  of  bond,  ........ 

Semi-annual  interest  for  10  years  at  5  per  cent, 
Interest  on  $2ol)  (the  semi-annual  interest  on  $10000) 
for  the  190  terms  of  ^  years  (the  series  of  1  to 
19  inclusive)  =  95  years  at  5  per  cent, 

Value  of  bond  at  maturity,  ... 


$10000.00 
5000.00 


1187.50 
$16187.50 


1267.     The  face  of  Bonds  or  Notes,  the  time  to  run,  and  the  rate  per  cent  of  yearly, 

half-yearly  or  quarterly  interest  that  they  hear,  given,  to  find  the  present  value  or 

worth  of  the  same  in  order  to  realize  from  the  interest  or  dividends  and  final 

payment  an  equivalent  to  a  given  rate  per  cent  per  annum  on  the  investment. 

1.  What  is  the  present  value  or  worth  of  a  $10000  bond  due  in  10  years,  with 
6  per  cent  semiannual  interest,  so  that  the  investment  may  be  equivalent  to  10  per 
cent  per  annum  payable  semi-annually,  allowing  simple  interest  on  all  the  semi- 
auuual  interests  on  the  bond  at  5  per  cent  ?  Ans.  $6540.41. 

OPERATION 

To  find  the  value  of  Bond  at  maturity. 

Face  of  bond, $10000.00 

Semi-annual  interest  for  10  years  at  5  per  cent,           .-...-..  5000.00 
Interest  on  $250  (the  semi-annual  interest  on  $10000)  for  the  190  terms  (the  series  of  1 

to  19  inclusive)  of  4  years  =  93  year.s  at  5  per  cent,          .        .        .        ^        .        .  1187.50 

Value  of  bond  at  maturity, $16187.50 

OPERATION 

To  find  the  value  of  a  $100  Bond  for  10  years,  at  10^^  semi-annual  interest,  the  investment  rate. 

Present  v.alue  assumed,      ..............  $100.00 

Semi-annual  interest  for  10  years  at  10  per  cent,             ........  100.00 

Interest  ou  $5  (the  semi-annual  interest  on  the  $100)  for  the  190  terms  (the  series  of  1  to 

19  inclusive)  of  ^  years  =  95  years  at  10  per  cent,       .......  47.50 

Value  of  $100  in  10  years  at  10  per  cent,  semi-annual  interest,  ...       $247.50 


I 


PROBLEMS    IN    STOCKS    AND    BONDS. 


693 


OPERATION 

To  find  the  present  value  of  the  maturity  value 
of  the  Bond. 


Explanation. — As  shoirn  in  the  operation,  we 
first  find  the  maturity  value  of  the  bond.  Then, 
in  order  to  produce  the  necessary  relationship 
numbers,  representing  present  and  maturity 
values  according  to  the  conditions  of  the  prob- 
lem, we  assume  |100  present  value,  to  which  we 
add  the  semi-annual  interest  and  the  simple 
interest  on  the  semi-annual  interest  for  the  10 
years  at  10  per  cent,  and  thus  obtain  $247.50, 
the  maturity  value  of  the  $100  assumed  value 

on  the  same  conditions  as  we  did  for  the  $10000  bond.     We  now  see  that  as  $100  present  value 

require  $L'47.50  maturity  value,  by  transposition  $247.50  maturity  value  require  $100  present  value. 

The  reasoning  for  the  line  statement  with  these  relationship  figures  is  so  simple  and  has  been  given 

so  often  that  we  here  omit  it. 


247.50 


9  ASSUJIED  VAL. 
100 

16187.50 
$6540.41 


THE   OPERATION   REVERSED  ' 

To  elucidate  the  problem  still  further,  we  present  the  figures  produced  by  reversing  the  work. 

Present  c.a.sh  value  of  bond, $6540.41 

Semi-annual  interest  on  same  for  10  years  at  10  per  cent,            ......  6540.40 

Interest  on  $327.02  (the  semi-annual  interest  on  $6540.41)  for  190  terms  of  |  vears  =  95 

years  at  10  per  cent, "...  3106.69 

Value  of  bond  at  maturity, $16187.50 

2,  What  is  the  present  value  of  a  $10000  bond  due  in  10  years,  with  5  per 
cent  semi-annual  interest,  in  order  that  the  investment  may  be  equivalent  to  10  per 
cent  per  annum  payable  annually,  allowing  simple  interest  on  the  semi-annual 
interest  or  dividend  of  the  bond  accruing  yearly  at  10  per  cent  per  annum  ? 

Ans.  $7040.82. 

OPERATION 

To  find  the  value  of  Bond  at  maturity. 

Face  of  bond, $10000 

Semi-annual  interest  on  same  for  10  years  at  5  per  cent,  5000 

Interest  on  $500  (the  sum  of  two  semi-annual  jiayments 
on  the  $10000)  for  45  terms  of  years  (the  decreasing 
series  from  9  to  1  inclusive)  at  10  jier  cent, 


OPERATION 
To  find  the  terms  and  years. 
9 
1 

10  X  4i  =  45. 


Value  of  bond  at  maturity,         .... 

OPERATION 

To  find  the  value  of  a  $100  Bond  for  10  years,  annual  interest  at  10  per  cent. 

Present  value  assumed,  .............. 

Annual  interest  for  10  years  at  10  per  cent,         .......... 

Interest  on  $10  (the  sum  of  two  semi-annual  jiayments  on  the  $100)  for  45  terms  of  years 
(the  decreasing  series  from  9  to  1  inclusive)  at  10  per  cent,     ...... 

Value  of  $100  in  10  years  at  10  per  cent  annual  interest,  ..... 

OPERATIOX 

To  find  the  present  value  of  the  maturity  value  of  the  Bond. 


2250 
$17250 


$100 
100 

45 

$245 


245 


100 
17250 

$7040.82    Ans. 

THE  OPERATION  REVERSED. 


Explanation. — The  operation  of 
this  and  the  preceding  problem  is 
so  explicit  that  a  special  explana- 
tion is  considered  unnecessary. 


Present  value  of  bond,  ............. 

Annual  interest  for  10  years  at  10  per  cent,        ......... 

Interest  on  $704.08  (the  annual  interest  on  $7040.82)  for  45  terms  of  years  as  above,  at 
10  per  cent,  .............. 


Value  of  bond  at  maturity, 


$7040.82 
7040.82 

3168.36 

$17250.00 


694  soule's  niiLosopiiic  practical  mathematics.  * 

3.  What  is  the  present  value  of  a  $10000  bond  clue  in  10  years,  with  5  per 
cent  semi-annual  interest,  in  oiiler  that  the  investment  may  be  equivalent  to  10  per 
cent  per  annum  payable  scmi-annualli/,  allowing  simjile  interest  on  all  the  semi- 
annual interests  of  tiie  bond  at  10  per  cent !  Ans.  $7020.20. 

OPERATION 

To  find  tlio  value  of  the  Bond  at  maturity. 

Face  of  bond, - $10000 

Semi-aunual  interest  for  10  years  at  5  per  cent,       ---------  5000 

Interest  on  $250  (tlie  semi-annual  interest  on  $10000)  for  the  190  terms  of  i  years  (the 

decreasing  series  from  19  to  1  inclusive)  =  95  years  at  10  i)er  cent,     ...        -  2375 


Value  of  bond  at  maturity,  .--..-.-..        $17375 

OPERATION 

To  find  the  value  of  a  $100  Bond  for  10  years,  semi-annual  interest  at  10  per  cent. 

I*resent  value  assumed,     ----- -..  $100.00 

Semi-aunual  interest  for  10  years  at  10  per  cent,              ----....  100.00 
Interest  on  $5  (the  semi-aunual  interest  on  the  $100)  for  the  190  terms  of  i  years  (the 

decreasing  series  from  19  to  1  inclusive)  :=  95  years  at  10  per  cent,             ...  47.50 


Value  of  $100  in  10  years  at  10  per  cent  semi-annual  interest,      ....        $247.50 

OPERATION 

To  find  the  present  value  of  the  maturity  value  of  the  Bond. 
$ 


247.50 


100 
17375 


$7020.20+    Ans. 

THE  OPERATION  UEVEKSED. 

Present  value  of  bond, $7020.20 

Semi-annual  interest  on  same  for  10  years  at  10  per  cent,              .-.-.-  7020.20 
Interest  on  $351.01  (the  semi-annual  interest  on  $7020.20)  for  190  terms  of  4  years  =  95 

years  at  10  per  cent,       ...-..-.-..-.  3334.60 

Value  of  bond  at  maturity, $17375.00 

4.  What  is  the  present  value  of  a  $10000  bond  due  in  10  years,  with  5  per 
cent  semi-annual  interest,  in  order  that  the  investment  may  be  equivalent  to  10  per 
cent  per  annum,  allowing  comi^ouiid  interest  on  the  semi-aunual  payments  of  inter- 
est at  10  per  cent  ?  Ans.  $7001.52. 

OPERATION 

To  find  the  worth  of  the  Bond  at  maturity. 

Face  of  Bond, $10000.00 

Semi-annual  interest  on  same  for  10  years  at  5  per  cent,  ......  5000.00 

Compound  interest  on  $250  (the  semi-annual  interest  on  $10000)  compounded  annually 
at  10  jier  cent  for  the  sum  of  the  ^  yearly  decreasing  series  of  terms  from  9}  to  | 
years  inclusive,     .----------.--  3167.93 

Value  of  bond  at  maturity,  ..--......        $18167.93 

OPERATION 

To  find  the  present  value  of  the  maturity  value  of  the  Bond. 
$2.5937425  )  $18167.9300000  (  $7004.52    Ans. 


117325000 

135753000 

60658750 
Wa  here  divide  the  matnrity  valnn  by  the  compound  amount  of  $1  for  10  years  at  10  pei;  cent» 


* 


PROBLEMS    IN    STOCKS    AND    BONDS.  695 


■which,  for  convenience  we  obtain  from  the  compound  interest  table  on  page  G12.     The  result  or 
<]Uotieut,  $71104.52,  is  the  present  value  of  the  bond. 

£xpla»tition. — In  the  first  part  of  the  operation,  to  find  the  compound  interest  on  the  $250  for 
the  sum  of  tlio  decreasing  scries  of  terms  of  i  years,  from  9i  to  +,  at  10  per  cent  annual  compound 
interest,  to  economize  time  ive  used  the  compound  interest  tables;  but  as  our  tables  are  calculated 
only  for  the  integral  or  rimnd  years,  ve  find  it  necessary  to  separate  the  |  yearly  decreasing  series 
into  two  yeariy  series,  viz. :  A  yearly  integral  series  from  9  to  1  iuchisive,  and  a  yearly  fractional 
series  from  OJ  to  li  inclusive,  and  also  the  final  i  year.  We  then  find  in  the  table,  in  the  10  per  cent 
column 

The  compound  amount  of  the  integral  series  from  1  to  9  years  to  be,         ...  $14.9374246 

From  which  we  deduct  the  sum  of  9  additions  of  $1  principal,  .        -        .        .  9. 

And  obtain,  compound  interest,     -.-- $5.9374246 

Then  to  find  the  amount  of  the  fractional  series  from  9i  to  1^ : 

We  multiply  the  amount  of  the  full  series  $14.9:!74246  by  $1.05 ; 

The  product  of  this  multiplication  is  $15.68429583,  from  which  we  deduct  $9,  the 

sum  of  9  additions  of  $1  principal,  and  obtain,  compound  interest,         -        -  6.68429583 

To  these  sums  we  add  the  comjiound  interest  on  $1  for  |  a  year  at  10  per  cent, 

which  is  ---......'.....  .05 

And  thus  obtain  the  total  compound  interest,      -....--.  $12.67172043 

This  we  multiply  by  $250,  the  semi-annual  interest  on  the  bond,       .        .        -        .  250 

And  produce,  as  the  total  annual  compound  interest  on  all  the  semi-anntial  inter- 
ests of  the  $10000  bond, $3167.9301075 

TO  FIND  THE    VALUE    OF    BONDS    UNDER    VARYING    CONDITIONS, 

STATED  IN  THE  PROBLEMS, 

1268.  1.  What  price  must  be  paid  for  a  $1000  bond  bearing  4  per  cent  inter- 
est payable  annually,  having  8  years  to  run,  so  that  the  investment  may  be  equal 
to  5  per  cent?  Ans.  $935,367. 

Remarks  1. — In  questions  of  this  kind,  in  wliich  there  are  no  conditions 
regarding,  simple,  annual  or  comijound  interest,  compound  interest  is  computed 
upon  tlio  annual  i)ayments  of  interest  on  the  bond.  (But  not  on  the  interest  of  the 
annual  interest  that  may  be  realized  on  the  investment). 

When  such  rate  is  not  specified,  the  rate  given  in  the  bond  is  taken  in  the 
solution  of  the  question. 

2.  The  value  of  a  bond  in  the  market  as  an  investment  at  a  given  rate, 
depend.s,  other  things  being  equal,  upon  the  rate  i)er  cent  interest  at  which  the 
re-investments  may  be  made  of  the  various  accruing  interest  payments. 

3.  "When  a  bond  is  bought  at  a  premium,  some  part  of  each  interest  install- 
ment received  on  the  bond  must  be  reinvested  at  compound  interest,  and  so  con- 
tinued until  the  maturity  of  the  bond.  The  accumulations  of  this  interest  investment 
must  amount  exactly  to  the  premium  paid  on  the  bond.  For  when  the  bond  is  paid, 
the  face  only  ■will  be  received. 

4.  When  a  bond  is  bought  at  a  discount,  by  the  same  reasoning  as  above,  the 
interest  installments  received  ■will  be  in  excess  compared  with  the  actual  amount 
paid  for  the  bond,  and  hence  there  -will  be  at  each  interest  paying  period  a  certain 
amount  that  may  be  exjiended  and  yet  leave  at  the  maturity  of  the  bond  the  exact 
amount  invested  therein. 

Note. — For  the  solution  of  problems  for  bond  investments,  tables  have  been  prepared  by 


696 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


different  authors,  and  aro  in  genoral  nso  \<y  hankers,  hriikers,  and  investors.  These  tahles  are 
prepared  on  ditiercnt  plans,  accordinj;  to  eertain  conditions  of  investing  and  re-investing,  and  hy 
their  nso  various  investing  results  are  shown  without  the  tedious  computations  iuvolving  the  prin- 
ciples of  compound  interest. 

In  the  first  solution  of  the  ahovo  prohlein,  we  will  make  use  of  the  Compound  Interest  Table, 
page  611,  and  in  the  second  solution  we  will  use  the  table  showing  the  Tresent  Worth  of  an 
annuity,  and  the  table  showing  I'reseut  Value  of  $1,  when  time  and  rate  per  cent  are  given,  (Se& 
Index,  Present  Worth). 


FIRST    SOLUTION. 

In  this  solution,  wo  consider  the  problem  as 
an  annuity  of  $40  per  annum  for  8  years,  with  a 
payment  of  $1000  at  the  expiration  of  tliat  time. 
And  since  the  present  worth  of  an  annuity  may 
he  found  by  dividing  the  payment  h.y  tlie  com- 
pound amount  of  $1  for  the  specified  time  and 
at  the  given  rate  per  cent,  hence  we  will  so 
divide  each  of  the  8  payments  ou  the  bond. 


Annuity 

Componnd 

Value  of 

Payments. 

Amuuut  of  $1. 

Annuity. 

1st       $40       - 

-      $1.05 

— 

$38.09 

2d          40       - 

1.1025 

= 

36.28 

3d         40      - 

1.1576 

= 

34.55 

4th        40      - 

1.2155 

= 

32.90 

5th        40      - 

1.2763 

= 

31.34-f- 

6th        40      - 

1.34 

=: 

29.85-f- 

7th         40       - 

1.4071 

=^ 

28.42 

8th     1010       - 

1.4775 

= 

703.93-1- 

Market  value  of  bond  ou  condi- 
tions named  in  the  problem. 


$935.36+ 


SECOND   SOLUTION. 

In  this  solution,  we  also  resolve  the  question 
into  an  annuity  as  in  the  lirst  solution.  Then 
since  the  present  worth  of  an  annuity  may  be 
found  by  multiplying  the  amount  of  an  annuity 
by  the  present  worth  of  an  annuity  of  $1  for 
tlie  time  specified  and  the  rate  per  cent  given, 
we  therefore  multiply  the  $40  annuity  by 
$6.463213,  which  is  the  present  worth  of  an 
annnity  of  $1  for  8  years  at  5  per  cent,  as  shown 
intheannuity tables,  whichsee.  Thusweobtain 
$258.52  as  the  value  of  the  $40  annuity. 

Then  to  find  the  present  worth  of  the  $1000 
bond,  we  multiply  the  same  by  $.676839,  which 
is  the  present  value  of  $1  for  8  years  at  5  per 
cent  as  shown  in  the  table  of  present  worth, 
which  see. 
By  this  we  obtain   as  the  present 

worth  of  the  bond,        -        -        -        $676.84 
Value  of  the  annuity  as  above         -  258.52-j- 


Value  of  bond  and  annuities, 


$935.36-(- 


Cost  of  bond,       -        .        -        . 
Interest  on  same,  1  ye.ar  at  5^^ 


Less  interest  on  bond,  1st  year. 


Balance,       -        -        .        -        - 
Interest  on  same,  2d  year  at  5,"q' 


Less  interest  on  bond,  2d  year, 


Balance,      .        .        .        -        - 

Interest  on  same,  3d  year  at  5% 


Less  interest  on  bond,  3d  year, 


Balance,       .        .        .        .        - 
Interest  ou  same,  4  years  at  5^^ 


Less  interest  on  bond,  4th  year, 
Balance  forward, 


$935.36 

46.77 


$982.13 
40.00 

$942.13 
47.11 

$989.24 
40.00 

$949.24 
47.46 

$996.70 
40.00 

$956.70 
47.84 

$1004.54 
40.00 

I  964.54 


Balance  brought  forward,  - 
Interest  on  same,  5th  year  5% 


Less  interest  on  bond,  5th  year, 


Balance,       .        .        -        .        . 
Interest  on  same,  6th  year  at  5% 


Less  interest  on  bond,  6th  year, 

Balance,       ,        -        -        -        . 
Interest  on  same,  7th  year  at  5% 


Less  interest  on  bond,  7th  year. 
Interest  on  same,  8th  year  at  5%' 
Less  interest  on  bond,  8th  year, 


Balance, 

Less  bond  redeemed, 


$  964.54 
48.23 

$1012.77 
40.00 

$  972.77 
48.64 

$1021.11 
40.00 

$  981.41 
49.07 

.48 
40.00 

$  990.48 
49.52 

$1040.00 
40.00 

$1000.00 
1000.00 


2.  What  price  should  be  paid  for  a  $500  bond  bearing  6  per  cent  interest, 
due  in  9  years,  bo  that  the  investment  will  produce  8  per  cent  income,  no  allowance 
to  be  made  for  the  interest  on  the  annual  interest  payments  ?         Ans.  $447.67-f . 


PROBLEMS    IN    STOCKS    AND    BONDS. 


697 


OPERATION. 


Face  of  bond,  $500.00 

Int.  for  9  years  at  Sj^^       -70.00 

Maturity  value,  §770.00 


Investment  of       -        -        -        $500.00 
Int.  ou  same  for  9  yrs.  at  8J'„         StiO.OO 

Value  of  investment,  -        $860.00 


8G0 


500 
770 


$447.6711  cash  val. 
of  bond. 


FIRST  PROOF. 


.  Cash  v.alue  of  bond, 
Maturity  value  of  bond, 

Gain  on  cost  of  bond  in  9  vears, 


$447.67iJ 
770.00 


447.67i?  I  $322.32y 
9      100 


322.32t| 

SECOXD   PROOF. 


*^"o  g^in  on  investment. 


Interest  on  $447.67  for  9  years  at  8  per  cent, 
Cash  value  of  bond,  .... 


Maturity  value  of  bond, 


$322.33 
447.67 

$770.00 


3.     If  in  tlie  above  problem,  interest  <at  0  per  cent  is  allowed  on  the  aimual 
interest  payments,  what  would  be  the  market  price  of  the  bond  to  make  the  invest- 

Aus.  $4S5.34ff. 


ment  equal  to  S  per  ceut  simple  interest  1 

OPERATION. 


Face  of  bond, $500.00 

Int.  for  9  yrs.  at  e^o,          -        -         -  270.00 

Int.  on  annual  int.  ($30)  for  36  yrs.  64.80 

Value  of  bond  at  maturity,      -        -  $834.80 


Investment  of, 
Int.  9  yrs.  at  S%, 

Value  of  investment. 


$500.00 
360.00 

,00 


860 


500 
834.80 

$480.3453  val. 
of  bond. 


FIRST  PROOF. 


Market  value  of  bond  as  above, 

Interest  on  same  for  9  years  at  8  per  cent. 

Maturity  value  of  bond,  ... 


$485.35 
349.45 

$834.80 


Cash  value  of  bond, 
Maturity  value  of  investment, 


SECOND  PROOF. 

$485.34Jf 
834.80 


Gain  on  cost  of  bond  in  9  years,        $349.45^ 


485.3411 
9 


349.45^j 
100 


S%  gain. 

4.  If  in  problem  2  above,  annual  interest  at  6  per  cent  is  allowed  on  the 
annual  interest  payments  on  the  bond,  and  annual  interest  at  8  per  cent  is  also 
allowed  on  the  $40  annual  interest  payments  on  the  investment,  what  would  be  the 
market  price  of  the  bond  to  make  the  investment  equal  to  8  per  cent  'i 

Aus.  $438.27—. 

OPERATION 


Face  of  bond,       ^ $500.00 

Interest  for  9  years  at  6  per  cent,        -  270.00 

Int.  ou  annual  int.  ($30)  for  36  yrs.    -  64.80 


Value  of  bond  at  maturity, 
975.20 


$500 
834.80 


$834.80 
investment. 


Investment  of, $500.00 

Int.  for  9  years  at  8  per  cent,      -        -  360.00 

Int.  on  annual  int.  ($40 )  for  36  yrs.  at  8%  115.20 

Val.  of  investment  at  the  close  of  9  yrs.  $975.20 


$428.02  value,  or  market  price  of  bond  on  conditions  named. 


698 


SOULE  S    PHILOSOPHIC    I'KACTICAL    MATHEMATICS. 


PROOF. 

Market  valno  of  liond, $428.02 

Interest  on  same,  for  9  years  at  8  ])('r  cent,      ---.-..---  308.17 
Interest  on  $34.L'4  (wliioh  is  tlie  interest  on  $428.02  for  1  year  at  8  x>er  cent)  for  36  years 

(tho  series  of  1  to  8  inclusive)  at  8  jier  cent,          --....-.  98.61 

Maturity  value  of  bond,            .............  $834.80 


PRACTICAL  MISCELLANEOUS  EXAMPLES  IN  STOCKS  AND  BONDS. 

1269.     1.     When  tlie  aiimial  dividend  of  stock  is  13  jier  cent,  and  the  interest 
of  money  is  10  per  cent,  at  what  iireniium  ought  the  stock  to  sell  1         Aiis.  205^. 

OPERATION   INDICATED. 


10 


100 
12 

$120  —  100  =  20j'o'    Ana. 


2.  What  would  be  my  income  by  investing  $4100  in  G  i)er  cent  .stock  at  82 
percent?  Ans.  $300. 

3.  I  bought  stock  at  6  per  cent  below  par,  and  sold  the  same  at  2J  per  cent 
above  par.  Paid  brokerage  J  per  cent  for  buying  and  J  per  cent  for  selling.  My 
gain  in  the  transaction  was  $750,  What  was  the  amount  of  my  investment,  includ- 
ing brokerage  for  buying,  and  what  was  the  brokerage  for  buying  and  selling? 

Aus.  $9450  investment.    $100  brokerage. 


OPERATION. 

6  %■  gain  on  pnreliase. 

2i"o  gain  on  sale. 

8i%  total  gain. 

1  %■  =  i%  Ijrok.  for  buying  and  i^^  for  selling. 

7^0  let  gain. 


15 


100 
2 

750 


$10000     par  of  stock. 
6,".f  discount. 


1800.00    discount. 
$10000  —  $600  =  $9400  +  $50  brokerage  for  buying  =  $9450  investment. 
Brokerage  on  $10000  at  l^^  for  buying  and  i%  for  selling  : 


4.    A  merchant  sells  6  per  cent  bonds  at  90,  and  invests  the  proceeds  in  8's 
at  par.    How  much  per  cent  does  he  increase  his  income?  Ans.  28%. 


$100    e^if  bonds. 
6 

$6.00  income  on  6"^  bonds. 


OPERATION. 

$96  8%  bonds. 
8 

$7.08  income  on  8%  bonds. 
6.00 


[6.00 


1.68 
100 


1.68  gain. 


PROBLEMS    IN    STOCKS    AND    BONDS. 


699 


5.  A  mercliant  invested  81372.5  in  Louisiana  6's  at  91J  ;  lie  held  tliem  iiutil 
they  rose  to  94,  then  sold  out  and  bought  New  Orleans  City  7  per  cent  bonds  at 
70i.     How  much  did  he  increase  his  annual  income  1!  Aus.  $500, 


OPERATION. 


183 


100 

2 

13725 

115000 


$15000 
94 

114100.00 


I    100 
141  I  2 

14100 


$20000 
7? 


1900.00  $1400.00     —$900  =  $500    Aus. 

The  free  Banking  Law  of  Louisiana  requires  that  the  stocks  deposited  Tvith  the  Auditor  of 
Public  Accounts  as  security  for  bank  note  circulation  "shall  always  be,  or  be  made  equal  to  stock 
of  the  State  bearing  not  less  that  6  per  cent  interest."  The  Auditor  is  prohibited  from  receiving 
"stock  or  bonds  at  a  rate  above  their  par  value,  or  above  their  market  value." 

6.    What  amount  of  circulating  uotes  could  a  bank  receive  on  stock  that 
produces  but  5  per  cent?  Aus.  S3J%  of  the  par  value  of  the  stock  deposited. 

OPERATION. 

$ 

I  100 
6    5 


r  =  83L°„''    Ans. 

7.  Suppose  in  the  above  example  that  the  stock  or  bonds  deposited  had  been 
7  per  cent,  and  the  market  value  00  per  cent,  Mhat  amount  of  circulating  notes 
would  the  bank  have  received  ?  Aus.  105%  of  par  value  of  stock. 


$100      par  value. 
90^0' 


OPERATION. 


$90.00 
'-0 

$6.30 


market  value. 


interest. 


100 
6.30 

$105  =  105;^'    Ans. 


8.     Suppose  again  that  the  stock  or  bonds  deposited  had  been  8  per  cent,  and 


the  market  value  110,  what  amount  of 
received  I 

OPERATION. 


circulating  notes  would  the  bank  have 
Ans.  133  J  %  of  i)ar  value  of  stock. 


100 


6    8 


I  S133i  =  1334,%-"    Ans. 

9.  In  the  last  question,  had  the  Auditor  been  allowed  to  receive  bonds  at 
their  market  value,  which  would  have  been  a  veiy  impolitic  transaction,  Mhat 
amount  of  circulating  notes  would  the  bank  have  received  1 

Ans.  110g%  of  par  value  of  stock. 

OPERATION. 


10 


par. 

=  10^^  premium. 


$110       market  value. 


$8.80     interest. 


100 
8.80 

$146|  =  146|%    Ans. 


700 


SOULE  S    I'lIILOSOPHIC    PRACTICAL    MATHEMATICS. 


10.  I  invested  $2598.75  in  4  per  cent  bonds  at  74J,  and  after  receiving  a 
year's  interest  sold  the  bonds.  My  whole  gain,  including  the  year's  Interest,  was 
$315 ;  at  what  price  did  I  sell  the  bonds  1  Ans.  79^%. 


297 


OPERATION. 


100 
4 

2598.75 

$3500       par  v.aluo  of  bonds  bouKlit. 


AVhoIe  gain, 
Interest, 

Gain  on  Rale, 
Cost  of  bonds, 


$315. 
140. 

$175. 
2598.75 


Selling  jirice  of  bonds,         $2773.75 


$140.00      interest. 

$2773.75  -H  $3500  =  $79i  =  79J»o  selling  price  of  $100  par  of  bonds. 

11.  A  merchant  owns  $40000  of  6  per  cent  railroad  first  mortgage  bonds, 
now  selling  at  85  per  cent,  lie  sells  and  invests  the  proceeds  in  C  i)er  cent  bonds, 
at  20  per  cent  discount.    What  will  be  his  increase  of  income  per  year  1 

Ans.  $150. 


TO  FIND  DIVIDENDS  AND  THE  VALUE   OF   STOCK  AFTEE  DECLAR- 
ING CASH  INSTALLMENT,  STOCK,  AND  FORCED  DIVIDENDS. 

1270.  1.  A  capitalist  subscribed  $25000  stock  in  a  bank,  the  capital  stock 
of  which  is  $1,000,000,  but  only  50  per  cent  paid  in.  A  cash  dividend  of  10  i)er  cent 
on  the  par  value  is  declared  and  iiaid.  What  amount  of  money  does  the  capitalist 
receive?  What  would  be  the  market  value  of  the  50  per  cent  paid  stock  to  make 
the  investment  equal  to  12  per  cent  interest  ? 

Ans.  $2500  dividend  received.     $83J  market  value  of  stock. 


FIRST  OPERATION. 


SECOND   OPERATION. 


1  year's  int.     1£ 


[100 
20       one  year's  dir. 


$1661  -^  2  = 


50 
10 

831-  market  value. 


2.  Suppose  in  the  above  example,  that  an  installment  dividend  of  10  per  cent 
had  been  declared  and  credited  to  the  stockholders  on  their  unpaid  stock,  and  that 
the  market  value  of  the  stock  before  the  declaring  of  the  dividend  was  $60,  what 
would  be  the  value  of  the  stock  after  the  dividend  had  been  declared  1 

Ans.  $72. 

OPERATION. 


50 


60 

60 


$72    Ans. 


PROBLEMS    IN    STOCKS    AND    BONDS. 


701 


3.  Suppose  again  tlmt  the  stock  with  50  per  cent  paid,  -was  Avortli  in  tlie 
market  60  per  cent  of  its  par,  and  that  a  stock  dividend  of  10  i)er  cent  be  declared 
and  delivered  to  the  stockholders,  Avhat  would  then  be  the  value  of  the  stock  ? 

Ans.  54-/f%. 

OPERATION. 


55 


60 

50 


110 


154-1^1-    Ans. 


120 

100 


$109,V  value  of  par  —  2  =  $54ft-    Ans. 


4.  Suppose  again  that  the  stock  with  50  per  cent  paid,  was  worth  in  the 
market  CO  per  cent  of  its  par,  and  that  a  forced  dividend  of  10  per  cent  be  declared 
and  paid  to  the  stockholders,  what  would  then  be  the  value  of  the  stock  ? 

Ans.  5i%. 


OPERATION. 


50 


60 

45 

$54  =  54?^    Ans. 


100 


120 

90 


1108  value  of  $100  par  —  2  =  $54  =  54%'. 


5.  Sections  15  and  16  of  the  United  States  National  Banking  law  require 
that  every  banking  association,  after  having  complied  with  the  pi-ovisions  of  the 
Banking  Act,  preliminary  to  the  commencement  of  banking  business  under  its 
provisions,  shall  transfer  and  deliver  to  the  Treasurer  of  the  United  States  any 
United  States  bonds  bearing  interest,  to  an  amount  not  less  than  one-third  of  the 
capital  stock  paid  in,  and  that  the  banking  association  making  such  deposit  shall 
be  entitled  to  receive  from  the  Controller  of  Currency  circulating  notes  equal  in 
amount  to  ninety  per  cent  of  the  current  market  value  of  the  United  States  bonds 
so  transferred  and  delivered,  but  not  exceeding  the  par  value  thereof,  if  bearing 
interest  at  the  rate  of  six  per  cent,  or  of  equivalent  United  States  bonds  bearing  a 
less  rate  of  interest. 

In  accordance  with  this  law  a  bank  deposited  with  the  United  States  Treasurer 
$800000  United  States  6  per  cent  bonds,  the  current  market  value  of  which  was 
]iar.  What  amount  of  circulating  notes  can  the  bank  receive  from  the  United 
States  Controller  of  Currency  ?  Ans.  $720000. 

OPERATION. 

$800000 
soJiT 


6. 


$720000.00    Au8. 
Suppose  in  problem  5,  that  the  current  market  value  of  the  United  States 


6  per  cent  bonds  deposited  with  the  United  States  Treasurer,  had  been  110,  what 
amount  of  circulating  notes  would  have  been  received?  Ans.  $792000. 


OPERATION. 


$110 
$6.60 


market  value, 
interest 


90 
6.60 


9.00  =  99% 


$110 
90 

$99.00  =  99% 


$800000 
$792000.00    Ans. 


702 


SOULE  S    riilLOSOPHIC    PRACTICAL    MATHEMATICS. 


7.  Suppose  iu  problfiu  5  that  tlie  United  States  bouds  deposited  bad  been 
United  States  4i  per  cent,  at  a  current  market  value  of  114,  what  would  Lave  been 
the  amount  of  circulating  notes  received?  Ans.  $615600. 


OPERATION. 


$114 


market  value. 


$5. 13       interest. 


90 
5.13 

$76.95 


1114 
90 

1102.60 
41 


$800000 
76.95 


$615600.0000 


6  )  $461.70 
$76.95 

8.  Suppose  again  in  problem  5,  that  the  United  States  bonds  deposited  had 
been  United  States  5  per  cent  bonds,  the  current  market  value  of  which  was  84  per 
cent,  what  would  have  been  the  amount  of  circulating  notes  received  ? 

Aus.  $504000. 

OPERATION. 

84   market  value. 
5"^   rate  of  interest. 


$4.20   interest. 


90 
4.20 


3.00  =  63;^' 


84 
90 

7560 
5 

6  )  37800 
$63.00  : 


90 

84 
5 


$800000 
63^ 

$504000 


Ans. 


63% 


1.00  =  63% 


9.  Suppose  again  in  problem  5,  that  the  $800000  United  States  bonds  depos- 
ited had  been  United  States  6  per  cent  bonds,  the  current  market  value  of  which 
was  $113i,  what  would  have  been  the  amount  of  circulating  notes  received  ? 

Ans.  $800000. 


OPBRATION. 

$113J      market  value.  $ 

6% 


$6.81      interest. 


90 
8.81 


:.15  =  102jife% 


Explanation. — In  this  problem  ■we  find  that 
the  6  i>er  cent  on  the  market  value  of  the 
bonds  is  equal  to  6i"j\,  per  cent  on  the  par,  and 
hence,  if  6  per  cent  par  bonds  give  90  per 
cent  issue  of  circulating  notes,  &f,f,s  per  cent 
will  give  102-,-,nj  per  cent  is-sue;  but  in  aecor- 


d.ance  with  the  law  governing  the  case,  as  shown  in  the  above  extract,  the  Controller  is  allowed  to 
issue  only  100  per  cent  of  circulating  notes  on  the  par  of  the  bonds  deposited. 

10.  A  merchant  subscribed  $10000  in  a  banking  incorporation,  the  paid  up 
capital  of  which  was  $400000.  A  cash  dividend  of  4  per  cent  on  the  par  value  of 
the  stock  was  declared.  The  paid  up  capital  being  40  per  cent  of  the  capital  stock, 
what  per  cent  gain  do«s  he  receive  on  his  investment,  and  what  i.s  tlie  capital  of  the 
bank?  Ans.  10%  gain.    $1,000,000  capital. 


*  PROBLEMS    IN    STOCKS    AND    BONDS.  7^3 

TO   COXSOLIDATE   COEPORATIONS,   WATER   THE    STOCK   AND    FIND 
THE  MARKET  VALUE  OF  THE  WATERED  STOCK. 

1271.  1.  Two  railroad  companies,  A.  with  a  capital  of  $25,000,000,  and  X. 
with  a  capital  of  $15,000,0(10  consolidated  nnder  tlie  corporate  name  of  A.  &  X.  R. 
B.  Co.    New  stock  was  issued  to  redeem  tlie  old  in  the  following  manner: 

(1).    For  every  four  shares  of  the  stock  of  the  A.  road,  five  shares  were  issued. 
(2),    The  stock  of  the  X.  road  was  watered  20  per  cent. 

What  was  the  capital  stock  of  the  Consolidated  Co.,  and  how  many  shares  oi 
watered  stock  were  added  ?       Ans.  $49,250,000  capital  stock  of  Consolidated  Co, 

92,500  shares  watered  stock. 

OPERATION. 

250000  =  shares  of  Co.  A.  150000  =  shares  of  Co.  X. 

62500  =  shares  of  watered  stock.  30000  =  20%  watered  stock. 


312500  =  shares  issned.  180000  =  shares  issued. 

180000 


492500  =  shares  at  $100  =  $49250000  capital  of  Consolidated  Co. 

62500  +  30000  =  92500  shares  -watered  stock. 
2.     In  the  above  problem,  suppose  the  market  value  of  the  stock  of  the  A. 
road  to  have  been  before  the  consolidation,  120,  and  that  of  the  X.  road  to  have  been 
85,  what  should  a  share  of  the  new  consolidated  company  be  worth,  in  the  market  ? 

FIRST  OPERATION. 

1  share  $100  of  pure  stock  of  Co.  A.  =  120  in  the  market. 

li  shares  S125  of  watered  stock  of  Co.  A.  =  120  in  the  market. 

Hence  ^'°  ^  ^°°  =  $96  the  value  of  1  share  watered  stock  of  Co.  A. 

125 

I  share  $100  of  pure  stock  of  Co.  X  =  85  in  the  market. 

II  shares  S120  of  watered  stock  of  Co.  X.  =  85  in  the  market. 
Hence  55-^-^  =  S70|  the  value  of  1  share  watered  stock  of  Co.  X. 

312500  shares  @  §96  =  §30,000,000 
180000  shares  @  S70J  =  812,750,000 
492500  )  42,750,000  (  $86.80  +  = 

Value  of  one  shai-e  of  consolidated  stock. 


SECOND  OPERATION. 


120 
25000000 


825,000,000  =  Capital  of  Co.  A.  ,.  o-q  qqq 
6,250,000  =  watered  stock  of  Co.  A.  ''     ' 

$31,250,000  =  new  capital  of  Co.  A. 

$15,000,000  =  Capital  of  Co.  X.  ^s  ^nn  nnn  I  '^■^ 

3,000,000  =  watered  stock  of  Co.  X.  18,0""."^  |  1500OOOO 


$96  value  of  one  share. 


$18,000,000  =  new  capital  of  Co.  X.  I  $70|  value  of  one  share. 

312500  shares  @  $96  =  $30,000,000 
180000  shares  @  $70J  =  $12,750,000 
492500  )  $42,750,000  (  $86. SO  +  = 

Value  of  each  share  of  consolidated  stock. 
3.    The  paid  up  capital  stock  of  a  newly  incorporated  company  is  $300000. 
The  business  of  the  company  gives  evidence  of  very  large  profits,  and  as  a  conse- 
quence the  market  value  of  the  stock  is  high,  say  $220  per  share,  and  the  demand 
greater  than  the  supply.    The  stockholders  thereibre,  in  order  to  supply  this  demand 


704 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


and  enrich  themselves,  clandestinely  resolve  to  ^^  water  the  stoek^^  to  the  extent  of 
$-'00000.  Accordingly  they  issue  and  divide  pro  rata  among  the  stockholders 
$L'0()000  new  stock,  tlius  increasing  the  capital  to  $500000.  What  would  be  the 
market  value  of  the  stock  after  it  was  thus  "  watered"  1  Aus.  shaa^. 


OPERATION. 


$100    original  stock  worth  $220. 
66|  watered  stock. 


$166f  present  stock. 


$300000  original  stock. 
200UU0   watered  stock. 


$500000  present  stock. 


I  220 
IGOJ  I  100 


I  220 
500000  300000 


$132  Ans. 


I  $132   Ans. 

4.  The  capital  stock  of  a  corporation  consists  of  $000000  regular,  $200000 
prefen-ed,  and  $200000  guaranteed  stock.  A  semi-annual  dividend  of  3  per  cent  is 
declared  on  the  regular  stock,  5  per  cent  on  the  preferred  stock,  and  G  i)er  cent  oa 
the  guaranteed  stock.  $7502.10  were  carried  to  reserve  fund.  What  was  the  net 
gain  of  the  comiJauy  during  the  six  months  business?  Ans.  $47562.10. 


TO  FIND  THE  MAEGIN  TO  CARET  STOCK. 

1272.  1.  A  speculator  deposited  $7000  margin  with  a  broker  and  instructed 
him  to  purchase  500  shares  of  X.  insurance  stock  and  150  shares  of  X.  bank  stock. 
The  margin  is  to  be  kept  good  at  10  per  cent.  The  broker  purchased  at  once  the  500 
shares  of  insurance  stock  at  90,  and  the  150  shares  of  bank  stock  at  110,  At  the 
close  of  six  days,  the  insurance  stock  had  depreciated  4  points  and  the  bank  stock 
had  advanced  2  points.  What  additional  sum  must  be  deposited  with  the  broker 
to  maintain  the  margin  on  market  value  at  10  per  cent.,  allowing  }£  per  cent,  coui- 
mission  for  buying,  ^  per  cent,  for  selling,  and  6  per  cent,  interest  ?    Ans.  $1579.66. 


Dr. 


operation. 

Speculator. 


Or. 


To  500  shares  at  90 
"   150  shares  at  110 
"  Commission  on  650 

shares  ®'J  per  cent, 
"  Int.  on  161662.50,  6  ds.  at  6%, 
"  Commission  for  selling, 

Total  indebtedness  to  date, 
Credit  as  per  opposite, 

Loss  to  date, 


$45000.00 
16500.00 

162.50 

61.66 

162.50 

$61886.66 
59807.00 

$2079.66 


By  Int.  on  .$7000  margin,  6  ds,  at  G^^ 
"   Value  of  stock  this  date : 
500  shares  at  86  -        .        .        . 

150  shares  at  112        -        .        -        . 


7.00 

$43000.00 
16800.00 

$59807.00 


650  shares  at  $100  par  =  $65000  at  10%  =  $6500,      margin  required. 
;  Loss  to  date.  2079.66  to  be  made  good. 


Margin  required, 
Margin  deposited, 

Additional  margin  required, 


$8579.66 
7000.00 

$1579.66 


Note  1. — In  m.aking  st.atements  of  this  kind,  the  commission  for  selling  is  considered  due  to 
the  broker  on  the  date  the  statenient  is  made.  For  in  case  the  required  additional  margin  was  not 
furnished  to  the  broker,  he  would  Lave  the  right  to  sell  the  stock  at  once  aud  thus  the  commission 
would  be  earned. 

Note  2. — Margin  is  estimated  on  the  par  value,  unless  otherwise  agreed.  The  margin  is  to 
protect  the  broker  in  case  the  stocks  should  decline  iu  value. 


PROBLEMS    IN    STOCKS    AND    BONDS. 


7o5 


ACCOUNTS  CUEEENT  AND  INTEEEST  ACCOUNTS  WITH  BROKEES. 


1273.  June  19,  1895,  Frank  Soule  placed  $5000  as  margin  -n-ith  H.  Lee,  liis 
broker,  and  instructed  liiui  to  purchase  the  following  stocks:  100  shares  of  Crescent 
€ity  E.  E.  stock;  200  shares  of  Metropolitan  Bank  stock,  and  50  shares  of  Sun 
Mutual  Insurance  stock.  June  21st  the  broker  bought  on  account  of  the  above 
■order,  70  shares  Crescent  City  E.  E.  stock  at  105  ;  115  shares  Metropolitan  Bank 
stock  at  165  and  30  shares  Sun  Mutual  Insui'ance  stock  at  112i. 

June  24,  he  bought  30  shares  of  Crescent  City  E.  E.  stock  at  104J ;  85  shares 
of  Metropolitan  Bank  stock  at  168 J  and  20  shares  of  Sun  Mutual  Insurance  stock 
at  111. 

June  25,  pursuant  to  instructions,  the  broker  sold  175  shares  of  Metropolitan 
Bank  stock  at  170^ ;  and  June  28,  he  sold  50  shares  Crescent  City  E.  E.  stock  at  104J. 

July  3,  a  settlement  was  made  with  the  broker.  IMake  the  bill  of  the  broker 
showing  balance  of  account,  allowing  interest  on  margin,  on  commission  and  ou 
pui'chases  and  sales  at  8  per  cent,  and  ^  i)er  cent  broker's  commission  ou  both 
pui'chase  and  sale  of  stocks. 

OPERATION. 

Dr.  Frank  Soule,  in  account  trith  H.  Lee,  to  July  3/95  at  8  per  cent.  Cr, 


Amount. 

Ds. 

lut. 

Amount. 

Ds. 

Int. 

1895 

1895 

Juue 

21 
21 

21 

21 
21 
24 
24 

70  sh.  C.  CR.  E. 
Btk.  ®  105, 

115  sh.  Metropol- 
itan Bk.  stk.  -a 
165, 

30  sh.  Snn  Insnr- 
aucestk.'all2i 

Com.  ou  215sh.'a 
4  °^ 

Int.'on  $29753.75 
12  ds.  -a  8»„', 

30  sh.  C.C.E.E. 
stk.  ®  104i, 

85  sh.   Metropol- 
itan Bk.  stk.  ® 

7350 

18975 

3375 

53 

3135 

00 

00 
00 
75 

00 

12 

79 

34 

June 

19 

25 

28 

Cash  deposited  as 

margin, 
Int.  on  same,  14 

ds.  ®  S%, 
175  sh.  Metropol- 
itan Bk.  stk.  'a) 

170i, 
Int.   on  same,    8 

ds.  ®  8%, 
50  sh.  C.  C.  R.  E. 

stk.  -S  104i, 
Int.   on  same,    5 

ds.  @  8%, 

Tot.  credit  int. 

5000 

29881 
5225 

74 

00 

25 
00 
49 

14 

8 
5 

15 

53 

5 

74 

56 

12 

81 
49 

168i, 

14322 

50 

July 

!1 

By  Balance, 

9459 

SG 

24 

20  sh.  Snn  Insur- 
ance stk.  ®  111, 

2220 

00 

24 

Com  ou  135  sh.  lai 

4^0' 

33 

75 

24 

Int.  on  119711.25 
9  ds.  ®  S%, 

9 

39 

42 

25 

Com.  on  175  sh. 

43 

75 

8 

08 

28 

Com.    50   sh.    ® 
Tot.  debit  int. 

12 
118 

50 
85 
10 

5 

118 

01 

85 

49640 

49640 

10 

Note  1. — Commission  is  due  as 
^m  Baid  time. 

Note  2. — See  Accounts  Current 


soon  as  a  purchase  or  sale  is  made,  and  will  draw  interest 
and  Interest  Acconnts,  in  thia  book,  pages  653  to  661. 


7o6 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


ACCOUNT  CUKRENT  WITH  BIIOKEKS,  WITU  "  SUOKT  SALES." 

1274.  1.  May  13,  L.  C.  Soiile,  deposited  with  liis  broker,  S.  Simon,  $10000  and 
instructed  biui  to  sell  sliort  S.  CO,  on  Lis  areount,  100  shares  Citizen's  Bank  stoek 
and  300  shares  Carrollton  li.  R.  stock.  May  15th,  said  stock  Mas  sold  at  the  follow- 
ing prices:  Citizen's  Bank  stock  at  85;  Carrollton  R.  R.  stock  at  109*.  July  10, 
uotice  was  given  to  the  buyer  and  stock  was  bought  to  "cover"  short  sales  at  the 
following  prices :  Citizen's  Bank  stock  at  81 ;  200  shares  Carrollton  R.  R.  stock  at 
107,  and  100  shares  of  same  at  105 J. 

Make  the  account  showing  balance  due  July  11,  allowing  8  per  cent  interest 
on  margin  and  on  commission,  but  not  on  short  sales,  and  com])ute  the  commissloa 
on  sales  and  purchases  each  at  ^  per  cent?       Ans.  $1245.92  bal.  due  L.  C.  Soul6. 

Account  which  the  student  will  verify : 


Dr.        i.  C.  Soule,  in  account  with  S.  Simon,  to  July  11/95,  at  8  per  cent. 


Cr. 


Amount. 

Ds. 

Int. 

Amount. 

Ds. 

Int 

1895 

1895 

May 

15 

Com.  on  400   sb. 

Int.   on   $100.    57 
ds.  'a)  S%, 

100 

00 

57 

1 

27 

May 

13 
15 

Cash   deposit   as 
margin, 

100   sb.  Citizen's 
Uk.  stk.  '&'  85, 

10000 
8500 

00 
00 

July 

10 
10 
10 

10 

100  sb.  Citizen's 

Bk.  stk.  &  81, 
200  sb.  Carrollton 

R.  R.  stk. -S;  107 
100  sb.  Carrollton 

R.    R.    stk.    'a 

105J, 
Com.  on400sb.'g. 

Int.  on  $4012.5,   1 
day  'a,  8%, 

To  Balance, 

8100 
21400 

10525 
100 

10 
1^45 

00 
00 

00 
00 

19 

1 

8 
10 

92 
19 

July 

15 

11 

300  sb.  Carrollton 
R.    R.    stk.   -a 
109i, 

Int.     on     $10000 
margin  59    ds. 

QOy- 
^X0> 

32850 
131 

00 
11 

11 

59 

131 
131 

11 
U 

41481 

11 

41481 

Note  1. — Brokers  do  not  allow  interest  on  "short  sales,"  as  there  is  no  payment  or  receipt, 
of  money. 

Note.  2. — It  is  the  custom  of  brokers  to  allow  interest  on  cash  sales,  and  to  charge  interest 
on  purchases  and  also  on  the  commission  for  buying  and  selling  regardless  of  the  condition  of  siile. 

Remark. — See  Accounts  Current  and  Interest  Accounts,  pages  653  to  661. 

2.  August  the  1st,  1895,  a  capitalist  placed  in  the  haiuls  of  Z.,  a  broker, 
$25000  for  investment  and  speculation  in  buying  and  selling  gold,  stocks,  and 
bonds,  for  60  days. 

Z.  immediately  purchased  200  shares  of  Texas  Railroad  stock  at  GO,  "b. 
26,"  no  margin  required,  and  sold  "short"  400  shares  of  New  Orleans,  Mobile  and 
Chattanooga  Railroad  stock  at  102J,  "b.  10."  Five  days  after  Z.  called  in  the 
Texas  Railroad  stock  and  sold  it  to  A.  at  61^;  August  5,  he  bought  $20000  of  Jive- 


PROBLEMS    IN    STOCKS    AND    BONDS, 


707 


twenties  at  109^ ;  at  the  same  time  the  party  to  whom  the  New  Orleans,  Mobile  and 
Chattanooga  Eailioatl  stock  was  sold,  called  it  in ;  Z.  paid  the  dilference,  the  stock 
being-  valued  at  lOl'i ;  August  24,  Z.  sold  the  five  twenties  at  llOi  cash,  and  at  the 
same  time  made  a  further  sale  of  $15000,  in  the  same  kind  of  bonds,  at  llOJ,  "  s. 
20, "  which  he  purchases  and  delivers  in  eight  days,  at  lOOJ ;  September  17,  Z.  sold 
$20000  in  gold  at  137J,  "  s.  10,"  which  was  not  delivered  at  the  expiration  of  the 
ten  days,  but  settled  at  137;  Z.  then  bought  "flat"  $25000  City  8's  at  91,  and  sold 
the  same  "flat"  at  95^. 

The  capitalist  and  broker  now  settled. 

By  the  terms  of  the  contract,  Z.  is  to  receive  ^  per  cent  brokerage  on  the  par 
of  actual  purchases  and  sales,  no  interest  is  to  be  allowed.  What  amount  of  money 
is  due  the  capitalist  ?  Ans.  $26375. 

OPERATION. 


Capitalist  in  account  with  Z. 


Dr. 


Cr. 


By  cash  deposited,  $25000.00 

To  200  shares  Texas  R.  R.  stock  at  60, $12000.00 

"  i°o  brokerage  on  .same,  ........  25.00 

By  400  shares  N.  O.  M.  &  Chattanooga  K.  R.  stock  at  102i,  short  h.  10, 
$41000. 
To  i^'o   hrokerage  on  same,  ........  50.00 

By  200  shares  Texas  R.  R.  stock  at  61|, 12375.00 

To  i%  brokerage  on  same,      .........  25.00 

"  $20000  U.  S.  5-20'8  at  1094, 21825.00 

"  4"o  brokerage  011  same,     .........  25.00 

"  loss  on  400  shares  N.  O.  M.  &  Chattanooga  R.  R.  stock,  bought 

at  102i  and  settled  at  102J,  $41150, 150.00 

By  $20000  IT.  S.  5-20'8  at  llOJ, 22050.00 

To  ^t  "J' brokerage  on  same,  ........  25.00 

By  $15000'  U.  S.  5-20'8  at  llOJ, 16518.75 

To  !?„'  brokerage  on  same,  ........  18.75 

"  $15000  U.  S.  5-20's  at  109|, 16481.25 

"  i^o  hrokerage  on  same,     ...-.•-..  18.75 

By  $20000  gold,  s.  10,  at  137i,  $27500. 

To  i%  brokerage  on  same,  ........  25.00 

By  gain  on  $20000  gold,  bought  at  137J  and  settled  at  137  ($27400),  -  100.00 

To  $2.5000  City  Warrants  at  91, 227.50.00 

"  i%  hrokerage  on  same,     .........  31.25 

By  $25000  City  Warrants  at  95^, 23812.50 

To  i%  brokerage  on  same,  .........  31.25 

Capitalist's  total  debits,  ........        $73481.25         

Capitalist's  total  credits, $99856.25 

73481.25 

Amount  due  the  capitalist,       -.«.....  $26375.00 

UNITED  STATES  STOCKS  AND  BONDS  IN  EUROPE. 


1275.     The  principal  markets  in  Europe,  for  United  States  stocks  and  bonds, 
are  London,  Paris,  Hamburg,  and  Frankfort. 

The  Euroi^ean  quotations  are  the  prices  of  $100  bonds  or  stocks  in  U.  S.  gold 
but  not  at  the  U.  S.  valuation,  as  will  be  seen  in  the  two  following  articles.  The 
accrued  interest  on  the  stock  or  bond  at  the  date  of  purchase  generally  constitutes 


7o8  soule's  rniLosuPHic  i-kactical  mathematics.  * 

a  part  of  its  value.     In  some  places  however  they  are  sold  "flat"  and  then  tli« 
accrued  iuterest  must  be  added. 

The  London  quotations  of  United  States  stocks  and  bonds  are  based  upon  an 
assumed  value  of  the  pound  sterling.  This  assumed  value  is  $."»,  inoditied  by  the 
rate  of  exchange.  Hence  to  find  the  actual  value  of  English  quotations  of  Ameri- 
can Securities,  we  must  multiply  the  quoted  price  by  the  rate  of  exchange.  Thus, 
if  Illinois  Central  11.  11.  stock  is  quoted  iu  London  at  103^  and  the  rate  of  exchange 
is  4.87 J  we  find  the  actual  value  of  the  quotation  by  the  following  operation : 

103.75  X  4.875 
5 

In  Paris  at  the  Bourse,  the  quotations  are  based  upon  the  conventional  value 
of  5  francs  for  $1,  while  the  actual  value  of  a  dollar  iu  exchange  transactions  is 
more  than  5  francs.  Hence  to  find  the  true  American  value  of  the  Paris  Bourse 
quotations,  we  must  deduct  therefrom  the  exchange  rate. 

The  German  price  of  American  Securities  is  the  marks  given  for  $100  par  of 
American  stocks  or  bonds.  The  rate  of  German  exchange  is  based,  by  American 
bankers,  upon  the  equivalent  value  of  4  marks  expressed  in  dollars  and  cents.  The 
exchange  par  of  4  marks  is  93^/;  the  rate  is  95|  more  or  less  according  as  premium 
is  charged  or  discount  allowed.     See  German  Exchange, 

To  ascertain  the  value  of  bonds  for  remittance  and  sale  in  Europe,  bankers 
base  their  comx)utations  upon  the  buying  rate  of  exchange. 

To  ascertain  the  net  i^roeeeds  of  American  stocks  or  bonds  sold  in  Europe, 
allowances  must  be  made  for  cost  of  remittance,  insurance  and  commission  |  and  if 
bills  of  exchange  are  drawn  against  the  net  proceeds,  allowance  must  also  be  made 
for  the  selling  rate  of  exchange. 


TO  FIND  THE  PRICE  OF  STOCKS  IE"  THE  FOTTED  STATES,  WHEN  THE 
ENGLISH  QUOTATION  AND  THE  EATE  OF  EXCHANGE  AEE  GIVEN. 

PROBLEMS. 

1276.     1.    Chicago,  Milwaukee  and  St.  Paul  Railway  stock  is  quoted  in  London 
at  63 J  and  the  rate  of  exchange  is  4.SCJ.     What  is  the  true  value  of  the  stock  ? 

Ans.  $62.03. 

OPEBATIOK    INDICATED. 


63.75  4 

4.865  or,  2 

5 


255  63.75  x  4.865 

973  or,         

5 


2.    Louisiana  4's  are  quoted  in  London  at  96J,  exchange  is  4.85J.    What  is 
the  equivalent  American  quotation  ?  Ans.  $94.02. 


*  PROBLEMS    IN    STOCKS    AND    BONDS.  709 

TO  FIND  THE   PEICE   OF   UNITED   STATES    SEOUEITIES   IN   LONDON, 

WHEN  THE  AMERICAN  QUOTATION  AND  THE  BATE  OF 

EXCHANGE  ARE  GIVEN. 

PROBLEMS. 

1277.     1.    New  York  Central  and  Hudson  River  Railway  stock  is  quoted  in 
Sgw  York  at  112J,  exchange  is  4.86§.    What  is  tlie  equivalent  price  in  London  ? 

Ans.  $115.20+. 

OPERATION    INDICATED. 


4.86625 


112.125          112.125  X  5.00 
5.00       or,    ■ 


4.86625 

2.  When  tbe  Soatfeern  Pacific  1st  mortgage  6  jier  cent  bonds  are  quoted  in 
New  Orleans  at  121|j  escbaage  feeing  4.87f,  what  should  be  the  equivalent  London 
quotation!  Ans.  $124.29+. 

©IPEKATIOSl    IKBICATED. 

I  121.25         121.25  X  5.00 
a.877S  !  5.00      or. — 


4,8775 

OPERATIONS  IN  FRENCH  AND  AMERICAN  QUOTATIONS. 

PROBLEMS. 

1278o  1.  If  the  Paris  Bourse  quotations  of  United  States  4  per  cent  bonds 
are  9SJ,  exchange  5.15,  what  would  be  the  equivalent  value  in  the  United  States 
market?  Ans.  $95.63+. 

OPERATtOS. 

The  $98i  Paris  Bourse  quotation  or  price  being  based  upon  the  conventional 
value  of  5  francs  for  $1,  which  is  too  high  a  value,  it  must  therefore  be  reduced  in 
the  ratio  of  5.15  francs  (the  exchange  value  of  $1)  for  5  francs. 


.Statement  to  thus  reduce  it. 
$ 
1 
5.15 


1                                    5  X  98.50 
5.00  or,         -T 


98.50  5.15 


$95,631  XJ.  S.  lu.irket  value  of  Bourse  quotation. 

2.    If  the  Paris  Bourse  quotation  of  United  States  bonds  is  1063,  exchange 
5.64J,  what  would  be  the  equivalent  value  in  the  United  States  market  1 

Ans.  $94,552+. 

OPERATION    INDICATED. 


11.29 

4 


1  106.75  X  5.00 

2  or. '■ 


5.00  5.645 

427 


$94,552+  U.  S.  marlcet  value. 

See  Foreign  Exchange  in  this  work  for  Extended  Operations  in  Exchange. 


1379.  Tlie  term  Exeliange  lias  various  significations  or  meanings,  according 
to  the  sense  or  manner  in  Avbich  it  is  used.  Tlie  general  acceptation  of  the  term  is 
the  act  of  giving  or  receiving  one  thing  for  another.  It  is  also  used  to  designate 
the  place  where  merchants,  brokers,  or  other  classes  of  business  men  assemble  to 
transact  business  at  certain  hoiu's  as  the  Cotton  Exchange,  the  Produce  Exchange, 
etc.  When  used  in  this  sense,  the  term  is  ofteu  contracted  into  "  change  " ;  as  at 
change  or  on  change,  meaning  at  the  exchange. 

Exchange,  when  the  term  is  used  by  commercial  men  and  bankers  in  con- 
nection with  transactions  of  finance,  signifies  the  method  or  system  of  paying  debts 
in  distant  cities  by  means  of  bills  op  exchange,  without  the  shipment  or  trans- 
mission of  money.    In  this  sense  we  here  use  the  term,  and  now  treat  the  subject. 

To  elucidate  the  Trorkings  and  to  eho-w  the  J)ractical  utility  of  exchange,  let  us  suppose,  1. 
That  the  sugar  and  cotton  merchants  of  New  Orleans  ship  theix  s^gar  and  cotton  to  New  York  for 
sale,  and  receive  in  payment  currency  or  coin,  which  is  forwarded  to  them  hy  express.  2.  Suppose 
that  the  dry  goods  and  grocery  merchants  of  JJew  York  ship  their  goods  to  New  Orleans  for  sale, 
and  receive  iu  payment  currency  or  coin,  which  is  forwarded  to  them  by  exjircss.  Now  it  is  clear 
that  if  the  sugar  and  cotton  purchasers  in  Kew  Y'oxk  had  paid  the  grocery  and  dry  goods  merchants 
there,  and  the  grocery  and  dry  goods  purchasers  in  New  Orleans  had  paid  the  sugar  and  cotton 
merchants  there,  the.  whole  business  smight  have  been  settled  without  the  expense  and  risk  that 
attend  the  shipping  of  currency  or  coin  from  one  place  to  the  other..  This  would  he  accomplished 
by  tho  sugar  and  cotton  merchants  drawing  drafts  or  bills  of  exchange  on  the  sugar  and  cotton 
buyers  in  New  Y'ork  in  favor  of  the  grocery  and  dry  goods  buyers  of  New  Orleans,  who  having 
bought  and  paid  for  these  drafts,  could  remit  them  to  the  grocery  and  dry  goods  sellers  in  New 
York,  in  pajmcnt  of  their  purchases.  On  presentation  to  the  sugar  and  cotton  buyers,  these  drafts 
would  be  paid,  and  in  this  manner  the  debts  due  from  New  Y'ork  to  New  Orleans  and  the  debts  due 
from  New  Orleans  to  New  Y'ork  are  both  paid  without  the  double  shipment  or  transmission  of 
currency  or  coin  from  each  place  to  the  other 

It  is  evident  that  to  make  this  system  general,  and  to  include  «ll  Kinds  of  trade  in  various 
places,  it  would  frequently  be  -very  difScult  for  those  having  bills  of  exchange  to  sell,  to  lind 
purchasers,  and  vice  versa.  To  obviate  this  difficulty,  exchange  brokers  and  exchange  banks  are 
established. 

Thej'  buy  bills  with  their  own  ca|)ital  from  tnose  who  nave  money  to  receive,  and  send  them 
forward  for  credit,  and  then  sell  their  own  bills  drawu  against  this  credit  in  such  amounts  as 
purchasers  may  require. 

The  exchange  broker  and  exchange  bank  continue  the  purchase  and  sale  of  bills  of  exchange 
in  this  manner,  charging  a  commission  only  for  their  services,  use  of  capital,  and  risk  incurred  in 
buying  bills  that  might  not  be  honored  when  presented,  until  the  demand  to  purchase  bills  becomes 
greater  than  the  supply  of  bills  offered  for  sale  or  rice  versa.  Then  they  either  sell  their  bills  at  a 
premium  or  par,  and  buy  at  par  or  discount. 

When  the  demand  is  in  excess  of  the  supply,  they  will  charge  not  only  a  commission,  but  a 
premium  sufficient  to  cover  the  expense  of  freight  and  insurance  to  ship  the  funds  to  meet  their 

(710) 


*  EXCHANGE.  7 II 

bills  at  maturity,  or  to  pay  the  interest  on  the  amount  of  their  over-drafts  until  the  commerce  of 
the  two  places  shall  have  adjusted  ths  difference.  They  -n-ould  also,  iu  this  case,  buy  at  a  premium. 
When  the  supplj-  is  in  excess  of  the  demand,  they  will  sell  at  a  discount  sufficient  to  cover  the 
expenses  of  shipping  their  funds  from  the  place  where  the  drafts  are  payable  to  their  own  place  of 
business. 

The  principal  cause  that  produces  the  excess  of  supply  or  demand  of  bills  of  exchange,  to 
which  we  have  here  made  mention,  is  the 

BALANCE  OF  TEADE. 

12S0.     Over-trading,  or  tlie  tjalance  of  trade,  is  in  itself  a  very  important 
subjei-t,  and  is  worthy  of  tbe  profoiiude.st  cousideratiou  by  business  men. 

In  this  connection  we  do  not  propose  to  discuss  tlie  subject,  but  for  the  infor- 
mation of  students  we  will  briefly  define  it.  By  tbe  "balance  of  TKADE"  is 
meant  tbe  monetary  excess  of  business  transactions  tliat  one  country  or  city  has 
over  another  with  which  she  deals.  For  example,  if  the  United  States  sells  more 
to  Enjiland  than  England  sells  to  the  United  States,  then  the  balance  of  trade  is  in 
favor  of  the  United  States.  If  the  reverse  is  the  case,  then  the  balance  of  trade  i.s 
against  the  United  States.  Or,  if  New  Orleans  sells  more  to  St.  Louis  than  she 
buys  from  St.  Louis,  then  the  balance  of  trade  is  in  her  favor;  but  if  the  reverse  is 
the  case,  then  the  balance  of  trade  is  iu  favor  of  St.  Louis. 

1281.  OEIGIN  OF  EXCHANGE. 

The  origin  of  negotiable  bills  of  exchange  has  been  a  subject  of  cliscussion  on  the  part  of 
legal  and  commercial  writers  for  ages,  and  it  is  yet  undecided  in  what  age  and  among  what  nation 
they  were  first  used.  Some  claim  that  they  were  known  among  the  Romans  and  Grecians,  but 
there  is  no  proof  to  substantiate  the  claim.  It  is  granted  however,  on  good  authority,  that  such  a 
transaction  as  a  request  by  A.,  in  Rome,  to  B.,  in  Alexandria,  to  pay  to  C.  on  his,  A's  account,  the 
money  that  B.  owes  A.,  was  of  common  occurrence ;  but  a  request  of  this  character,  and  a  com- 
pliance therewith,  do  not  constitute  a  bill  of  exchange. 

In  1272  bills  of  exchange  were  in  use  in  Venice,  but  there  is  no  evidence  to  show  that  they 
were  then  negotiable.  There  are,  however,  records  of  them  at  an  earlier  date,  and  various  theories 
have  been  presented,  on  questionable  evidence,  to  account  for  their  origin.  The  French  writers 
ascribe  their  origin  to  the  Jews  in  1181,  when  oppressively  and  wicliedly  exjielled  from  their 
homes,  in  the  kingdom  of  France,  by  Philip  Augustus.  But  this  date  and  circumstance,  or  cause  of 
origin,  is  questioned  by  other  writers.  Mr,  Eeddie,  in  his  Historical  View  of  the  Law  of  Maritime 
Commerce,  says,  "that  if  we  inquire  what  cause  has  led  to  the  invention,  the  true  answer  is,  the 
necessities  of  commerce  ;  hut  that  if  we  inquire  who  were  the  inventors,  in  what  position  and  by 
whom  these  necessities  were  most  strongly  felt,  and  what  persons,  experiencing  the  urgency  of 
these  necessities  in  the  most  lively  manner,  produced  the  thing  invented,  it  would  be  absurd  to 
call  the  extension  of  commerce  the  inventor,  for  this  would  be  to  confound  the  mover  with  the 
agent.  It  is  also  highly  probable  that  the  Jews,  being  in  these  ages,  as  we  h.ave  seen,  the  chief 
money  lenders,  persecuted  from  mistaken  religious  views  and  on  account  of  their  alleged  pecuniary 
extortions,  scattered  over  the  European  kingdoms,  yet  in  a  manner  forced  to  keep  up  a  pretty  con- 
stant communication  with  each  other,  clever  and  acute  naturally,  and  comparatively  skillful  in 
such  business  from  having  been  trained  to  it  for  generations,  were  really  the  first  inventors  of  biJIs 
of  exchange  in  a  rude  state.  "  But  he  concludes  that  there  is  no  evidence  that  the}'  made  the 
invention  at  the  precise  time  and  by  reason  of  their  expulsion  from  France. 


7i2  soule's  philosophic  practical  mathematics.  * 

1282.  THE  IMPORTANCE  OF  EXCHANGE. 

Money  ■was  the  first  great  invention  by  or  through  ■nhich  coninicrcial  transactions  were 
extended  and  facilitated. 

It  answered,  in  the  language  of  Mr.  Parson,  in  his  treatise  on  Notes  and  Bills,  "all  the 
purposes  for  which  it  was  ^Yanted,  for  many  ages.  It  satisfied  all  the  re(iuireBicut8  of  social  life, 
and  of  commerce,  through  the  early  Eastern  empires,  and  those  of  Greece  and  Kome.  It  is  said, 
hut  upon  somewhat  doubtful  authority,  that  some  kind  of  paper  money  was  used  in  Tartary,  or 
China,  or  Jajiau,  a  thousand  years  ago ;  but  nothing  is  known  certaiuly  about  this.  In  Europe, 
gold  and  silver  money  were  the  only  circulating  medium,  and  were  sufficient;  but  five  or  six  hun- 
dred years  ago,  the  discovery  was  made  of  a  new  circulating  medium,  bills  of  exchange,  of  which 
it  is  the  characteristic  quality  that  it  represents  that  which  represents  everything  else. 

"The  use  of  money  enlarged  hum-an  intercourse,  or  so  much  of  it  as  may  be  included  in  the 
widest  sense  of  the  word  commerce.  It  made  interchanges  possible  and  easy,  which  would  other- 
wise have  been  very  difficult,  if  not  impossible.  AVe  cannot  imagine,  for  example,  the  whole 
commerce  of  Greece  and  Kome,  or  a  hundredth  part  of  it,  carried  on  by  actual  barter  of  commodi- 
ties. Precisely  in  the  eame  way,  the  invention  and  use  of  paper  to  represent  money  gave  a  new 
enlargement  to  commercial  intercourse,  and  greatly  increased  its  facilities  and  its  possibilities. 
For  we  could  not  now  suppose  the  commercial  intercourse  between  America  and  Europe,  for 
example,  to  be  carried  on  wholly  by  actual  exchange  of  the  precious  metals,  as  must  be  the  case 
if  bills  and  notes  were  abandoned,  without  a  cost  and  hindrance  which  would  bo  fatal  to  a  very 
large  part  of  it  The  invention  and  nse  of  money  conferred  upon  mankind  the  vast  benefits  which 
have  ever  flowed  therefrom,  because  money  represented  all  other  commodities,  and  for  no  other 
reason  whatever.  He  who  had  any  superfluities  on  hand  was  no  longer  obliged  to  take  the  trouble 
of  storing  them  and  the  risk  of  their  destruction,  or  to  save  them  by  exchanging  them  for  the 
superfluities  of  some  accessible  neighbor,  whether  these  were  precisely  what  he  wanted  or  not. 
For  BOW  he  might  sell  his  superfluities,  and  their  value  was  then  invested  in  somethiug  easily 
preserved,  and  which  could  always  be  exchanged  for  the  very  article  he  wanted,  as  soon  as  he 
found  it  within  his  reach.  But  after  a  time,  this  exchange  was  to  be  made  in  such  quantities,  and 
at  such  distances,  that  it  cost  too  much  in  time  and  trouble  to  be  profitable;  and  here  is  .a  natural 
limit  to  commerce  by  mere  money,  which  seems  to  have  been  reached  by  the  nations  of  Europe 
some  few  centuries  ago. 

"Beyond  this,  therefore,  it  is  plain  that  commerce  could  not  have  grown,  unless  new  facili- 
ties, by  means  of  new  instruments,  had  been  provided  for  it;  and  this  was  done  by  the  invention 
and  use  of  bills  of  exchange.  Just  as,  by  the  help  of  money,  a  hundred  oxen  could  be  exchanged 
for  a  hundred  pieces  of  cloth,  at  distances  which  would  have  made  the  actual  transfer  of  the  oxen 
and  the  cloth  too  onerous  to  be  advantageous,  so  now  commercial  transactions  which  would  have 
required  large  bags  or  boxes  of  money  to  be  sent  back  and  forth  at  great  cost,  both  of  time  and 
money,  and  with  much  trouble  and  some  hazard,  can  be  carried  into  full  effect,  with  equal  prompti- 
tude, safety,  and  facility,  by  exchanging  small  pieces  of  paper.  And  these  two  inventions,  one 
made  at  the  beginning  of  human  society  and  the  other  but  a  few  centuries  since,  are  useful  for 
precisely  the  same  reason:  Money  represents  all  commodities,  and  so  prevents  the  necessity  of  an  actual 
exchatige  of  commodities ;  and  liills  and  notes  represent  money,  and  so  prevent  the  necessity  of  an  actual 
transfer  of  money. 

"Moreover,  as  gold  and  silver  were  first  used  by  weight,  and  it  was  a  distinct  though  very 
speedy  improvement  to  coin  them  into  money,  so  bills  of  exchange  were  first  used  only  for  the 
benefit  of  a  specified  payee,  but  were  soon  perfected  into  the  indispensable  instrument  of  commerce 
which  they  now  are  by  being  made  negotiable.  For  this  adds  an  entirely  new  element  to  their 
utility.  By  means  of  indorsements,  which  may  be  extended  indefinitely,  negotiable  paper  not 
merely  makes  money  in  one  place  or  at  one  time,  but  becomes  money  in  another  place  or  at  another 
time,  without  actual  transfer;  and  not  merely  makes  credits  tho  equivalent  of  money,  but  it 
represents  and  carries  with  it  the  accumulated  credit  of  all  who  become  parties  to  the  paper. " 


EXCHANGE. 


713 


BILLS  OF  EXCHANGE. 

1283.  A  Bill  of  Exchange  is  a  written  order  for  the  payment  of  a  specified 
sum  of  money  unconditionally.  (The  bill  may  be  printed  with  the  exception  of  the 
signature  of  the  drawer,  which  must  be  written). 

Bills  of  exchange  are  divided  into  two   classes,   Domestic  or  Inland,   and 
Foreign, 

A  domestic  or  inland  hill  is  one  drawn  and  made  payable  in  the  same  country. 
A  foreign  hill  is  one  drawn  in  one  country  and  made  payable  in  another. 

Note. — It  is  gcnorally  understood  that  a  bill  dra-nn  in  any  city  and  State  of  the  United 
States  and  made  payable  in  any  other  city  and  State  of  the  United  States,  is  a  domestic  or  inland 
bill;  but  it  lias  been  decided  by  onr  couits,  that  a  bill  draw  n  in  one  State  and  made  payable  in 
another,  is  a  foreif^n  bill,  for  the  reason  that  each  State  is  a  sovereign  government  in  many 
particulars.  But  nothwithstanding  this  legal  decision,  in  commercial  jiarlance,  domestic  exchange 
means  exchange  on  any  of  the  States  and  territories  of  the  United  States. 

For  the  form  of  Bills  of  Exchange,  Drafts,  Notes,  etc.,  and  for  an  explanation  of  the  parties 
to  bills  of  exchange,  and  for  other  forms  of  commercial  instruments  of  -writing,  see  pages  549  to 
557  of  this  book. 


1284. 


SET  OF  rOEEIGX  BILLS  OP  EXCHAls^GE. 


£1800.9.7. 

Ni 

•w  Okleans,  April  22,  1895. 

Sixty  days  after 

sight 

of  this  F 

rst   of  Exchange   (second   and   third 

unpaid),  pay  to  the  order  of  J. 

M.  Bntchee&Co. 

,  Eighteen  Hundred  Pounds, 

Nine  Shillings  and  Seven  Peuc 

.>,  Sterling. 

To  Messrs.  Quealy  & 

Harvey, 

Liverpool,  Eng. 

MEAD  &  "WELSH. 

XISOO.9.7.  Nkw  Okleans,  April  22,  1895. 

Sixty  days  after  sight  of  this  Second  of  Exchange  (first  and  thinl 
unpaid),  pay  to  the  order  of  J.  M.  Eutchee  &.  Co.,  Eighteen  Hundred  Pounds, 
Nine  Shillings  and  Seven  Pence,  Sterling. 

To  Messrs.  Quealy  &  Harvey, 

Liverpool,  Eng.  AIEAD  *  AVELSH. 


£1800.9.7.  New  Orleans,  April  22,  1895. 

Sixty  days  after  sight  of  this  Third  of  Exchange  (first  and  second 
unpaid),  pay  to  the  order  of  J.  M.  Biitchte  &,  Co.,  Eighteen  Hundred  Pounds, 
Nine  Shillings  and  Seven  Pence,  Sterling. 

To  Messrs.  Quealy  &  Harvey,' 

Liverpool,  Eng.  MEAD  &.  AVELSH. 


Note. — Exchange  is  drawn  in  sets  in  order  to  secure  greater  safety  in  transmission  through 
the  mails.  All  the  bills  of  a  set  are  of  the  same  tenor  and  date,  and  are  sent  by  different  mails 
and  \vhen  either  is  received  and  juiid  the  other  or  other.s  become  void.  In  domestic  exchange,  it  is 
a  recent  custom  of  many  banks  to  fill  but  one  exchange,  and  if  that  .should  be  lost,  then  to  fill  the 
duplicate.  In  foreign  exchange,  many  merchants  mail  by  different  Bteamera  two  of  the  set  and 
hold  the  third. 


714 


SOULE  S    rillLOHOPHIC    I'KACTICAL    MATHEMATICS. 


LETTERS  OF  CREDIT. 

1285.  A  Letter  of  Credit  is  a  written  instrument  drawn  by  a  banker, 
instructing  any  of  iseveial  cture.spondent  banks  in  ditierent  cities  and  countries,  to 
pay  to  tlie  holder  sueli  .sums  of  money  as  be  may  wisii,  not  exceeding  the  aiut)unt 
sjjecitic'.l  in  the  Letter  of  Credit,  and  promising  to  le-jjay  the  sums  ,so  advanced  to 
tlie  i)erson  or  persons  advancing  tliem,  or  to  honor  his  or  tlieir  bills  drawn  for  such 
amounts  as  they  may  have  individually  I'urni.slied. 

Note  1. — Letters  of  Credit  are  drawn  in  dift'ereiit  iorniR  nreording  to  the  ideas  of  tbe  parties 
issuing  tliem,  and  tbe  re<iniienien1»  of  persons  ai)i)lyinf;  fo»  tlieni. 

NoTK  2. — A  Letter  of  Credit  difters  from  a  bill  of  exebanfic  in  tbat  it  is  payable  at  different 
places,  at  different  times  and  in  different  sums.  Wbile  a  bill  of  excbauge  is  payable  at  a  certain 
place,  at  a  certain  time  and  for  a  speciticd  amount. 

Letters  op  Credit  are  of  great  convetiience  to  travelers  by  enabling  them 
to  draw  such  sums  at  such  times  and  at  such  iilaces  as  they  may  wish. 

To  procure  a  Letter  of  Credit,  tlie  i>arty  is  required  to  make  a  deposit  of  cash 
or  securities  with  a  banker  who  is  ^jrejiared  to  issue  them,  for  the  amount  of  the 
Letter  of  Credit. 

An  Interest  Account  is  kept  by  the  bank  or  banker  issuing  the  Letter  of  Cre- 
dit with  the  holder  thereof,  and  at  the  close  of  the  journey,  an  account  is  rendered 
and  settlement  made,  of  the  interest  and  of  the  unexpended  balance,  if  any.  The 
interest  allowed  on  the  deposit  varies  from  IJ  to  3  per  cent,  and  the  interest 
charged  on  the  payments  is  the  current  rate. 

When  the  holder  of  a  Letter  of  Credit  wishes  to  draw  any  money,  lie  presents 
the  same  to  any  banker  named  therein  or  on  the  listof  the  drawer's  correspondents, 
who  will  pay  him  the  amount  he  wishes,  in  the  currency  of  the  country  in  which  he 
may  be,  at  the  regular  rate  of  exchange  then  existing,  and  either  take  his  receipt 
or  a  draft  on  the  drawing  bank  for  the  amount.  This  receipt  or  draft  is  remitted 
to  the  drawing  bank  and  is  charged  to  the  account  of  the  payee  or  holder  of  the 
Letter  of  Credit. 

Foreign  bankers  who  make  payment  on  I^etters  of  Credit,  usually  charge  the 
payees  or  holders  a  small  commission,  varying  according  to  amount  jjaid,  from  ^  to 
1  per  cent. 

Form  of  a  Letter  of  Credit. 


Office  of  Haunky,  Cadv  &  Co.,  Biinkers, 
New  Yokk,  July  1,  1895. 


No.  504. 

For  £350. 

To  our  Correspondents  : 

Gentlemen. — We  bave  opened  a  credit  for  Three  Hundred  Fifty  Founds 
Sterling,  in  favor  of  j1/r.  A.  X.  Green,  wbose  name  appears  below  in  bis  own 
band-writing,  and  we  resjiectfully  request  that  you  honor  bis  drafts  on  us,  at 
sight,  for  such  funds  as  be  may  require,  not  exceeding  the  .available  extent  of 
this  letter,  indorsing  all  such  drafts  paid  by  you  on  this  credit,  which  will 
continue  in  force  until  January  1,  1896, 
Very  Respectfully, 

(Signature  of  tbe  holder). 


i^fe 


^    ^-^. 


'lee-n,. 


KoTE. — The  purchaser  of  this  Letter  of  Credit  is  supplied  with  a  list  of  Harney,  Cady  & 
Go's  correspondents  and  their  addresses,  to  whom  be  is  autliorized  to  present  it.  This  list  is  often 
printed  on  the  Letter  of  Credit. 


i 


*  EXCHANGE.  7 1 5 

AMERICAN  EXPEESS  COMPANY  TRAVELERS'  CHECKS. 

1286.  As  an  equivalent  of  the  Bankers'  Letters  of  Credit,  the  American 
Express  Co.  now  issue  travelers'  checks,  payable  in  various  parts  of  nearly  all  the 
countries  of  the  world.  Like  Letters  of  Credit,  these  checks  are  a  great  service  to 
the  traveling  public.  They  may  be  used  by  different  members  of  a  family  or  party, 
are  of  convenient  form,  are  easily  utilized  as  money,  offer  a  safe  and  a  cheap  way 
of  carrying  money  when  traveling  in  our  own  or  in  foreign  countries,  and  they  may 
be  purchased  in  hundreds  of  cities  where  Letters  of  Credit  cannot  be  obtained. 

Form  of  an  Express  Company  Travelers'  Check. 


WHEN  COUNTERSIGNED  BELOW  ^^tSen«ilEspressfJ;>--.  000000 

WITH  THIS  SIGNATURE.  i^'    Trailers Cheqae      ''«»?, 


/S9. 


T 


™  HmmcEH  FijiFess  f  miiipanii 


WiU  pay  to  tbe  Order  of -|J20^. 


^,IB  Cnited  Stat«  and  Canad.-i  S^rj^-^s^^ 

FRANCE     „,_,,, .,,.|    ITALY 

NORWAY 

Sw«.  Dn. 

HOLLAND  X. 

^ 

^^d  Twenty  ^onars.\i\ I  ? 

Fr.  1  Ct.    Mks. 
102 1 60     83 

Pf.|l,.re  Ct. 
soil 03  50 

Kr.  Ore 
73    39 

Fir.   Ct.  fi 
49    03fc 

-• 

COUNTERSIGNED :  (ses  sroiiATon.s  above). 

Tkeas. 

THE  FACE  AXD  CHARACTER   OF  FOREIGN  BILLS. 

1287.  1.  Foreign  Bills  of  Exchange  arc,  witli  rare  exceptions,  drawn  in  the 
monetary  unit  or  money  of  account  of  the  country  where  tbej-  are  made  payable  or 
in  which  the  drawee  resides.     See  Monetary  Units  of  difl'erent  nations. 

2.  Foreign  Bills  of  Exchange  are  accei>ted  and  bear  grace  whether  drawn 
at  sight  or  on  time. 

Bills  are  sometimes  drawn  with  "  acceptance  waived, "  in  which  case  they  are 
not  subject  to  protest  for  non-acceptance,  but  if  not  paid  at  maturity,  would  then  be 
protested  for  non-payment. 

The  most  of  our  foreign  exchange  is  effected  through  a  few  of  the  most 
prominent  mercantile  and  financial  centers  of  Europe,  such  as  Loudon,  Paris, 
Hamburg,  Amsterdam,  Frankfort,  Bremen,  Antwerp,  Vienna,  Berlin,  and  Leipsic. 

In  mercantile  parlance,  we  have  Short  Sight  Bills,  Long  Bills,  Bankers'  Bills, 
Commercial  Bills,  Clear  Bills,  Bill  of  Lading  Bills,  or  Documentary  Exchange,  A 
1  Bills,  and  Prime  Bills. 


7i6  soule's  philosophic  practical  mathematics.  * 

A  Sliort  Sight  Bill  is  a  bill  tlrawii  at  .'J  days  sight.  The.  most  of  English 
bills,  and  also  other  foreign  bills,  are  drawn  at  60  days  sight,  and  are  called  Long 
Bills. 

Bankers'  Bills  are  those  sold  by  bankers. 

Commercial  Bills  are  those  sold  by  merchants. 

Clear  Bills  are  those  tliat  luive  no  collateral  security,  but  are  negotiated 
solely  \\\Mn\  the  lespousibility  of  their  makers  and  indorseis. 

Bill  of  Lading  Bills  or  Documentary  Exchange  are  those  secnred  or 
protected  by  the  assignment  and  translei-  of  the  bill  of  lading,  and  policy  of  insur- 
ance, of  some  merchandise  or  commodity  consigned  for  sale  under  order  at  the 
maturity  and  for  account  of  the  bill. 

Bills  are  sometimes  secured  by  other  collaterals  than  bills  of  lading  and 
policies  of  insurance. 

A  1  Bills  are  either  commercial  or  bankers'  bills  that  are  considered  by 
reason  of  the  wealth,  capacity  and  integrity  of  the  makers  and  drawees,  pei'fectly 
good. 

Prime  Bills  are  generally  regarded  as  synonymous  with  A  1  bills,  but  by 
some  they  are  considered  a  shade  better.  We,  however,  cannot  conceive  /tow  they 
can  be  better  than  A  1,  when  A  I's  are  perfectly  good. 

PAR  OP  EXCHANGE. 

1288.  The  Par  of  Exchange  is  the  established  value  of  the  monetary  unit 
of  one  country  in  that  of  another,  and  it  is  either  intrinsic  or  commercial. 

The  Intrinsic  Par  of  Exchange  is  the  value  of  the  monetary  unit  of  one 
country  expressed  in  that  of  some  other,  and  is  determined  by  comparing  the 
weight  and  purity  of  the  coins  of  different  countries.  Thus  the  intrinsic  par  with 
England  is  now,  1895,  li.SGCS;  with  France  it  is  $.193,  and  with  Germany  it 
is  $.238. 

The  Commercial  Par  of  Exchange  is  the  market  value  of  the  coin  of  one 
country  when  bought  or  sold  in  another;  and  this  value  varies  somewhat  according 
to  the  law  of  supply  and  demand. 

See  English,  French,  German,  and  other  Foreign  Exchange  for  the  quotations 
of  Foreign  Exchange. 

THE  COMMERCIAL  RATE  OR  COURSE  OF  EXCHANGE. 

1289.  The  Rate  of  Exchange  is  the  variable  price  that  is  paid  for  bills. 
There  are  three  conditions  on  which  bills  of  exchange  can  be  bought  and  sold  by 
first  parties ;  they  are,  par,  premium,  or  discount. 

To  understand  fully  the  reasons  that  combine  to  produce  these  three  different 
rates,  requires  careful  study  of  the  whole  subject  of  exchange.  For  domestic 
exchange,  the  following  reasons  are  the  principal  ones  governing  the  rate :  The 
« balance  of  trade"  and  the  probability  of  its  remaining  for  some  time  in  favor  of 


*  EXCHANGE.  717 

or  against  the  city  ■where  the  bills  are  drawn  ;  the  opportunity  or  advantages  pre- 
sented to  the  seller  of  the  exchange  to  use  his  money  in  his  own  city  or  where  the 
exchange  is  made  payable ;  the  rate  of  interest  that  the  seller  has  to  pay  on  his 
over-drafts;  and  the  exjiense  of  shipping  money  to  meet  his  over-drafts,  or  of 
shijiping  his  surplus  funds  in  another  city  to  his  own. 

The  rate  of  exchange  was  formei-ly  based  exclusively  upon  per  cent.  But 
during  the  past  few  years  many  bankers  throughout  the  United  States  base  the  rate 
for  domestic  exchange  upon  the  M.  (one  thousand.)  We  shall  therefore  use  both 
methods  in  this  work. 

BUSINESS   CUSTOMS  TO   BE    OBSERVED    IN   BUYING   AND    SELLING 

EXCHANGE. 

1290.  When  exchange  is  bought  or  sold  through  agents,  it  is  the  custom,  in  the 
absence  of  a  special  contract,  to  allow  the  agent  a  commission  ujion  the  face  of  the 
bill,  j)/«s  the  banker's  premium,  or  minus  the  banker's  discount.  And  when  bought 
through  an  agent  and  a  broker,  to  allow  both  the  commission  and  brokerage  on  the 
face  of  the  bill,  plus  the  banker's  ijremium,  or  minus  his  discount. 

When  money  is  invested  at  a  pi'emium  or  discount,  it  is  the  custom  of  the 
banker,  founded  on  equitable  princijjles,  to  charge  a  premium  or  allow  a  discount 
on  the  face  of  the  bill  that  the  money  will  buy,  instead  of  calculating  the  same  ou 
the  amount  invested. 

PROBLEMS  IN  DOMESTIC  OR  INLAND  EXCHANGE. 

Problems  in  huying  on  our  own  account. 

1291.  1.     What  cost  a  bill  for  $15000,  bought  at  par  ?  Ans.  $15000. 

2.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  premium  ? 

Ans.  $15150. 

3.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  discount  ? 

Ans.  $14850. 

4.  What  cost  a  bill  for  $15000,  bought  at  $3.25  per  M.  premium? 

Ans.  $15048.75. 

OPERATION  FOE  THE  IST  PROBLEM. 
$15000  face  aud  cost  of  bill. 


OPERATION  FOR  THE  2d  PROBLEM. 

$15000  face  of  bill. 
150  =  l^Q  premium. 


OPERATION  FOR  THE  3d  PROBLEM. 

$15000  face  of  bill. 
150  =  1  ^„  discount. 


$14850  cost. 


$15150  cost. 

OPERATION  FOR  THE  4TH  PROBLEM. 

$15000  $15000 


3J  M.  3.25 

45.000  '         $48.75000 

3.75 


or,       $15000  X  1003.25  =  $15048.75000  cost  of  bill. 


$48,750  =  premium. 


7i8 


SOULE  S    nilLOSOPHIC    PRACTICAI.    MATHEMATICS. 

l'rublcm.1  in  huyhuj  through  an  Afjcuf. 


5.  What  cost  a  bill  for  $15000,  bought  at  i)ar,  paying  the  agent  1  per  cent 
commission?  Ans.  $15150. 

G.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  premium,  paying  the 
agent  1  per  cent  commission  ?  Ans.  $15301.50. 

7.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  discount,  paying  the 
agent  1  per  cent  commission  ?  Ans.  $14998.50. 

8.  What  cost  a  bill  for  $15000,  bought  at  $3.25  per  M.  discount,  paying  the 
agent  1  per  cent  commission  ?  Ans.  $15100.76 

OPERATION  FOR  THE  5Tn  PROBLEM. 

$15000  face  of  liill. 

150  =  1°J  commission. 

$15150  cost. 

OPERATION  FOR  THE  6Tn  PROBLEM. 

$15000       face  of  liill. 

150       =  1  °„  jiremium. 


$15150       face  +  premium. 
151.50  =  1  °o  commission. 


OPERATION  FOR  THE  7x11  PROBLEM. 

$15000       face  of  Ijill. 
150       =  1  "„'  discount. 


$14850       face  —  discount. 
148.50  =  1  ^a  commission. 


$15301.50  cost.  I  $14998.50  cost. 

OPERATION  FOR  THE  STH  PROBLEM. 


$15000       face  of  bill 

48.75  =  $3.25  per  M.  disconut. 

$11951.25  face  —  discounii. 


$14951.25 

149.51  =  1  %  commission. 


$15100.76.cost. 


Explanation. — In  solving  questions  of  this  character,  in  conformity  to  business  custom,  we 
allow  the  agent  a  commission  on  the  face  of  the  bill,  plus  the  haulier's  premium  or  minus  the 
banker's  discount. 

Prohlems  in  buying  through  an  Agent  and  a  Broler. 

9.  What  cost  a  bill  for  $15000,  bought  at  par,  paying  the  agent  1  per  cent 
commission  and  the  broker  J  per  cent  brokerage  ?  Ans.  $15225. 

10.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  premium,  paying  the 
agent  1  per  cent  commission  and  the  broker  i  per  cent  brokerage? 

Ans.  $15377.25. 

11.  What  cost  a  bill  for  $15000,  bought  at  1  per  cent  discount,  paying  the 
agent  1  per  cent  commission  and  the  broker  J  per  cent  brokerage  ? 

Ans.  $15072.75. 

12.  What  cost  a  bill  for  $15000,  bought  at  $3.25  per  M.  premium,  paying  the 
agent  1  per  cent  commission  and  the  broker  J  per  cent  brokerage  ? 

Ans.  $15274.49. 


EXCHANGE. 


719 


OPERATION  FOE  THE  !)TII  PROBLEM. 

$15000  face  of  bill. 

150  =  1  J!(|'  commission. 
75  =  i  %  brokerage. 


$15225  cost. 


I 


OPERATION  FOR  TIIK  IOtH  PROBLEM. 

$15000       face  of  bill. 

150       =  1  ^^  premium. 

$15150       face  +  premium. 
151.50  =  1  3'o  commJKsiou. 
75.75  ^  i  "0  brokerage. 


$15377.25  cost. 


OPERATION  FOR  THE  llTIl  PROBI.EM. 

$15000       face  of  bill. 


50       face  —  discount. 
118.50  =  1  "„'  oonimissioH. 
74.25  =  i  %  brokerage 


$15072.75  cost. 


OPEKATION  FOE  THE  IStH  PKOHLEM. 


I 


$15000       face  of  bill. 

48.75  =  $3.25  per  M.  premium. 

$15048.75  =  face  +  premium. 


$15048.75 

150.49  =  1  J'l^  commission. 
75.25  =  i  ^^  brokerage. 


$15274.49  =  cost. 


Explanation. — Where  tbe  purchase  of  exchange  is  made  through  both  an  agent  and  a  broker, 
or  where  a  charge  is  made  for  commission  and  brokerage,  in  conformity  to  custom,  in  the  absence 
of  a  special  agreement  to  the  contrary,  we  allow  both  the  commission  and  brokerage  on  the  face 
of  the  bill,  plus  the  premium  charged  or  minus  the  discount  allowed. 

Problems  in  selling  on  our  own  account. 

13.  What  will  be  realized  for  $15000  sold  at  par'?  Ans.  $15000. 

14.  What  will  be  realized  for  $15000  sold  at  1  per  cent  premium  ? 

Ans.  $15150. 

15.  What  will  be  realized  for  $15000  sold  at  1  per  cent  discount? 

Ans.  $14850. 

16.  What  will  be  realized  for  $15000  sold  at  $3.25  per  M.  premium  ? 

Ans.  $15048.75. 

Note. — For  the  solution  of  the  above  four  problems,  see  the  solution  of  jiroblems  1,  2,  3  and 
4,  page  717. 

Problems  in  selling  through  an  Agent 

17.  What  will  be  realized  for  $15000  sold  at  par,  paying  1  jier  cent  commis- 
sion ?  Ans.  $14850. 

18.  What  will  be  realized  for  $15000  sold  at  1  per  cent  premium,  paying 
1  per  cent  comiiiissioii  ?  Ans.  $14998.50. 

19.  What  will  be  realized  for  $15000  sold  at  1  per  cent  discount,   paying 
1  per  cent  commission?  Ans.  $14701.50. 

20.  What  will  be  realized  for  $15000  sold  at  $3.25  per  M.  discount,  paying 
1  per  cent  commission  ?  Ans.  $14801.74. 


720 


SOULE  S    PH/LOSOI'inC    PRACTICAL    MATHEMATICS, 


OPEKATION  FOR  THE  17TH  PROBLEM. 

115000 

150  =  1  "  '  commission. 


114850  amount  realized. 


OPERATION  FOR  TlIK  18X11  ruOULKM. 


$1.5000 
150 


1  /'„'  premium. 


$15150       face  +  premium. 
151.50  =  1  %  commission. 


$14998.50  amount  realized. 


OPERATION  FOR  TlIF,  19rH  PROBLEM. 


$15000 
1.50 


=  1  "o  discount. 


50       face  —  discount. 
148.50  =:  1  5n  commission. 


$14701.50  amount  realized. 


$15000 

48.75  =  $3.25  per  M.  discoimt, 

$14951.25  face  —  discount. 


OPERATION  FOR  THE  20TH  PROBLEM. 

$14951.25 


149.51  =  1  ?o'  commission. 


$14801.74  :=  amount  realized. 


Problems  in  selling  ihrovgli  an  Agent  and  a  Broker. 

21.  "What  will  be  realized  for  $15000  sold  at  par,  paying  1  per  ceut  commis- 
sion aiid  J  per  cent  brokerage  ?  Ans.  $14775. 

22."  What  will  be  realized  for  $15000  sold  at  1  per  cent  premium,  paying 
1  per  cent  commission  and  i  per  cent  brokerage  f  Ans.  $14922.75. 

23.  What  will  be  realized  for  $15000  sold  at  1  per  cent  discount,  paying 
1  per  cent  commission  and  A  ]ier  cent  brokerage'/  Ans.  $14627.25. 

24.  Vt^hat  will  be  rerUized  for  $15000  sold  at  $3.25  per  M.  premium,  paying 


1  per  cent  commission  and  h  jter  cent  brokerage? 

OPERATION  FOR  THE  21ST  PROBLEM. 

$15000 
1  PjJ'  commission  $1.50  )  q,,- 

4  %  brokerage        75  J  ""'"' 


Ans.  $14823,01. 


Amount  realized 
OPERATION  FOR  THE  22d  PROBLEM. 


1  %  premium, 


1  %  commission,  | 
i  %  brokerage. 

Amount  realized. 


151.50 ) 
75.75  S 


$15000.00 
150.00 

$15150.00 
227.25 

$14922.75 


$14775 
OPERATION  FOR  THE  23d  PROBLEM. 


1  ^^  discount, 


1  %■  commission,  $148.50 
i  %  brokerage,        74.25 

Amount  realized, 


OPERATION  FOE  THE  24TH  PROBLEM. 


$15000 

48.75  =  $3.25  per  M.  premium. 

$15048.75 

225.74  =  commission  and  brokerage. 


1  %  commission  = 
i  %  brokerage  =: 


$15000.00 
150.00 

$14850.00 
222.75 

$14627.25 


$150.49 
75.25 


$225.74 


$14823.01  =  amount  realized. 

Note. — In  the  foregoing  24  problems  we  have  omitted  extended  explanations,  believing  that 
the  operations  alone  will  render  them  easily  understood.     The  student  must  be  careful  to  observe 


k 


*  EXCHANGE.  721 

that  the  agent  and  broker  are  allowed,  ■when  buying  or  selling  exchange,  a  commission  and  brokerage 

oil  the  face  of  the  iill  2>his  the  premium  charged  or  minus  the  discount  alloived. 

We  now  a]>])roaeh  exchange  ^vork  that  is  generally  considered  very  difficult,  and  although 
by  our  ]>hilo8ophic  system  it  is  rendered  comparatively  easy  and  simple,  yet  it  will  require  the 
exercise  of  unimpaired  brains  and  a  remembrance  of  the  business  customs  pertaining  to  exchange 
operations,  to  fully  comprehend  it.  We,  therefore,  here  solicit  the  student's  earnest  attention  and 
elosest  reasoning. 

Problems  in  investing  money  in  Exchange. 

25.  What  amount  of  bill  can  be  bought  for  $15000  at  par?      Ans.  $15000. 

26.  What  amouut  of  bill  can  be  bought  for  $15000  at  1  per  cent  premium  I 

Ans.  $14851.49. 

27.  What  amount  of  bill  can  be  bought  for  $15000  at  1  -per  cent  discount  ? 

Ans.  $15151.52. 

28.  What  amount  of  bill  can  be  bought  for  $15000  at  $3.25  per  M.  premium  ? 

Ans.  $14951.41. 

OPERATION  FOR  THE  26th  PROBLEM.  Explanation. — Since  by  business  custom,   and 

in  this  case  strict  justice,  the  banker  is  allowed 

$100  amount  of  bill  assumed.  a  premium  only  on  the  face  of  the  bill  that  the 

1=1%  premium.  money  will  buy,  after  paying  the  premium,  it  is 

clear  that  we  cannot  compute  the  premium  on 

$101   cash  cost  of  $100  bill.  the  S15000,  for  were  we  to  do  so,  we  would  be 

computing   premium  upon  premium,   which  is 
$  not  allowed.     To  illustrate  this  we  will  make 

I  100  the  figures,  thus:    1  per  cent  of  §15000  is  $150, 

101  I  15000.00  (  $14851.49  -which  deducted  from  the  $15000  leaves  $14850, 

I as  the  face  of  the  bill,   and  as  1  per  cent  of 

I  $14850  is  but  $148.50,  it  is  plain  that  it  is  incor- 

rect to  work  on  the  face  of  the  sum  to  be  invest- 
ed. Having,  Wnce,  no  figures  npon  the  face  of  which  we  can  work,  we  are  obliged  to  assume 
some  number  to  represent  the  amount  or  face  of  the  bill,  and  u])on  which  we  may  work  according 
to  the  conditions  of  the  problem  and  produce  relationship  uumliers  of  face  and  cost  of  bill,  with 
which  and  the  $15000  we  may  make  a  proportional  solution  statement.  We  then  assume  $100  as 
the  face  of  the  bill,  and  compute  and  add  thereto  the  premium  charged,  and  thus  produce  $101  as 
the  cash  cost  of  the  $100  bill.  This  gives  us  the  required  relationship  numbers  with  which  and 
the  $15000  we  make  the  solution  statement. 

In  making  this  statement  we  first  observe  that  as  a  $100  bill  cost  $101  in  cash,  by  transposition 
$101  in  cash  will  buy  a  $100  bill,  and  hence  we  can  buy  as  many  $100  bills  as  $15000  cost  is  equal  to 
$101  cash;  but  to  make  the  operation  clear  and  philosophic  we  place  the  $100  assumed  face  on  the 
statement  line,  and  reason  thus:  If  $101  cash  will  buy  a  $100  bill,  $1  in  cash  will  buy  the  101st 
part,  and  $15000  in  cash  will  buy  15000  times  as  much,  which  is  $14851.49,  answer. 

Note. — In  all  work  of  this  character,  in  actual  practice,  the  assuming  of  the  $100,  and  the 
finding  and  adding  thereto  the  premium,  or  the  deducting  therefrom  the  discount,  should  be  per- 
formed mentally. 

The  foregoing  work  gives  the  only  exact  method  of  solving  problems  of  this  character,  but 
some  accountants  solve  them  approximately  by  first  deducting  the  premium  from  the  amount  to  be 
invested,  then  add  to  the  remainder,  the  premium  on  the  first  premium,  and  thus  continue  the 
alternation  of  subtracting  and  adding  until  the  desired  degree  of  accuracy  is  obtalusd.  To 
illustrate  this  method,  we  will  solve  the  foregoing  problem  by  it. 

OPERATION. 

$15000.00    sum  to  be  invested. 

150.00    =  1  %  premium  deducted. 


$14850.00 

1.50    =:  1  %  on  $150  premium  added. 

$14851..50 

.015  =  1  %  on  $1.50  premium  subtracted. 


$14851.485    Ans. 

In  some  cases,  in  which  the  sum  to  be  invested  and  the  rate  per  cent  premium  or  discount  is 
even  numbers,  this  method  can  be  used  to  advantage,  but  generally  the  exact  method  is  preferable. 


722 


SOULE  S    I'lIILOSOPIIIC    I'RACTICAL,    MATHEMATICS. 


orKUATioN  rou  tiik  Jnii  problem. 


$100 
1 


f;ice  of  l)ill  assumed. 
=  1  ^'i^  discount. 


$99   casli  cost  of  $100  Ijill. 


Explanation. — Agcaiu  liaviiig  no  nnmlier  repre- 
Bentiiig  the  I'ai-o  of  the  bill,  on  ■which  we  can 
"work,  we  therefore  assume  |100  in  order  to  work 
upon  and  produce  tlie  necessary  relationship 
numbers,  as  explained  in  the  preceding  problem, 
to  solve  the  question.  On  this  assumed  $100  w© 
tompute  and  deduct  therefrom  the  1  per  cent 
discount,  and  thus  obtain  $09  as  the  cash  equi- 
valent, or  as  the  cash  cost  of  the  $100  bill.  We 
then  observe  by  the  exercise  of  our  reason  that 
as  a  $100  bill  cost  $99,  by  transposition  $99  cash 
will  buy  $100  bill,  and  hence  we  can  buy  as 
many  $100  bills  of  exchange  as  $15000  are  equal 
to  99.  But  working  more  philosophically,  we  place  the  $100,  assumed  face  of  bill,  on  our  state- 
ment line,  and  reason  thus :  If  $99  cash  will  buy  a  $100  bill  of  exchange,  $1  will  buy  the  99th  l>art, 
and  $15000  will  buy  15000  times  as  much,  which  is  $15151.52,  answer. 

OPERATION  POR  THE  28TH  PROBLEM. 


99 


100 
15000 

$15151.52 


$1000       face  of  bill  assumed. 

3.25  =  $3.25  per  M.  vreminm. 


$1000    =  face  of  bill  .assumed. 
3i  ==  $3.25  per  M.  premium. 


$1003.25  =  cost  of  $1000  bill. 


1000 
1003.25     15000.00 


$1003i 


4013 


1000 
4 

15000 


Ans. 


$14951.41    Ans.  $14951.41 

Frohlems  in  investing  through  an  Agent. 

29.  What  amount  of  bill  can  be  bought  for  $15000  at  par,  paying  the  agent 
1  per  cent  cominissiou?  Ans.  $14851.49. 

30.  What  amount  of  bill  can  be  bought  for  $15000  at  1  per  cent  premium, 
paying  the  agent  1  per  cent  commission  ?  Ans.  $14704.45. 

31.  What  amount  of  bill  can  be  bought  for  $15000  at  1  per  cent  discount, 
paying  the  agent  1  per  cent  commission  ?  Ans.  $15001.50. 

32.  What  ainoiuit  of  bill  can  be  bought  for  $15000  at  $3.25  per  M.  premium, 
paying  the  agent  1  iier  cent  cominissiou  ?  Ans.  $14803.38. 

OPERATION  FOR  THE  29th  PROBLEM.  Explanation. — In  this  problem  the    agent    is 

entitled  to  a  commission  on  the  face  of  the  bill ;. 

$  and  as  there  is  no  ])reminra  or  discount  to  con- 

I  100  sider,  the  operation  and  reasoning  is  the  Banie> 

101  I  15000  (  $14851.49)  as  in  the  21st  jiroblem  of  investing,  where  the 

b.anker  was  entitled  to  a  premium  on  the  face 
of  the  bill. 

FIRST  SOLUTION  OF  THE  30TH  PROBLEM. 


OPEIIATIOK 

To  take  out  the  Agent's  commission. 

$100  amount  invested  assumed. 
1  =  1^^  commission. 

$101  cash  required  to  invest  $100. 

$ 
I  100 
101  I  15000  (  $14851.49  invested). 

$14851.49 

$148.51       commission. 


OPERATION 

To  take  out  the  Banker's  premium  and  find 

face  of  bill. 

$100  face  of  bill  assumed. 
1  =  1  J!„'  premium. 

$101  cash  cost  of  $100  bill. 

$ 
I  100 
101  I  14851.49  (  $14704.45)     Ans. 

$14704.45 
1  "^ 


$147.04        premium. 
Explanation . — As  it  is  the  business  custom,  iu  exchange  transactions,  to  allow  the  ageni  • 


EXCHANGE. 


723 


commission  on  tlio  face  of  the  hill  or  drafts,  plus  the  premium  charged  or  minus  the  discount  allowed  by 
the  banker  or  seller  of  the  exchange,  we  therefore  tirst  take  the  agent's  commission  ont  of  the 
amoiHit  to  be  invested.  To  do  this  we  work  and  reason  the  same  as  wo  did  in  finding  the  face  of 
the  hill  ill  problem  Ko.  Sfi,  under  the  head  of  Investing.  The  result  of  this  statement  is  $14S51.49 
to  bo  invested,  and  $148.51  is  the  agent's  commission. 

Then,  having  tlie  amount  to  be  invested,  we  invest  the  same,  as  shown  by  the  second  state- 
ment, with  the  banker  at  1  ]>er  cent  jiremiuni.  The  reasoning  for  this  second  statement  is  the  same 
as  that  given  in  problem  No.  26,  under  the  head  of  Investing. 

Note.- — By  finding  1  per  cent  on  the  amount  invested,  or  by  finding  the  difference  between 
the  $15000  and  the  amount  invested,  we  have  the  agent's  commission  ;  and  by  finding  1  per  cent  on 
the  face  of  the  bill,  or  by  finding  the  difference  between  the  $14851.49  and  the  face  of  the  bill, 
$14704.45,  we  have  the  banker's  premium. 

To  add  the  agent's  commission  and  the  banker's  premium  together,  and  make  but  one  pro- 
portional statement  thus, 

100 
102     15000 


would  give  an  incorrect  result,  for  the  reason  that  the  agent  is  entitled  to  a  commission  on  the  face 

of  the  bill2}lus  the  banker^s premium, 

SECOND  SOLUTION  OF  THE  30TH  PROBLEM. 

Instead  of  making  two  separate  line  statements,  we  can  state  it  thus: 


101 
101 


10201 


100 

15000 

100 


150000000  (  $14704.45)     Ans. 


FIRST  SOLUTION  OF  THE  31ST  PROBLEM. 


OPERATION 

To  take  out  the  Agent's  commission. 

$100  amount  invested  assumed. 
1  =  1%  commission. 

$101  cash  required  to  invest  $100. 


101 


100 

15000  (  $14851.49) 


Ans. 


OPERATION 

To  find  face  of  bill. 

$100  face  of  bill  assumed. 
I  =  X%  discount. 

$99  cash  cost  of  $100  bill. 

$ 
I  100 
99  I  14851.49  (  $15001.50)     Ans. 


Explanation. — The  operation  of  this  problem  is  so  nearly  the  same  as  that  of  the  preceding 
one,  and  the  reasoning  for  the  statements  being  precisely  the  same,  we  therefore  omit  repeating 
^hem. 


SECOND  SOLUTION  OF  THE  31ST  PROBLEM. 


OPERATION. 


1  100 
101  I  15000 
99  I  100 


9999  I  150000000  (  $15001.50)     Ans. 


Explanation. — To  save  time  and  figures  we 
prefer  the  one  line  statement  in  all  cases  where 
it  can  he  made.  But  in  case  the  agent  takes 
his  commission  out  before  going  to  the  bank  to 
purchase  exchange,  as  he  would  naturally  do, 
then  separate  statements  would  necessarily  have 
to  bo  made. 


101 


100 
15000 


OPERATION  FOR  THE  31 
I  1000 
1003.25  I  14851.49 


D-  PROBLEM. 


$14851.49 


$14803.38 


101 
1003.25 


100 

15000.00 

1000 


724  soule's  philosophic  practical  mathematics.  * 

Prohlcmn  in  investing  through  an  Agent  and  a  BroJcer. 

33.  What  amount  of  bill  can  be  bought  for  $15000  at  par,  paying  the  agent 
1  per  cent  commission  and  the  broker  i  ]>er  cent  brokerage?  Ans.  $14778.33. 

34.  What  amount  of  bill  can  be  bought  for  $15000  at  1  per  cent  premium, 
paying  the  agent  1  jjer  cent  commission  and  the  broker  J  per  cent  brokerage  ? 

Ans.  $14632.01. 

35.  What  amount  of  bill  can  be  bought  for  $15000  at  1  per  cent  discount, 
paying  the  agent  1  i)er  cent  commission  and  the  broker  J  per  cent  brokerage  ? 

Ans.  $14927.61. 

36.  What  amount  of  bill  can  be  bought  for  $15000  at  $3.25  per  M.  discount, 
paying  the  agent  1  per  cent  commission  and  the  broker  J  per  cent  brokerage  ? 

Ans.  614826.52. 

SOLUTION  OF  THE  33d  PROBLEM. 


OPERATION. 

00  assumed  as  amount  invested, 
1.00   =  1  }^  commission. 
.50   =  i  %  brokerage. 


$101.50   cash  cost  of  $100  bill. 


I  100 
101.50  I  1.5000.00  (  $14778.33)    Ans. 


Explanation. — In  this,  as  in  the  above  jirobloms 
in  Investing,  and  for  reasons  therein  given,  we 
assnme  and  work  upcin  $100.  And  as  it  is  the 
bu.siness  cnstom  to  allow  both  the  commission 
and  brokerage  on  the  face  of  the  bill  plus  the 
Ijanker's  premium  or  minus  his  discount,  we 
compute  as  shown  in  the  o])eration,  and  add  to 
the  $100  the  commission  and  brokerage,  and 
thus  obtain  $101.50  as  the  cash  cost,  or  cash 
equivalent,  of  a  $100  bill  bought  at  1  per  cent 
commission  and  J  per  cent  brokerage. 

The  reasoning  for  the  line  statement  is  the 
same  as  given  in  preceding  problems  whei'e 
commission  only  w.as  charged. 


FIRST  SOLUTION  OF  THE  34TH  PROBLEM. 


OPEKATION 

To  take  out  commission  and  brokerage. 

$100.00  assumed  as  amount  invested. 
1.00  =  1  l'^  commission. 
.50  =  i  %  brokerage. 


$101.50  cash  required  to  invest 


I  100 
101.50  I  15000.00  (  $14778.33    Ans. 


OPERATION 

To  take  out  premium  and  find  face  of  bill. 

$100  assumed  face  of  bill. 
1  =  1%  i)remium. 

$101  cash  cost  of  a  $100  bill. 


$ 
1  100 
101  I  14778.33  (  $11632.01    Ans. 


The  reasoning  for  these  statements  is  the  same  as   in   the  preceding  problenm   and   hence 
is  omitted. 


PROOT'"  BY  REVERSING  THE  OPKRATION. 


Face  of  bill,  -        .        .        . 

Banker's  premium  at  1  "„  added, 


Face  of  bill  plus  premium. 
Commission  at  1  %  (on  $14778.33), 
Brokerage  at  i  %  (on  $14778.33), 


Total  amount. 


$14632.01 
146.33 

$14778.33 

147.78 

73.89 

$15000.00 


* 


I 


EXCHANGE. 
SECOND  SOLUTION  OF  THE  34TH  PROBLEM. 


725 


10L50 
101 


100 

15000.00 

100 

S14632.01     Ana. 


203 
101 


100 
2 

15000 
100 


$14632.01     Ans. 
FIRST  SOLUTION  OF  THE  35TH  PROBLEM. 


OPERATION 

To  take  out  commission  ami  brokerage. 

1.00  assumed  as  amount  invested. 
1.00  =  1  %  commission. 
.50  =  i  ^Q  brokerage. 


$101.50  cash  required  to  invest  $100. 


I  100 
101.50  I  15000.00  (  $14778.33    Ans. 


OPERATION 

To  find  face  of  bill. 

$100  assumed  face  of  bill. 
1  =  1%  discount. 

$99  casli  cost  of  $100  bill. 


I  100 
99  I  14778.33  (  $14927.61    Ans. 


SECOND  SOLUTION  OF  THE  35TH  PROBLEM. 


101.50 
99 


100 

15000.00 

IOC 


$14927.61     Ans. 


203 
99 


100 

2 

15000 
100 

$14927.61    Ans. 


PROOF  BY  REVKRSING  THE  OPERATION. 

Face  of  bill,             $14927.61 

Banker's  discount  at  1  %  deducted,      - ....  149.28 

Face  of  bill  minus  discount,          ....---..---  $14778.33 

Commission  at  1  »„'  (on  $14778.33), 147.78 

Brokerage  at  i  %  (on  $14778.33), 73.89 

Total  amount, $15000.00 

OPERATION  FOR  THE  36TH  PROBLEM. 


$100       as,sumed. 

1       =  1  5^5  commission. 
.50  ^  i  °^  brokerage. 


$101.50 


I  100 
101.50    15000.00 


778.33 


$1000       assumed  face  of  bill. 
3.25  =  $3.25  per  M.  discount. 


.75  cost  of  $1000  bill. 


1000 
996.75  I  14778.33 


$14826.52    Ans. 


7^6  soule's  niiLOsopiiic  practical  mathematics.  * 

MISCELLANEOUS    PROBLEMS,    INVOLVING    THE    FOREGOING    PEIN> 

CIPLES. 


1292.    1.    Find  the  cost  of  $2850  New  York  exclistnge  at  |  per  cent  premium. 

Ans.  $28GU.09. 

2.  Find  the  cost  of  $6540  exchange  on  Boston  at  $2.75  per  M.  premium. 

Ans.  $0557.99. 

3.  Buy  $16250  exchange  on  San  Francisco  at  $4.50  per  M.  premium  and  find 
the  cost.  Ans.  $16323.13. 

4.  Buy  $30400  exchange  on  New  York  at  $1.12^  per  M.  discount  and  find 
the  cost.  Ans.  $30365.80. 

5.  What  cost  a  bill  for  $22368.40  bought  at  |  per  cent  discount  ? 

Ans.  $22228.60. 

6.  Find  the  cost  of  $5862.70  exchange  bought  through  an  agent  at  $4.50  per 
M.  premium  juaying  the  agent  J  per  cent  commission.  Ans.  $5918.53. 

7.  Find  tlie  cost  of  $7250.80  exchange  at  J  per  cent  premium  paying  an 
agent  1  per  cent  commission  and  a  broker  J  per  cent  brokerage.      Ans.  $7396.30. 

8.  Bought  $24910.60  exchange  at  $3.70  per  M.  discount,  and  paid  J  per 
cent  commission,  and  J  per  cent  brokerage.     How  much  did  I  pay  ? 

Ans.  $25004.57.     • 

9.  "What  cost  a  bill  for  $38215  bought  at  par  paying  an  agent  J  per  cent 
commission  and  a  broker  J  per  cent  brokerage  ?  Ans.  $38501.61. 

10.  Invested  $8500  in  exchange  at  $5  per  M.  premium.  What  was  the  face 
of  the  bill?  Ans.  $8457.71. 

11.  Invested  $9000  in  exchange  at  $4.50  jier  M.  discount.  What  was  the 
face  of  the  bill  I  Ans.  $9040.68. 

12.  Exchange  is  quoted  at  2J  per  cent  premium,  how  much  can  be  bought 
for  $12500?  Ans.  $12195.12. 

13.  Exchange  is  quoted  at  1|  jjer  cent  discount,  how  much  can  be  bought 
for  $40490?  Ans.  $47257.94. 

14.  If  you  pay  1  per  cent  commission  and  J  per  cent  brokerage  and  buy 
exchange  at  $1.25  per  M.  discount,  how  much  exchange  can  you  buy  for  $15000  T 

An.s.  $14796.83. 

15.  Bought  the  following :  Excliange  $4700  at  ^  per  cent  premium ;  $0250  at 
f  per  cent  premium;  $5600  at  ^  per  cent  discount;  $800  at  $1.10  per  M.  premium; 
$16420.05  at$l  per  M.  discount,  and  through  a  broker  $12286.45  at  par;  paid  the 
broker  $5  per  M.  on  the  last  exchange.  How  much  exchange  did  I  buy  and  what 
did  it  cost?  Ans.  $46057.10,  bot.     $46141.75,  cost. 

16.  You  owe  a  New  York  merchant,  for  goods  bought  of  him  $18400,  which 

you  wish  to  remit  in  exchange.    The  rate  is  $2.50  per  M.  discount.    What  is  the 

cost  of  the  bill  ?  Ans.  $18354.00 

Note. — The  student  of  book-keeping  may  make  the  Cash  Book  entry  for  both  firms  for  this 
transaction. 


*  EXCHANGE.  727 

17.  Yon  owe  :i  Xew  York  merchant,  for  goods  sold  for  him,  in  yovir  capacity 
as  a  commission  merchant,  $18400,  with  which  you  wish  to  buy  aud  remit  exchange. 
The  rate  is  $2.50  per  M.  discount.    What  is  the  face  of  the  bill  ?   Aus.  $18446.12. 

Note. — The  student  of  'book-keeping  may  make  the  Cash  Book  entry  for  Ijoth  firms  for  this 
transaction. 

18.  Jones  of  ifew  Orleans  owes  Sniith  of  Chicago,  a  balance  of  $1018  for 
which  he  buys  exchange  at  $1  per  M.  premium,  and  remits  to  Smith.  What  did 
the  bill  cost?  Ans.  $1019.02. 

Note. — The  student  of  liook-keeping  may  make  the  Cash  Book  entries  in  both  Jones  and 
Smith's  books,  for  this  transaction. 

19.  A.  of  Galveston  owes  B.  of  Dallas,  for  net  proceeds  of  cotton,  $8232.38 
which  he  invests  in  New  York  Exchange  at  $1  per  M.  premium.  "What  is  the  face 
of  the  bill?  Ans,  $8224.16. 

Note. — The  student  of  book-keeping  may  make  the  Cash  Book  entries  in  the  books  of  A.  and 
B.  for  this  transaction. 

20.  A  commission  merchant  sold  a  consignment  of  goods  for  $28900.  The 
charges  were  $2140,  commission  2i  per  cent.  With  the  net  proceeds  he  purchased 
exchange  at  1^  per  cent  premium.    What  was  the  face  of  the  bill  ? 

Ans.  $25652.71. 

Note  1. — ^The  student  of  boot-keepiag  may  make  the  Cash  Book  entry  for  the  purchase  of 
the  bill  by  the  consignee  and  for  its  collection  by  the  consignor. 

Note  2.— See  .Souli^'s  New  Science  and  Practice  of  Accounts,  pages  178,  179,  272  and  273, 
edition  of  1895  or  1896,  for  the  jiroper  entries  for  such  transactions. 


TO  FIND  THE  EATE  OF  EXCHANGE. 

1293.     1.    Paid  $2436  for  a  bill  of  $2400.    What  was  the  rate  per  cent  ? 

Aus.  1J%. 

OPERATION. 

$2436  $ 

2400  I  36 

2400  I  100 

$36  =  premium.        I 

1      U°^    Ans. 

2.    The  cost  of  exchange  was  $3484.60 ;  the  face  of  the  bill  was  $3500.    What 
was  the  rate  per  M.  and  the  rate  per  cent  discount  ? 

Ans.  $4.40  per  M.  or,  .44%  discount. 


$3500 
2484.60  3500 

^15.40  =  discount. 


OPERATION. 

15.40 
1000 

3500 

15.40 

100 

$4.40  per  M.  discount. 

.44^0  or  '^%% 

728 


soule's  philosophic  practical  mathematics. 


3,    Exchange  cost  $4050.15,  face  of  bill  $4000 ;   paid  brokerage  for  buying 
f  per  cent.     What  was  the  rate  per  M.  premium  and  the  brokerage  ? 

Ans.  $5  per  M.  premium.     $30.15  brokerage. 


I  100 
100.75    4050.13 


OPERATION. 

$4020 
4000 


$4020.00  =  face  of  bill  +  premium. 
4050.15  =  cost  of  bill. 


|20  premium. 


4000 


20 
1000 

$5      per  M.  premium. 


$30.15  =  brokerage. 

4.  A  bill  of  exchange  cost  $0,504.51  J,  the  face  was  $0500;  paid  commission 
1  per  cent  and  brokerage  J  per  cent.  What  was  the  rate  i)er  cent  discount,  the 
commission  and  the  brokerage  ?  Ans.    See  operation. 


$100 

n 

$101i 


OPERATION. 


101} 


100 
6564.51J 

$6467.50    =  face  of  bill  —  i  °^  discount. 


$6500 
e467.50 

$32.50  =  discount  on  $6500. 


$6467.50  at  1  «^  =  $64.67}  commission. 
$6467.50  at  }  %  =  $32.33f  brokerage. 


6500.00 


32.50 
100 


i%  discount 


TO  FI]SrD  THE  FACE   OF  BILLS  WHEN  THE   COST  AND    RATE    ARE 

GIVEN. 

1294.     1.     The  cost  of  a  bill  was  $9024.75;  the  rate  was  $2.75  per  M.    What 
was  the  face  of  the  bill  ?  Ans.  $9000. 

OPERATION. 

$ 

I  1000       assumed  face  of  bill. 
Cost  of  assumed  face,  1002.75  |  9024.75  cost  of  bill. 

I  $9000     Ans.     Face  of  bill. 

2.     Paid  $3244.14  for  a  bill  of  exchange  bought  at  J  per  cent  premium,  bro- 
kerage i  per  cent.     What  was  the  face  1  Ans.  $3200. 

OPERATION. 


100.50 


100         assumed  face. 
3244.14 


807 


I  $322S.   =  face  -f-  premium. 


100 

8 

3228 

$3200  Ans. 


100.50 


1001 


100 

3244.14 

100 


$3200   Ans. 


EXCHANGE. 


•29 


3.  The  cost  of  a  bill  was  $15377.25.     Paid  1  per  cent  premium,  1  per  cent 
commission  and  J  per  cent  brokerage.    What  was  the  face  1  Ans.  $15000. 

4.  Invested  throrigh  a  broker  $15000.     Paid  brokerage  1  per  cent ;   face  of 
bill  was  $14803.38.     What  was  the  premium  per  M  ?  Ans.  $3.25. 


101 


t)rF.i;ATi<>x. 

$ 

100 

15000 

$14851.49  =:  face  of  Ijill  -f  banker's  iireiiihiiu. 
14803.38  =  face  of  bill. 


14803.38 


$ 

48.11 

1000 

i  $3.25  per  M.  premium. 


$48. 11  =  banker's  premium. 

Note. — In  -n-orking  the  statement  to  find  the  $3.25  per  M.  premium,  the  last  dividend  which 
gives  the  final  figure  5  is  a  very  little  too  small  to  produce  even  5.  This  results  mainly  from  the 
fact  that  the  divisor  $14803.38  is  slightly  in  excess  of  the  true  amount,  which  is  $14803. 379-t-. 

5.  The  face  of  a  bill  for  a  sum  invested  at  1  per  cent  discount,  paying  an 
agent  1  per  cent  commission  and  a  broker  J  per  cent  brokerage,  is  $14927.61.  What 
was  the  amount  invested  ?  Ans.  $15000. 

Note. — See  problem  35,  page  725  for  aid  in  the  solution  of  this  problem. 

6.  Invested  in  exchange  at  IJ  per  cent  discount  and  received  a  bill  for 
$4100.    What  was  the  sum  invested  ?  Ans.  $4038.50. 

OPERATION. 
$4100 

61.50  =  li  %•  discount. 


$4038.50  ^=  sum  invested. 


TO  FEND  THE  BATE  OF  EXCHANGE  ON  TIME  BILLS. 

1295.  1.  What  would  be  the  equivalent  rate  per  cent  premium  or  discount 
on  a  30  day  bill,  money  being  worth  6  per  cent  and  sight  exchange  on  the  same 
place  being  2  per  cent  discount?  Ans.  2^^%  discount. 


OPERATION. 


$100  assumed  to  work  on. 


$100 


60 


100 
33 

$2.00    discount  due  the  purchaser 

.55    interest      "      "            " 

55c.  interest. 



$2.55    sum  of  dis.  and  int.  due  the  purchaser,  and  which 
being  on  the  $100,  is  hence  equal  to  2ii  ^^  dis. 


730 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.     What  would  be  the  equivalent  rate  per  cent  premium  or  discount  ou  60 
day  bills,  money  worth  8  per  cent  and  sight  exchange  1  per  cent  premium  'I 

Ans.  f  %  discount. 

OPERATION. 


$100  .assumed  to  work  on. 

$ 

100 
45    63 

$1.40  int.  (Ine  the  purchaser. 
1.00  premium. 


$100 

l"o'   premium  due  the  bauker. 


$1.00    premium. 


.40  excess  of  interest,  which  being  ou  the  $100  and  in  favor  of  the  purchaser  is  equal 
to  I  j!^  discount. 

3.     What  is  the  equivalent  per  cent  ]>reniium  or  discount  on  a  15  day  bill, 
money  being  worth  4  per  cent  and  sight  exchange  J  per  cent  premium  ? 

Ans.  |5%  premium. 

OPERATION. 

$100  assumed  to  work  on. 


90 


$ 

100 

18 

20c.  interest  in  favor  of  the  purchaser. 


$100 


=  87ic.  prem.  in  favor  of  the  hanker. 
20c.    interest. 


67ic.  excess  of  premium. 

Explanation. — By  these  statements,  we  see  that  there  is  an  excess  of  tlie  premium  of  OTJc,  and 
as  it  is  on  the  $100,  and  in  favor  of  the  banker,  it  is  therefore  the  per  cent  premium  on  15  day 
bills.     By  reduction  we  find  it  equal  to  |5  per  cent. 

4.  Sight  exchange  is  $4  per  M.  discount,  and  money  is  worth  5  per  cent. 
What  is  the  rate  of  exchange  per  M.  for  90  day  bills  allowing  grace,  btit  not  discount 
day  1  Ans.  $16.91§  per  M.  discount. 

OPERATION. 

$1000  assumed  to  work  on. 


4     :=  $4  per  M.  discount. 


$  4.000    =  discount  in  favor  of  buyer. 
12.91f    =  interest  in  favor  of  buyer. 


1000 
'Xi 

$12.91|  interest  in  favor  of  buyer. 


$16.91j 


interest  and  discount  in  favor  of  buyer,  and   which   being   on   the   $1000   is, 
therefore,  the  rate  of  discount  i)er  JI. 


TIME    EXCHANGE. 


1296.  1.  What  is  the  cost  of  a  $15000  bill  drawn  at  60  days  sight,  money 
being  worth  8  per  cent  and  sight  exchange  being  1  per  cent  premium,  allowing  for 
3  days  of  grace  ?  Ans.  114940. 


FIRST    OPERATION. 


$15000  face  of  bill. 

150  =  1  %  premium  for  sight. 

$15150  cost  of  a  sight  lull. 

210  int.  on  $15000  for  63  days  at  8  %. 

$14940  cash  cost  of  bill. 


$  Explanation. — As  the  cherk  or  bill  has 

15000  63  days  to  run  after  presentation,    it   is 

45    63  clear  th.at  we  pay  it  63  days  before  due, 

and  hence  we  find  and  deduct  from   the 

$210  sight  cost  the  banker's  discount  ou   the 

face  of  the  bill,  $15000,  for  the  63  diiys 
at  8  per  cent,  and  thus  obtain  the  present 
cash  cost. 

Note.— In  practice,  10,  30,  60,  and  90  day  exchange  is  generally  drawn  at  a  certain  per  cent 
^reminiu  or  discount,  which  renders  work  of  this  character  of  but  little  use. 


EXCHANGE.  73 1 


SECOND    OPERATION. 

$100.00  assumed  cost.  $ 

1.00  =  1  ;„'  premium  on  sight  bill.  I  100 

45  I  63 

$101.00  cost  of  sight  bill.  

1.40  int.  on  $100  ass.  for  63  ds.  at  8  »^.  )  $1.40 


$99.60  cost  of  $100,  60  day  bill. 

15000. 


$14940.0000   cost  of  $15000,  60  day  bill. 

THIRD    OPERATION. 

Premium  as  above  on  $100  =:  $1 
Interest  as  above  on    100  =    1.40 

.40c.  discount  on  the  $100  or  |  "o  discount. 
$15000  $15000 

.40  60 


^60.0000  =  discoimt.  $14940    =  cost  of  bill. 

2.  "What  is  the  cost  of  $8000  exchange  on  Baltimore,  drawn  at  30  days 
sight,  money  being  worth  6  per  cent,  sight  exchange  being  2  per  cent  discount"? 

Ans.  $7790. 

OPERATION. 

$8000  face  of  bill.  $ 

160  =  2  » '  discount.  I  8000 

60  I  33 

$7840  cash  cost  of  a  sight  bill.  

44  interest.  |  $44.00  interest. 

$7796  cash  cost  of  $8000,  30  day  bill. 

3.  A  merchant  owes  a  balance  of  $2511.25,  which  he  wishes  to  pay  with  a 
draft  on  St.  Louis  at  30  days  sight.  Allowing  money  to  be  worth  6  per  cent,  and 
sight  exchange  or  drafts  to  sell  at  1  per  cent  premium,  for  what  amount  must  he 
draw  his  draft  so  that,  when  sold,  it  will  net  the  amount  that  he  owes  ? 

Ans.  $2500. 


OPERATION. 


I 


$100.00  face  of  draft  assumed.  $ 

1.00  =  1  °a  premium.  [  100 


$101.00  cash  value  of  $100  sij^ht  draft. 
.55  int.  on  |100  for  33  davs  at  6  «i. 


60 


33  100.45 

55c.  interest. 


100 
2511.25 

$2500     Aus. 


45  cash  value  of  $100,  3C  day  draft. 
To  sell  this  draft  the  following  figures  are  produced: 

OPERATION. 

$2500  00  face  of  draft. 

25.00  =  1  ^0  premium  for  sight  drafts. 

60 

$2525.00  value  of  $2500  sight  draft. 

13.75  interest. 


2500 
33 


i.75  interest. 


$2511.25  cash  xalue  of  30  day  draft. 


732 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


excha:n"ge  and  cash  notes  combined. 

1297.  1.  I  have  a  balance  clue  S.  S.  Packard  &  Co.,  New  York,  of  $6500.  He 
instructs  me  to  purchase  sight  exchange  and  remit  to  him.  Sight  checks  are  J  per 
cent  discount;  I  charge  J  per  cent  commission  for  investing,  and  give  my  note,  with 
collaterals,  for  such  an  amount  as,  ■when  discounted  at  8  per  cent  for  34  days,  will 
net  the  amount  due  the  banker  for  the  exchange.  What  is  the  commission,  the 
face  of  the  note,  and  the  face  of  the  bill  ? 

Ans.  $32.34  commission,    $6510.90  face  of  note.     $6500.10  face  of  bill. 


OPERATION 

To  take  out  i  %  commission. 


I  100 
201  I  2 
6500 


$6467.66 


SOLUTION. 

OPERATION 

To  invest  $6467.66  at  i%<lisc't. 

$ 
I  100 
199  I  2 

I  6467.66 

$6500.16 


OPERATION 

To  find  the  face  of  the  note. 

$  assumed  note. 
I  4500 
4466    6467.66 


$6516.90    Ans. 


2,  Bought  of  the  Merchants'  Bank  the  following  exchange :  A  check  on 
New  York  for  $2500,  at  |  per  cent  discount;  a  check  on  St.  Louis  for  $2000,  at  J 
per  cent  premium ;  and  a  gold  check  on  Galveston  for  $5000,  at  J  per  cent  discount, 
gold  being  10  per  cent  premium.  For  the  cost  of  these  three  checks  I  gave  $4000 
currency  and  my  cash  note  at  30  days,  at  8  per  cent,  secured  by  collaterals.  What 
is  the  face  of  the  note,  allowing  grace  and  discount  day?  Ans.  $6016.71. 


OPERATION   INDICATED. 


Gold 


$2500      at  }  %  discount  =  |18.75  =  $2481.25       cost. 
2000      at  I  »„'  premium  =     17.50  =    2017.50       cost. 


1    ^500  I  at  J  »„' discount  =     27.50=    5472.50       cost. 


$ 
I  4500 
4466    5971.25 


$6016.71    Ans. 


$99.71.25  =  total  cost. 
4000.00  =  cash  paid. 

$5971.25 
DISCOUNT  ON  TIME  DEAFTS. 

1298.  When  time  drafts  or  bills  of  exchange  are  negotiated  before  maturity, 
they  are  subject  to  discount  on  the  same  conditions  as  promissory  notes.  In  trans- 
actions of  this  character,  the  premium  or  discount,  and  the  interest  are  computed 
on  the  face  of  the  draft  or  bill. 

EXAMPLES. 

1. 

$8000.  New  Orleans,  July  10,  1895. 

Sixty  days  after  sight,  for  value  received,  pay  to  the  order  of  Geo.  Fernandez, 
Eight  Thousand  Dollars,  and  charge  the  same  to  our  account. 

To  Baab  &  Maspero,  SCEIBNEE  &  WAGNEE. 

St.  Louis,  Mo. 


EXCHANGE. 


733 


"What  are  the  net  proceeds  of  this  draft  if  discounted  the  same  day  that  it 
was  drawn  at  8  per  cent,  the  rate  of  exchange  being  1  jjer  cent  premium  ? 

Aus.  $7900.22. 

OPERATION. 


$8000.00   face  of  draft. 
80.00  =1%  premium. 


I  80.00 
45    64 


00  value  of  draft. 
113.78  interest. 

$7966.22  net  proceeds.     Ans. 


$113.78  interest. 


2.    What  is  the  value  of  a  draft  for  $10000,  payable  15   days  after  sight, 
exchange  f  per  cent  discount,  interest  8  per  cent?  Ans.  $9882.78. 


OPERATIOX. 


$10000.00   face  of  draft. 
75.00  =  J  ^o'  discount. 

$9925.00  value  of  draft. 
42.22  interest. 

2.78    Ans. 


$ 
I  100.00 
45     19 


I  $42.22  interest. 


To  buy,  sell,  and  invest  in  exchange  and  pay  commission  and  hrolierage  on  the  face  of 

the  bill  only. 

1299.     1.     What  will  a  bill  for  $15000  cost,  bought  at  1  per  cent  premium, 
paying  1  per  cent  commission  and  J  per  ceut  brokerage  on  the  face  of  the  bill  ? 

Ans.  $15375. 


OPERATION'. 


Face  of  bill, 
1  ?o  premium, 
1  5^  commission, 


Cost  of  bill, 


$150 
150 


$15000 


$15375 


Explanalion. — As  tlie  commission  and  bro- 
kerage are  here,  as  ■well  as  the  premium, 
charged  ou  the  face  of  the  bill,  they  could 
have  been  added,  making  2J  per  cent,  and 
the  correct  result  olitained  by  adding  to  the 
$15000  the  amount  of  2i  jier  cent  of  itself. 


2.     What  will  be  realized  for  a  $15000  bill  sold  at  1  per  cent  premium,  paying 
1  per  cent  commission  and  i  per  cent  brokerage  on  the  face  of  the  bill  ? 


Aus.     $14925. 


OPERATION. 

Face  of  bill, $15000 

1  ^  premium,  -------  150 

Value  of  bill, _-        $15150 

1  %  commission,         -----        $150 
i  9fl  brokerage,  ...        -        -  75 


$14925 


734  SOLLE  S    THILOSOPHIC    rKACTlCAL    MATHEMATICS.  * 

3.    What  anioant  of  bill  can  be  bought  for  $15000  at  1  per  cent  premium, 
paying  1  per  cent  comDiission  and  i  per  cent  brokerage  on  the  face  of  the  bill  f 

Aus.  ^14034.15. 

OPFRATIOX. 

JlOO.OO  assumed  face  of  bill. ,  S  Explanation. — The    reasoning 

l^  =  }  ^V"'^'}"".-  ,n.,vii}^nn  for  the  line  statement    Is    the 

1.00   =  1  *^  commission.  lOJ.oO  |  loOuO.OO 

.50  ^  i  ;„'  brokerage.  j same  as  given  in  the  26th  prob- 

$14634.15     Ans.  ,  ,  ,  .  .^    , 

$102.50   cash  cost  of  $100  bill.  lem,  and  hence  is  omitted. 

IfoTE. — By  the  foregoing  ■work.  Tve  have  illnstrated  39  different  transactions  in  purchafinii, 
telling,  and  inreating  in  domestic  exchange  according  to  the  custom  governing  exchange  transac- 
tions, and  also  several  different  problems  where,  by  special  agreement,  the  commission  and  brokerage 
are  charged  on  the  face  of  the  bilL 

This  classification  is  entirely  new,  mnch  more  extended  than  ever  before  presented,  and  folly 
Olostratea  'work  that  has  always  confused  students  and  the  great  majority  of  bosiuess  men. 


jnSCELLA>T:OUS  PKOBLEMS  IN  DOMESTIC  EXCHANGE. 

1299.  1.  "What  is  the  amount  of  the  banker's  premium  on  $25000,  yew- 
Orleans  exchange  at  |  per  cent  premium;  what  is  the  agent's  commission  at  J  per 
cent,  and  what  is  the  cost  of  the  billt  Ans.  $156.25,  banker's  premium. 

•?125,78,  agent's  commi-ssion. 

.$25282.03,  cost  of  bill. 

2.  A  merchant  invested  through  his  correspondent  $7S40..50  in  exchange  on 
Philadelphia  at  f  per  cent  discount.  The  correspondent  charged  for  investing  1 
per  cent  commission  and  paid  a  broker  J  per  cent  for  transacting  the  business. 
WTiat  was  the  face  of  the  bill,  what  the  banker's  discount,  what  the  correspondent's 
commission,  and  what  the  broker's  brokerage  I  Ans.  $7802.22,  face  of  bilL 

$58.52,  discount. 
$77.44,  commission, 
819.36,  brokerage. 

3.  Bought  exchange  at  1  per  cent  discount  and  paid  $2385.90 ;  what  was  the 
face  of  the  bill !  An.s.  $2410, 

OPEEATIOX    IXDICATED. 

$100  —  $1  =  99.      ($2385.90  X  100)  —  99  =  $2410. 

4.  A  merchant  in  New  Orleans  owes  a  merchant  in  St,  Louis  $15000. 
Exchange  on  St.  Louis  is  1  per  cent  premium,  and  exchange  in  St.  Louis  on  New 
Orleans  is  1  per  cent  discount;  making  no  allowance  for  the  use  of  money  or 
telegraph  expenses,  which  is  it  better  for  the  Xew  Orleans  merchant  to  do,  buy  in 
New  Orleans  at  1  per  cent  premium  and  remit,  or  to  instruct  the  St.  Louis  merchant 
to  draw  on  him  for  such  a  sum  as  when  sold  at  1  per  cent  discount  will  net  the 
$15000  !  Ans.  $1.52  better  that  the  !New  Orleans  merchant  buy  and  remit. 

KoTE. — See  solution  for  the  following  problem. 


EXCHANGE. 


735 


5.  In  tlie  above  prnbleTn.  estimating:  that  money  is  ■worth  to  the  Xew  Orleans 
merchant  8  i)er  cent,  and  that  lie  instructs  the  St.  Louis  merchant  by  telegiam 
costing  $2.40,  to  draw  on  him  at  sight  for  such  an  amount  as  vhen  sold  at  1  per 
cent  discount  will  net  the  SloOOO,  and  that  it  will  require  4  days  for  the  St.  Louis 
bill  to  reach  the  Kew  Orleans  merchant,  what  will  be  the  gain  to  each  merchant  by 
this  mode  of  settling,  over  that  of  buying  a  bill  in  Kew  Orleans  at  1  per  cent 
premium,  money  being  worth  6  jier  cent  per  annum  in  St.  Louis,  and  4  days  time 
required  to  transmit  the  exchange  if  bought  in  2sew  Orleans  ? 

Ans.  $9.55  gain  to  the  New  Orleans  merchant, 
gain  to  the  St.  Louis  merchant. 


SOLUTION. 


OPERATIOK 

Showing  the  cost  of  §15000  exchange  bonght  in 
Kew  Orleans  at  1  per  cent  premium. 

$15000  face  of  bill. 
150  =  1  ^^  preminm. 

$15150  cost. 


OPERATION 


Showing  face  of  hill  drawn  in  St.   Louis  and 
sold  at  1  per  cent  discount  to  net  $15000. 


99 


100 
15000 

$15151.52  face  of  Ijill. 


Face  of  St.  Louis  bill. 
Cost  of  New  Orleans  bill. 


Loss  to  the  New  Orleans  merchant  by  paying  the  St.  Louis  bill, 
To  which  we  add  telegraph  charges,  ..... 


§15151.52 
15150.00 

§1.52 
2.40 


And  produce  a  loss  of  .............  §3.92 

Thus  far  we  see  that  the  Kew  Orleans  merchant  has  lost  $3.92;  but  by  reason  of  the  St.  Louis 
merchant  drawing  on  him,  he  has  had  the  use  of  §15150,  the  co.st  of  a  §15000  bill  in  New  Orleans, 
for  4  days  at  8  per  cent ;  and  the  St.  Louis  merchant  has  had  the  use  of  §15000,  the  proceeds  of  the 
St.  Louis  bill,  for  4  days  at  6  per  cent. 


OPERATION 

To  find  the  worth  of  §15150  for  4  days  at  8  ^^. 

§ 

I  15150 
45     4 


I  §13.47  worth  or  int. 


OPERATION 

To  find  the  worth  of  §15000  for  4  days  at  6  %. 


60 


$ 

15000 
4 


§10 


worth  or  int. 


By  these  last  two  operations  we  see  that  the  St.  Louis  merchant  gains  by  the  use  of  §15000, 
for  4  days  at  6  per  cent,  §10;  and  that  the  New  Orleans  merchant  gained  by  the  nse  of  §15150,  for 

4  days  at  8  per  cent,  - •*        §13.47 

From  which  we  deduct  the  losses,  as  shown  in  the  first  part  of  the  operation,    ...  3.92 


And  thus  obtain  a  net  gain  of 


§9.55 


v^^I^^P?"^^/ 


/^ 


y^^^' 


rbitration  of  Exchange. 


^ 


1300.  Arbitration  of  Exchange  is  the  process  of  findings  the  proportional 
cost  of  exchange  between  two  cities  or  countries,  by  means  of  one  or  more  interme- 
diate exchanges. 

The  object  of  this  system  of  exchange  is  to  ascertain,  by  comparison  of  the 
costs  or  rates  of  exchange,  between  several  places,  the  most  advantageous  channel 
through  which  to  remit  bills. 

The  subject  is  divided  into  simple  and  compound  arbitration. 

Simple  Arbitration  is  the  process  of  finding  the  cost  or  rate  of  exchange 
between  two  cities  through  a  third. 

Compound  Arbitration  is  the  process  of  finding  the  cost  or  rate  of  exchange 
between  two  cities  through  several  others. 

Exchange  of  different  articles  of  merchandise,  and  different  weights  and 
measures  of  different  nations,  may  be  arbitrated  in  the  same  manner  as  bills  of 
exchange  and  currencies. 

PROBLEMS. 

TO  FIND  THE   COST  OF  A  BILL  THAT  WILL  PAY  A  CERTAIN  SUM 

m  ANOTHER  PLACE. 

1301.  1.  A  merchant  of  New  Orleans  owes  a  balance  of  $5000  in  New  York, 
•which  lie  wishes  to  pay.  Direct  exchange  on  New  York  is  1  per  cent  premium; 
exchange  at  the  time  on  St.  Louis  is  i  per  cent  discount,  and  St.  Louis  exchange  on 
New  York  is  J  per  cent  premium.  Allowing  ^  per  cent  brokerage  for  reinvesting 
on  the  jirHnvestment  or  banker's  cost  of  the  bill  in  St.  Louis,  which  is  the  better 
way  for  the  New  Orleans  merchant  to  remit,  and  how  much  will  he  gain  by  remitting 
the  better  way  ?  Ans.    Through  St.  Louis  is  the  better  way. 

$37.03,  gain. 

OPEHATION 

For  direct  Exchange  on  New  York. 

$5000  face  of  bill. 

50   ^  1  °^  premium. 

$5050  cost  of  bill. 

(736) 


ARBITRATION    OF    EXCHANGE.  737 

FIKST    OPERATION 

For  finding  the  cost  of  Exchange  by  remitting  through  St.  Louis. 

s 

Bill  on  St.  Louis,  100  |  99.50  cost  of  hill  on  St.  Louis  at  ^  %  discount. 

Bill  ou  New  York,  100  |  100.50  cost  of  hill  on  New  York  at  ^  %  premium. 

Face  of  hill  on  N.  Y.      100     100.25  cost  of  hill  on  New  York  at  |  %  brokerage. 

5000  dollars  invested. 


SECOND   OPERATION. 


1  $5012.37    Ans. 

Statement  of  costs. 


$5000.00  face  of  bill. 

25.00   =  i  j",,  premium  in  St.  Louis.  Cost  by  direct  exchange,  -        -        $5050.00 

Cost  through  St.  Louis,  -        -  5012.37 

$5025.00   St.  Louis  cost.  

12.56  =  i  "q  brokerage  iu  St.  Louis.  $37.63  gain. 


$5037.56  cost  including  brokerage. 

25.19   :=  i  %  discount  in  New  Orleans  on  St.  Louis  bill. 

$5012,37  cost  of  $5037.56  iu  New  Orleans. 

2.  Suppose  in  the  above  problem,  that  the  \  per  cent  brokerage  Lad  been 
charged  on  the  face  of  the  bill,  instead  of  the  net  investment  or  banker's  cost, 
what  would  have  been  the  cost  of  the  exchange  through  St.  Louis? 

Ans.  $5012.31. 

FIRST    OPERATION. 


Bill  on  St.  Louis,     100 
Bill  on  N.  Y.  100 


99.,50     cost  of  bill  on  St.  Louis  at  i  %  discount. 

100.75   cost  of  bill  on  N.  Y.  at  ^  %  premium  and  J-  %  brokerage. 

5000      dollars  invested. 

$5012.31    Ans. 

SECOND    OPERATION. 

$5000.00  face  of  bill. 
25.00   =  i  %  premium. 
12.50   =  i  %  brokerage. 

$5037.50  cost  of  bill  in  St.  Louis  including  premium  and  brokerage. 
25.19  ^  i  %  discount,  N.  O.  exchange  on  St.  Louis. 

$5012.31   net  cost  in  New  Orleans.    ■ 

3.  Suppose  again,  that  in  the  above  problem  brokerage  had  been  charged  on 
the  irhoJe  amount  for  investing,  by  the  St.  Louis  agent,  what  would  have  been  the 
cost  of  the  exchange  through  St.  Louis  !  Ans.  $5012.40. 

FIRST    OPERATION. 

$ 

Bill  ou  St.  Louis, 100  |  99.50     cost  of  bill  in  St.  Louis  at  i  %  dis. 

Bill  on  New  York,  ....  100  |  100.50   cost  of  bill  on  N.  Y.  at  i  %"  premium. 

Proceeds  of  am't  invested,  brok.  deducted,  99.75     100        amount  invested  in  N.  Y.  exchange. 

5000      dollars  invested. 


99.75 


$5012.40    Ans 

SECOND    OPERATION. 

$5000 
25 

face  of  bill. 

=  i  ?o  premium  on  New  York. 

$5025 
100.00 

St.  Louis  cost. 

$5037.59  total  cost  including  brokerage  on  same  for  investing. 
25.19   =  i  %  discount  on  St,  Louis  exchange  in  New  Orleans. 


$5012.40  net  cost  in  New  Orleans. 


738 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


TO  FIND  TEE  TROCEEDS  OR  FACE  OF  A  BILL  FOR  A  CERTAIN  SUM 

INVESTED. 

1302.  1.  A  merchant  of  New  Orleans  wishes  to  invest  and  remit  $5000  to  a 
New  York  correspondent.  Exchange  on  New  York  is  1  i)er  cent  2)remium ;  exchange 
on  St.  Louis,  at  the  same  time,  is  i  jjcr  cent  discount,  and  St.  Louis  exchange  on 
New  Y^ork  is  ^  per  cent  premium,  allowing  ^  per  cent  brokerage  on  the  net  amount 
invested  by  the  agent,  which  is  the  better  way  to  remit  and  how  much  will  he  gain 
by  remitting  the  better  way  1 

Ans.    Through  St.  Louis  is  the  better  way.    $37.16  gain. 


For  direct  exchange  on  New  York : 


SOLUTIONS. 
OPERATION. 


I  100   face  of  bill  on  New  York. 
Cost  of  bill    101  I  5000 

I  $4950.50   face  of  bill.    Ans. 

OPERATION. 


Cost  of  bill    . 

99.50 

Cost  of  inv. 

100.25 

Cost  of  bill  on  N.  Y. 

100.50 

For  remitting  through  St.  Louis : 

I 

100.00  face  of  bill  on  St.  Louis. 

100.00  amount  invested  by  agent  in  N.  Y.  exchange. 

100.00  face  of  bill  on  New  York. 

5000       dollars  invested. 

I  $4987.66   proceeds  or  face  of  bill. 
4950.50  proceeds  by  direct  exchange, 

$37.16  gain  as  above. 

2.  Suppose  that  in  the  same  problem  the  agent  had  charged  J  per  cent 
brokerage  on  the  whole  amount  invested,  what  would  have  been  the  proceeds  on  the 
face  of  the  bill  1  Ans.  $4987.62. 

OPERATION. 

$ 

Cost  of  bill  on  St.  Louis,      99.50  I  100  face  of  bill  on  St.  Louis. 

Cost  of  investment,  100.      |  99.75  .amount  invested  by  .agent  in  N.  Y.  exchange. 

Cost  o*  bill  on  New  York,  100.50     100.00  face  of  bill  on  New  York. 

5000  dollars  invested. 


I  $4987.62  proceeds  or  face  of  bill. 

3.  Suppose  again,  in  the  above  problem  that  the  agent  had  charged  J  per 
cent  brokerage  on  the  /ace  of  the  bill,  what  would  have  been  the  proceeds  or  face 
of  bill?  Ans.  $4987.72. 

OPKRATION. 


Cost  of  bill  on  St.  Louis,  99.50 

Cost  of  N.  Y.  exchange,  100.75 

Invested  at  i  %  prem.  and  J  %  brok. 


100.00  face  of  bill  on  St.  Louis. 
100.00  .amt.  invested  in  N.  Y.  ex. 
5000      dollars  invested. 

$4987.72  proceeds  or  face  of  bill. 


ARBITRATION    OF    EXCHANGE. 


739 


In  the  foregoing  problems,  we  liave  elucidated  the  three  different  methods  of 
agents  of  charging  commission  or  brokerage  for  reinvesting,  wheu  the  cost  of  a 
bill  for  a  certain  sum  is  required,  and  also  "when  the  proceeds  or  face  of  a  bill  for  a 
certain  sum  invested  is  required. 

There  is  no  statute  la\v  regulating  the  method  of  making  this  charge;  and 
as  the  business  custom  relating  thereto  is  not  uniform,  it  is,  therefore,  a  matter  of 
special  agreement. 

In  the  arbitration  of  exchange,  brokerage  or  commission  is  charged  at  each 
intermediate  i)lace  through  which  it  is  remitted. 

In  the  arbitration  of  foreign  exchange,  the  commission  or  brokerage  for 
reinvesting  is  sometimes  included  in  the  rate  of  exchange;  but  when  not  thus 
included,  a  separate  charge  is  made. 

FOEMULA  OF  STATEMENTS. 

1303.  To  elucidate  the  operation  of  charging  commission  or  brokerage  on  the 
net  investment  or  face  of  bill,  or  ichole  amount  for  investment,  when  buying  or 
investing,  we  present  the  following  ibrmula  of  statements: 


When  luying  a  certain  amount  of  exchange. 

To  charge  commission  or  brokerage  ou  the 
net  cost  of  bill,  the  statement  is  made  thus: 


100    100.50 

To  charge  commission   or  brokerage  on  the 
face  of  bill,  the  statement  is  made  thus: 


100 


100.50  +  banker's  premium  or 
—  banker's  dis. 


To  charge  commission  or  brokerage  on  the 
whole  amount,  the  statement  is  made  thus: 


99.50 


100 


When  iiivcsling  a  certain  sum  in  exchange. 

To  charge   commission  or  brokerage  on  the 
net  cost  of  bill,  the  statement  is  made  thus: 


100.50  I  100, 


To   chiirge  commission  or  brokerage  on  the 
face  of  bill,  the  statement  is  made  thus : 


100.50  ■ 

the  banker's  premium 
or  —  banker's  dis. 


100, 


To  charge  commission  or  brokerage  on  the 

whole  amount,  the  statement  is  made  thus : 

I 
100  I  99.50. 


Explanation. — In  these  statements,  we  have  assumed  }  per  cent  as  the  commission  or  brokerage 
charged. 


By  Net  Investment  is  meant  the  sum  invested  after  the  agent's  brokerage 
or  commission  has  been  deducted. 

By  "Whole  Amount  of  Investment  is  meant  the  sum  invested,  or  to  be 
invested,  before  deducting  the  agent's  brokerage  or  commission. 

By  Face  of  Bill  is  meant  tlie  amount  of  the  bill  that  is  obtained  froni  the 
banker,  after  paying  the  agent's  commission  or  brokerage,  and  the  banker's  premium, 
or  allowing  for  the  banker's  discount. 


740  soule's  philosophic  practical  mathematics.  * 

To  find  the  equivalent  rate  of  direct  exchange  to  correspond  with  the  arbi- 
trated rate,  we  would  omit  from  the  statements  the  last  amount  on  the  increasing 
side;  or,  we  would  divide  the  final  result  of  the  statement,  or  answer,  by  the  last 
amount;  or,  if  the  last  amount  is  of  a  different  unit,  as  it  often  is  in  foreign, 
exchange,  we  would  divide  bj'  its  equivalent,  reduced  to  the  same  unit. 
To  elucidate  this,  we  present  the  third  examplo  above;  thus; 

99.50  I  100.00 
100.75  I  100.00 

I  .9975+%  (lis.     Ans. 

ENGLISH  EXCHANGE. 

1301.  In  the  introduction  of  exchange  we  defined  it  to  be,  iu  a  commercial 
sense,  a  system  of  paying  debts  in  distant  places  by  means  of  bills  of  exchange, 
■without  the  transmission  or  shipment  of  money.  By  the  term  English  Exchange, 
•we  mean  the  system  of  paying  debts  in  England  and  her  dependent  Provinces 
without  the  transmission  of  money. 

1305.  HISTORY  OF  THE  ENGLISH  MONETARY  POUND. 


The  English  monetary  pound  is  of  French  origin.  In  the  year  A.  D.  768,  Charlemagne,  or 
Charles  the  great,  ascended  the  throne  of  France,  and  during  hi.s  reign  he  wielded  the  scepter  with 
siicli  consummate  skill  and  wisdom  as  to  win  the  plaudits  of  the  world  and  weave  around  his  name 
a  brighter  luster  than  was  radiated  from  the  jewels  in  his  crown.  In  the  year  800  he  was  crowned 
by  the  Pope  of  Rome  as  "  Charles  Augustus,  Emperor  of  the  West,  "  and  during  the  reign  of  this 
great  king  and  emperor,  which  continued  till  816,  he  accomplished  what  had  been  undertaken  by 
his  imperial  predecessor,  Augustus  the  First,  the  unillcation  of  tlie  money  of  his  empire.  He 
ordained  that  the  French  livre  or  po-.iDd  weight  of  silver  should  constitute  the  money  unit  of  value 
of  the  French  kingdom;  and  this  monetary  pound  of  Charlemagne,  reduced  in  weight,  lived  in 
daily  use  through  the  vicissitudes  of  ten  centuries  in  France,  until  1792  when  the  French  Nation 
adopted  the  "metric  system"  of  weights  and  measures,  and  the  Franc,  decimally  divided,  as  the 
monetary  unit.    It  then  disappeared  from  the  money  drawers  of  France. 

In  A.  D.  1066,  this  Livre  or  pound  was  carried  across  the  channel  into  England  by  William  of 
Normandy,  and  as  conqueror  he  imposed  it  on  the  English  people.  The  "Tower  Pound"  of  A.  D. 
1066,  the  first  coined  in  England,  actually  contained  one  pound  weight  of  silver.  But  the  word 
has  long  since  ceased  to  possess  any  truthful  significance,  for  the  English  monetary  pound  Jias  been 
for  five  centuries,  since  the  year  1200,  dwindling  iu  weight,  until  now  the  20  shillings  into  which 
it  is  divided  weigh  hut  S^^g  ounces  troy. 

In  1620,  when  America  was  colonized,  the  people  brought  with  them,  or  soon  adopted,  the 
English  pound  as  their  uuit  of  money.     Some  of  the  colonies  still  further  reduced  the  weight  of 


ENGLISH    EXCHANGE. 


741 


the  English  twenty  shillings,  and  some  coined  smaller  ilenominatioils,  with  sub-divisions  in  copper. 
This  they  continued  until  1660,  -svhen  Charles  the  second  rcascended  the  throne  of  England,  and 
issued  his  royal  order  prohibiting  any  further  coinage  by  the  colonies.  The  pound  then  remained 
the  monetary  unit  until  1783  when  England  recognized  the  political  independence  of  the  United 
States,  and  further,  until  the  adoption  of  the  present  xmit  of  money,  the  dollar.  Ou  the  Gth  of 
July,  1785,  the  Continental  Congress  passed  the  memorable  monetary  ordinance  reported  by  the 
"grand  Committee  of  thirteen,  "  by  which  they  rejected  the  denominations  of  pounds,  shillings, 
pence  and  farthings,  and  established  the  dollar  as  the  permanent  monetary  unit  of  the  United 
States,  with  the  proviso,  that  it  should  be  decimally  divided.  The  precise  ■weight,  however,  was 
not  fixed  until  August  8th,  1786.  The  weight  and  value  of  the  American  silver  dollar  was  then 
made  to  conform  in  value,  or  very  nearly  so,  to  the  value  of  the  old  Spanish  Carolus  jiillar  dollar 
■which,  by  the  usage  of  bankers  for  ages  had  been  the  standard  of  value  for  the  monetary  pound 
during  its  many  changes  in  ■weight. 

As  a  matter  of  history,  ■we  ■would  here  remark  that  this  Spanish  dollar  was  not  original  with 
,  Spain,  she  having  borrowed  it  from  Austria,  during  her  union  with  Austria  under  the  empire  of 
Charles  the  Fifth.  The  true  ancestor  of  the  American  dollar  is  the  "Joachim's  thaler,"  first 
coined  in  the  mines  of  the  Bohemian  valley  of  Saint  Joachim  (or  James). 

From  1786  to  1834,  the  commercial  value  of  the  English  monetary  pound,  compared  with  the 
United  States  dollar,  was  $4.44*,  or  £9  sterling  was  ecjual  to  $40,  and  rice  rersa.  With  this 
equivalent  of  value  between  the  monetary  units  of  the  two  nations,  we  have  but  to  multiply 
Bounds  by  40  and  divide  the  product  by  9,  to  find  the  equivalent  of  pounds  in  dollars,  and  vice 
versa  to  find  the  equivalent  of  dollars  in  pounds. 


Thus,  to  reduce       £270      to  dollars. 
40 

9  )  10800 


And  to  reduce       $1200   to  pounds. 
9 


40  )  10800 


$1200 


£270 


But,  by  Act  of  Congress,  taking  efl'ect  August  1st,  1834,  the  weight  of  the  American  coin  was 
reduced.  The  ten  dollar  gold  eagle,  which  then  weighed  270  grains  troy,  |i  pure,  was  reduced  in 
weight  12  grains,  and  all  other  denominations  of  gold  coin  were  reduced  in  like  proportion.  Con- 
gress, also,  in  1837,  further  reduced  the  coin  by  increasing  the  alloy  from  -jV  to  -i\i.  By  reason  of 
this  decrease  of  ■n-eight  and  increase  of  alloy,  the  relative  value  of  pure  gold  in  the  United  States 
was  enhanced,  and  the  jnfrinsic  value  of  the  pound  sterling,  the  English  gold  sovereign,  was  thus 
enhanced  or  increased  in  like  proportion.  This  increased  value  of  the  sovereign  or  pound  sterling 
is  ascertained  by  calculating  the  actual  number  of  grains  of  pure  gold  contained  in  the  standard 
coin  of  Great  Britain,  and  estimating  the  same  according  to  the  United  States  valuation  of  gold. 


The  following  -work  gives  this  increased  value  of  the  poUnd  sterling,  elucidates 
the  manner  of  obtaining  it,  and  furnishes  the  data  for  the  basis  of  the  bankers' rate 
of  English  exchange,  from  1S34  to  1874,  -which  -was  nearly  9J  per  cent,  more  or  less, 
(according  as  premium  -was  charged  or  discount  allowed  in  selling  the  exchange) 
premium  on  the  old  $4.44^  value  of  £1 : 


742  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

From  the  original  weight  of  tlio  11.  S.  eagle  (.$10  gold  piece)  wliicli  was     '       -        -  270.0   gra. 

We  iirst  deduct        --...-...---..  12.0   grs. 

And  have  in  the  roinaiuder 258.0   grs. 

its  present  weight,  i'„-  of  which  is  pure  gold. 
From  this  we  deduct  iV  alloy,        ..--..-....  25.8   grs. 

And  have  in  tho  remainder  ...........  232.2    grs. 

of  pure  gold. 

By  Act  of  Parliament,  one  pound  weight  of  standard  {\i  pure)  gold  is  coined  into 

£46  14s.  Gd.,  which  gives  the  sovereign,  the  monetary  pound,  a  weight  of     -         123.2744    grs. 
From  which  we  deduct  i^  alloy     ...--.....-  10.2728  grs. 

And  have  in  the  remainder,  of  pure  gold,  ........        113.0016   grs. 

Having  now  the  weight  of  pure  gold  in  the  monetary  unit  of  England,  and  of 
one  of  the  denominations  of  the  American  monetary  unit,  to  ascertain  the  value  of 
the  English  pound  in  the  American  unit  of  value,  we  have  but  to  measure  or  value' 
its  contents  in  pure  gold  by  the  United  States  valuation  of  gold. 

OPERATION  !■      7         .•  «  T         J.     ^^       r  ■  i       „ 

Explanation. — According  to  the  foregoing  work,   we 

To  find  the  value  of  the  English  see  that  $10  American  gold  coin  contains  232.2  grains 

,  ,  of  pure  gold,  and  hence,  by  transposition,  232.2  grains 

mone  ary  poun   .  are  worth  $10;  with  these  facts  and  premises  we  place 

J,  the  $10  on  the  statement  line,  and  reason  as  follows: 

?«  Since  233.2  grains  of  pure  gold  are  legally  worth  $10, 

„„„  r,    iiQnoif!  1  grain  is  worth  the  232. 2d  part,  and  113.0016  grains 

'"     ^ (the  weight  of  the  sovereign  in  pure  gold)  are  worth* 

S4  8rr5  113.0016  times  as  much,  which  worked,  gives  $4.8665. 

By  deducting  from  this  $4.8665  the  old  par  value  of  the  pound,  $4.44:44,  we 
have  a  gain  of  42.21  cents,  which  gives  nearly  9J  jicr  cent  gain  on  the  old  par 
value,  or  the  value  before  the  changes  in  the  American  coin  in  1834  and  1837.  Thus, 
(.4221  X  100)  -f-  4.4444  =  9f|if|. 

This  value,  $4.8665,  is,  by  Act  of  Congress  of  March  3d,  1873,  which  went 
into  effect  January  1st,  1874,  the  present  value  of  the  pound  sterling  and  the  basis 
of  English  exchange.  By  the  same  act,  bankers  and  exchange  dealers  were 
required  to  base  their  exchange  transactions  with  England  upon  this  value,  and 
allowing  for  premium,  or  discount,  quote  the  rate  of  exchange  in  the  U.  S.  dollar 
and  cent  equivalent  of  the  English  monetary  pound.  In  pursuance  of  this  law, 
bankers  and  exchange  dealers,  have  adopted  $4.86 J  as  the  par  of  exchange  and 
auote  the  rate  $4.86J^  more  or  less  according  as  a  premium  or  discount  is  declared. 


MONETARY   TABLE. 


1306. 


TABLE  OF  FOREIGN  COINS 


Shoiciiuj  the  t'alueit,  in  United  States  f/old,  of  the  ]}ure  Gold  or  Silver  representing  the 

monetary  units,  respectively,  of  Foreign  Countries. 

Prepared  by  the  director  of  the  Philadelphia  Mint,  January  1,  1895. 


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J  Silver  the  iiumiual  standard.    Paper  the  actual  curreucy,  the  depreciation  of  wliicU  ia  measured  by  the  yold  standard. 


744 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 

THE  EATE  OF  EXCHANGE. 


Note. — For  the  Face  and  Character  of  Foreign  Bills,  see  Article  1287,  page  715. 

1307.  The  basis  of  English  Exchange,  as  above  stated,  is  $4.8GJ  more  or  less, 
according  as  premium  or  discount  is  declared. 

By  custom 'of  exchange  dealers,  this  rate  is  generally  increased  or  decreased 
by  a  varying  scale  of  i  cent,  or  i  point,  whicli  is  very  nearly,  and  for  practical 
convenience,  considered  and  quoted  i^;,  per  cent.  The  exact  per  cent  figures  for  the 
^^  increase  or  decrease  on  the  intrinsic  value  of  the  pound  sterling  are  ^^§  per 
cent  or  decimally  expressed  .102743+  per  cent. 

lu  some  cases,  the  rate  varies  by  J  and  J  cent  in  the  quotations. 

TO  FIND  THE  RATE  OF  EXCHANGE  WHEN  A  PREMIUM  IS  CHARGED. 

1308.  1.  Considering  the  balance  of  trade,  the  use  of  money,  etc.,  the  seller 
of  Englisii  Exchange  wishes  to  charge  a  premium  of  1  per  cent.  What  would  be 
the  rate  of  bills?  Ans.  $4.91  J. 

OPERATION 


Based  on  the  intrinsic  value  of  the  pound. 

$4.8605      =  value  of  pound  sterling. 
.0486+  =  1  "q  premium  charged. 

$4.9151      =  (practically  $4.91i). 


OPERATION 

Based  on  the  exchange  value  of  the  pound. 

$4,865  =  banker's  basis  of  exchange. 

.05    =  prem.  charged,  allowing  ^-„  ^q  to  equal 

ic.  increase. 

$4,915  =  (practically  $4.91i). 


What  is  the  rate  of  exchange  when  J  per  cent  discount  is  allowed  ? 

Ans.  $4.84. 


FIR.ST  OPERATION. 

By  subtracting  J  per  cent  to  the  dollar  and  cent 
value  of  j£l. 

$4.8665      =  value  of  £1. 
.0243+  =  i  ;'„'  discount. 

$4.8422      =  (practically  $4.84)    Ans. 


SECOND  OPERATION. 

By  .allowing  -j^ij  per  cent  to  equal   |c.   decrease 
on  the  l>auker'8  basis  of  exchange. 

$4,865  =  banker's  basis  of  exchange. 
.025  =  discount. 

$4,840  =  Ans. 


Explavation. — In  the  second  operation  since,  -^n  per  cent  equals  4c.  i  per  cent,  which  is  -,\  per 
cent,  equals  |c.  which  is  2ic.  or  2.5  cents  discount. 

TO   FIND   THE    PREMIUM   CHARGED   OR  THE    DISCOUNT    ALLOWED 
WHEN  THE  EATE  OF  EXCHANGE  IS  GIVEN. 


1309. 


1.    The  rate  of  exchange  is  $4.91  i,  Avhat  was  the  premium  charged  ! 

Ans.  1%. 


FIRST  OPERATION. 


$4,915     rate  of  exchange. 
$4.8665   value  of  pound  sterling. 

.0485    =  the  premium. 


4.8665 


.0485 
190 

lSiS?%  practically  1%    Ans. 


4.865 


SECOND  OPERATION. 
$4,915 

$4,865 


.050 


.050 
100 


§^  %  practically  1  %    Ans. 


ENGLISH    EXCHANGE. 


"45 


2.     Tbe  rate  of  exchange  is  $4.84,  what  was  the  discount  allowed  ? 

Ans.  i%. 


FIRST    OPERATION. 

$4.84      =  rate  of  exchange. 
4.8665  =  value  of  jCl. 


.0265  =  fliscount. 

I  .0265 
4.8665    100 


I  ma   (practically  i  %). 


SECOND   OPERATION. 

$4.84    =  rate  of  exchange. 
4.865  =  basis  or  jiar  of  exchange. 


.025  =  di-scount. 

I  .025 
4.865    100 


5-igg    (practically  }%). 


THE  RATE  AND  THE  PAR  OF  EXCHANGE  GIVEX  TO  FIND  THE   PER 
CENT  PREMIUM  OR  DISCOUNT  DECLARED. 


1310.     1.     The  rate  of  exchange  is  4.87^,  the  par  cf  exchange  is  4. SOi.     What 
is  the  per  cent  premium  charged?  Ans.  .25(>9%  premium. 


4.871 
4.86i 

lie 


4 
4.865 


OPERATION. 

C.  C 

4 


1000  or,        973 

.2569  % 


2 
100 


2.     The  rate  of  exchange  is  4.80;   the  par  is  4.86i.     What  is  the  per  cent 
discount  allowed  !  Ans.  1.33607%  discount. 


PROBLEMS  IN  ENGLISH  EXCHANGE 


TO   REDUCE   THE  ENGLISH  MONETARY   UNIT   TO   THAT   OF   THE   UNITED   STATES. 


1311.     1.     What  is  the  cost  of   a  bill  of    exchange    on   London  for   £.5000; 
exchange  4.87.^?  Ans.  $24375. 


OPERATION. 

£5000 
4.87i 


2435000 
2500 


$24375.00    Ans. 


Explanation. — Since  the  rate  of  exchange  is  the 
exchange  value  of  jCl,  It  is  clear  that  by  multiplying 
the  rate  by  the  number  of  pounds  we  will  obtain  the 
cost. 


2.     What  Avill  £1480  lis.  8d.  cost  in  currency,  exchange  at  4.85i  ? 

Ans.  $7217.36. 


746 


SOULE  S    raiLOSOPHIC    PRACTICAL    MATHEMATICS. 


SOLUTION. 

OPERATION 

To  reduce  the  lis.  Sd.  to  the  decimal  of  a  pound, 
n  8 

.55  +    .033J-  =  £.583i. 


OPERATION' 

To  reduce  pounds  to  dollars. 

£1486.5833 

4.85^  rate  of  exchange. 

72099'29005 
7432916 

$7217.361921     Ans. 


Note. — See  Article  869,  page  450,  for  an  explanation  for  reducing  shillings  and  pence  to  tho 
decimal  of  a  pound. 

3.    What  will  £21408  3s.  London  exchange  cost  at  5.32 1 


Ans.  $114210.56. 

OPERATION 

To  reduce  the  pounds  to  dollars. 

£21468.15 
5.33 


OPERATION 

To  reduce  the  3s.  to  the  decimal  of  a  pound. 

3s.  X  .05  =  .15  pounds. 

or,  3s.  —  20  =  .15  jjounds. 

$114210.5580    Ans. 

Note.— In  this  .and  m.any  of  the  problems  in  exch.ange  operations,  we  use  the  multiplicand 
ks  the  multiplier.     This  we  do  in  conformity  with  custom  aud  for  practical  convenience. 

4.  A  mercliant  imports  an  invoice  of  goods  from  England,  amounting  to 
£1723  14s.  lid.  What  is  it  worth  in  United  States  money  at  the  legal  value  of  the 
fcovereigu?  Ans.  $8388.61. 


FIRST   SOLUTION. 
OPERATION 

To  reduce  14s.  lid.  to  the  decimal  of  a  pound. 


14 
5 

.70 


11 

.045J 
.70 

£    .745* 


OPERATION 

To  reduce  pounds  to  dollars. 

£  1723.746 

$4.8665  legal  v.ilue  of  £1. 


$8388.6099090    Ans. 


SECOND  OPERATION. 

£1723.746 

$4 


i  of  $6894.984 

1^  of    1378.996 

114.916 


amount  considering  1  pound  worth  $4.    $4.00 
:  ^  of  $4  valuation  of  1  pound,  .80 

:  1^  of  ^  of  $4  valuation  of  1  pound,  6j 


.896  =  amount  considering  £1  to  be  =  to 


$4.86J 


Explanation. — This  opera- 
tion  is  based  upon  the  pre- 
sumption that  £1.  =  $4.86}- 
instead  of  $4.8665;  hence  the 
slight  excess,  of  28+c.  in  the 
.answer.  It  is  a  shorter  /.leth- 
od  than  the  first  and  is  used 
by  many  account.ants.  As 
sliowu  in  the    operation    we 


first  assume  the  pound  sterling  to  be  worth  $4,  then  add  to  that  value  ^  of  the  $4  valuation,  and 
also  Vi  of  the  ^  of  the  $4  value.  . 

5.    A  merchant  imported  an  invoice  of  English  goods,  amounting  to  £12400 
2s.  8d.    The  importation  duty  is  40  per  cent  on  the  invoice  price.    What  is  the  cost 


*  ENGLISH    EXCHANGE.  747 

of  tlie  invoice  iu  United  States  money,  including  tlie  duty  but  no  otlier  importing 
charges?  Ans.  $84483.35. 

OPERATION   INDICATED. 

£12400.1333 

4.8665  , 


$60345.25       cost  of  invoice. 
24138.10        =  40  J'o' duty  on  cost. 


$84483.35       importing  cost. 

6.  By  telegram,  we  learn  tliat  medium  cotton  is  selling  iu  Liverpool  at  7Jd. 
Sterling  exchange  is  5.29.  The  freight  on  cotton  to  Liverpool  in  .sailing  bottoms 
is  /ed.  per  pound. 

From  these  figures,  making  no  allowance  for  the  insurance,  interest  on 
money,  and  the  charges  to  eflect  sales,  what  is  the  equivalent  Aalue  of  cotton  of 
like  classification  in  U.  S,  money?  Ans.    Practically  16^/. 

OPERATION, 

7id.       Liverpool  price. 

■r^d.       freight  cliarges  deducted. 

7ftd.  =  7.3125d.  wliich  —  240  =  .030469  pounds. 
.030469  X  5.29  the  rate  of  exchange  gives  $.16118101,  practically  lejc.     Ans. 

d.  d.  Explanation. — As  many  producers 

7.3125  or  thus:        16  1117  of   cotton    unac(]uainted    with    the 

240    5.29  24U  |  5.29  customs  and  workiiig.s  of  commerce, 

•  I look  upon  and  regard  these  figures 

$.1611797    Ans.  |  $.1611797    Ans.  as  the  price  of  cotton   iu  our  own 

market,  we  will  remark  that  t< 
closely  approximate  even,  from  data  of  this  character,  the  price  of  cotton  in  our  market,  we  musj 
allow  for  the  insurance  to  ship  cotton  to  Liverpool,  the  various  charges  to  effect  sales  in  Liverpool 
the  interest  on  money,  and  in  addition  to  this  we  must  take  into  consideration  the  probability  oj 
the  increase  or  decrease  iu  the  Liverpool  jirice  of  cottou,  the  rate  of  exchange,  and  lastly  th« 
margin  for  profit  to  the  buyer  must  be  considered. 

7.  Good  middling  cotton  is  quoted  in  Liverpool  at  5Jd.  "What  is  the  eqniva 
lent  price  in  United  States  money,  the  rate  of  exchange  being  4.90  and  no  allow 
ancesmade?  Ans.  $.10718+,  practically  10|^/. 

A  NEW  METHOD. 

1312.     The  following  new  method  may  be  often  used  with  advantage  when  th( 
amount  is  in  pence  and  the  rate  of  exchange  is  near  4.86. 

1.    What  is  the  equivalent  in  United  States  money  of  8d.,  exchange  4.88? 

Ans.  $.1626§,  practically  16J/. 

Explanation. — By  this  method  of  work  we  first  multiply 
the  pence  by  24'i,c.   which   is  the    value    of   Id.    when 
exchange  is  4.86 ;  then  to  that  part  of  the  product  obtain- 
ed by  multiplying  by  ^5,  we  add  to  or  subtract  i^r  part 
$.1626f  of  the  same  for  every  jc.  increase  or  decrease  of  tlie  rate 

of  exchange  from  4.86.  In  this  example  the  rate  ol 
exchange  being  4.88,  the  difference  between  the  same  and  4.86  in  half-cents  is  4,  hence  we  multiply 
the  .2c.  by  T^.  . 


3PERATI0N. 

8d 

X 

2c. 

=  16c. 

8d 

X 

-Ac 

=  00.2 

2c. 

X 

A 

=  0006J 

748 


SOULE  S    rilll.OSOPHIC    rRACTICAI,    MATHEMATICS. 


2.     What  is  the  cquiviileut  m  United  States  gold  of  J)id.,  cxcliaiige  4.87i? 

Ans.     Practically  10|/. 


OPEKATION 

By  the  new  niothod. 

9Jd   X  24^0.  =  in.2375c. 
.2375c.  X    Ac.  =  00.0593+ 


19.2968+    Ails. 


OPERATION 

By  the  first  method. 

d. 
2  I  19 
240    4.875 


I  .19296+0.     Ans. 

3.    What  is  the  equivalent  iu  cents  of  2s.  Gd.,  exchange  at  4.90? 

Ans.  61i/. 

OPEliATION 


By  the  new  method. 

2s.  6d.  =  30d.  X  2,i,c.  =  60.75c. 
.75c.   X    A     =      .5 


61.25c.     Ans. 


OPERATION 

By  the  first  method. 

d. 

I  30 
240  I  4.90 

I  61ic.     Ans. 


PROBLEMS. 

1.  What  is  the  cost  of  a  bill  for  £05241.3.  11,  exchange  $4.88J  ? 

Ans"  $318703.24. 

2.  What  is  the  imjiorting  cost  of  an  English  invoice  of  goods  amounting  to 
JE65241  3.  11  ?  Ans.  $317496.28. 

3.  Exchange  on  Liverpool  is  quoted  at  490,  and  wheat  in  Liverpool  is  quoted 
at  42s.  8d.  i)er  quarter,  what  is  the  equivalent  value  per  bushel  in  the  United  States 
monetary  unit?  Ans.  $1.30§. 

OPERATION   INDICATED. 


240 

480 


512 
4.90 

00 


Note. — 480  pounds  is  one  q^uarter.     See  Table  of  Weights  and  Measures. 

4.  A  merchant  imported  from  J.  Wood,  London,  goods  amounting  to  £460 
14.  10,  on  which  there  is  a  duty  of  50  per  cent.  What  is  the  cost  in  United  States 
money?  Ans.  $3363.30. 

Note. — Coninicrcial  students  should  make  the  Journal  and  Cash  Book  entries  for  the  above 
.  transactions. 

5.  A  merchant  bought  and  remitted  to  J.  Wood,  London,  a  bill  for  £460 
14s.  lOd.  when  exchange  was  4.88.     What  was  the  cost?  Ans.  $2248.42. 

Note. — Commercial  students  should  make  the  Cash  Book  entries  for  the  above  transactions. 
TO  EEDUCE  THE  UNITED  STATES  MONETARY  UNIT  TO  ENGLISH. 


1313.     1.    What  amount  of  English  exchange  can  I  buy  for  $24375;  the  rate 
of  exchange  being  4.87i  ?  Ans.  £5000. 


OPERATION. 

4.875)  $24375.000  (£5000.     Ans. 
24375 


Explanation. — Since  the  rate  of  exchange  is  the  dollar  and  cent 
exchange  equivalent  of  £1  Sterling,  it  is  evident  that  we  may 
buy  as  many  pounds  as  $24375  is  times  equal  to  $4.87^. 


.*' 


ENGLISH     EXCHANGE. 


749 


2.  What  amount  of  English  exchange  can  be  bought  for  $40250.65,  the  rate 
of  exchange  being  5.32  ?  Ans.  £7565  ISs.  3d. 

OPERATION. 

$40250.65  —  0.32  =  £7565.9116. 

20 

8.  18.  2320 
12 

d.  2.  784 

3.  What  will  be  the  face  of  a  60  day  bill  of  exchange  on  Liverpool,  that  can 
be  bought  for  $7217.36,  exchange  at  $4.S5J  ?  Ans.  £1486  lis.  8d, 

4.  A  merchant  owes  a  London  correspondent  $8119.53  currency,  which  he 
desires  to  invest  and  remit.  The  rate  of  exchange  is  4.85J,  and  gold  is  12i  per 
cent  premium.    What  is  the  face  of  the  bill?  Ans.  £1480  lis.  8d. 

SECOND   SOLUTION. 

$4,855         gcl'l  rate  of  exchange. 
.606875  =  12i  %  premium  gold. 

5.461875   currenov  rate  of  exchange. 
$8119.53  —  5.461875  =  £1486.5829 

20 

s.  11.6580 
12 


FIRST 

SOLUTION. 

112} 

4.855 

$ 

100 
8119.530 

£1486.5829 
20 

8.  11.6580 
12 

d.  7.896 

Operation  to  buy  gold. 

$ 

100 
112}     8119.53 

$ 

7217.36  gold. 

d.  7.896 


THIRD   SOLUTION. 


4.855)  $7217.36  (£1486.5829 
20 


s.  11.6580 
12 


d.  7.896 

5.  A  Manchester  merchant  consigned  to  us  an  invoice  of  goods  which,  pursu- 
ant to  his  order,  were  sold  for  his  account.  The  net  i^roceeds  are  $11240.20  which 
we  wish  to  remit;  exchange  is  5.41.    What  is  the  face  of  the  bill  ? 

Ans.  £2077  13s.  5d. 

6.  I  have  a  balance  of  $4210.12  belonging  to  a  Liverpool  merchant,  which  I 
am  instructed  to  invest  in  60  day  Engli.sh  exchange  and  remit.  The  rate  of 
exchange  is  $4.84^;  I  charge  J  per  cent  commission  for  investing;  what  is  my 
commission  and  what  is  the  face  of  the  bill  ? 

Ans.  $20.95  commission.     £865  Is.  8d.+  face  of  bill. 


PER  CENT  AND  INTEREST  ON  ENGLISH  MONET. 


1314.  For  the  method  of  computing  per  cent  on  English  money,  see  Article 
868,  page  449,  and  for  the  method  of  computing  interest  on  English  money  see 
Article  1135,  page  569.  For  accounts  current  and  interest  accounts  in  English 
money,  see  Article  1209,  page  661. 


y5o  soule's  riiii.osoPHic  practical  mathematics.  * 

TO  FIND  TUE  EATE  OF  EXCHANGE  WHEN  THE  FACE  OF  THE   BILL 

AND  THE  COST  AliE  GIVEN. 

1315.  1.  The  face  of  a  bill  on  Londou  was  £5000,  tbe  cost  was  $24375. 
What  was  tlio  rate  of  exchange  ?  Ans.  $4.87J. 

OPERATION  INDICATED. 

5000)  24375  ($4,875    Ans. 

2.     Tbe  cost  of  £4G0  14s.  lOd.  was  $2248.42.     What  was  the  rate  of  exchange  ? 

Ans.  $4.88. 

PEACTICAL  MISCELLANEOUS  PEOBLEMS, 

1316.  1.  A  Liverpool  cotton  factor  sold  200  bales  of  cotton,  weighing  90000 
pounds,  at  SJd.  per  pound.  He  charges  2J  per  cent  commission  for  selling.  What 
is  his  commission,  and  what  are  the  net  proceeds'? 

Ans.  £82  Os.  7Jd.  commission. 

£3199  4s.  4Jd.  net  proceeds. 

riE«T   OPERATION. 

90000       -weiKht.  Commission  reduced  to  £.  s.  and  d. 


8id.  price. 


810000 


12  )  19687.50d.   commission  as  above. 


22500  20  )     1640. 7 Jd.   =  1640s.   7id. 

787500a.    value  in  d.  £82 

2i%  rate  of  commission. 


19687.50d.  commission  in  d. 
787500.00     value  as  above. 


£82  Os.  7Jd.   commission. 


12  )  767812.50d.  ~l  proceeds  in  d. 

reduced  to  £. 

20  )     63984.4id.   J-     s.  and  d. 

£3199.4s.      J 

SECOND    OPERATION. 

90000  lbs.  weight  of  cotton.  £3281  5s. 

8|d.    price  per  pound.  2i?iJ'   rate  of  commission. 


810000  £82.  Os.  7Jd.   commission. 

22500  3281.  5     0        sales  of  cotton. 


12  )  787500d.   "lvalue   in    d.  £3199.  4s.  4id.   net  proceeds. 

I      reduced  to 

20  )     65625s.      }     £.  aud  s. 

£3281.5s.J 
Note. — To  compute  2J  per  cent  commission,  consider  the  -whole  amount  100  per  cent  and 
divide  by  40,  as  2^  per  cent  is  -^i;  of  100.     Or  first  compute  the  commission  at  5  per  cent  and  divide 
by  2. 

2.  A  merchant  in  New  Orleans  has  a  balance  of  $5000  currency  due  to  his 
Liverpool  correspondent,  and  is  instructed  to  invest  the  same  in  60  day  sterling 
exchange  and  remit.  The  rate  of  exchange  is  $4.86 ;  the  New  Orleans  merchant 
charges  1  per  cent  commission  for  investing.    What  is  the  face  of  the  bill  ? 

Ans.    £1018  12s.  5d. 


ENGLISH     EXCHANGE. 


75i 


KIRST   OPERATION 

To  take  out  1  per  cent  commission. 

$100  investment  assumed. 
1   ^  1  %  commission. 

$101   cask  required  to  invest  $100. 


I  100 
101     5000 


SECOND   OPERATIOir 


To  convert  dollars  to  pounds. 


4.86 


4950.50 


£1018.6214 
20 


12.4280 
12 


5.136 


I  $4950.50 

3.  What  is  the  amount  of  duty  on  the  folio-wing  invoice,  the  United  States 
tariff  rates  being  IJ/  per  pound  on  the  hoop  iron,  and  10/  per  pound  ou  the  files 
not  over  10  inche.s  long,  and  30  per  cent  on  the  cost  of  same;  and  6/  on  the  files 
over  10  inches  long,  and  30  per  cent  on  the  cost  of  same  ?  Ans.  $270.98. 

BiKMiNGHAM,  17  September,  1895. 
Messrs.  Eand  &  Bercegeay. 

Bot.  of  Jno.  Ludlow  &  Sons. 


N.  O. 


Weight. 

C.  Q.  L. 
12    8 


Weight. 


C.   Q.   L. 

12    0 


7.'>  buDdlea  Elephaut  Hm>p  Iron 

40 

30 

40         "  ••  ••         ■• 

50 


.  235  Bundles. 

IJ  by  18 
H  by  17 
l|by  16 
2J  by  14 
3    by  13 


19 
0 
13 


At  £7  15a.  per  ton, 
Delivered  free  in  Liverpool. 


1 
1 

0 
5 

0 
0 

0 
0 

5 

17 

2 

0 

.  Cask   348. 


48  doz.  W.  &.  S.  Butchers  hand  saw  Taper  Files  3J  in.  3/4 

60     ••                   "                          ■•                   ■'  4    in.  3/9 

75     "                     "                            •■                    ••  4Jin.  4/3 

6     "                     "                    mill                        •'  10    in.  ll/'J 


Discount,  35  per  cent.  ...... 

doz.  W.  &  S.  Butchers  mill  saw  Taper  Files  12    in.      "JO/e 

14    in.        23/ 


Cask, 


Discount,  32i  per  cent, 
7/9  ;  carriage  tu  Liverpool,  5/3, 


EECAPITDLATION. 


Amount  of  235  bundles, 
AmouDt  of  cask  348, 


CHAItGES. 

12    9  114 

Commission,  2^  per  cent  on  £45  10s.  7d. ;  3  per  cent  on  £35  9s.  9d. 


Shipping,  port  charges  and  bill  lading. 
Consul's  certificate,      .        .        .        - 


Due  in  cash  this  day, 


£. 

S. 

d. 

£ 

s. 

45 

10 

7 

45 

10 

8 

11 

5 

15 

18 

9 

i 

10 

U 

38 

14 

3 

13 

10 

11 

25 

3 

7 

8 

C 

6 

18 

14 

6 

6 

4 

13 

1 

9 
35 

13 
13 

9 

45 

10 

35 

9 

81 

0 

2 

4 

K\ 

4 

1 

0 
11 

84 

15 

JUG.  LUDLOW  &  SOJfS. 


752 


souLE  s  rniLosopiiic  practical  mathematics. 


OPEUATION. 

Weight  of  iron,  5  t.  17  c.  2  qr.  0  lb,  =  13160  lbs.  at  lie 

"       "   files  uot  over  10  in.  long,  1  c.  2  qr.  8  lbs.  =  176  lbs.  at  10c. 
"       "   liles  over  10  in.  long,  1  v.  2  qr.  0  lb.  =  168  lbs.  at  6e. 

Cost  of  files,  i"35  9s.  9il.  at  $4.8665  =  $172.70 ;  30%  on  $173.  is    - 

Amount  of  duty  ....... 


.*17.60 
10.08 
51.90 


$197.40 


79.58 


$276.98 


TO  FIND  COST  OF  DIFFEEENT   ARTICLES   OF   IMPORTED  GOODS. 


1317.  1.  To  ascertain  the  cost  of  the  different  articles  of  iinx)orted  goods,  the 
amount  of  all  the  charges  is  first  found,  and  then  the  invoice  price  of  each  article  is 
increased  in  the  proportion  that  the  charges  bear  to  the  invoice  price  of  the  goods. 

When  the  cost  is  thus  found  the  profits  are  then  added. 

In  foreign  invoices  it  is  frequently  convenient  to  first  find  the  per  cent  that 
the  charges  are  of  the  invoice,  and  then  increase  the  cost  of  each  article  according 
to  that  per  cent. 

2.  A  merchant  imported  from  Birmingham,  England,  an  invoice  of  goods 
amounting  to  £1723  14s.  lid.  In  the  invoice  were  12  dozen  knives  which  cost  £28 
8s.  Gd  ;  allowing  the  duty  to  be  40  per  cent  and  the  freight  to  be  $234.90,  what  was 
the  per  cent  of  freight  on  invoice  and  duty,  and  what  was  the  retail  selling  price 
per  knife  to  gain  25  per  cent  on  the  imjiortation  cost  ? 

Ans.  2%.    $1.71  retail  selling  price. 


OPKRATION. 

£1723.14.  11  =  £1723.746  x  $4.8665  $8388.61 

40%  duty  on  $8388.61  •  3355.44 

Cost  of  whole  invoice,  $11744.05 

£28  8s.  6(1.  £28.425  X  $4.8665 
40%'  duty  on  $138.33, 


Cost  of  12  dozen  knives, 
2%  freight  on  knives, 


$138.33 
55.33-f 

$193.66+ 
3.87+ 


N.  O.  cost  of  12  doz.  including  frt.  $197..53+ 
25%  gain  on  N.  O.  cost. 


Price  of  12  doz.  knives, 
Price  of  1  doz.  knives. 
Retail  jirice  of  1  knife. 


49.38+ 
12  )  246.92— 


12  )     20.57+ 

$1.71+    Ans. 


OPERATION 

To  find  the  rate  per  cent  of  freight. 

I  234.90 
11744.05  I  100 

I  2.00016%'',   practically  2%. 


NoTK  1. — See  pages  456  and  457,  for  Practical  Work,  Marking  Goods  and  Computing  the  Rate 
Per  Cent  of  Freight. 

Note  2. — .See  Customhouse  Business  for  Importing  Goods,  pages 503  to  508  for  other  problems 
in  importing  goods. 


ENGLISH    EXCHANGE. 


753 


COMPUTATIONS  OF  FREIGHT  IN"  ENGLISH  MONEY. 


1318.     1.     What  is  the  freight  on  540  bales  cotton,  -weighing  2430S4  pounds,  at 


|d.  per  pound  from  New  Orleans  to  Liverpool  ? 


Ans.  £G3o  Os.  8d. 


FIRST   OPERATION. 

243084 


8  )  1215420 
12  ) 
20  ) 


12  )     151927.5d. 

12660   =  8d. 


£633  Os. 


SECOND  OPERATION. 

D. 
8 
12  243084 
20 


£633.0313 
20 


Os.  6260 
12 

7d.  512 


2.    What  is  the  freight  and  primage  on  68  bales  cotton,   weighing  31416 
pounds,  at  -ffd.  per  pound;  primage  5  per  cent  ?  Ans.  £122  14s.  4id.,  freight. 

£6  2s.  9d.,  primage. 


FIRST   OPERATION. 


) 

31416 

lid. 

16 

471240 

12 

) 

29452.5d. 

20 

) 

2454     = 

4id. 

£122        = 

148. 

£122  14s.  4id.     amount  of  freight. 
5  %    rate  of  primage. 


£6.13  .11  .101     amount  of  primage. 
20 


2s.71 
12 

8d.62i 


NoTE.^ — Primage  is  a  charge  in  addition  to  the  freight; 
originally  it  was  a  gratuity  to  the  captain  of  the  vessel  for  his 
particular  care  of  the  goods.  It  now  belongs  to  the  owners  or 
freighters  of  the  vessel,  unless  by  agreement,  it  is  allowed  in  whole 
or  in  part,  to  the  caiitain. 


SECOND   OPERATION 
To  find  freight. 


16 
12 
20 


D. 

15 
31416 


£122.7188 
20 

14s.  376 
12 


4d.  512 


3.  What  will  be  realized  in  United  States  money  from  the  freight  and 
primage  of  21000  bushels  of  wheat  from  New  Orleans  to  Liverpool;  freight  12id. 
per  bushel,  primage  5  per  cent;  exchange  being  5.40?  Ans.  $6201.56. 


FIRST   OPERATION. 


12id.    freight. 
|d.   =  5  %  primage  on  12id. 

131d,   freight  and  primage  per  bushel. 


12 
20 


D. 

105 

21000 

5.40 

$6201.56    Ans. 


Explanation. — We  here  reduce  and  place  the 
rate  of  freight  and  primage  IS^d.  on  our  state- 
ment line,  and  multiply  the  same  by  21000,  the 
number  of  bushels  on  which  freight  and  primape 
are  charged.  This  gives  us  the  amount  of 
freight  and  primage  in  pence,  which  we  reduce 
to  pounds  by  dividing  by  12  and  20,  or  240. 
Then,  as  the  result  thus  obtained  is  pounds,  or 
pounds  and  decimals  of  pounds,  we  reduce  the 
same  to  United  States  money  by  multiplying  by 
the  rate  of  exchange  5.10. 


754 


SOULE  S    rHILOSOPIIIG    I'RACTICAL    MATHEMATICS. 


SECOND    OPERATION. 


21000  biishols. 

12Jil.  rate  of  freight. 

262500d.  freight. 

13r25d.  =  5  ^'j  primage. 


£1148  8s.  Oil.  =  £1148.1375+ 
5.40 


J;G201.56L'50 


Ans. 


12  )  275G25(1.      freight  and  primage. 
20  )     22968  9d. 
£1148  8s. 

4.  Wliat  is  tlie  freight  and  primage  in  United  State.s  money  on  25000 
bnsliels  of  corn  from  New  Orleans  to  Liverpool,  at  24s.  i)er  imperial  quarter  of  4S0 
pounds;  primage  5  jier  cent,  and  exchange  4.8G?  Ans.  $17860.50. 


riHST   OPERATION. 


480 


100 
•20 


S. 

24 

56 

2r)000 

105 

4.86 

|;17860.50  Ans. 


SECOND   OPERATION. 

25000  X  56  =  1400000  pounds. 


480 
20 


1400000 
24 

70000s. 


£3500   amount  of  freight. 

175   =  5  ^0  primage  on  £3500. 

£3675     freight  &  prim,  in  LiverpooL 
4.86     rate  of  exchange. 


$1786.50    Ans. 


TO  FIND  THE  PUECHASE  PKICE  OF  ARTICLES  TO  FILL  THE  ORDERS 

OF  ENGLISH  MERCHANTS,  AT  A  SPECIFIED   PRICE,  FREE  OF 

ALL  CHARGES  WHEN  DISCHARGED  IN  BRITISH  PORTS. 

1319.  1.  A  Liverpool  correspondent  instructs  me  to  purchase  and  ship  to 
him  7200  centals  of  white  wheat,  on  condition  that  the  total  cost,  including  all 
expenses  for  purchasing,  shipping,  etc.,  shall  not  exceed  lis.  8d.  per  cental  when 
placed  on  the  dock  at  Liverijool.  The  freight  is  25s.  per  imperial  quarter,  shipping 
expenses  and  commission  for  purchasing  estimated  at  9  per  cent;  exchange  $4.90. 
What  is  the  maximum  price  to  be  paid  per  bushel  ?  Ans.  Sl-^^^^. 

SOLUTION. 

Statement  to  iind  the  freight  on  1  Cental. . 

8. 

I  25 
480  I  100 

58.  2Jd.  freight  on  1  cental. 


11.S.  8(1.  Liverpool  price  per  cent.al. 
5    2^      freight  expenses. 


Cs.  51d.  ^  77id.  =  ^-pd.  to  pay  for  wheat  and  remaining  charges. 


ENGLISH    EXCHANGE. 


755 


STATEMENT 


To  find  tlm  purchasing  price  of  wheat. 


0 

155 

100 

60 

:.'40 

4.90 

109 

100 

87-4^sC.    Ans. 


Explanation. — We  first  find,  as  Bhown  in  the 
first  statement,  the  freifjht  on  100  jiounds,  and 
tlien  deduct  the  same  from  tlio  Liveri)Ool  limit 
of  price,  and  thus  obtain  6s.  D^d.  =  Hid.,  ■«  ith 
^vhich  ve  are  to  buy  the  wheat  and  pay  purchas- 
ini^  and  shipping  charges.  Then  to  find  the 
price  that  this  sum  will  pay  per  bushel,  after 
being  reduced  to  American  money  at  the  rate 
of  exchange  given,  and  paying  all  charges,  we 
place  the  -f-d.  (771)  on  our  statement  line,  and 
reason  thus:  If  100  pounds  (1  cental)  cost  -"-l^d., 
1  pound  will  cost  the  100th  part,  and  60  pounds 
or  1  bushel  will  cost  60  times  as  much.     Then 


to  convert  the  pence  to  pounds,  we  divide  by  240;  to  convert  the  English  monetary  unit  to 
American,  we  mnlti]ily  by  the  rate  of  exchange  4.90,  then  to  deduct  the  shipping  expenses  and  our 
commission,  botli  being  on  the  same  value,  we  multiply  by  100  and  divide  by  109.  The  final  result 
of  the  entire  statement  is  the  sum  that  we  can  pay  for  one  bushel  of  wheat  and  comply  with  the 
conditions  of  the  problem. 

2.  A  London  correspondent  advises  tlie  purchase  and  shipment  of  5,000 
bbls.  of  Extra  Flour  at  32  s.,  6d.  per  bbl.  C.  I.  F.  (Tliat  is  with  Cost,  Insurance, 
Freight  and  all  other  charges  to  deliver  the  same  on  the  wharf  at  London,  paid.) 

The  freight  is  2  s.  8  d.  per  bbl. ;  primage,  5%  ;  all  other  shipping  expenses, 
including  Insurance,  Freight,  Drayage,  Labor,  etc.,  are  estimated  at  4%  on  first 
cost;  Commission  for  purchasing,  5%  ;  Exchange,  $4.85.  What  price  can  be  paid 
for  the  flour  in  the  American  market.  Ans.    $0.5954+ . 


2s.   8d. 

coy 


freight, 
primage. 


2s.  8d.     freight. 
1|       primage. 


1.60  =  lid.   primage. 


2s.  9|d. 


OPERATION. 

D. 

32s 

5 

1782 

2 

240 

4.85 

104 

100 

29s 

105 

100 

6d.      Liverpool  price. 
9*        deducted. 

8id.   =  3561d.  =  ^^A. 


6.595+    Ans. 

To  reverse  the  question  and  operation,  and  find  the  London  cost  of  floiif 
bought  here  at  $0.5954+,  with  all  the  specified  charges  added,  we  produce  the 
following  figures : 


100 

100 

4.85 


FIRST    OPERATION. 

$ 

6.5954  cost  of  1  barrel  in  New  Orleans. 

104 

105 


£1.485  =  £1 


9s. 
2 


8.4d. 

9.6       freight  and  primage. 


£1     12s.     6d.       London  cost. 
SECOND    OPERATION. 


Ans. 


6.5954  purchasing  price  of  flour  in  New  Orleans. 
.2638  =  i%  shipping  expenses  on  $6.5954. 


$6.8592   cost  of  flour  with  4%  shipping  charges  added. 
.3429   =  5j.'(,'  commission. 

$7.2021   cost  of  flour  in  London  in  U.  S.  money,  freight  and  primage  unpaid. 


4.85  I  7.2021 

£1.485    ■■ 
20 


:£1 


9.700 
12 


9s.     8.4 

2       9.6  freight  and  primage  added. 


£1    12s.    6d.  London  cost.    Ans. 


8d.  4 


^56 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


3.  American  tallow  is  quoted  in  London  at  44s.  Cd.  per  cwt.  Allowing  $12^ 
per  American  short  ton  for  freight,  5  per  cent  for  primage,  3  per  cent  of  prime  cost 
for  shipping  exjjenses,  and  5  per  ('ent  of  sales  for  London  charges,  what  is  the 
maximum  price  to  be  paid  in  New  Orleans  when  exchange  is  4.80,  in  order  to  be 
sold  at  the  London  quotation,  and  a  gain  of  10  jier  cent  on  prime  cost  realized  ? 

Ans.  $.0743599. 


44s. 


6d.        London  selling  price. 
5%       rate  of  Loudon  charges. 


2s.    2.'o<i-   commission. 


OPERATION. 

44s.     6d.        London  selling  price. 
2       2i"(j       London  charges. 


4l2s.     Si'jd.    =  507i'od.  London  net  proceeds,  per  cwt. 


Statement  to  find  the  net  proceeds  of  1  pound  iu  U.  S.  mouey. 


10 
240 
112 


5073 

4.80 


$.0905892-1-   net  proceeds  of  1  pound. 
Operation  to  find  the  freight  and  primage  on  1  pound. 


I'reiglit  on  1  ton,     - 
Primage  at  5%  on  $12^, 


Freight  and  primage  on  2000  pounds,  ...  .        .        . 

Freight  and  primage  on  1  pound, -        - 

Which  we  deduct  from  the  London  proceeds  of  1  pound, 

N.  O.  cost,  including  3%  shipping  charges  and  105^  gain  on  prime  cost, 


$12.50 
.625 


2000  )  $13,125 


0065625 
.090,J892 

$.0840267 


Operation  to  deduct  the  3  per  cent  shipping  charges  and  10  per  cent  gain. 

$ 
I  100 
113  I  .0840267 

I  f.0713598    Ans.     N.  O.  prime  cost,  or  the  maximum  N.  O.  purchasing  price. 


EXCHANGE  BETWEEN  LONDON  AND  VAEIOUS  OTHEE  COUNTEIES. 


1320.     London,  England,  has  been  called  the  great  clearing  house  of  the  world. 

The  following  table  shows  the  rates  of  exchange  of  recent  dates  between 
London  and  the  cities  and  countries  named  in  the  list. 

It  should  be  remembered  that  these  rates  do  not  show  the  intrinsic  value  of 
the  difl'ercnt  monetary  units,  and  that  they  are  subject  to  a  limited  increase  or 
decrease  through  the  same  causes  that  govern  the  rate  of  exchange  in  our  own 
country.  They  are  presented  as  average  rates,  for  general  information  concerning 
exchange. 


PROBLEMS   INVOLVING   ENGLISH   EXCHANGE. 


A  New  Orleans  house  received  a  cable  message  offering  Sd.  for  100  bales  of 
cotton,  good  middliag,  good  color  aud  staple,  f.  i.  c.  and  6%  tare  in  Manchester- 
What  would  be  tlie  total  New  Orleaus  selling  price  aud  the  price  per  pouud,  allow- 
ing freight  at  2Sc-  per  100  lbs.,  and  insurance  at  J%,  exchange  at  4.80  f 

Aus.  $7400.72  selling  price.     Price  per  lb.  $.1480144. 

FIRST   OPERATION. 

100  B/C  =      50000  lbs. 

Less  ()%  tare,  3000    " 


47000  lbs.  net  -S)  Sd.=376000d.  at  4.86  = 


Operation  to  Reduce 
pence  to  dollars,  exchange  486 
d. 
376000 
240       4.86 


$7614.00 
Less  freight  on  50000  lbs.  ®  28/  per  C.  lbs.      140.00 


$7614.  +  10%  ($761.40)  =  $8375.40  the 

amount  to  insure. 
5%  insurance  on  $8375.40  = 


$7614.00 


240 


PROOF. 

8d. 
486 


Selling  price, 
$7400.72  -. 

8d.  ®  486 


50000  lbs.  =  $.14801+ 

$.162  =  price  of  1  lb. 
.00972  =  6%  tare. 


$7474.00 


73.28 


$7400.72 


$.162 


.15228  =  net  price. 
.0014657  =  J=  insurance. 


Operation  to  find  the  amount 
to  insure. 
$  .15228  =  net  price. 
.015228  =  10 7o  increase. 


.1508143  =  price  less  insurance. 
.0028  freight  at  28/  per  C.  pounds. 


$.1480143  net  price  of  1  pound. 
60000  =  lbs. 


$  .167508  =  anit.   to  insure  ® 
^1%  =  $.0014657. 


$7400.715  =  purchasing  price. 


SECOND     OPERATION. 

The  following  inaccurate  but  approximate  method  is  used 
by  some  cotton  house  accountants. 

100  B/C  =  50000  lbs. 
Less  6%  =    3000  " 


47000  'a>  8d.  = 
Less  freight  ®  28/  per  100  lbs.,  $140 
(at  Exchange,  480) 

Less  J%  ins.  on  £1566  plus  10% 


£1566—13—4 
29—  3—4 

£1537-10-0 
15—  1—5 


PROOF. 

8.d. 
.48  =  6% 


7.52 
.0724  =  ins. 


7.4476 
.14  =  freight,  Ex.  480. 


7.3076d. 
(7.3070d.  X  4.86) 
—  240  =$.1479789  per  lb. 
X   50000  lbs..  =  $7398.95 
Difference  caused  by 
the  decimals,        -        -   6 


Selling  price, £1522—8—7 

Exchange  ®  486  =         -        -        -        -     $7399.01 

$7399.01 

Note  1. — The  inaccuracy  of  this  method  consists  in  using  480  as  the  value  of  the  pound 
sterling,  in  calculating  the  freight  and  in  omitting  the  shillings  and  pence  on  amount  insured. 

Note  2. — The  first  operation  is  strictly  accurate  and  much  shorter  and  easier,  as  we  work 
in  dollars  and  cents,  instead  of  pounds,  shillings  and  pence,  after  the  first  item. 


756K 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


A  firm  iu  Manchester,  Eug.,  instructs  its  correspondent  in  New  Orleajis  to 
buy  100  bales  of  good  middling,  good  color  and  staple  cotton  on  the  following 
conditions:  8d.  per  pound  f.  i.  c.  and  6%  tare. 

What  price  per  pound  may  the  New  Orleans  house  pay,  allowing  $1.10  per 
bale  expenses  iu  New  Orleans,  50  cents  per  bale  commission,  30  cents  per  100  lbs. 
freight,  1%  insurance,  estimating  each  bale  to  weigh  500  1 

Ans.  for  1st  operation,  $.1427. 


"       "   2d 


.1448. 


FIRST   OPERATION. 


8d.  ®  Ex.  480  = 
1%  insurance  = 

.16  cents 
.0016 

6%  tare 

.1584 
9504 

30/  frt.  per  100  lbs. 
=  .3/  on  1  lb. 

.1489 
30 

50/  com,  on  500 
lbs.  =  .1/  on  1  lb. 

.1459 
10 

110/  exp.  on  500 
lbs.=.22/onllb. 

.1449 

22 

N.  O.  buying  price     .1427 


SKCOND   OPERATION. 

8d.  Ex.  48605  =  .1622+ 

1  fo  insurance  =  .0016 


6%  tare 

.1606 
96+ 

30/  frt.  per  100  lbs. 
=  .3/  on  1  lb. 

.1510 
30 

50/  com.  on  500  lbs. 
=  .1/  on  1  lb. 

.1480 
10 

110/  exp.  on  500  lbs. 
=  .22/  on  1  lb. 

.1470 

22 

N.  O.  buying  price 


.1448 


Note  1. — It  is  the  custom  of  cotton  buyers  in  problems  of  this  kind  to  base  their  calcula- 
tions on  a  480  valuation  of  the  pound  sterling,  thus  facilitating  the  operation  to  some  extent. 

Note  2.— The  difference  between  Exchange  480  and  48665  is  .0665c.  =  1.3854%'  or,  practi- 
cally 1^%.  Hence,  by  adding  If  %  to  the  480  Exchange  price  we  produce  the  48665  Exchange 
price  nearly. 

Note  3. — By  multiplying  the  .480  Exchange  price  by  the  current  rate  of  exchange  and 
dividing  the  product  by  480  we  would  produce  the  price  at  the  current  rate  of  exchange. 


ENGLISH   EXCHANGE.  757 

EXCHANGE  BETWEEN  LONDON  AND  YAEIOUS  OTHER  COUNTRIES. 

1.320.     Loudon,  England,  Las  been  called  the  great  clearing  house  of  the  world. 

The  following  table  shows  the  rates  of  exchange  of  recent  dates  between 
Loudon  and  the  cities  and  countries  named  iu  the  list. 

It  should  be  remembered  that  these  rates  do  not  show  the  intrinsic  value  of 
the  difierent  monetary  units,  and  that  they  are  subject  to  a  limited  increase  or 
decrease  through  the  same  causes  that  govern  the  rate  of  exchange  iu  our  own 
country.  They  are  presented  as  average  rates,  for  geueral  information  concerning 
exchange. 


Exchange  on  Paris 

"  "  Amsterdam 

"  "  Brussels 

"  "  Rome 

"  "  Rome 

"  "  Berlin 

"  "  Athens 

"  "  Vienna 

"  "  Madrid 

"  "  Lisbon 

"  "  St.  Petersburg 

"  "  Hong  Kong 

"  "  Shanghai 

"  "  Calcutta 

"  "  Rio  de  Janeiro 

"  "  Copenhagen 

"  "  New  York 


and  other  cities  in  France  25.25,  fr.  and  cts.  for  j£l. 

"  •'  Holland  12.3,  fl.  . and  stiv.  for  £1. 

"         "  "  "  Belgium  25.30,  fr.  and  cts.  for  £1. 

"         "  "  "  Italy  (gold)  25.40,  lire  and  cent  per  £L 

"         "  "  "  Italy  (paper)  28.10,  lire  and  cent  per  £1. 

"         "  "  "  Germany  20.55,  mks.  and  pf.  for  £1. 

"        "  "  "  Greece  28.25,  dr.  and  cts.  for  £1. 

"        •'  "  "  Austria  23.90,  crowns. 

"        "  "  "  Spain  25.25,  peseta  and  cts.  for  £1. 

"        "  "  "  Portugal  o2d.  for  1  Milreis. 

"        "■  "  "  Russia  30d.  for  1  Rouble. 

'•         "  "  "  China  46d.  for  1  Dollar. 

"  "  "  China  62d.  for  1  Tael.   . 

"         "  "  "  Inilia  22d,  for  1  Rupee. 

"        "  "  "  S.  America  26d  for  1  Milreis  of  Brazil. 

"         "  '•  "  Denmark  13d.  for  1  Crown. 

"         "  "  "  United  States  49d.  for  1  Dollar. 


Note  1. — The  milries  of  Portugal  is  equal  to  52id.  or  $1.08  which  is  a  much  larger  nnit  than 
the  milries  of  Brazil,  which  is  equal  to  54.6  cents.    The  table  for  Portugal  money  is  as  follows: 
1  milreis  =  2J  crusados  =  25  reales  =  1000  reis. 
1  =  10      "      =    400     " 

1       "      =     40     " 
Note  2. — By  the  decree  of  October  19,  1868,  the  money  of  Spain  was  assimilated  to  that  of 
France,  and  the  Peseta  of  reales  was  considered  as  equivalent  to  a  franc  and  was  taken  as  the 
monetary  unit. 

Note  3. — The  new  monetary  unit  of  Denmark  is  the  Krona  or  Crown  which  equals  100  Ore. 
1  Crown  =  26.8  cents  or  13id. 

Note  4. — The  money  of  account  of  the  Netherlands  (Holland)  is  the  florin  of  100  cents, 
worth  40.2  cents  United  States  gold.     The  florin  was  formerly  divided  into  20  stivers. 

London  Exchange  on  Paris. 

2.     What  is  the  cost  of  a  bill  on  Paris 
for  fr.  lOSlO.SO.  exchange  25.25  ? 

Ans.  £417  8s.  lid. 


London  Exchange  on  2few  Torlc. 

1.    What  is  the  cost  of  a  bill  on  New 
Tork  for  $5000,  exchange  at  49id.  ? 
Ans.  £1026  Os.  lOd. 

OPERATION. 

$500        face  of  bill. 
49id.  exchange. 

12  )  246250d. 

20  )    20520    lOd. 

£1026    Os. 


OPERATION. 

25.25)  10540.50  (£417.4455 
20 


9100 
12 


lOd.  92 


758 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


London  Exchange  on  Calcutta. 
3.    What  is  the  cost  of  a  bill  on  Calcutta  for  2000  company  rupees  4  annas  j 

Ans.  £191  13s.  lOd. 

Explanation. — Tlie  money  of  ac- 
count in  Calcutta  is  the  rupee,  -which 
is  divided  into  IG  annas,  and  each 
anna  into  12  pice. 


exchange  23d.  ? 

OPERATION. 

2000  Company  rupees  4  annas  =  2000.25       Company  rupees. 

23d. 

4     annas  :=  

lV  C.  K.  =  .35  12  )  4()005.75d. 


20  )     3833 
£191 


9M. 

138. 


4. 


London  Exchange  on  Shangliai. 
What  is  the  cost  of  a  bill  for  1250  taels  5  maces ;  exchange  58.Jd.  ? 


FIRST    OPERATION. 


1250  taels  5  maces  =  12505  maces  ^  125050 
candarines  =  1250500  cash. 


1250500 

5»i 


1000 
12  )  73154.250d. 


20  ) 


6096. 
£304 


2.250d. 


16s. 


2Jd. 


SECOND    OPERATION. 

1250  taels  5  maces  =  1250.5 

1250.5  X  oU  =  73154.25d. 

73154.25  reduced  to  £.  s.  and  d.  =  £304  16s. 

2id.  as  in  the  first  operation. 


Ans,  £304  ICs.  2id. 

Explanation. — Accounts  are  kept  in  China  in 
tael,  maces,  candarines  and  cash.  Tlio  table  is 
10  cash,  1  candarine,  10  candarines,  1  in.ace  and 
10  maces,  1  tael.  Tlie  tael  of  Shanghai  is  equiv- 
alent to  $1.14  -while  the  tael  of  Haik-wan  or 
customs  tael,  is  equivalent  to  $1.27. 

There  are  no  national  gold  and  silver  coins  in 
China.  The  only  money  made  is  the  cash,  and 
it  is  not  coined  but  cast.  It  is  composed  of  6 
parts  of  copper  and  4  of  lead.  It  is  circular  in 
form,  marked  on  one  side  only  and  has  a  squ.are 
hole  in  the  center,  by  -which  means  they  are 
carried  like  beads  on  a  string  or  wire.  Foreign 
coins  are  in  general  circuhation  in  the  ports 
open  to  the  commerce  of  the  world  and  they 
are  considered  as  bullion  rather  than  coin,  and 
are  usuilly  taken  by  weight.  Silver  ingots  are 
also  ui,ed  as  money,  being  weighed  by  the  unit; 
tael.     See  Exchange  on  China. 


TJOTTED  STATES  YALUE   OP  UNITED   STATES   STOCKS  AND  BOKDS, 

WHEN  QUOTED  IN  LONDON,  AND  THE   LONDON  VALUE  OP 

SAME  WHEN  QUOTED  IN  THE  UNITED  STATES. 


PROBLEMS. 

1321.     1.     United  States  4  per  cent  bonds  are  quoted  in  London  at  115J;  -what 
is  the  equivalent  quotation  in  the  United  States,  sterling  exchange  being  4.84? 

Ans. 


112.26f. 


OPERATION. 

4.86  X  115i  =  56133  —  5  ■. 


■■  112.26  J. 


KoT.E. — All  American  stocks  and  bonds  are  quoted  in  London  on  an  assumed  value  of  the 
pound  sterling.  This  assumed  value  is  $5,  modified  by  the  rate  of  exchange.  See  Stocks  and  Bonds 
in  this  book,  pages  708  and  709. 

2.  The  quotation  of  American  stocks  in  New  York  is  93J,  exchange  is  4.8CJ. 
What  is  the  equivalent  London  quotation  ?  Ans.  95.84-. 

OPERATION. 

93.25  X  $5  =  46625  —  4.86i  =  95.8+. 

3.  Louisiana  4  per  cent  consols  are  quoted  in  New  Orleans  at  101,  sterling 
exchange  is  4.85.     What  is  the  equivalent  London  quotation  ?  Ans.  104.12+, 


FRENCH    EXCHANGE. 
FRENCH  EXCHANGE. 


759 


KoTE, — See  French  Monetary  Table,  page  235. 

1323.  The  intrinsic  value  of  the  franc  and  the  United  States  legal  value  for 
Customhouse  computations,  is  $.193  or  19 1%/. 

The  rate  or  course  of  exclianfje  011  France  is  always  the  variable  number  of 
francs,  or  francs  and  centimes,  allowed  for  a  dollar. 

The  basis  for  the  computations  of  the  rate  of  exchange  is  5.20  francs  for  #1 
gold,  and  it  is  quoted  5.20  more  or  less  according  as  premium  is  charged  or  discount 
allowed. 

Note. — In  English  exchange  the  rate  is  the  dollar  and  cent  value  of  £1,  according  as 
premium  is  charged  or  discount  allowed.  But  in  French  exchange  the  rate  is  tlie  franc  and 
centime  value  of  SI.  Hence  in  English  exchange  the  higher  the  rate  the  dearer  is  exchange.  But 
in  French  exchange  the  higher  the  rate  the  cheaper  is  exchange,  and  the  lower  the  rate  the  higher  is 
exchange. 


TO  FIND  THE  BATE  OP  EXCHANGE  WHEN  A  PEEMIUM  IS  CHARGED. 


1333.  1.  Taking  into  consideration  the  balance  of  trade,  the  worth  of  money, 
etc.,  the  banker  or  seller  of  French  bills  determines  to  charge  say,  2  per  cent  i)re- 
mium.  What  would  be  the  gold  rate,  and  if  gold  is  15  per  cent  premium,  what 
would  be  the  currency  rate  for  French  bills  ?         Ans.  Fr.  5.09f ,  gold  rate. 

Fr.  4.433+,  currency  rate. 


OPERATION. 

Fr.  5.20  =  $1.00  par  of  exchange. 

.02  =  2  ^0  premium  added. 

$1.02  cost  of  5.20  exchange. 

Fr. 
I  5.20 
1.02     1.00 


Fr.  5.09^4  practically  fr.  5.09i. 


Explanation. — By  the  operation,  we  see  that 
the  cost  of  fr.  5.20  exchange,  at  2  per  cent 
premium,  is  $1.02.  The  reasoning  for  the  line 
statement  is  as  follows:  If  $1.02  huy  fr.  5.20, 
ic.  -will  huy  the  102d  part,  and  $1,  or  100c.  will 
huy  100  times  as  much,  which  is  practically  fr. 
5.09i. 


To  find  the  currency  rate  of  exchange  at  2  per  cent  premium,  gold  at  15  per 
cent  premium,  we  make  the  following  statements : 


FIRST    STATEMENT. 

Fr. 

5.20 
1.02  1.00 
115    100 


Fr.  4.433    Ans. 
Practically  fr.  4.43f. 


SECOND    STATEMENT. 

5.20  =  $1.00     p.ar  of  exchange. 
.02     =  2^0  premium. 

11.02     gold  cost  of  fr.  5.20. 
.1.53   =  155(,  premium  on  gold. 

$1,173  currency  cost  of  fr.  5.20  exchange. 

Fr. 
I  5.20 
1.173     1.000 


Fr.  4.433    Ans. 


76o  soule's  niiLosui'iiic  practicai.  mathematics.  * 

TO  FIND  THE  KATE  OF  EXCHANGE  WHEN  A  DISCOUNT  IS  ALLOWED. 

1324.     1.    Wliiit  would  be  the  gold  rate  if  tlie  banker  or  seller  allowed  1  percent 
discount,  and  if  gold  was  12^  per  cent  premium,  what  would  be  the  currency  latel 

Ans.  Fr.  5.252,  gold  rate.     Fr.  4.GCS9,  currency  rate* 

OPEUATIOX. 

Fr.  .5,20  =  $l.nO   ]i:ir  o.f  pxclian<;e.  Fr. 

.01    =  IjJj  discount.  I  o.'JO 

■ .99  I  1.00 

|.99   cost  of  fr.  5.20  exchange.  I  — 

I  Fr.  5.252-)-     Aiis.     Practically  fr.  5.25i. 

Operation  to  find  the  currency  rate. 

Fr.  Fr.  5.20  =  .fl.OO  Fr. 

5.20  .01  I  5.20 

1.00  1.11375  I  1.00000 

100  or,  .99  

.12375  =  12L"oPr<""-  |  Fr.  4.6689     Acs. 


.99 
112i 


Fr.  4.66S9     Ans. 


1.11375   cost  of  fr.  3.20. 


TO   FIND   THE    PREMIUM   CHARGED   OR   THE    DISCOUNT   ALLOWED, 
WHEN  THE  RATE  OF  EXCHANGE  AND  THE  PAR  ARE  GIVEN. 


1325.     1.     The  rate  of  exchange  is  5.09|},  the  jiar  is  5.20,  what  was  the  rate 
per  cent  premium  charged  1  Ans.  2%. 


OPERATION. 

.    .5.20       par  of  $1. 

G. 

Or,  with  the 

5.09^   =  rate  of  exchange  deducted. 

lOi? 

fractions  re- 



5.0911 

100 

duced  thus: 

.lOJJ   gain  on  fr.  5.09H. 

2% 

Ans. 

51 
26000 


G. 

5.20 
51 
100 


2%     Ans 

Note. — It  will  he  ohserved  that  the  seller  gains  on  his  investment,  of  5.09J}  francs  auil  not 
on  the  jiiir  of  5.20. 

2.  The  currency  rate  of  exchange  is  4.4.33,  the  premium  on  gold  is  15  per 
cent,  what  wa.s  the  rate  per  cent  premium  charged,  or  discount  allowed  by  the  seller 
of  the  exchange?  Ans.  2%  premium. 

OPERATION. 

Fr.   4.433  currency  rate.  G. 

.665  =  15^(^  j)remium  on  gold.  1.102 

■ 5.098  I    100 

Fr.  5.098  gold  rate.  

5.20  par  of  exchange.  I  2+      practically  2^^  premium. 

.102   gain  on  fr.  5.098. 


*  FRENCH    EXCHANGE.  761 

3.     Tlie  rate  of  exchange  is  5.25J,  and  tlie  i)ar  5.20,  \rliat  "was  the  rate  of  dis- 
count allowed  by  the  seller  of  the  exchange?  Aus.  1%  discount. 


OPERATION. 


It.   5.20     5>arof$l.  I,. 


b.25i  rate  of  exchange.  4 

. ■  2101 

.05^   loss  Ijy  the  exchange. 


21 

4 

100 

84o2    practically  1^0  ilisconiit. 


Note. — It  will  l>e  ohserveil  that  the  seller  loses  on  his  investment  of  5.25i  francs  and  not  oa 
the  imv  of  5.20. 

We  have  omitted  the  reasoning  in  the  line  statements  of  the  foregoing  ])rob- 
lems,  because  so  often  given  in  previous  operations. 

Pages  437  to  448  give  full  explanations  of  similar  percentage  line  statements. 

TO  COISrVERT  UNITED  STATES  MONEY  INTO  FRANCS. 

1326.     1.     What  is  the  equivalent  Aalue  in  francs  of  $5000,  exchange  5.15J? 

Ans.  Fr.  25775. 


OPERATION. 


Vr.  5.15J    exchange  value  of  $1. 
5000 


2575000 
2500 


Explanation. — As  the  rate  of  exchange  is  the 
nnmber  of  francs  and  centimes  given  for  $1, 
and  as  Ihe  rate  is  here  fr.  5.15^,  it  is  clear  that 
for  $5000  we  can  buy  5000  times  the  fr.  5.15^; 
hence  we  have  but  to    inulti|)ly    the    rate    of 


,,     o^-7r  on     A  exchange  by  dollars  to  convert  dollars  to  Iraucs. 

2.     What  amount  of  exchange  can  I  buy  for  $1410.60,  exchange  at  4.45J? 

Ans.  Fr.  G280.Gt>65. 

OPERATION. 

$1410.60  X  4.45i  =  fr.  G280.6965     Ans. 

TO  CONVERT  FRENCH  MONET  INTO  UNITED   STATES  DOLLARS. 

1327.     1.     What  is  the  equivalent  value  in  dollars  of  fr.  25775,  exchange  5.15 J  T 

Ans.  $5000. 

OPERATION.  Explanation. — As  the  fr.  5.15}  is  the  exchange 

equivalent  of  $1,  it  is  clear  that  in  fr.  25775  we 
5.15i  =  5.155  )  25775.000  (  $5000    Ans.  have  as  many  dollars  as  it  is  times  equal  to  fr. 

5.15*.     Hence  we  divide  the  francs  by  the  rate 
of  exchange. 

2.     What  is  the  cost  of  fr.  50500.40J,  exchange  fr.  4.05^  ? 

Ans.  $12446.18. 

OPERATION. 

2  I  101000.81  or,     4.0575  )  50500.403  (  12446.18    Aus. 

1623    4 


$12446.18    Alls. 


762  soule's  philosophic  practicai.  mathematics.  *■ 

TO    IXVEST    IK    FRANCS    FOR    A    CORRESPONDENT    AND    CHARGE 

COMMISSION. 

132S.     1.     Invested  $10000  in  French  exchange  at  0.20.     My  commission  on  the 
amount  invested  is  J  per  cent;  what  is  the  face  of  the  bill  bought? 

Alls.  Fr.  51741.L'-4,  face  of  bill. 

OPEKATION. 

$ 
100 


201 


10000 
5.20 


Fr.  51741.24     Ans. 

TO    rURCHASE    FRENCH    EXCHANGE    ON    ORDERS    FROM    DISTANT 

CITIES. 

1329.  1.  A  banker  in  New  York  instructs  his  correspondent  in  New  Orleans 
to  purchase  fr.  200000.  Ho  limits  the  price,  including  all  expenses,  at  5.20.  New 
Orleans  sight  exchange  on  New  Tork  at  the  time  is  f  per  cent  discount.  The  cor- 
i-espondcnt  who  makes  the  purchase  charges  J  per  cent  commission  on  prime  cost. 
What  is  the  cost  of  the  bill,  what  is  the  commission  and  discount,  and  what  is  the 
maximum  purchasing  rate  of  exchange  ? 

Ans.  $38461.54,  cost. 

$189.93,  commission. 

$284.90,  discount. 

$5.2GJ  maximum  purchasing  rate  of  exchange. 

SOLUTION. 

Operation  to  find  the  cost  of  the  200000  francs  at  5.20,  the  limited  price. 

Fr.  200000.00  -^  5.20  =  $;38461.54  cost,  including  4"^  commission  and  i%  discount. 

Operation  to  deduct  l^o  commission  and  f,"^  discount,  and  thus  produce  the  prime  cost. 

$100.00  prime  cost  assumed.  $ 

.50   =  iy'o  commission.  I  100 

.75   =  S,"o'  discount.  101.25  |  38461.54 


$101.25   =  cost  including  the  commission  and  discount.  |  $37986.71   prime  cost. 

189.93   =  i%  commission. 
284.90  =  j;^  discount. 

$38461.54     Ans.     Total  cost. 
Operation  to  find  the  maximum  purchasing  rate  of  exchange. 
37986.71  )  200000  (  5.2649,  practically  5.26J  maximum  purchasing  rate  of  exchange. 

2.  Suppose  in  the  above  problem,  the  price  had  been  5.25  and  the  rate  of 
New  Orleans  exchange  on  New  York  had  been  ^  per  cent  i)remium,  instead  of  | 
per  cent  discount,  what  would  have  been  the  full  cost  of  the  200000  francs  ? 

Ans.  $38381.67. 


*  •  FRENCH    EXCHANGE.  763 

SOLUTION. 

Operation  to  find  tlie  cost  of  tlie  200000  francs  at  5.25,  the  limited  price,  ^vitliout  considering  the 
f  per  cent  premium  or  the  i  per  cent  commission, 

Fr.  200000.00  -H  5.25  =  $38095.24  cost. 

This  cost  includes  the  J  per  cent  commission  but  is  exclusive  of  the  allowance 
to  be  made  for  the  f  per  cent  premium.  The  following  operation  will  allow  for  the 
I  per  cent  premium,  and  exclude  or  deduct  the  J  per  cent  commission. 

$100.00  prime  cost  assumed.  $ 

.50  =  i°^  commission.  I  100 

'  99.75  I  38095.24 

$100.50  cost  with  ^"5  commission  included.  I 

.75   =  f°(,' premium  on  |100  deducted.  |  $38190.72   prime  cost. 

190.95   =  4^5  commission  added. 

$99.75  cost  of  $100  including  i^^  commission 

and  allowing  for  the  $,"„  premium.  $38381.67    Ans.     Full  cost. 

286.43   =  4%  premium  deducted. 

$38095.24  net  cost  after  paying  i°^  com- 
mission and  receiving  J% 
premium  on  exchange. 

MISCELLANEOUS  PEACTICAL  PROBLEMS. 

1330.     1.     What  cost  fr.  2500.50,  exchange  5.33  ?  Ans.  $470.02. 

2.  Find  the  value  of  6400  francs  at  5.2U.  Ans.  $1227.23. 

3.  Find  the  value  of  5940.75  francs  at  5.1S3.  Ans.  $1145.20. 

4.  How  many  francs  can  be  bought  for  $10000  at  5.22J  1  Ans.  Fr.  62250. 

5.  Invest  $4000  in  francs  at    5.25i.  Ans.  Fr.  21010. 

6.  What  are  the  proceeds  of  a  draft  on  Paris  for  32500  francs,  exchange 
5.30,  and  brokerage  on  exchange  J  per  cent  ?  Ans.  $0124.40. 

OPERATION. 

32500  -H  5.30  =  $6132.07.  !,"„'  on  $6132.07  =  $7.07.  $6132.07  —  $7.67  =  $6124.40  Ans. 

7.  What  is  the  cost  of  a  draft  on  Paris  for  32500  francs,  exchange  5.30, 
brokerage  on  exchange  ^  per  cent  ?  Ans.  $0139.74. 

8.  What  will  be  the  face  of  a  bill  on  Havre  that  is  bought  for  $5000, 
exchange  being  5.35|?  Ans.     Fr.  2(;70.S.75. 

9.  Sold  exchange  for  12000  francs  on  Paris  through  a  broker;  exchange 
5.23,  brokerage  ^  per  cent.     What  were  the  proceeds  of  the  bill  ?    Ans.  $2288.72. 

10.  A  French  invoice  amounts  to  5S40.G0  francs ;  the  duty  on  same  is  50  per 
cent,  the  freight  is  $480.     What  is  the  importing  cost  ?  Ans.  $2170.85. 

OPERATION. 

Fr.  5840.60  X  $-193  =  $1127.24  invoice  +  SO^o'  duty  =  $1690.85,  +  $480  freight  =  $2170.85  import- 
ing cost. 

11.  What  cost  a  bill  for  the  above  fr.  5840,00,  exchange  at  5.23  ? 

Ans.  $1116.75. 


764  SOULe's    MIIIOSOI'HIC    IRACTICAI.    MATHEMATICS.    ■  * 

12.  Iiuporteil  from  France,  iiicrcliaudise  amouutiiig  to  fr.  8542.50.     Wliat  is 
the  valne  in  Uiiitod  States  inonoy  ?  Ans.  $1048.70. 

13.  Eouylit  tliroiigh  a  broker  a  bill  lor  i'r.  25800.     Exchange  5.10,  brokerage 
J  per  cent.     What  was  the  cost  of  the  bill  ?  Ans.  $5012.50. 

Note.— 25800.00  H- 5.16  =  §5000.    i,%' on  .$5000  =  $12.50.     $5000  +  $12.50  =  $5012.50. 

14.  Invested  tlirong;li  a  broker  $7400  in  French  exchange  at  5.24,  paid  ^  per 
cent  brokerage.     What  Wius  the  face  of  the  bill!  Ans.  Fr.  38727.58+. 

Note.— 17400.00  ^  lOOi  =  $7390.76  x  5.24  =  fr.  38727.5824. 

15.  Received  $1425  for  a  French  bill  that  was  sold  throngh   a   broker  at 
5.25 ;  the  brokerage  was  J  per  cent.    What  was  the  face  of  the  bill  ? 

Ans.  Fr.  7500. 

OPERATION    INDICATED. 
$1425  ^  99i  =  $1428.57+  X  5.25  =  fr.  7499.99+,  practically  fr.  7500. 

1331.     Paris  Exchange  on— 

Fr.  C. 
London,  short  and  3  months,  ...  25.20  per  £1  Sterling. 

Belflium,     "       "     3  '  ...  99.50  "    100  Francs. 

Italv,  <i       <i     3         .<  ...  89,50  "    100  Lire  (paper). 

Netherlands,  3  months,  ....  211.50  "    100  Florins. 

Germany,         3       "  ....  122.50  "    100  Marlis. 

Austria,  3       "  ....  105  "    lOO  Crowns. 

Spain,  3       "  ....  505  "100  Pesos. 

Portugal,         3      "  .        .        .        .  55O  "    100  Milreis. 

Russia,  3       "  -        -         -        -  312  "    100  Roubles. 

Note. — These  are  average  rates  and  may  he  more  or  less  according  to  circumstances  govern- 
ing exchange. 

EXCHANGE   ON  BELGIUM,    SWITZERLAND,   ITALY,    SPAIN,    GEEECE, 

AND  FINLAND. 

1332.  The  monetary  unit  of  all  these  nations  is  the  same,  in  value,  as  that  of 
France,  and  hence  the  operations  of  exchange  would  be  the  same  as  in  French 
exchange. 

Note  1. — In  Italy  the  monetary  unit  is  called  the  lire,  of  100  oentesimi,  in  Spain  the  peset.n, 
of  100  centesimos;  in  Greece,  the  drachnie  of  lOOlepta;  and  in  Finland,  the  mark  of  100  pfennigs. 

Note  2. — The  old  monetary  unit  of  Spain  was  the  hard  dollar  or  duro,  valued  at  99c.  con- 
taining 20  reales,  each  leale  containing  34  maravedis, 

GERMAN  EXCHANGE. 

Note. — See  German  Monetary  Table,  page  236. 

1333.  The  present  money  of  account  throughout  the  German  Empire  is  the 
mark  (Eeichsmarks). 

Note.— Following  the  German  and  French  war,  came  the  formation  of  the  German  Empire 
and  the  unification  of  her  iloneiartj  Units,  the  Thaler,  P'lorin  and  Marc  Banco.  liy  acts  of  the 
German  Parliament  taking  effect,  one  Dec.  4,  1871,  and  one  July  9,  1873,  the  "Mark"  was  adopted 
as  the  monetary  unit  of  the  German  Empire. 

The  German  Monetary  Law  of  Dec.  4,  1871,  ordains  Ist,  That  "there  shall  be  coined  an 


GERMAN    EXCHANGE. 


765 


imperial  gold  coin,  ISStJ-  Jiieccs  of  -nhich  phall  contain  one  'ponnd  of  pnre  golil."  2il,  Tliat  the 
"tenth  of  this  yold  coin  shall  be  called  mark,  an<l  shall  be  divided  into  100  pfennigs."  The  same 
law  also  provides  that  "besides  the  imperial  gidd  coin  of  10  marks,  there  shall  be  coined  im])erial 
gold  coins  of  20  marks,  of  which  69f  pieces  shall  contain  one  ]ioun(l  of  pure  gold."  By  act  taking 
■eli'ect  Jnly  9th,  ISTo,  5  mark  gold  coin  are  also  anthorized,  279  of  which  must  contain  one  pouncl 
of  pnre  gold.  By  this  same  act  the  following  coins  are  also  authorized  :  1st.  Silver  coins,  5  mark, 
2  mark,  1  mark,  oO  ])eniiy  and  20  penny  ]]ieces.  2d.  Nickel  coins,  10  penny  and  5  penny  pieces.  3d. 
Copper  coins,  2  ]ienny  and  1  penny  pieces.  A  ponnd  of  tine  silver  is  coined  into  20  jive  mark,  50 
tioo  mark,  100  otie  mark,  200  fifitj  penny  and  500  tweiity  penny  pieces. 
The  gold  and  silver  coins  are  all  -j^u  pure  and  -^  alloy. 

The  intrinsic  and.  the  United  States  Customhouse  value  of  the  mark  in  U.  S. 
gold  is  23.8  cents ;  which  is  a  smaller  value  than  any  of  the  previous  German 
monetary  units;  and  because  of  its  small  value,  exchange  dealers  and  bankers 
find  it  more  convenient  to  base  the  rate  of  German  exchange  upon  the  equivalent 
value  of  4  gold  marks  exi)ressed  in  United  States  money.  Thus  23.8/  x  i  =  95.2 
cents,  practically  for  exchange  operations,  95;J/  gold,  which  is  quoted  more  or  less 
according  as  a  premium  is  charged  or  a  discount  is  allowed.  To  find  the  currency 
or  silver  rate,  the  premium  on  gold  is  added  to  the  gold  rate. 


TO  COlSrVERT   GEEMAN  MO]!TET  INTO  AN  EXCHANGE   EQUIVALENT 
OF  UNITED  STATES  DOLLAES. 


133-lr.     1.     What  is  the  gold  cost  of  15000  marks,  exchange  at  94? 

Ans.  $3525. 

OPERATION. 

Explanation. — The  rate  of  exchange  being  the 
price  of  4  marks,  we  reason  thus  :  Since  4  marks 
equal  94  cents,  1  mark  is  equal  to  the  4th  part, 
and  15000  marks,  15000  times  as  much. 


94 

15000 


$3525.00     Ans. 

2.     Wliat  will  a  bill  of  exchange  for  24832  marks  cost;  exchange  at95.J,  gold 
11  per  cent  premium  ?  Ans.  $6580,79. 


OPERATION. 

191 

24832 

■  j;5928.64  gold. 

652.15   =  ll°o  premium  on  gold. 

$6580.79    Ans. 


Explanation. — Here  we  reason  thus  :  Since  4 
marks  cost  ^^  cents,  1  mark  will  cost  the  J 
part,  and  24832  will  cost  24832  times  as  much, 
which  is  $5928.64  gold,  to  which  we  add  the 
premium  on  gold  and  i)roduce  the  correct  result 
$6580.79. 


3.  Received  instructions  from  a  correspondent  to  purchase  German  exchange 
to  the  amount  of  4189.  marks  aud  45  pfennigs.  Exchange  is  107i.  What  is  the 
cost?    '  Axis.  $1125.91. 

OPERATION  INDICATED. 


C. 

2  1215 
4    4189.45 


766 


SOULE  S    nilLOSOPHlC    PRACTICAL    MATHEMATICS 


TO  CONYEKT  UXITED  STATES  DOLLARS  liSTTO  AN  EXCHANGE  EQUIV- 
ALENT OF  MARKS. 

1335.     1.     What  amount  of  German  marks  can  be  bouglit  for  $10000   gold, 
exchange  being  96?  Ans.  41G6G.G7  marks,  or  416C6.  marks  and  07  x)fennigs. 


OPERATION. 

M. 

I  ^ 
96  I  10000.00 

I  41666.661  m.arks. 


Explanaiion. — In  this  proljlem,  we  reason  tliiis : 
Since  96  cents  \my  i  marks,  1  cent  will  buy  the 
96th  part,  and  1000000  cents  will  buy  1000000 
times  as  many. 


2.    Invested  $1260.55  silver  in  German  exchange  at  lOoJ.    What  is  the  face 
of  the  bill  ?  Ans.  4790.69  ms. 

NoTE.^Tbe  rate  of  exchange  in  this  and  the  two  following  problems   is  based  upon   the 
supposition  that  gold  is  at  a  premium  over  silver. 


4.21 


FIRST  OPERATION. 

M. 

4 
4 
1260.55 


4790.69   marks.     Ans. 


SECOND  OPERATION. 


M. 

I  * 

105.25     1260.65 


4790.69   marks.     Ans. 


3.  A  merchant  invested  for  a  correspondent  $5000  silver,  in  German 
exchange.  The  rate  of  exchange  was  104;  he  charged  1  per  cent  commission  on 
the  silver  invested.     What  amount  of  marks  did  he  buy  ?  Ans.  19040.37  ms. 


riRST    OPERATION. 

$ 
I  100 
101  I  5000.00 
1.04    4 


19040.37   marks. 


SECOND   OPERATION. 


I  100 
101     5000 


1  14950.495  invested. 


•M, 

I  * 
104  I  49.50.495 

19040.37   marks. 


4.  Invested  for  a  correspondent  $20000  silver,  in  German  exchange  at  108f. 
How  many  marks  did  I  buy,  and  how  much  does  my  correspondent  owe  me,  allowing 
1  per  cent  commission  on  the  $20000  for  investing?  Ans.     73563.22  marks. 

$20200.  he  owes. 

1336. 


IIamuithg  Exchange  on — 


London, 
France, 

Belgium,  " 

Netherlands,         " 
New  York,  " 

Austria,     3  months, 
Italy,         3        " 
Portugal,  3        " 
Spain,        3        " 


short  and  2  months, 

"       "    3        " 

"       "     3 
u       ,<    3 

"       "    3        " 


Ms.  Ps. 
20.42  per  £1  Sterling. 


81. 

80.25 
169.50 
413 
175 

75 
450 


100  Francs. 
100  Francs. 
100  Florins. 
100  Dollars. 
100  Florins. 
100  Lire. 
100  Milreis. 
100  Pesetas. 


Note. — These  are  average  rates  subject  to  an  increase  or  decrease  according  to  circumstances 
governing  exchange. 


*  AUSTRIAN    EXCHANGE.  767 

MISCELLANEOUS  PRACTICAL  PEOBLEMS. 

1337.  1.  Imported  from  E.  Gretzner  of  Dresden,  goods  amounting  to  85420.75 
marks.  If  the  duty  is  60  per  cent  and  the  freight  $3-10.30,  what  is  the  importing 
cost!  Ans.  $32868.52. 

OPKRATlOJf. 

Marks  85420.75  x  $.238  =  $20330.14,  Dresden  cost  of  iuvoice;   60^"^  duty  =  $12198.08  $20330.14  + 
$12198.08  +  $340.30  =  $32868.52. 

Note. — The  commercial  student  should  make  the  proper  entries  for  the  above  transactions. 

2.    Remitted  to  E.  Gretzner,  a  bill  of  exchange  for  the  above  invoice  of 
85420.75  marks,  exchange  96J.    What  was  the  cost  of  the  bill  ?     Ans.  $20607.76. 


OPERATION   INDICATED. 
4 


85420.75 
.965 


Note. — The  commercial  student  should  make  the  proper  entries  for  the  above  transactions. 

3.  Imported  from  A.  T.  Kohn,  Hamburg,  Germany,  merchandise  amounting 
to  4212.15  marks.    What  is  the  importing  value  in  American  money? 

Ans.  $1002.49. 

4.  Remitted  to  A.  T.  Kohn,  a  bill  for  4212.15  marks  bought  at  94.  What 
did  it  cost?  Ans.  $989.86. 

Note. — The  commercial  student  should  make  the  proper  entries  for  the  remittance  of  this 
liill. 

See  pages  248,  249,  270,  271,  272  and  273,  of  Sould's  New  Science  and  Practice  of  Accounts,  for 
the  proper  Invoice  and  Cash  Book  entries  for  such  transactions  as  the  above. 

5.  Invested  for  a  correspondent  $20000  in  German  exchange,  at  95J ;  charged 
J  per  cent  brokerage.    What  was  the  face  of  the  bill  and  what  was  my  brokerage  ? 

Ans.  83571.65  marks  face  of  bill.     $99.50  brokerage. 

6.  Bought  for  a  correspondent  20000  marks  at  95^,  and  charged  ^  per  cent 
brokerage.  What  was  the  cost  of  the  marks,  what  was  my  brokerage,  and  what 
■was  the  total  cost  ?  Ans.  $4762.50  cost  of  marks. 

$11.91  brokerage. 
$4774.41  total  cost. 

AUSTRIAN  EXCHANGE. 

1338.  The  kingdom  of  Austria-Hungary,  by  law  of  August  2,  1892,  changed 
its  monetary  standard  from  silver  to  gold,  and  by  the  same  law  it  changed  its 
monetary  unit  from  the  florin  to  the  krone  (Crown)  which  is  divided  into  one 
hundred  "hellers." 

Under  this  new  law  the  coin  of  Austria-Hungary,  is  as  follows : 
(1).    20  and  10  gold  crown  pieces.     (2).    1  crown  silver  pieces.     (3).    20  and 
10  uickle  and  heller  pieces.    (4).    2  and  1  bronze  heller  pieces. 


768  soule's  riiiLosoPHic  practical  mathematics.  * 

The  valup  of  tlie  gold  crown  is  20.3  cents  American  gold.  At  this  valuation 
the  United  States  Customhouse  estimates  merchandise  imi)orted  from  Austria.  See 
Table  of  Foreign  Coins  for  the  value  of  the  former  monetary  units. 

Notwithstanding  the  new  monetary  unit  and  the  new  monetary  standard 
adopted  by  Austria  in  1892,  American  and  European  bankers  continue  (August, 
1895,)  to  base  Austrian  exchange  on  the  value  of  the  florin  as  it  existed  prior  to 
1892.  This  gold  value  of  the  florin  is  48.2  cents;  the  silver  value  is  38  cents  more 
or  less,  according  to  the  United  States  mint  value  of  silver. 

The  ratio  value  between  the  new  unit,  the  crown,  and  the  old  unit  the  florin, 
is  practically  100  crowns  equals  42  florins  gold  ;  or,  100  florins  equals  237.10  crowns. 

The  e.rchange  value  of  the  florin  varies  according  to  the  premium  or  discount 
declared,  and  according  to  the  value  of  silver  in  the  United  States. 

Note. — The  Metric  System  of  Weights  and  Measures  is  compulsory  in  Austria-Hungary. 

PROBLEMS. 

TO  CONVERT  AUSTRIAN   MONEY   TO   THE   EXCHANGE   EQUIVALENT 
OF  UNITED  STATES  DOLLARS. 

1339.     1.    What  is  the  cost  of  1480.45  florins,  exchange  at  41J  cents  ? 

Ans.  $614.39. 

OPERATION. 

ri.  1480.45 
41J 


.3867i     Ans. 

2.    Bought  5862.70  florins,  exchange  45,  and  paid  h  per  cent  commission. 
What  was  the  cost?  Ans.  $2051.41. 


TO  CONVERT  UNITED  STATES  DOLLARS  TO  THE  EXCHANGE  EQUIV- 
ALENT OF   AUSTRIAN  MONET. 

1340.     1.     How  many  florins  can  be  purchased  for  $9703.45,  exchange  at  48J 
cents?  Ans.  Fl.  20007.11  +  . 

OPE  RATION'. 

I  9703.45 
.485).$9703.450(F1.  20007.11+     Ans.  or  thus:  97  |  J 

2.     Invested,  through  a  broker,  $20000  in  florins,  exchange  40  cents,  broker- 
age \  per  cent.     What  was  the  face  of  the  bill  ?  Ans.  Fl.  49875.30. 

BRAZILIAN   EXCHANGE. 

EXCHANGK  ON  RIO  JANEIRO. 

13-11.     The  money  of  account  of  Brazil  is  the  rets,    1000   of  which   make   a 
milreis. 


*  BRAZILIAN     EXCHANGE.  769 

The  Intrinsic  value  of  the  milreis  in  United  States  money  is  54. G  cents. 

Tlie  exchange  value  is  oi.C  cents  for  1000  reis,  more  or  less,  according  as 
premium  or  discount  is  declared. 

In  the  notation  of  accounts,  the  milreis  are  separated  from  the  reis  by  a  sign 
thus  ©,  or  thus  $,  called  cifrao;  and  the  milreis  from  the  million  by  a  colon  thus: 
Es.  4  :  500  $  200,  is  read  four  thousand  tive  hundred  milreis,  two  hundred  reis. 

Note. — The  French  Metric  System  of  AVeights  ami  Measures  is  the  etandard,  and  has  heen 
compulsory  in  Brazil,  since  1872.  But  the  ancient  ivei;;;hts  and  measures  are  still  partly  eniployeil 
by  business  men.     The  old  Libra  ^  1.012  lbs.  avoirdupois,  and  the  arroba  =  32.38  lbs.  avoirdupois. 


PROBLEMS. 

TO  CONVEET  BRAZILIAN  MONEY  TO  THE   EXCHANGE   EQUIVALENT 
OF  UNITED  STATES  DOLLAES. 

1342.     1.     What  is  Es.  2  :  430  $  300  worth,  exchange  at  55 J  ? 

Ans.  $1348.85. 

OPERATION.    ■ 
Rs.  2430.360  x  SoJ  =  |;1348.&498. 

2.     What  will  Es.  4C7  $  000,  Braziliau  exchange  cost,  exchange  53J  ? 

Ans.  $249.85. 


TO  CONYEET  UNITED  STATES  DOLLAES  TO  THE  EXCHANGE  EQUIVA- 
LENT OF  BEAZILIAN  MONEY. 

1343.     1.     How  many  milreis  can  be  purchased  for  $2400,  exchange  at  51 1 

Ans.  Es.  4  :  705  $  882  + . 

OPERATION. 

Es. 

I  1000 
.51     2400.00 


Es.  4  :  705  $  882 


2.     Invested  $20000  in  exchange  on  Eio  Janeiro,  exchange  at  52J.     What  is 
the  face  of  the  bill  ?  Ans.  Es.  38  :  095  $  238. 

EXCHANGE  ON  BRAZIL  IN  ENGLISH  MONEY. 

Note. — Most  of  the  exchange  on  Brazil  by  United  States  Bankers  and  Merchants  is  drawn  in 
Pounds,  Shillings,  ami  Pence. 


T^Q  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

TO  KEDUCE  BRAZILIAN  MONEY  TO  ENGLISH. 

PROBLEMS. 

1344.     1.    An  invoice  of  coffee  amouuted  to  Rs.  77  :  426  $  100,  exchange  at 
21  |d.  per  milreis,  or  per  1000  reis.    What  was  the  face  of  the  bill  in  sterling? 

Ans.  £6976  8s.  3d. 

OPERATION. 

Es.  77  :  426  $  100  x  Sljd.  =  1674339.412id. 
1674339. 4125d. 


12 
20 


1395288.  3d. 
£6976.  8s. 


2.     What  is  the  face  of  a  bill  for  Rs.  62  :  418  $  610,  exchange  at  24 Jd  ! 

Ans.  £6274  7s.  5d. 


TO  REDUCE  UNITED  STATES  MONET  TO  ENGLISH. 

1346.     1.     What  is  the  face  of  a  bill  for  $27646.98,  exchange  at  $4.85  ! 

Ans.  £5700  8s.  2d. 

OPERATION. 

4.85)$27646.98(£5700.4082 
20 


8.1640 
12 

1.968  ■ 

2.  An  invoice  of  coffee  amounts  to  $8490.16.  What  is  the  face  of  a  sterling 
bill  for  the  amount,  exchange  at  4.86  ?  Ans.  £1746  18s.  lid. 

Note. — For  invoices  of  coffee  as  they  are  made  in  Rio  de  Janeiro,  and  for  tile  exchange,  and 
the  accounts  current  and  interest  accounts  resulting  from  the  importation  of  coffee  from  Brazil,  see 
Customhouse  Uusiuess,  pages  505  to  508. 

EXCHANGE  ON  PORTUGAL. 

1346.  The  monetary  unit  of  Portugal  is  the  milkeis,  the  intrinsic  and 
Customhouse  value  of  which  is  $1.08.     (53Jd). 

The  exchange  value  is  $1.08  more  or  less,  according  as  premium  or  discount 
is  declared.  The  operations  of  exchange  are  the  same  in  form  as  those  in  Brazilian 
exchange. 

The  monetary  table  is  as  follows : 

1  Milreis  =  2\  Crusados  =  25  reales  =  1000  Eeis. 
1  "        =10     "       =    400     " 

1     "        =40     " 

The  almude  of  Lisbon  =  3.7  gallons.    The  almude  of  Oporto  =  5.6  gallons. 


*  EXCHANGE  ON  VARIOUS  COUNTRIES.  7/1 

EXCHANGE  ON  HONDURAS,  COSTA  RICA,  GUATEMALA,  NICARAGUA, 

SALVADOR  AND  COLOMBIA, 

1347.  The  monetary  uTiit  of  all  these  countries  is  the  silver  peso,  or  dollar  of 
100  centavos,  the  value  of  vhich  is,  January  1,  1S95,  45.5  cents  United  States  gold. 
This  value  of  the  silver  peso,  is  determined  by  the  value  of  silver  at  the  United 
States  Mints. 

The  rate  of  exchange  is  the  cents  charged  for  a  peso,  and  varies  according 
to  the  general  principles  of  exchange,  the  political  and  civil  conditions  of  these 
countries  and  the  facilities  of  trade  between  them  and  the  United  States. 

EXCHANGE  ON  PERU,  ECUADOR  AND  CHILI. 

1348.  The  monetary  unit  of  Peru  is  the  silver  soL,  valued  at,  January  1, 
1895,  45.5  cents  gold. 

The  monetary  unit  of  Ecuador  is  the  silver  SUCRE,  the  value  of  which  is, 
January  1,  1895,  45.5  cents  gold. 

The  monetary  unit  of  Chili  is  the  gold  and  silver  peso  of  100  centavos,  valued 
at  91.2  cents  gold. 

The  rate  of  the  exchange  is  the  cents  charged  for  a  sol  or  sucr6,  or  peso. 
Note. — See  remarks  above  on  Honduras  Exchange. 

EXCHANGE  ON  MEXICO  AND  THE  ARGENTINE  REPUBLIC. 

1349.  The  monetary  unit  of  Mexico  is  the  silver  peso  or  dollar  of  100 
centavos,  the  value  of  which  is,  January  1,  1895,  49.5  cents  gold.  The  gold  dollar 
or  peso  of  Mexico,  is  worth  98.3  cents. 

The  rate  of  exchange  is  the  cents  charged  for  a  peso. 

The  monetary  unit  of  Argentine  Republic  is  the  gold  and  silver  PESO  of  100 
centesimos,  the  intrinsic  value  of  which  is  9G.5  cents  gold. 
The  rate  of  exchange  is  the  cents  charged  for  a  peso. 

CUBAN  EXCHANGE. 

1350.  The  monetary  unit  of  Cuba  is  the  gold  and  silver  peso  of  100  centesi. 
mos,  the  intrinsic  value  of  which  is  92.6  cents. 

The  rate  of  exchange  is  the  cents  charged  for  a  pes' 
Note. — See  Customhouse  Business,  page  504,  for  problems  in  importing  goods  from  Cuba. 

RUSSIAN    EXCHANGE 

EXCHANGE  ON  ST.  PETERSBURG. 

1351.  The  money  of  account  of  Russia  is  the  ruble  of  100  kopecks. 


772  soule's  philosophic   practical  mathematics,  * 

The  silver  ruble  is  the  legal  unit  aiul  it  contains  17.99G1  grams,  and  is  worth 
January  1,  181)5,  in  American  gold  30.4:  cents.     Tlie  gold  ruble  is  worth  January 
1,  1895,  in  American  gold  77.2  cents. 

The  exchange  value  is  30  or  77  cents,  more  or  less,  according  to  the  premium 
or  discount  declared,  and  according  as  tlie  exchange  is  for  silver  or  gold  rubles. 

Russia  has  an  imperial  gold  coin  worth  $7,718,  and  a  half  imperial  gold 
coin  worth  $3,859. 

Paper  rubles  are  in  general  circulation  in  Russia,  and  their  value  is  much 
below  that  of  the  silver  ruble. 

Since  the  reign  of  Peter  the  Great,  ten  changes  have  been  made  in  the 
weights  and  fineness  of  Russian  silver  coin. 

Note. — In  Russia  the  Jiili.in  calendar,  or  old  style,  is  still  retained.  This  calendar  is  now, 
and  till  the  year  )9()0  will  remain,  12  days  later  than  the  new  style,  and  in  leap  years  13  days  later. 
Thus,  January  15  in  the  United  .States,  is  January  3il  in  Russia.  From  the  year  1900  to  2000,  the 
diU'erence  will  he  13  days,  and  in  leap  years  14  days.  This  deviation  commences  March  1,  each 
year. 

peoblems. 

TO    CONVERT    RUSSIAN    MONEY    TO    THE    EXCHANGE    EQUIVALENT 
OF  UNITED  STATES  DOLLARS. 

1352.     1.    What  is  the  exchange  equivalent  or  cost  of  940  rubles,  28  kopecks, 
exchange  at  09^  ?  Ans.  $051.14. 

OPERATION. 

940  rubles,  28  kopecks  =  940.28   rubles. 

69i 


$651.1439 


2.    What  cost  31428.65  rubles,  exchange  at  73J,  brokerage  for  buying  J  per 
cent?  Ans.  $23157.8L 

TO  CONVERT  UNITED  STATES  DOLLARS  INTO  AN  EXCHANGE  EQUIV- 
ALENT OF  RUSSIAN  RUBLES. 

1353.  1.    What  amount  of  exchange  on  St.  Petersburg  can  I  buy  for  $840.10, 
exchange  at  70  ?  Ans.  1208.71  rubles. 

OPE  RATION. 

846.10  —  70  =  1208.71?. 

2.     Invested  $5000  in  exchange  on  Moscow  at  68.    What  was  the  face  of  the 
bill  t  Ans.  7352.94  rubles. 

OPERATION. 

$5000  —  68  =  7352.94+. 

JAPANESE  EXCHANGE. 

1354.  The  monetary  unit  of  Japan  is  the  gold  yen  or  dollar  which  contains 
100  sens.    It  is  worth  intrinsically  99.7  cents  gold.     The  value  of  the  silver  yen 


*  EXCHANGE    ON    VARIOUS    COUNTRIES.  ']'J'^ 

is,  January  1,  1895,  49.1  cents  American  gold.     The  gold  coins  consist  of  20,  10,  5, 

2,  and  1  YEN  pieces.     The  silver  coins  are  1  yen  and  50,  20,  10  and  5  sen  pieces. 

The  cojjiier  coins  are  2  sens,  1,  J  and  n,-  SEN  jiieces.     Paper  currency  is  iu  general 

circulation,  of  which  there  are  various  denominations,  corresponding  to  those  in 

coins. 

The  rate  of  exchange  is  the  cents  charged  for  a  yen  ■which  varies  according 

to  the  general  principles  governing  exchange,  and  according  as  the  exchange  is  for 

gold  or  silver  yens. 

Note. — The  metallic  value  of  silver  in  foreign  coins  depends  upon  the  mint  market  value  of 
bullion  silver  iu  the  United  States. 

problems. 

1.  What  cost  a  bill  on  Tokio,  for  Y.  2530.40,  exchange  at  98iC.  ? 

Ans.  $2492.44. 

OPERATION. 

2530.40  X  98i  =  2491.4440. 

2.  How  many  yens  can  be  bought  for  $5000,  exchange  at  $1.02i  ? 

Ans.  487S.04+  yens. 

OPERATION. 

$5000  —  1.021  =  4878.04  +  yens. 

3.  An  invoice  of  silk  goods  imported  from  Yokohama,  amounts  to  4210.60 
YENS.  If  the  Customhouse  value  is  99.7  cents  per  yen  and  the  rate  of  exchange  is 
$1,  -what  is  the  importing  cost  of  the  goods,  and  what  is  the  cost  of  the  bill  of 
exchange,  to  remit  in  payment  of  the  invoice?  Ans.  $4197.97  cost  of  goods. 

4210.G0  cost  of  bill. 

OPERATION. 

4210.60  X  90.7c.  =  $4197.96820 
4210.60  X  $1.   =  4210.60 

4.  The  United  States  duty  on  the  above  invoice  is  50  per  cent  ad  valorem. 
Supposing  the  freight  and  insurance  to  be  10  per  cent  on  the  invoice,  what  would 
be  the  total  cost  to  the  importer  ?  Ans.  $6716.75. 

Note. — Commercial  students  should  make  the  Journal  and  Cash  Book  entries  for  .nil  the 
tTansactions  in  the  above  3d  and  4th  problems.  See  Soule's  New  Science  and  Practice  of  Accounts, 
pages  249  and  270  to  271,  for  the  proper  entries. 


EXCHANGE  ON  SWEDEN,  NORWAY  AND  DENMAEK. 

1355.     The  monetary  unit  of  Sweden,  Norway  and  i)enmark,  is  the  goxd  KRONE 
or  CROWN  of  100  ore,  the  intrinsic  value  of  which  is  26.8  cents  gold. 

The  rate  of  exchange  is  the  cents  charged  for  a  crown,  or,  the  value  of  100 
crowns  in  dollars. 


774  SOULE  S    PHILOSOPHIC    TRACTICAL    MATHEMATICS.  •" 

CHINESE  EXCHANGE. 

1356.  The  money  of  account  in  China  consists  of  the  tael,  mace,  candarink 
and  CASH.  The  only  official  coinage  of  China  is  the  copper  cash,  of  which  about 
1600  to  1700  equal.s  1  Haikwan  or  customs  tael. 

Large  payments  are  made  l)y  weight  of  silver  bullion,  the  standard  or  unit 
of  weight  being  the  liang  or  tael.  Hence  the  tael  is  a  unit  of  weiglit  rather 
than  a  monetary  unit.  By  treaty,  it  is  equal  in  weight  to  1 J  ounces  avoirdupois, 
and  an  Haikwan  tael  is  this  weight  of  ])ure  silver. 

The  following  is  the  table  of  weight: 

10  Sze  =  1  Hu. 

10  Hu  =  1  Hao. 

10  H.io  =  1  Li  (nominal  rash). 

10  Li  =  1  Fun  (Candareu). 

10  Fun  =  1  Tsien  (Mace). 

10  Tsien  =  1  Liang  (Tael)  IJ  oz.  aToirdnpois  by  treaty. 

16  Liang  =  1  Kin  (Cattv)  U  lbs.  "  "         "; 

100  Kiu  =  1  Tan  (Piciil)  133i  lbs.         "  " 

Note. — The  Chinese  standards  of  weight  and  length  vary  throughout  the  Empire. 

The  intrinsic  value  of  the  silver  tael  of  Shanghai  is,  January  1,  1895,  Gl.3^ 
American  gold;  and  that  of  Haikwan  or  customs,  is  74.9  cents  gold. 

Note. — By  an  imperial  decree,  issued  in  1890,  the  silver  dollar  coined  at  the  new  Canton 
Mint  is  made  current  all  over  the  Empire.  This  dollar  is  equal  in  value  to  the  United  States  and 
Mexican  silver  dollar  and  to  the  silver  Yen  of  Japan,  taken  at  the  United  States  Mint,  market 
value. 

The  rate  of  exchange  is  the  variable  dollar  and  cents  charged  for  a  tael. 

PROBLEMS. 

1.  What  is  the  cost  of  a  bill  for  2400  taels,  exchange  at  67 J  cents  f 

Ans.  $1620. 

OPERATION. 

2400  X  .67i  =  $1620. 

2.  What  cost  1650  taels  840  cash,  exchange  at  75  cents?      Ans.  $1238.13. 

OPERATION. 

1650.840  at  .75  =  $1238.13. 

3.  Invested  $3000  in  taels,  exchange  at  72  cents.  What  was  the  amount 
bought  1  Ans.  4166.6G6§  taels,  or  4106  taels  and  666J  cash,  or  4166 

taels,  6  mace,  0  candarine,  and  6J  cash. 

OPERATION. 

$3000  -i-  72c.  =  4166.666J. 
ARBITRATION  OF  FOREIGN  EXCHANGE. 

1357.  1.  A  merchant  in  New  Orleans  wishes  to  remit  10000  marks  to 
Hamburg,    direct    exchange    is    96/.     Exchange    on    Loudon    is    4.87;    London 


y^ 


ARBITRATION    OF    FOREIGN    EXCHANGE. 


775 


excbaiijre  on  Hamburg  is  20.50  ni.  for  £1  ;  New  Orleans  exchaTige  on  Paris  is  fr. 
ii.lO;  I'aris  exchange  on  London  is  fr.  L'o.U),  and  London  excliange  on  Hamburg  is 
20.50  m.  for  £1.     Wbicli  is  tlie  most  advantageous  way  to  remit  ? 

Ans.  Through  Londou. 

SOLUTION. 

First  operatiou  to  tiiid  the  cost  of  direct  exchange, 
c. 

I  90 
4  I  10000 

I  $2400.00   cost  of  direct  exchange. 
Second  operation  to  find  the  cost  of  exchange  through  London. 

$ 
I  4.87  =  1£. 
1£  =  m.  20.50  I  10000        marks. 

I  $2375.61   cost  through  London. 
Third  operation  to  find  the  cost  of  exchange  through  Paris  and  London. 


Francs            -        5.10 

1.00 

dollar. 

Pound  sterling     1 

25.10 

francs. 

Marks             -      20.50 

I 

pound  sterling 

10000 

marks. 

12400.77   cost  through  Paris  and  London. 

2.  I  owe  a  correspondent  in  Taris  fr.  25000,  which  I  wish  to  settle  by- 
remitting  exchange.     Direct  exchange  on  Paris  is  5.10. 

But  before  making  a  i)archase  1  find  on  inquiry  that  exchange  on  London  is 
4.87,  and  London  exchange  on  Paris  is  £1  for  fr.  25.20. 

I  also  learn  that  exchange  on  New  York  is  ^  per  cent  discount,  and  that  New 
York  exchange  on  London  is  4.S6i,  and  London  exchange  on  Paris  as  above,  fr. 
25.20.  Allowing  i  per  cent  comniission  on  the  whole  amount  invested  at  each 
intermediate  place" for  re-investing,  which  is  the  most  advantageous  way  to  remit? 

Ans.  Through  New  York  and  Loudon. 

SOLUTION. 
First  operation  to  find  the  cost  hy  direct  exchange. 

Francs    5.10  I  1 

I  25000.00   francs. 

I  $4901.96   cost,     Ans. 
Second  operation  to  find  the  cost  through  London. 


Proceeds  of  amount  invested      99.50 
Francs  given  for  £1  25.20 

Pound  sterling  1 


100.00  assumed  investment. 

1  pound  sterling. 

4.87  exchange  value  of  1£. 

25000  francs. 


99.50  1 


100 
25000.00 


$4855.64    cost.     Ans. 

Or  thus : 

25.20  I  ?  =  francs.  ?  =  £. 

4.87 


?  =  francs.  |  ?  =  £.  ?  =  14855.63    Ans. 

Third  Operation  to  find  the  Cost  through  New  York  and  London. 


Assumed  face  of  New  York  bill  100 
Amount  invested  in  New  York  99.50 
Amount  invested  in  London  99.50 
Francs  given  for  £1  25.20 

Pound  1 


99.25  cost  of  New  York  bill. 

100  cost  of  investment  in  New  York. 

100.00  cost  of  investment  in  London. 

1  pound  sterling. 

4.865  exchange  valup  of  1£. 

25000  francs. 

$4838.46  cost.     Ans. 


776  soule's  niiLosoPHic   practical  mathematics.  * 

3.  Suppose  ill  tlie  operation  throug;li  New  York  and  London  that  commission 
for  reinvesting  liad  been  allowed  only  on  the  net  amount  invested,  what  would  have 
been  the  cost  of  exchange  t  Aus.  $4838.21. 


Assumett  faro  of  Now  York  1)111  100 

Assiiiiiod  iiivostniont  in  Now  York      100 
Assumecl  iuvestmoiit  in  Louilou  100 

Francs  given  for  XI  ,  25.20 

Pound  Bterliug  1 


OPKRATION. 

99.25  cost  of  New  York  bill. 

100. .">0  coat  of  New  York  inve.stinent  at  J"j(  commission. 

100.50  cost  of  investment  in  London  at  |j„'  commission. 

1  ]iound  sterliii;;. 

4.865  exoUauge  value  of  IjC. 

25000  francs. 


$4838.21    Ans. 

THE  UNIFICATION  OF  THE  MONEY  OF  THE  WORLD. 

1358.     The  importance  of  a  uniform  monetary  measure  for  the  world  is  so 
great  that  we  cannot  close  our  work  ou  exchange  without  referring  thereto. 

The  great  variety  of  monetary  units  of  the  different  nations  of  the  world,  and 
the  consequent  comi)lications  and  intricacies  of  exchange,  as  well  as  the  loss  of  time 
and  increase  of  labor  that  are  thereby  entailed  upon  commercial  men,  proves  most 
clearly  the  need  for  the  uuiticatiou  of  the  money  of  the  world. 

Since  the  downfall  of  the  Roman  empire,  the  various  countries  of  Europe 
Lave  been  afflicted  with  a  heterogeneous  coinage,  and  for  ages  past  the  loss  of  time 
and  money,  resulting  from  this  cause,  has  been  seriously  felt  by  travelers  and 
business  men  of  all  classes  and  climes. 

Napoleon  the  First,  looking  down  on  the  world  from  the  rock  of  St.  Helena, 
declared  that  what  Europe  most  needed  was  a  common  law,  a  common  measure,  and 
a  common  money. 

This  solemn  and  wise  utterance  was  a  legacy  not  only  to  Europe  but  to  the 
■whole  family  of  nations.  In  1821,  the  American  Secretary  of  State,  John  Quincy 
Adams,  in  his  celebrated  report  to  the  Congress  of  the  United  States,  pointed  out 
the  incalculable  advantages  of  a  common  measure  and  a  common  money,  which 
should,  in  his  own  comprehensive  language,  "  overspread  the  globe  from  the  equator 
to  the  i^oles." 

With  clear  political  sagacity,  he  saw,  and  said,  that  the  object  could  oidy  be 
accomplished  "  by  a  general  convention  of  nations  to  which  the  world  shall  be 
parties,"  and  "in  which  the  energies  of  opinion  must  precede  those  of  legislation." 
In  18G7,  the  delegates  of  nineteen  nations,  the  United  States  being  one,  representing 
320  millions  of  people,  met  at  Paris  to  consider  the  important  question  of  unifying 
the  money  of  the  different  nations,  and  in  1892  an  international  monetary  conference 
was  held  in  Brussels,  at  which  the  question  of  a  single  monetary  unit  for  the 
leading  nations  of  the  earth  was,  with  other  financial  matters,  discussed;  and 
although  much  good  has  resulted  from  these  conferences,  the  great  object  remains 
unaccomplished. 

We,  therefore,  urge  the  commercial  student  and  general  reader  to  labor  U 
behalf  of  this  cause  until  the  dollar  of  the  United  States,  the  Pound  of  England 
the  Franc  of  France,  the  Mark  of  Germany,  the  Crown  of  Austria,  the  Ruble  c 
Russia,  the  Milries  of  Brazil,  the  Tael  of  China,  the  Y'en  of  Japan,  and  the  othf 
monetary  units  of  the  world  are  all  of  the  same  metal,  weight,  and  purity.  Tb 
names  need  not  be  the  same,  but  the  value  and  subdivisions  should.  This,  the  ap 
demands,  commerce  requires,  and  common  sense  approves ;  and  but  for  the  prej- 
dice,  the  stupidity,  bigotry,  and  the  false  national  pride  and  fear  of  a  few  crowul 
heads,  it  would  have  been  accomplished  years  ago. 


bXORAGE. 


1359.  Storage  is  the  keeping  of  goods  in  a  store,  -warehouse,  or  other  place  of 
deposit,  until  required  for  use  or  shipment. 

1360.  Storage  is  also  the  price  or  charge  made  for  the  keeping  of  goods  in  a 
storehouse. 

1361.  Cold  Storage  is  storage  in  refrigerating  chambers  or  other  places 
artificially  cooled,  to  preserve  articles  liable  to  be  damaged  by  heat. 

1362.  Bonded  Warehouses  are  Government  storehouses  in  which  imported 
goods  on  which  duties  have  not  been  paid,  are  stored. 

1363.  A  Grain  Warehouse,  in  which  grain  is  stored,  is  called  an  elevator. 

1364.  Cash  Storage  is  storage  that  is  comjiuted  and  i>aid  on  each  withdrawal 
or  delivery  of  goods. 

1365.  Storage  Receipts  or  Warehouse  Receipts  are  receipts  showing  the 
amount  of  goods  in  the  storehouse,  and  are  frequently  bought  and  sold,  esiiecially 
those  for  grain  and  other  produce,  and  are  used  as  collaterals  in  discount  and 
exchange  transactions. 

1366.  Credit  or  Time  Storage  is  storage  on  one  or  several  lots  of  goods 
that  may  be  received  and  delivered  in  different  quantities  at  different  times,  and 
the  charges  made  periodically,  or  at  the  final  withdrawal. 

1367.  The  Rates  of  storage  charges  are  generally  fixed  by  the  Boards  of 
Trade  or  Chambers  of  Commerce  of  the  different  cities,  and  are  most  usually  made 
by  the  month  of  30  days,  per  box,  barrel,  bale,  sack,  etc.  No  deductions  are 
allowed  when  merchandise  or  property  is  withdrawn  before  the  close  of  the  month. 
A  fractional  part  of  a  month  is  counted  as  a  full  month. 

In  some  cities  and  towns,  if  the  property  is  taken  out  within  fifteen  days 
after  the  expiration  of  the  first  month,  a  half  month  is  charged,  but  if  after  fifteen 
days  a  whole  month. 

The  owners  or  agents  of  property  pay  all  expenses  incurred  at  the  store  or 
warehouse  for  receiving,  storing  and  delivering  the  property. 

(777) 


778  soule's  philosophic  practical  mathematics.  * 

136S.  Average  Storage.  Independent  of  the  regular  storage  charges  goods 
are  sometimes  received,  and,  by  special  agreement,  charges  are  made  for  the  actual 
time  they  were  stored.  In  cases  of  this  kind  it  is  customary,  in  computing  the 
storage,  to  average  the  time,  and  charge  a  certain  rate  per  month  of  thirty  days. 

PROBLEMS. 

TO  FIKD  THE  AMOUNT  OP  STORAGE  PEE  AETICLE  PER  MONTH. 

1369.  1.  What  is  the  cost  of  storage  on  500  barrels  molasses  from  November 
3,  to  December  12,  1895,  at  6/  per  barrel  i^er  month  1  Ans.  $60. 

OPERATION. 

500  bbls.  from  Nov.  3,  to  Tiec.  3,  =  1  month at  6c.  =         $30.00 

500  bbls.  from  Dec.  3,  to  Dec.  9,  =  fraction  of  month     -        -        -        at  6c.  =  30.00 

Total  amonnt, $60.00 

ETplanaiion. — Tho  general  ou.stom,  as  above  stated,  is  to  charge  for  a  whole  month  whenever 
the  goods  are  on  storage  a  fraction  of  a  month. 

2.  What  would  be  the  cost  of  storage  on  300  bbls.  flour  received  Oct.  10,  and 
delivered  Oct.  28;  400  bbls.  received  Oct.  16,  and  delivered  Nov.  15,  and  600  bbls 
received  Nov.  18,  and  delivered  Nov.  30,  at  5/  per  bbl.  per  month  ?         Ans.  $65. 

OPERATION. 

300  bbls.  Oct.  10,  to  Oct.  28,  fraction  of  a  month  -        -        -        at  5c.  =        $15.00 

400  bbls.  Oct.  16,  to  Nov.  15,  1  month at  5c.  =  20.00 

600  bbls.  Nov.  10,  to  Nov.  30,  fraction  of  a  month  -        -        -        at  5c.  =  30.00 


$65.00 


TO    FIND    THE    AMOUNT    OP    STORAGE    POR    AVERAGE    PERIOD    OF 

THIRTY  DAYS  PER  ARTICLE. 

1370.  1.  What  is  the  cost  of  storage  on  500  bbls.  molasses  received  Nov.  3^ 
and  delivered  Dec.  12,  1895,  at  6  cents  per  barrel?  Ans.  $39. 

OPERATION. 

1895. 
Nov.  3.     500  bbls.  for  39  davs  =  19500  bbls.  for  1  dar.  6c.  X  650  =  $39. 

19500  4-  30  =  650  bbls.  for  30  days,  or  1  month. 

TO   FIND    THE    AMOUNT   OF    STORAGE    PER   AVERAGE   PERIOD    OF 

THIRTY  DAYS  PER  ARTICLE  WHEN  DELIVERIES 

HAVE  BEEN  MADE. 

1371.  1.  From  the  warehouse  books  we  find  the  following  receipts  and  deliv- 
eries of  flour,  for  the  account  of  F.  W.  Ahsen  &  Co.  By  contract,  storage  is  charged 
for  the  actual  time  the  flour  was  on  storage,  at  the  rate  of  5/  per  barrel  per  month: 


*  STORAGE    DAILY    BALANCE    METHOD.  779 

Receipts :  Kov.  4,  1895,  200  bbls. ;  Xov.  5,  700  bbls. ;  Nov.  15,  1300  bbls. ; 
Nov.  29,  3800  bbls. ;  ISox.  30, 1200  bbls. 

Deliveries  :  Nov.  7, 1895,  1000  bbls. ;  Nov.  11,  500  bbls. ;  Nov.  18,  600  bbls. ; 
Nov.  20,  100  bbls.;  Nov.  27,  1000  bbls.;  Dec.  5,  1400  bbls.;  Dec.  7,  1200  bbls,; 
Dec.  13,  800  bbls. ;  Dec.  16,  150  bbls. ;  Dec.  21,  300  bbls. ;  Dec.  24,  500  bbls. ;  Dec, 
27,  300  bbls.  What  is  the  amount  of  storage  due  January  1,  1896,  and  how  many 
barrels  remain  in  the  warehouse  ?  Aus.  $234.15. 

1150  barrels  iu  warehouse. 


FIRST  OPERATION  BY  THE  DAILY  BALANCE  METHOD. 


F.  W.  Ahsen  &  Co. 


^'% 

No.  baiTels  stored  for 

1895. 

lHo.  bbla.  received. 

>Jo.  bbls.  delivered. 

No.  bbls.  ill  -w-arehonse. 

3i 
1 

oue  day. 

Nov. 

4 

2000 

2000 

2000 

5 

700 

2700 

2 

5400 

7 

1000 

1700 

4 

6800 

11 

500 

1200 

4 

4800 

15 

1300 

2500 

3 

7500 

18 

600 

1900 

2 

3800 

20 

100 

1800 

7 

12600 

27 

1000 

800 

2 

1600 

29 

3800 

4600 

1 

4600 

. 

30 

1200 

5800 

.5 

29000 

Dec. 

5 

1400 

4400 

2 

8800 

7 

1200 

3200 

6 

19200 

13 

800 

2400 

3 

7200 

16 

150 

2250 

5 

11250 

21 

300 

1950 

3 

5850 

24 

500 

1450 

3 

4350 

27 

300 

11.50 

5 

5750 

9000 

7850 

1 
30 

)  140500 

9000  bbls.  received. 
7850     "      delivered. 


Practically  4683  barrels. 
5c. 


4683i 


1150 


in  warehouse. 


$234.15.    Ans. 


Explanation. — By  the  operation  of  the  -work,  we  see  that  there  were  140500  barrelB  on  storage 
for  1  day,  and  allowing  30  days  for  the  month,  there  were  4683J  (practically  4683)  barrels  on 
storage  for  one  month,  which,  at  5c.  per  barrel  per  month,  amounts  to  $234.15. 

This  method  of  equating  the  time  and  making  up  the  account  for  storage  is 
applicable  also  to  various  other  departments  of  business. 

Butchers  and  live  stock  dealers  frequently  hire  their  stock  pastured  or  fed, 
with  the  privilege  of  entering  and  withdrawing  stock  as  they  may  require,  and  pay 
for  the  actual  time  that  stock  was  pastured. 


78o 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


SECOND  OPERATION'  BY  THE  PRODUCT  METHOD. 


Br. 

Date. 


Kfc'i'ived. 


Days 


I'liiducts 


Hate. 


Deliveries 


Days. 


Gr. 

Products. 


1895 

1895 

Nov. 

4 

20on 

58 

lUiOOO 

Nov. 

7 

1000 

55 

5500O 

5 

700 

57 

39900 

11 

500 

51 

2.5500 

n 

15 

1300 

47 

61100 

18 

600 

44 

26400 

ii 

:!9 

3800 

33 

125400 

20 

100 

42 

4200 

11 

30 

1200 

32 

38400 

27 

1000 

35 

35000 

Dec. 

o 

1400 

27 

37800 

9000 

7 

1200 

25 

30000 

7850 

13 

800 

19 

1520O 

16 

150 

16 

240O 

1150  bills,  im  liaml. 

21 

300 

11 

330O 

24 

500 

8 

4000 

Total  Dr.  pi 

27 

300 
7850 

5 
Tot.  C 

150O 

'oducts,   380800 

r.prod't  240300 

"   Cr. 
Bal.  of  Dr. 

240300 

products,  140500 

140500  -:-  30  =  4683^  practically,  4683  barrels.  5c.  X  4683  =  $234.15  storage. 
Explanation. — By  this  method,  we  as.siime  that  all  items  of  receipt  were  in  storage  from  the 
date  received  till  January  1,  and  tind  tlie  equivalent  storage  on  1  barrel  for  such  time  for  each  item 
of  receipt,  which  gives  380800  days  storage  on  1  barrel.  Then,  since  there  were  deliveries  made, 
we  tind  the  number  of  days  that  each  item  of  delivery  was  made  before  January  1,  and  the 
equivalent  storage  on  I  barrel  for  snch  time,  for  each  item  of  delivery,  which  gives  240300  days 
storage  on  1  barrel.  This  is  deducted  from  the  total  debit  product  and  gives  a  net  debit  of  140500 
day.s  storage  on  1  barrel,  which  we  divide  liy  30,  the  storage  period,  and  obtain  the  number  of 
barrels  ou  which  storage  is  due  for  one  mouth  at  5  cents. 

2.  The  proprietor  of  a  pasture  received  and  delivered  cattle  for  account  of 
John  Gunn,  as  follows: 

Received  :  Jan.  1, 1895,  900  head ;  Jan.  5,  .350  head ;  Jan.  21,  1100  head. 

Delivered:  Jan.  2,  1895, 120  liead;  Jan.  3,  120  head ;  Jan.  4,  150  head;  Jan. 
14,  360  head;  Jan.  17,  200  head;  Jan.  23,  250  head;  Jan.  27,  400  head;  Jan.  28, 
120  head;  Jan.  31,  100  head.  What  will  be  the  cost  for  pastnring  these  cattle 
from  Jan.  1,  to  Feb.  1,  1895,  counting  the  actual  time  that  the  stock  was  jiastured, 
at  75  cents  per  head  per  week,  (7  days)  1  Aus.  $2556.42f . 

OPERATION  BY  THE  DAILY  BALANCE  METHOD. 

Account  of  Pasturage  of  Cattle,  at  seventy-five  cents  per  lieadj)er  tceek  (seven  days),  for 

John  Gunn,  from  January  1,  to  February  1,  1895. 


Number  head 

Number  head 

Number  liead  in 

No.  days. 

Number  head  pastured 

1895. 

received. 

delivered. 

]).i.Hturaffe. 

p.i-stured. 

Jor  one  day. 

Jan. 

1 

900 

900 

1 

900 

?, 

120 

780 

1 

780 

.S 

120 

660 

1 

660 

4 

150 

510 

1 

510 

5 

350 

860 

9 

7740 

14 

360 

500 

3 

1500 

17 

200 

300 

4 

1200 

«1 

1100 

1400 

•7 

2800 

aa 

250 

1150 

4 

4600 

27 

400 

750 

1 

750 

9» 

120 

630 

3 

1890 

31 

100 

530 

1 

530 

2350 

1820 

23860 

Number  received 
Number  delivered 


2350 
1820 


Number  in  pasture     530 


)  23860     number  pastured  for  one  day. 

3408J   average  number  pastured  for  one  week. 
75c.  X  3408*  =  $2556.425. 


STORAGE. 


781 


TO  FIND  THE  STOEAGE  ON  GOODS  RECEIVED  AND  DELIVEEED  AT 
DIFFERENT  DATES,  THE  STORAGE  TEEM  AND 


1372. 

warehouse : 


RATE    BEING    GIVEN. 


1.     The  following  are  the  receipts  aud  deliveries  of  coffee  by  a  general 


Received. 


1895.     July     7.   800  bags. 
"      28,   250     " 
Aug.   13,   300     " 


Delivered. 

1895.     Jnlv  24;  700  bags. 

Aug.     4,  100     " 

8,  200     " 

"      26,  350     " 


How  nnicli  is  due  for  storage  Angu.st  26,  at  the  rate  of  3  cents  per  bag  for 
a  period  of  10  days,  or  any  part  of  10  daysT  Ans.  $85.50. 


OPERATION. 

Received. 

Delivered. 

Storage  Charges. 

July     7,   800 

July   24,    700 

700  at  6c.     -    $42.00 

"     28,    2.^0 

Aug.     4,    100 

100  at  9c.     -        9.00 

Aug.  13,   300 

■'       8,   200 

200  at  6c.     -       12.00 

"     26,   350 

50  at  9c.     -        4.50 
300  at  6c.     -      18.00 

1350  1350  $85.50  total  storage. 

Note. — All  goods  delivered  are  deducted  from  the  oldest  receipts  in  the  warehouse. 

Explanatiort. — I?y  inspection,  we  see  that  the  first  delivery  of  700  bags,  July  24,  was  made  17 
days  after  the  first  rt-ceijit  of  800  bags,  which  is  one  term  aud  a  fraction  of  a  term  ^  2  terms. 
Hence  a  charge  is  made  on  the  70U  bags  for  2  terms  at  6  cents  per  bag  =  $42.  The  remaining  lOO' 
bags  received  July  7,  were  delivered  Aug.  4,  which  is  28  days,  =  2^,^  terms  =  3  terms.  Hence  a 
charge  is  made  on  the  100  bags  for  3  terms  at  9  cents  per  bag  =  $9.  Aug.  8,  200  bags  were  delivered 
which  are  applied  to  tlie  250  received  July  28,  thus  making  11  days  storage  =  !,'„  term  =  2  terms 
=  6  ceuts  storage  on  the  200  bags  =  $12.  Aug.  26.  350  bags  were  delivered.  50  of  these  bags  were 
applied  on  the  receipt  of  July  28,  and  300  on  the  receipt  of  Aug.  13.  The  50  bags  owe  storage 
from  July  28,  to  Aug.  26,  =  29  days  =  2;'„-  terms  =  3  terms  or  9  cents  per  bag  =  $4.50.  The  300- 
bags  owe  storage  from  Aug.  13,  to  Aug.  26,  =  13  days  =  li'o  terms  =  2  terms  at  6  cents  per  bag  ^ 
$18. 

2.     The  following  are  the  receipts  and  deliveries  of  corn  by  a  grain  elevator  : 

Reccivcil.  Delivered. 


1895.     May     4,  2000  bushels. 

"      28,  1500 

June    4,  1000 

July  10,  3000 


1895.     May  31,   1200  bushels. 
June20,   2000         " 
Julv  15,    4300         " 


The  storage  on  the  above  corn  was  3  cents  per  bushel  for  the  Hrst  15  days- 
or  fraction  thereof,  and  2  cents  per  bushel  for  each  subsequent  15  days  or  fractional 
part  thereof.     What  is  due  for  storage  July  15, 1S95  ?  Ans.  $379. 


OPERATION. 

ptriflge  due  on  1200  bu.  Mav  4,  to  Mav  31,  ^=  27  days  =  2  terms  at  5  cents  :=  ... 

Storage  due  on  (2000-1200  =)  800  bu.  M.ay  4,  to  June  20,  =  47  days  =  4  terms  at  9  cents  ^    - 

Storage  due  on  1200  bn.   (amount  taken"  from  receipt  of  May  28,  and  added  to  balance  of 

May  4,  to  make  delivery  of  June  20,)  from  May  28,  to  June  20,  =  23  days  =  2  terms 

at  5  ceuts  =:  .....-.-.-.--.- 

Storage  due  on  300  bu.  (balance  of  1500-1200)  from  May  28,  to  July  15,  =  48  days  =  4  terms 
at  9  cents  =  .........------ 

Storage  due  on  1000  bu.  from  June  4,  to  July  15,  =  41  days  =  3  terms  at  7  cents  = 
Storage  due  on  3000  bu.  from  July  10   to  July  15,  =  5  days  =  1  term  at  3  cents  = 


$60 

72 


60 

27 
70 
90- 


782 


soule's  philosophic  practical  mathematics. 


TO    FIND    THE    STORAGE    ON    GRAIN    SOLD    IN   THE   ELEVATOR    OR 

WAREHOUSE. 


1373.     1.    A  g^rain  wareliouse  received,  stored  and  delivered  wlieat  as  follows: 


Received. 

1895.    May     3,  7000  bushels. 

"       20,  ()(XI0         " 

"      2fi,  9000         " 

Jime     8,  8000         " 

30000 


1895. 


Delivered. 

May   13,    6000  liusliela. 

June  9,  7000  " 
"  13,  8000  " 
"      26,    9000         " 


30000 


Compute  the  total  and  tlie  extra  storage  on  this  account  based  on  the  follow- 
ing conditions:  The  seller  to  pay  2  cents  per  bushel  for  storage  for  the  first  20 
days,  or  part  of  20  days ;  the  buyer  to  pay  J  cent  per  bushel  for  each  subsequent 
10  days  or  part  of  10  days;  5  days  are  to  be  allowed  the  purchaser  without  extra 
storage  during  which  time  the  wheat  is  to  be  delivered. 


OPERATION. 


Keceipta 

With. 

Storage  on 

Days. 

^.  Eate. 

Withdraw- 

Withdraw- 

From    1       To 

Anit. 

Total. 

Extr». 

buahela. 

drawals. 

als  from 
receipt8. 

als  from 
halaucea. 

1895 

Mar 

3 

7000 

1000 

Miiy 

3',June 

U 

42 

3^ 

35 

00 

15 

00 

13 

6000 

6000 

" 

3  May 

1       "^ 

18 

15 

2 

120 

00 

It 

20 

6000 

June 

9 

7000 

6000 

it 

20 

June 

14 

25 

n 

150 

00 

30 

00 

May 

26 

9000 

1000 

tt 

26 

July 

1 

36 

3 

30 

00 

10 

00 

Juno 

13 

8000 

8000 

a 

26'Juue 

18 

23 

2^ 

200 

00 

40 

00 

(( 

8 

8000 

it 

26 

9000 

8000 

June 

8 

July 

1 

23 

-'* 

200 

00 

$735 

00 

40 
$135 

00 

Explanation. — First  find  storage  on  6000  l>u.  delivered  May  13,  from  May  3,  to  May  18,  (May 
13,  +  5  days)  =  15  days  or  1  term  at  2  cents  a  liu.  =  $120.00. 

Then  since  the  balance  of  the  7000  —  6000  =  1000  bu.  remained  iu  storage  from  May  3,  to  the 
next  sale  of  7000  bu.  on  June  9,  we  must  find  storage  on  same  1000  bu.  from  May  3,  to  June  14, 
(June  9  +  5  days)  =  42  days  or  4  terms  at  3J  cents  =  $35.  Balance  of  sale  of  June  9,  must  Lave 
been  taken  out  of  the  receipt  of  6000  bu.  on  May  20,  hence  we  find  storage  on  6000  bu.  from  May  20, 
to  June  11,  (June  9  +  5  days)  =  25  days  or  2  terms  at  21  cents  =  $150.  Now  find  storage  due  on 
the  withdrawal  of  June  13,  8000  bu.  from  May  26,  to  June  18,  (June  13  +  5  days)  =  23  days  or  2 
terms  at  2i  cents  =  $200. 

Since  the  withdrawal  of  8000  bu.  on  the  13th  of  June,  left  a  balance  of  1000  bu.  of  the  receipt 
of  May  26,  we  find  storage  on  that  1000  bu.  from  May  26,  to  July  1,  (5  days  credit  added  to  with- 
drawal of  June  26, )  =  36  days  or  3  terms  at  3  cents  =  $30. 

The  final  work  is  to  find  the  storage  on  the  last  receipt  of  8000  bu.  on  June  8,  from  June 
8,  to  July  1,  (June  26  +  5  days)  =  23  days  or  2  terms  at  2^  cents  =  $200.  Thus  producing  $735 
total  storage  and  $135  extra  storage,  $735  —  $135  s  $600  storage  due  by  the  seller  and  $135  due  by 
the  buyer. 


STORAGE. 


783 


2.  What  is  the  storage  on  the  foHowiiig  receipts  and  deliveries  of  flour  at  1 
cent  per  barrel  for  the  first  10  days,  or  fraction  thereof,  and  ^  cent  for  each  subsequent 
5  days  or  fraction  thereof?  Ans.  $14.25. 


Receipts. 

1895.     Feb.    3,  500  barrels. 

"      21,  250         " 

"     24,  300        " 

"     28,  150         " 

1200 


Deliveries. 

1895.     Feb.  11,    100  barrels. 

"      18,    200  " 

"     22,    225  " 

"     24,    300  " 

Mar.    5,    375  " 

1200 


Keceipta 

Witlnlniw. 

Stora 

•;o  oil 

To 

l)av8. 

Kate. 

Date. 

AVitbdrawals 

1 

From 

Amt. 

Total. 

barrels. 

als  grusa. 

from 
receipts. 

Balances. 

1895 

1 

Fob. 

3 

500 

400 

ti 

11 

100 

100 

Feb. 

■.i 

Feb. 

11 

8 

1  c. 

1 

00 

ii 

18 

200 

200 

200 

3 

it 

18 

15 

lie. 

3 

00 

ti 

21 

250 

200 

ti 

3 

it 

22 

19 

2c. 

4 

00 

it 

22 

225 

25 

225 

it 

21 

H 

22 

1 

Ic. 

25 

ti 

24 

300 

225 

li 

21 

tt 

24 

3 

Ic. 

0 

25 

it 

24 

300 

75 

ti 

24 

tt 

24 

0 

225 

n 

24 

Mar. 

5 

9 

Ic. 

0 

25 

it 

28 

150 

Mar. 

5 

375 

150 

tt 

28 

a 

0 

0 

Ic. 

1 

50 

14 

25 

3.    Elevator  "A"  of  the  Mississippi  Elevator  Co.,  received  and  delivered  for 
B.  Milmo  &  Co.,  Avheat  as  follows : 

1894.  Kov.   18,  received  5450''^<^  busliels,  which  were  delivered  Jan.  31,  1895. 

"  "  "      31, 

"  '  "     31, 

1895.  Jan.    12,         "         6920"^         "  "  "  "  "      31, 

"  "  "      31, 

"  "  "    [31, 

Total  bnshels,    21647^^ 


Dec. 

6, 

4740ia 

a 

11, 

3265'^ 

Jan. 

12, 

6920"^ 

it 

23, 

750<!^ 

tt 

26, 

520^^1 

According  to  the  Eules  and  Tariff  of  the  Elevator  Co.,  J  cent  for  each  10 
days  or  part  of  10  days  is  charged  for  storage,  and  1  cent  per  bushel  is  charged  for 
delivering.  No  charge  is  made  on  the  fractions  of  a  bushel.  What  is  the  amount 
due  for  storage  and  delivery?  Ans.  $483.30. 

The  student  will  make  the  bill  for  the  above,  as  per  the  following  form : 


784 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


FORM  OF  BILL. 


ELEVATOR     "A." 

ISTew  Orleans,  January  31,  1895. 


Messrs  B.  Milmo  &  Co., 

For  storage  charges  on  wheat. 


To  Mississippi  Elevator  Co.  Dr., 


Wlieat. 

Corn. 

5450 

SQ 

4740 

IQ 

3265 

QO 

6920 

li 

750 

QQ 

520 

Ml 
ii 

21647 

bushels. 

Ex.  El.  A  Bins. 


In 

Out 

1894 

1895 

Nov. 

18 

Jan. 

31 

Dec. 

6 
11 

., 

1895 

Jan. 

12 
28 

" 

ti 

26 

Sbippetl  on 


S.  S.  Pendarves, 


AmonDt. 


$118 

103 
6 

|483 


50 

80 
50 
30 


Received  payment. 


Note. — In  this  bill  the  student -niU  find  the  extensions  of  the  first,  third  and  fifth  items  of 
delivery,  and  the  rate  of  cents  to  charge  for  the  second,  fourth  and  sixth  items. 


uCt 
'•^?- 


anufacturing  on  Shares  and  Tollage. 


1374.  Manufacturing  on  shares  means  that  the  manufacturer  receives  a 
specified  per  cent  or  part  of  the  articles  or  things  manufactured  in  comi)ensation 
for  services  rendered  in  manufacturing  the  material  supplied  to  him.  This  method 
of  paying  for  services  is  quite  common  in  certain  lines  of  business. 

By  tollage  is  meant  the  taking  of  a  specified  portion  of  goods  or  articles  in 
compensation  for  services  bestowed  upon  the  goods  or  articles,  or  upon  the  owner's 
part  thereof,  after  taking  the  toll  out ;  as  the  grain  taken  by  the  miller  in  payment 
for  grinding.  ' 

PROBLEMS. 

1.  A  planter  sends  to  a  miller  to  have  ground  80  bushels  of  corn.  The 
miller  charges  J  of  the  whole  number  of  bushels  for  grinding.  How  much  is  his 
toll  ?  Ans.  10  bushels. 


OPERATION. 

8  )  80  bushels. 

10  bushels.     Ans. 


Explanation. — Since  the  miller  charges  |  of 
the  whole  number  of  bushels  for  grinding,  he  is 
entitled  to  as  many  bushels  as  the  80  is  times 
equal  to  8,  which  is  10  times.  Hence  we  see 
that  he  charges  10  bushels  for  grinding  70  for 
the  planter  or  1  bushel  toll  for  every  7  bushels 
ground  for  the  planter. 


2.  Suppose,  in  the  preceding  example,  that  the  miller  had,  by  request,  ground 
the  whole  80  bushels  for  the  jilanter,  and  that  the  planter  had  agreed  to  send  the 
miller  his  toll  the  next  day.     How  much  corn  should  he  send  ? 

Ans.  llf  bushels,  =  11  bush.  1  pk.  5  qts.  If  pts. 


FIRST    OPERATION. 
8   )   80 

10.     Toll  for  grinding  {. 

TOLL. 

110 

7  I    8 

I  lis   bushels.     Ans. 


Explanation. — By  the  exercise  of  our  reason, 
■we  see  that,  as  tlie  miller's  toll,  10  bushels,  is  j 
of  the  whole  quantity,  he  actually  charges  the 
toll  for  grinding  |  of  the  corn  for  the  planter, 
and  hence  it  is  clear,  as  shown  by  the  line 
statement,  that  if  the  grinding  of  |  of  the  corn 
for  the  jilanti-r  is  worth  10  bushels  toll,  the 
grinding  of  i  is  worth  the  seventh  part,  and  f 
or  the  whole,  is  worth  8  times  as  much,  which  is 
llj  bushels. 


(785) 


786 


SOULE  s  PHiLosorinc  i-ractical  mathematics. 


SECOND   oriCRATION. 


8  )  80   Imsliols  corn. 
10    bushels  toll. 

70  bushels  {rroiind  corn. 


70 


Explanation. — In  this  Bolntinn  -we  use  nnmbers 
representing  the  actuiil  iiinnber  of  bushels  of 
corn  brought  to  the  mill  and  gronud,  iustead  of 
the  fractional  iiunihers  as  in  the  lirst  solution, 
and  in  order  to  show  clearly  the  relationship 
iiunibers,  and  the  manner  of  obtaining  them, 
■\V6  jiresent  in  tlio  first  part  of  the  operation  the 
iigures  showing  the  bushels  of  ioJl  and  ground 
corn  or  meal  that  the  miller  and  planter  would 
liave  resi)ecti  vely  received,  had  the  usual  custom 
of  taking  toll  from  the  amount  of  corn  brought 
to  the  mill  been  observed.  By  these  figures  we  see  that  70  bushels  of  ground  corn  require  10 
bushels  of  toll,  and  as  by  the  agreement  the  planter  was  to  receive  HO  bushels  of  ground  corn,  it  is 
clear,  as  shown  by  the  line  statement,  that  if  70  bushels  ground  corn  refpiiro  10  bushels  toll,  1 
bushel  will  require  the  seventieth  part,  and  80  bushels  80  times  as  much,  which  is  11?  bushels. 

THIRD    OPERATION. 


TOLL. 

I  10 

I  80 

11}   bushels  corn,     Ans. 


100    bushels  assumed  as  the  quantity  ground. 
12}  is  i  of  same  for  grinding. 

87J  is  left  after  deducting  toll. 


B. 

I  100 
87J  I    80 

I  9I7  bushels  required  to  grind  and  deliver 
80,  and  pay  toll,  and  80  deducted 
from  91',  gives  llj. 


Or, 

To  find  the  toll  on  91J  bushels,  we  operate  thus : 
8  )  91? 

113.     Ans. 

3.  A  planter  agrees  to  gin  and  bale  the  cotton  of  Lis  neighbor  for  iV;  ^"^ 
under  this  agreement  lie  ginned,  baled  and  delivered  20  bales,  weighing  9200 
pounds.    How  many  pounds  of  cottou  is  due  him  for  Lis  labor? 

Ans.  1022f  pounds. 


FIRST    OPERATION. 

1^  =   /„  =  9200  pounds   the 


pound  ^   1 S 

owner's  share  of  an  unknown  quantity  to  be 

ginned. 


Hence  the  following  statement: 

I  9200 
9  I  10 

I  10222|  pounds  =  |5  the  total  quantity  to  gin 
as  both  the  owner's  and  the  ginner's 
shares. 

Then  10222^  —  9200  =  1022?  pounds  the  ginner's 
share,  or  the  quantity  yet  to  be  ginned. 


Explanation. — By  the  conditions  of  the  prob- 
lem, the  planter  who  gins  and  bales  the  cotton 
is  entitled  to  /„  of  the  whole  quantity  ginned, 
and  had  he  taken  his  pay  out  of  the  9200 pounds 
he  would  have  received  920  pounds.  But  as  the 
whole  9200  pounds  were  delivered  to  the  owner, 
the  planter  must  now  gin  such  an  additional 
quantity  as  will  pay  him  according  to  the  con- 
tract. By  the  terms  of  the  contract  he  is  to 
receive  iV  of  all  that  is  ginned,  and  hence  as 
shown  by  the  o))eration,  for  each  pound  ginned, 
1^,1  is  for  himself  and  -{-'a  for  the  owner.  There- 
fore the  9200  pounds  is  ,'''rt  of  the  total  quantity 
that  must  bo  ginned,  on  which   ,^,i  may  be  com- 


jiuted  as  the    share    or   compensation    of  the 
planter.     Hence,  since   ,'„  =  9200  pounds  |8  = 
10222J  pounds;  iV  of  which  is  1022^  pounds  yet  to  be  ginned  by  and  for  the  planter. 

SECOND    OPERATION. 

1  pound  ginned,  —  -^  pound  ginner's  share,  =  ^\j  pound  owner's  share. 

9200  pounds  ginned  for  owner.     ,V  of  9200  =  920  pounds  ginner's  share,  to  be  paid  in  cotton 
to  be  ginned.     9200  pounds  —  920  pounds  =  8280  pounds. 

Then  the  following  proportional  statement : 

8280  :  9200  :  :  920  :  10225  pounds  yet  to  be  ginned. 
Note. — The  explanation  for  this  operation  is  similar  to  that  of  the  second  solution,  problem  2. 


*  STORAGE.  'j'&'J 

THIRD    OPERATION. 

9200  —  10  =  920  pounds  due  for  ginning  9200  pounds. 

1  pound  =  •{  S  —  1^1?  =  V'lT  pounds  quantity  ginned  for  tlie  owner  to  earn  -j'rr  pound.  Hence  for 
each  pound  ginned  the  planter  receives  -^„  jiound  and  tlie  owner  r'(i.  'Since  it  is  worth  920  pounds 
to  gin  V'f  of  tlie  cotton  for  the  owner,  to  gin  ^^  it  is  worth  the  ^th  part,  and  to  gin  \%  it  is  worth 
ten  times  as  much,  which  is  IO225  pounds 

4.  A  planter  contracts  ■with  his  laborers  to  give  them  \  of  all  the  corn  they 
can  raise  on  his  plantation.  The  division  to  be  made  on  the  .stalk.  When  the  corn 
becomes  ripe,  he  contracts  to  pay  them  -.h-  of  his  share  for  gathering  the  same. 

The  entire  croiJ  amounted  to  14800  bushels.     What  was  the  planter's  share? 

Aus.  7104  bushels. 

OrERATION. 

\  )  14800   total  number  of  bushels. 


•j'j-  )  7400     bushels  ^  planter's  \. 

296     bushels  ==  ^V  paid  for  gathering. 


7104     bushels  =  jilanter's  share. 

5.  A  rice  planter  contracts  to  have  his  rice  harvested,  thrashed  and  sacked 
for  \  or  25  per  cent  of  the  yield. 

The  contractors  harvested,  thrashed,  sacked  and  delivered  to  the  jilanter 
1G500  sacks  of  1G2  lbs.  each  of  rough  rice.  It  is  then  agreed  that  they  shall 
continue  to  harvest,  thrash  and  sack  enough  more  rice  to  pay  them  according  to 
the  contract,  and  that  the  planter  shall  purchase  the  same  at  $1.80  per  sack.  What 
sum  will  the  planter  owe  the  contractors  ?  Ans.  $9900. 

FIRST    OPERATION. 

1  sack  —  1  =  J  =  1(!500  sacks  the  owner's  proportion  or  share  of  an  unknown  number  to  be 
haryested,  thrashed  and  sacked.     Hence  the  following  statement : 

I  16500 
3  I  4 

I  22000   sacks  to  be  harvested,  thrashed  and  sacked.     \  of  22000  =  5500  sacks,  the  share  of  the 
contractors  or  harvcstors. 

5500  X  $1.80  =  $9900     Ans. 

SECOND    OPERATION. 

leSOO  —  4  =  4125  sacks  due  contractors. 

1  — J  =  i  14123 

3|4 

I  5500   sacks  to  be  harvested,  thrashed  and  sacked  for  the  contractors. 
5500  X  $1.80  =  $9900     Ans. 
Note. — See  second  and  third  operations  of  problem  3. 

THIRD    operation. 

i  =  25^'';;';  100^^  —  25^^  =  75?^  ;  then  16.500  sacks  =  75^'^  of  a  number;  hence  1?^  will  be  the 
75th  part,  and  100"^,  or  the'number  will  be  100  times  as  much.  Thus,  as  75  :  100  :  :  16500  :  22000. 
25^0'  of  22000  is  5500  X  $1-80  =  $9900. 

\ote. — Kice  that  has  been  harvested  is  thrashed  and  sacked  for  8  to  10  cents  per  sack  or 
barrel  of  162  pounds. 


788  socle's  i'hilosopiiic   practical  mathematics.  * 

C.  Tlio  owner  of  timbercil  laud  engaged  laborers  to  make  and  bunch 
sliinglcs;  and  for  their  services  they  were  to  receive  J  of  what  they  manufactured. 
They  manufactured  240000  shingles.  IIow  many  of  the  number  are  they  entitled 
to  according  to  the  terms  of  the  agreement?  Ans.  80000. 

7.  Suppose,  ill  the  preceding  example,  that  the  laborers  had  been  required 
to  manufacture,  in  excess  of  the  240000,  the  shingles  to  Avhich  they  were  entitled. 
How  many  would  they  have  manufactured  for  themselves  in  order  to  make  a  just 
settlement?  Ans.  120000. 

8.  A  planter  agrees  to  gin  his  neighbor's  cotton,  and  receive  in  payment  for 
bis  labor  and  use  of  machinery  xs  of  the  cotton  ginned.  lie  received  70000  pounds 
of  seed  cotton,  which  made  20000  pounds  of  ginned  cotton.  How  many  pounds 
of  ginned  cotton  is  he  entitled  to?  Ans.  2000  pounds. 

9.  The  same  planter  referred  to  in  the  preceding  problem  agreed  with 
another  neighbor  to  gin  his  cotton,  and  receive  in  ])ayment  for  his  services  and  use 
of  machinery  seed  cotton  to  the  amount  of  -^g-  of  the  seed  cotton  ginned.  He 
received  70000  pounds  of  seed  cotton.  How  many  pounds  must  he  deduct  before 
ginning  to  pay  him  according  to  contract,  and  how  many  pounds  of  ginned  cotton 
did  he  deliver  to  his  neighbor,  allowing  3i  pounds  of  seed  cotton  to  make  one  jiound 
of  ginned?  Ans.  G3G3i\  iiounds  to  be  deducted. 

18181 1\  pounds  he  delivered. 

OPERATION. 

lV  +  1  or  Id  =  ji  3i)  63636 iV  pounds  to  gin. 

11)  70000      total  amount  received.  • 

6363 iV  pounds  of  seed  cotton  due  the  ginner.  18l81i'-V  pounds  ginned  cotton    delivered    to       j 

tlie  planter.  1 

Eiplanalion. — In  this  operation  we  see,  by  the  exercise  of  our  judgment,  that,  as  tlie  planter  f 
gins  the  cotton  for  i^,-  of  the  amount  ginned,  we  cannot  compute  the  -,^|,-  on  tlie  70000  pounds 
received,  for  tliat  would  he  charging  -,V  on  the  anuiunt  ginned,  and  on  the  cotton  cliarged  for 
ginning.  By  careful  thought  we  nee  that  if  -,^1,  of  what  is  ginned  is  charged  for  ginning,  tlien  the 
amount  received  to  gin  is  equal  to  -^-fj  plus  1,  or  jg,  whicli  is  -|A,  .and  hence  if  the  amount  received  is 
■fi  of  the  amount  ginned,  -,V  is  the  amount  to  he  deducted  for  ginning. 

SECOND    OPERATION. 
100  pounds  assumed  to  gin.  p. 


10   pounds  due  for  ginning. 

110 

110  pounds  of  seed  cotton  required  to  gin  100. 


100 
70000 

63636,*!    pounds  to  gin. 
70000 

6363i',    pounds  of  S.  C.  due  the  ginner. 

3i  )  63636,V 


18181,''V  pounds  of  ginned  cotton. 

Explanation. — The  principles  of  this  second  solution  have  been  so  repeatedly  explained  in 
Per  Centum,  Commissiou  and  Brokerage  operations,  that  we  deem  it  unnecessary  to  occupy  space 
with  them. 

10.  A  housekeeper  wishes  to  supply  his  family  with  30  bushels  of  flour.  His 
miller  charges  -j\-  toll  for  grinding.  How  many  bushels  of  wheat  must  he  send  to 
mill  in  order  to  obtain  the  exact  30  bushels  of  flour  ? 

Ans.  32-i^f  bush.  =  32  bu.  2  p.  7  qts.  1^  pts. 


I 


LLIGAXION. 


1375.  Alligation,  which  signifies  a  tie  or  binding,  is  a  process  for  solving 
questions  rehitiiig  to  the  mixing  or  compounding  of  different  kinds  of  ingredients, 
or  of  combining  different  qualities  or  values. 

It  is  generally  treated  under  two  heads  or  divisions,  viz:  Alligation  Medial 
find  Alligation  Alternate. 

1376.  Alligation  Medial,  or  Medial  Proportion,  is  the  process  of  finding 
the  mean  or  average  jirice  or  rate  of  a  mixture  of  different  kinds  or  qualities  of 
ingredients,  when  tlie  quantity  and  value  of  each  kind  are  given. 

PROBLEMS. 

1.  A  merchant  mixes  400  lbs.  of  sugar  worth  9  cents  jier  lb.,  and  600  lbs. 
vorth  14  cents,  Avith  1000  lbs.  worth  15  cents ;  what  is  the  mixture  worth  per  pound  T 

Ans.  13J  cents. 

OPERATION. 

400  lbs.  .it    9c.  =:    $36.00  Explanation. — In  tliis  example,  we  have   only 

600      '•     '■  14c.  =      84.00  to  niultijily  each  given  quantity  liy  its  price  to 

1000      "     "  15c.  =    150.00  obtain  its  iliill  value;  thi-n  it  is'eviilent  that  the 

•                              sum   of    these   values,   |;l'70.0(I,    divided   l>y   the 

2000   lbs.                =  !f270.00  whole   number   of  pounds,   2000,    will   give   the 

mean  value,  or  average  price  per  pound  of  the 

Hence  1  lb.  ^     Ans.         IS^c.  mixture. 

2.  If  I  mix  .50  lbs.  of  tea  at  60  cents  per  lb.,  with  40  lbs.  at  75  cents  and  100 
lbs.  at  80  cents,  what  is  the  average  value  of  the  mixture  per  pound  ? 

Ans.  73||  cents. 

3.  What  per  cent  of  Alcohol  in  a  mixture  of  40  gallons,  85  per  cent  strong; 
CO  gallons,  90  per  cent  strong;  50  gallons,  98  per  cent  strong;  and  30  gallons,  95 
l)er  cent  strong f  Ans.     91j^%. 

OPKBATH)S   INDICATED. 

40  gals.  <a)  S5?„'  =  3400 

60  ■'   "  90%  =  5400 

50  "   "  98?^  =  4900 

30  "   "  95^0  =  2850 

180  )  16550  ( 

4.  "What  is  the  purity  of  a  mixture  of  20  pwt.  of  gold,  14  carats  fine;  30 
pwt.  16  carats  fine ;  50  pwt.  12  carats  fine ;  40  pwt.  18  carats  fine ;  and  10  pwt.  pure 
gold  1  Ans.  15^  carats  fine. 

ALLIGATION   ALTERNATE. 

1377.  Alligation  Alternate  is  a  kind  of  Indeterminate  Reciprocal  Proportion, 
and  is  the  process  of  finding  the  ratios  of  the  quantities  of  the  articles,  whose 
values  or  qualities  are  given,  that  must  be  taken  to  form  a  compound  or  mixture 
of  a  given  mean  rate  of  value  or  quality. 

(789) 


790  SOULe's    rHILOSOPHlC    PRACTICAL    MATHEMATICS.  * 

It  is  the  reverse  of  Alligation  Medial,  and  may  be  proved  by  it. 

The  solution  of  question.s  in  Alligation  Alternate  is  based  on  the  principle 
of  an  equality  of  gains  and  losses;  that  is,  in  the  composition  of  a  mixture,  the 
gains  on  the  articles  whose  values  are  less  than  the  mean  value,  must  be  equal  to 
the  losses  on  the  articles  whose  values  are  greater  than  the  mean  value.  Thus,  if 
we  mix  1  pound  of  sugar  worth  10  cents  with  1  pound  worth  12  cents,  we  will  have 
2  pounds  worth  22  cents,  or  an  average  of  11  cents  per  pound.  We  here  see  that 
the  gain  of  1  cent  on  the  first  pound  is  just  equal  to  the  loss  on  the  second  pound ; 
the  amounts  used  being  equal,  the  gain  and  loss  are  relatively  equal.  lint  suppose 
we  Lad  1  pound  worth  10  cents  that  we  wished  to  mix  with  sugar  worth  13  cents, 
so  as  to  make  a  compound  worth  12  cents  per  pound,  we  see  that  there  will  be  a, 
gain  of  2  cents  on  1  pound  of  the  first  kind,  and  a  loss  of  1  cent  on  1  pouTid  of  the 
second  kind,  and  hence  to  produce  an  equilibrium  of  gains  and  losses,  we  must 
take  enough  of  the  kind  at  13  cents  to  give  a  loss  of  2  cents,  that  is  2  pounds. 
Here  we  see  that  the  quantity  required  of  e.ach  kind  is  inversely  i)roportional  to 
the  gain  or  loss  on  1  pound  of  that  kind,  as  compared  with  the  other,  and  if  there 
were  several  other  kinds  used,  the  same  reasoning  would  apply,  as  shown  by  the 
following  illustrations : 

The  subject  is  treated  under  four  cases  or  conditions. 

Case  I. — Analytic  Method. 

TO    FIND    THE    PROPORTIONAL    PAKTS    TO    BE    USED    WHEN    TUB 

MEAN  OR,  AVERAGE  PRICE  OF  A  MIXTURE  AND  THE  PEICES 

OF  THE  DIFFERENT  ARTICLES  AEE  GIVEN. 

1378.     1.     What  relative  quantities  of  sugar,  at  5  cents  per  lb.  and  8  cents 

per  lb.,  must  be  used  to  make  a  compound  worth  (>  cents  per  pound  ? 

Ans.  2  lbs.  at  5  cents.     1  lb.  at  8  cents. 

OPEUATioN.  Explanation. — If  a   compound  of  two  grade* 

of  BUgar,  one  worth  5  cents  and  the  other  worth 
8  cents,  be  made  iu  certain  priiportions,  and 
the  mixture  be  priced  or  valued  at  6  cents,  there 
will  bo  a  gtiin  on  the  5  cent  grade  .and  a,  loss  oa 
Ans.  the  8  cent  grade.      This   gain  and   lo.ss   in   the 

whole  mixture  must  be  exactly  equal,  and  hence 
we  must  take  Kuch  quantities  of  each   as   will 
PnooF.  give  a  gain  and  a  loss  of  equal  amount  in  the 

unit  of  value,  here  1  cent;   the  result  will   be 

2  lbs.  at  5c.  =  10c.  the  correct  quantities  to  be  compounded. 

1    lb.    at  8c.  =    8c.  By   selling  1  pound  of  sugar  worth  5  cents  for 

. ■  6  cents,  there  will  be  a  gain  of  6  —  5  =  1  cent; 

3  lbs.  at  Gc.  =  18c.  hence  we  place  1  pound  opposite  the  5. 

By  selling  1  ]>ound  of  sugar  worth  8  cents  for  6  cents,  there  will  be  a  loss  of  8  —  6  =  2  cents, 
and  to  lose  1  cent  will  require  i  of  a  pound,  hence  wo  place  i  pound  Ojiposite  the  8.  Therefore,  1 
pound  at  5  cents  and  A  pound  at  8  cents  are  tho  proportional  quantities  for  the  compound  or  mixture. 
Now,  since  tho  gain  and  loss  are  equal  in  these  proportional  quantities,  it  is  clear  that  they  will  be 
equal  in  any  number  of  times  the  same.  Therefore,  wo  may  multiply  the  proportional  quantities 
by  any  number  and  produce  other  proportional  numbers. 

In  this  problem,  we  multiply  by  2,  the  least  common  multiple  of  their  denominators,  and 
thus  obtain  2  pounds  at  5  cents  and  1  pound  at  8  cents,  for  the  mixture. 

Note. — See  page  159,  for  the  Least  Common  Multiple. 


6^ 


I    8      } 


:i 


ALLIGATION. 


791 


2.  "Wliat  proportions  of  tea  at  48  cents,  60  cents,  60  cents,  and  72  cents  a 
pound,  must  be  mixed  together  so  that  the  mixture  shall  be  worth  64  cents  per 
pound?  Aus.  1  lb.  at  48  cents,  2  lbs.  at  06  cents, 

1  lb.  at  CO  cents,  3  lbs.  at  72  cents. 


(iS 


6ii 


;  60 


66 


I  72 


FIRST   OPKnATION. 


1 

2 

3 

4 

5 

,', 

1 

1 

i 

1 

1 

* 

0 

2 

i 

2 

2 

Ans. 


PROOK. 

1    111.    at  48c.  =       48c. 

1  •■      '■  60c.  =        60c. 

2  lbs.   "  66c.  =$1.32 
2     ■'      "  72c.  =    1.44 


"  64c.  =  $3.84 


Explanation. — To  maintain  the  equality  of  i/ains  and  losses,  we  must  always  compare  tlie 
prices  of  two  articles  or  ingredieDts,  one  that  is  greater  and  one  that  is  less  than  the  average  or 
mean  price,  and  treat  each  pair  or  couplet  thus  compared  as  a  separate  problem.  In  the  first 
operation,  we  formed  two  couplets  and  compared  48  and  72,  and  60  and  66.  In  the  second  operation, 
we  compared  48  and  66,  and  60  and  72. 

In  the  first  operation,  we  find  1,  By  selling  1  pound  of  tea  worth  48  cents,  for  G4  cent.s,  there 
will  be  a  gain  of  16  cents;  and  to  gain  1  cent,  would  reijuire  ,\-  of  a  pound,  which  wo  write  in  tho 
first  column  opposite  the  48  cents.  2.  By  selling  1  pound  of  tea  worth  72  cents,  for  64  cents,  there 
is  a  loss  of  8  cents;  and  to  lose  1  cent,  would  require  -J  of  a  pound,  which  is  written  in  the  first 
column  opposite  the  72  ceuts.  3.  By  selling  1  pound  of  tea  worth  60  cents,  for  64  cents,  there  is  a 
gain  of  4  cents;  and  to  gain  1  cent,  would  require  i  of  a  pound,  which  we  write  in  the  second 
column,  opposite  the  60  cents.  4.  By  selling  1  jiound  of  te.a  worth  66  cents,  for  64  cents,  there  is  ;i 
loss  of  2  cents;  and  to  lose  1  cent  would  require  ^  of  a  pound,  which  we  write  in  the  second 
column,  oppo.site  the  66  cents. 

Kow  having  finished  with  the  couplets  and  produced  the  correct  proportional  quantities,  we 
reduce  the  fractional  quantities  to  integers  or  whole  numbers,  which  is  done  by  multijilying  the 
quantities  in  the  respective  columns  of  gain  and  loss  proportionals,  by  the  least  common  multiple 
of  their  respective  denominators.  Thus,  -i\.  and  |  were  multiplied  by  16,  giving  1  and  2  pounds, 
which  we  write  in  the  third  column.  The  J  and  i  were  multiplied  by  4,  giving  1  and  2  pounds 
■which  we  place  in  the  fourth  column.  The  proportional  quantities  are  then  arranged  in  column 
five. 

Note  1. — \Vben  the  prices  of  the  ingredients  and  the  mean  rate  are  whole  numbers,  the 
intermediate  steps,  as  shown  in  columns  1  to  4,  may  bo  omitted,  and  the  same  results  obtained  by 
taking  the  dift'erence  between  each  article  used  as  a  couplet  and  the  mean  rate,  and  placing  it 
opposite  the  other  number  of  the  couplet.     See  the  linking  method  a  few  pages  beyond. 

Note  2. — There  will  be  as  many  columns  of  proportional  quantities  as  there  are  couplets. 

SECOND    operation. 


r48 

^*]66 


1 

2 

3 

4 

iV 

1 

i 

2 

* 

8 

i 

1 

il 


Ans. 


Note. — This  operation  shows  that  by  com- 
paring different  prices,  dili'erent  jiroportional 
quantities  are  obtained. 


Note  1. — '\^1len  there  is  an  odd  quantit.v,  compare  the  same  with  one  that  has  been  already 
compared  and  take  the  sum  of  the  two  numbers  opposite  tho  one  used  twice. 

Note  2. — A  Common  Factor  may  be  used  in  any  couplet  or  omitted  from  it  without  changing 
the  proportional  parts.  Hence  it  is  obvious  that  there  may  be  any  number  of  answers  in  tho  same 
proportion. 


792  SOULE  S    rillLO.SOPHIC    I'RACTICAI.    MATHEMATICS. 

1379.  LOSS  AND  GAIX  METHOD. 


Tiiiiii)  on 

:HATI<)N 

ANIJ 

r.atio. 

PUOOF. 

ProportioBal 
quantities. 

(1  lb.  at  48c. 
„,  1  lb.  at  60i'. 
•'■'l  1  lb.  at  66c. 

.1  lb.  at  72c. 

=  +  16c. 

=  +    4c. 

1  = 

20c.  gain 

=  2 

to 

=  1 

1  lb.    at  48c.  =    48c. 
1  lb.    at  6()c.  =    60c. 

=  —  2c. 
=  —    8c. 

10c.  loss 

2  llis.  at  66c.  =  132c. 
2  lbs.  at  72c.  =  144c. 

6  lbs.  at  64c.  =  384c. 

Explanation. — In  this  operation,  we  first  'write  the  prices  of  the  given  commodities  in  a 
column  in  rcgnlar  order,  beginning  with  the  smallest,  and  the  mean  price,  64  cents,  on  the  left,  as 
a  basis  of  comjiarison.  We  then  see  tluit  "when  we  use  1  pound  of  tea  at  48  cents,  in  a  mixture  to 
lie  worth  64  cents,  we  have  a  gain  of  16  cents;  we  place  this  to  tlie  right  of  the  )>rice  that  produced 
It,  and  mark  it  j)/«s,  to  indicate  a  gain.  We  tlien  compare  tlie  next  ]iricc,  60  cents,  with  the  mean 
jirice,  64  cents,  as  in  the  first  case,  and  find  a  gain  of  4  cents,  wliich  we  ]ilace  as  before  to  the  right 
of  its  producing  term,  60,  and  mark  it  plus  also.  Then  jirocceding  with  the  next  term,  66,  we  find 
by  comparing  it  with  the  mean  64,  that  in  using  1  pound  there  will  be  a  loss  of  2  cents:  we  place 
tliisto  the  right  of  66,  and  mark  it  minus  ( — )  to  indicate  a  loss.  Comjiaring  the  fourth  term,  72, 
in  the  same  manner,  we  tinil  a  loss  of  8  cents,  wliich  we  place  and  mark  as  m  the  last.  Having 
now  found  the  gains  and  losses  on  all  the  terms  taken  singly,  as  coni])ared  with  the  mean  term  on  a 
basis  of  1  lb.  of  each  kind,  we  next  add  all  the  gains,  marked  plus,  together,  giving  a  total  gain  of 
20  cents,  and  set  to  the  right.  We  then  add  all  the  losses,  marked  minus,  giving  a  total  loss  of  10 
cents,  and  place  to  the  right.  Then,  as  these  two  amounts  have  .a  connuon  measure,  10,  we  divide 
them  both  by  10,  and  set  the  quotients  still  to  the  right.  These  results  have  the  same  ratio  to  each 
other  as  the  whole  gain  and  loss,  and  hence  we  have  the  ratio  of  the  gain  to  the  less  as  2  is  to  1,  or 
a  relative  gain  of  2  cents  to  a  loss  of  1  cent  in  the  admixture  of  1  pound  of  each  kind  of  tea.  But 
as  the  gains  and  losses  in  the  whole  mixture  must  be  the  same,  we  must  equalize  them  by  changing 
the  relative  ]>roportions  of  the  ingredients  used ;  in  order  to  do  this,  we  must  multiply  the  quanti- 
ties used,  1  pound  of  each,  by  the  reciprocal  ratio  of  its  gain  or  loss.  Now,  the  ratio  of  gain  to 
loss  is  as  2  to  1,  and  their  reciprocal  ratio  is  as  1  to  2.  Therefore  we  multiply  the  given  quantities, 
1  pound  of  each  ingredient,  that  gave  a  gain,  or  that  stands  opjiosite  the  sigapliis,  by  the  relative 
number  or  ratio  of  loss,  1,  anil  ))lace  the  jiroduct  to  tlie  right  in  a  c<diimu  headed  "Proportional 
Quantities  ;"  then  we  multijily  the  units,  1  pound,  opposite  the  minus  sign,  by  the  relative  number 
or  ratio  of  gain,  2,  and  set  the  results  in  the  column  of  proportionals.  These  proportionivls  are  the 
qiuantities  reqtiired  to  he  taken  of  each  kind  of  the  ingredients  at  the  prices  opposite  to  which 
they  stand. 

To  prove  the  correctness  of  the  work,  multiply  these  proportional  quantities  by  their 
respective  prices,  and  set  the  products  to  the  right.  The  sum  of  the  products  divided  by  the  sum 
of  the  proportionals  will  give  for  a  quotient  the  mean  price,  as  in  the  example.  By  this  method 
the  proportional  quantities  of  all  the  commodities  that  give  a  gain  are  always  equal,  and  all  those 
that  give  lo-sses  are  equal.  These  proportionals  may  be  multiplied  by  any  number,  and  heuce,  an 
indefinite  number  of  answers  having  the  same  ratios  will  be  found. 

LINKING  METHOD. 

1380.  We  Tvill  now  present  another  method  by  which  the  reciprocation  of 
ratio.s  is  more  readily  effected,  and  the  results  may  be  different  from  tliose  of  the 
first  two  methods,  thoujjli  botli  are  correct.  Tlie  reason  however  is  not  so  apparent, 
though  this  method  is  often  used,  and  has  jriveu  the  name  to  this  branch  of  Arith- 
metic Alligation,  as  the  process  is  indicated  by  alligatiiig,  or  linking,  one  rate  that 
is  less  than  the  meau  rate  with  one  that  is  greater,  as  shown  below : 


OPKRATION  AND  PROOF 

J  60 


'48- 

RA  I   601 

6"     66J 
[t2_ 


8  lbs.  at  48c.  =:    $3.84  Explanation. — We  here  write  the  simples  as  in 

2  lbs.  at  60c.  =      1.20  the  first  method,  and  link  one  that  is  less  with 

4   lbs.  at  66c.  =:      2.64  one  that  is  greater  than  the  mean  rate;  that  is, 

16   lbs.  at  72c.  =    11.52  we  link  48  with  72,  and  60  with  66.     We  then 

—                               take  the  first  couplet,  48  and  72,  and  comparing 

30  lbs.  at  64c.  =  $19.20  each  term  with  the  mean  rate  64,  we  find  there 

will  he  a  gain  on  1  pound  costing  48  cents,  when 


ALLIGATION. 


'93 


used  in  a  compound  valued  at  64  cents,  of  16  cents,  and  a  loss  of  8  cents  on  1  pnnnil  costing  72 
cents.  Now,  as  the  gain  is  to  the  loss  as  16  is  to  H.  in  this  conjilet  where  1  pound  of  each  is  used, 
and  as  the  mixture  must  give  an  equal  coniljinatiou  of  gains  and  losses  from  each  couplet,  we  know 
from  reason  and  previous  explanations,  that  the  quantities  used  must  be  taken  in  au  inverse  or 
reciprocal  ratio  to  the  gain  and  loss,  tliat  is,  as  8  is  to  16.  We  therefore  place  the  diti'erence 
betweeu  64  and  48  =  16,  not  ojiposite  the  48,  to  represent  the  gain  as  iu  the  second  method,  hut 
opposite  the  7'2,  to  rejiresent  the  relative  quantity  to  be  taken  at  that  price;  and  for  the  same 
reason  we  place  the  difference  between  64  and  72  =  8,  opposite  the  48.  The  terms  of  the  other 
couplet,  60  and  66  linked  together,  are  treated  iu  the  same  manner,  by  placing  the  difference 
between  the  mean  rate,  64,  and  each  term,  opposite  to  the  one  w  ith  which  it  is  linked.  These 
differences  thus  arranged  constitute  the  jiroportional  quantities  that  are  to  be  taken  of  the  several 
ingredients  that  form  tlie  compound,  at  the  prices  opposite  to  which  they  stand.  The  proof  is  the 
same  as  iu  the  first  method.  When  there  are  more  than  three  terms,  in  which  two  or  more  are 
less,  and  two  or  more  are  greater  than  the  mean,  they  can  be  linked  in  various  ways,  each  variation 
giving  a  new  set  of  ratios,  and  as  each  set  of  ratios  admit  of  indefinite  multiples,  the  answers  to 
questions  of  this  character,  with  those  conditions,  may  be  infinite.  When  any  rate  has  several 
others  linked  with  it.  each  alligated  term  will  produce  separate  differences,  and  their  sum  will  be 
the  proportional  required  for  that  rate. 

PEOBLEMS  WOEKED  BY  THE  ANALYTIC  METHOD. 

1381.  1.  A  wine  niercliant  lias  -wine  worth  $1.10,  $1.80,  $2.50,  per  gallon, 
which  he  wishes  to  mix  with  water  so  that  the  couipouiid  will  be  worth  $1.50  jjer 
gallon.    What  will  be  the  proportional  quantities  of  wine  and  water  ? 

Aus.  10  gallons  water. 

3  "        wine  at  $1.10. 

4  "  "      "  $1.80. 
15       "  "      "  $2.50. 

OPERATION. 


$1.5p. 


1 

*) 

3 

4 

0.00 

-h 

10 

1.10 

i 

3 

1.80 

h 

i 

2.50 

1 
1  0 

15 

10 
3 
4 

15 


AlKS. 


2.  A  grocer  has  three  grades  of  coffee  worth  respectively  10,  11,  and  15/  per 
pound.  In  what  proportions  must  they  be  mixed  so  that  the  compound  would  be 
worth  12/  per  pound  1  Ans.  In  equal  quantities. 


OPKRATION. 


1 

2 

3 

4 

5 

i 
i 

1 

i 

3 

2+ 

3 
1 

3 
3 
3 

Ans. 


PROOF. 

3  lbs.  at  10c.  =  i 
3  lbs.  at  lie.  = 
3   lbs.  at  15c.  = 


.30 
.33 
.45 


9  lbs.  at  12c.  =  $1.08 


Explanaiion. — Here  we  have  two  articles  less  than  the  mean  rate,  and  one,  greater.  First 
compare  10  and  15  as  the  first  couplet  with  12,  thus :  10  from  12,  2,  4  ;  12  from  15,  3,  J.  Second  com- 
pare 11  and  15  as  the  second  couplet  with  12,  thus  :  11  from  12,  1 ;  12  from  15,  3,  \.  Then  multiply 
the  proportional  numbers  iu  column  1  by  6,  and  those  in  column  2  by  3.  The  6  and  3  being 
respectively  the  least  common  multiple  of  the  denominators  of  the  numbers  in  the  respective 
columns.     Then  arrange  these  results  iu  the  5th  column. 


794 


soule's  rniLOSoPHic  tractical  mathematics. 


3.  A  planter  has  oats  -worth  40  cents,  corn  worth.  60  cents^  and  barley 
\rorth  70  cents  ]ivv  bushel.  lie  wishes  to  mix  them  so  as  to  make  a  compouTul  worth 
fiO  cents  per  but^licl.     AVhat  proportions  must  ho  use!       Ans.  3  bushels  of  oats. 

1  husliel  of  corn. 
1  bushel  of  barley. 


OPKRATIOX. 


50  s'  (id 
^70 


1 

o 

3 

4 

T^,r 

-'+ 

1 

V,F 

1 

■^'^ 

1 

i! 


Ans. 


F.J-phiiuit'uin. — TTero  -wc  have  two  articles  greater 
than  the  mean  rate  and  (ino  less.  First  conijiaro  40 
and  70,  as  the  first  ciinplet,  witli  50,  tlie  mean  price. 
Second  compare  40  and  CO,  as  the  second  couplet, 
with  50.     Then  proceed  as  in  the  previous  problems. 


4.     What  relative  qualities  of  flour  worth  $4|,  $5,  and.  $7^  per  barrel,  must 
be  sold  to  realize  an  average  price  of  $6J  ? 

Ans.  4  bbls.  at  $4J ;  4  bbls.  at  $5 ;  and  12  bbls.  at  $7^. 

Explanation. — First,  use  4J-  and  7J  as  the  first  couplet, 
■which  gives  1}=J  =  J-  and  1. 

Second,  use  5  .and  7i  as  the  second  couplet,  ■which 
)  gives  1J=5  =  |  and  1.     Then  multiply  the   proportional 

/  Ans.  numbers  in  column  1  by  7  and  tho.'ie  in  column  2  by  5. 
)  See  problem  1,  page  790,  and  problem  2,  page  793. 


OPKRATIOX. 


6J 


hi 


1 

2 

3 

4 

f 

4 

* 

4 

1 

1 

7+ 

5 

5.     If  I  mix  wines  worth  respectively  60  cents,  80  cents,  $1.00,  $1.30,  and 
$1.50  per  galloii,  how  many  gallons  of  each  kind  will  be  required  to  form  a  mixture 
worth  $1.10  per  gallon  ? 
Ans.  4  gals,  at  CO/,  2  gals,  at  80/,  2  gals,  at  $1,  4  gals,  at  $1.30,  5  gals,  at  $1.50. 
And  other  answers  by  different  coupling. 


110 


r   61 

8( 
l.Oi 


60- 
80- 
00] 


1.30 
l..=)0 


1 

2 

3 

4 

5 

G 

1 

S'.T 

4 

O 

■JlT 

V-n- 

2 

^ 

■Ar 

3+ 

1 

A 

5 

Ans. 


Oats. 

Com. 

r.ye. 

Parley. 

Wheat. 

]  bu. 

2  bn. 

4  bn. 

2  bn. 

1  bu. 

9bu. 

0  Im. 

10  bu. 

10  bn. 

10  bu. 

6bu. 

3  bu. 

4bu. 

4bu. 

Gbu. 

6.  Mixed  oats  at  40  cents,  corn  at  50  cents,  rye  at  75  cents,  barley  at  SO 
cents,  and  wheat  at  $1  per  bu.shel.  How  many  bushels  will  it  take  to  make  a 
mixture  worth  70  cents  per  bushel  1 

Ans.  By  the  1st.  or  reciprocal  method, 
Ans.  iJy  the  loss  and  gain  method, 
Ans.  By  the  linking  method. 

And  other  ratios  bj^  different  methods  of  linking. 

7.  A  goldsmith  would  mix  gold  12  carats  tine,  with  some  of  13,  14,  16,  IS, 
and  20  carats,  and  alloy  valued  at  nothing.  How  much  of  each  kind  will  be 
required  to  make  a  compound  of  15  carats  fineness? 

Ans.  By  the  1st  or  reciprocal  method,  5  parts  of  12  carats;  3  parts 
of  13  carats;  1  part  each  14  and  IG  carats^  2  parts  of  IS 
carats ;  and  3  parts  of  20  carats. 

And  other  ratios  by  different  method  of  work. 


ALLIGATION. 


795 


Case — II. 
1383.     Wlien  tlie  quantity  of  one  of  tbe  commodities  composing  a  mixture  oi 
a  given  mean  value  is  given,  to  find  the  quantity  of  each  of  the  others. 

1.  A  merchant  has  molasses  -worth  35  cents,  52  cents,  and  70  cents  a  gallon. 
He  ■wishes  to  form  a  mixture  ■worth  CO  cents  per  gallon,  and  to  have  the  same 
contain  IL'O  gallons  of  the  35  cents  per  gallon  molasses.     How  many  gallons  of  each 


grade  of  molasses  must  he  take  ? 


Ans.  See  oi)eration. 


OPERATION'. 


(35 

60  .;  52 
(»70 


1 

2 

3  1 

4 

A 

2 

i 

0 

A 

1-0 

5+ 

i 

2  X  60  =  120  ) 
5  X  60  =  300  } 
9  X  60  =  540  S 


Ans. 


Explanation. — ^We  first  find,  as  in  case  1, 
article  1378,  the  proportional  (juantities  to  be  2 
gallons  of  the  35  cents,  5  gallons  of  the  52  cents, 
and  9  gallons  of  the  70  cents  molasses.  But 
since  120  gallons  of  the  35  cents  molasses, 
instead  of  2,  are  to  be  used,  we  must  take  ■'■|^ 
:=  60  times  each  of  the  other  grades  or  priced  molasses,  in  order  that  the  gai:i  and  loss  may  be 
equal.  Hence  -we  have  2  x  60  =  120  gallons  at  35  ceuts,  5  X  60  =  300  gallons  at  52  cents,  and  9  X 
60  =  540  gallons  at  70  cents. 

2.  A  grocer  Las  100  pounds  of  coffee,  -worth  10  cents  per  pound,  which  he 
■wishes  to  mix  ■with  other  grades  of  coffee  ■worth  respectively  12  cents,  14  cents,  15 
cents,  16  cents,  and  18  cents  per  pound,  .so  that  the  mixture  ■will  be  worth  15  cents 
per  jjound.     How  much  of  each  kind  Avill  be  required  ?  Ans.  See  operation. 

OPKRATIOX. 

7 


15  < 


1 

2 

3 

4 

5 

6 

rio — 

^ 

3 

12— 

i 

1 

14-, 

15 

16— 

1 

1 

1 

1 

3+ 

1 

U«_ 

— 

i 

5 

3  X  33 it  =  100    pounds. 
1  X  33i  =    33i 

1  X  33i  =    33J- 

any  quantity  at  15c. 

4  X  33;^  =  133jt  pounds, 

5  X  33  j  =  166 J 


Ans. 


Explanation. — We  first  find  the  relative  or  proportional  quantities,  as  in  case  1,  these 
quantities  are  sho-n-n  in  column  7.  Then,  since  100  jiounds  of  cotfeo  -n-orth  10  cents  per  pound  are 
to  be  used,  we  divide  100  by  3,  the  relg,tive  number  for  the  10  cents  coffee,  and  in  the  quotient  33J 
we  have  the  ratio  with  which  to  mtiltiply  each  of  the  proportional  quantities,  as  shown  in  column 
7,  to  obtain  the  quantity  required  of  each. 

Note. — -The  15  cent  grade  of  coffee  being  of  the  same  value  as  the  mean  rate,  it  is  not 
considered  in  the  operation.     It  may  be  omitted  from  the  list  of  prices,  if  desired. 

3.  A  merchant  has  280  pounds  of  coffee,  ■worth  16  cents  per  pound,  Avhich 
he  ■n'ishes  to  mix  ■with  other  qualities  worth  IS  cents,  21  cents,  22  cents,  and  24 
cents  respectively.  How  much  of  each  additional  kind  will  be  required  to  compose 
a  mixtm-e  that  will  be  worth  20  cents  per  pound  ? 

Ans.  By  1st  method,  280  lbs.  each  at  16,  22,  and  24  cents,  and  560 
lbs.  each  at  18,  and  21  cents. 
And  other  ratios  by  different  methods  of  work  and  different  Unkings. 

Operation  by  the  1st  method. 
4       5       6 


20  <^ 


1 

2 

3 

16 

J. 

18 

+ 

+ 

21 

1 

22 

+ 

24 

i 

1+ 


X  280  —  280 
X  280  =  560 
X  280  =  560 
X  280  =  280 
X  280  =  280 


280  -^  1  =  280  ratio  number  to  multiply  each  of  the  proportional  quantities. 


796 


SOULE  S    PHILOSOPHIC     PRACTICAL    MATHEMATICS. 


4.  A  stock  broker  has  140  sliarcs  of  E.  II.  stock  that  cost  him  $00  per  share. 
How  many  more  shares  must  ho  buy  at  $S0  per  share,  so  that  he  can  sell  the  wliole 
at  $75  per  share  and  gain  20  per  cent  on  liis  investment  7     Ans.  20  shares  at  '■ 


OPERATIONS. 


Statement  to  fiiiil  average  cost 


I  100 
120    75 


$62  i   cost. 


Bv  tlie  1st  niethoj. 


m{\ 


1 

o 

60 

i 

It 

80 

k 

2 

1  X  20  =.  20    Ans. 


140  -7-  7  =  20  ratio  iuiml)cr  by  which 
to  iniilti|>ly  proportioual  quantity 
of  railroad  stock. 


By  the  linking  method. 

„,,  (60    17i  =  '/=7 
"-'  (  80'     2i  =   I   =  1 

1  X  20  =  20     Ans. 


By  th»  loss  and  gain  method. 

.oi  <  fiO    2^  gain  =  1  =  7 
"-*  ^80  17ilo8S    =7  =  1 

1  X  20  =  20     Ans. 


Case— III. 


1383.     When  the  quantity  of  two  or  more  of  the  ingredients  or   articles   is 
given  or  limited,  to  find  the  quantity  of  each  of  the  others. 

1.  How  many  pounds  of  Naples  walnuts  at  30  cents,  and  of  almonds  at  27 
cents,  must  be  mixed  with  300  pounds  Tenn.  peanuts  at  5  cents,  200  pounds  of 
Italian  chestnuts  at  8  cents,  150  pounds  of  Brazil  nuts  at  9  cents,  and  250  pounds 
of  Louisiana  pecans  at  16  cents,  so  that  the  average  price  of  the  mixture  may  be 
worth  15  cents  ? 


OPERATION. 


300  lbs.  Tenn.  peanuts         at    5c.  =  $15.00 

200  ll>s.  Italian  chestnuts  at    8c.  =    16.00 

150  lbs.  Brazil  nuts  at    9c.  =    13.50 

250  lbs.  Louisiana  pecans  at  16c.  =    40.00 


900  lbs. 


$84.50  (  97,c.  =  average  or  mean  price  of  the  given  articlea. 


(   9.^ 
15  •(27 
^30 


27U 

101 


+216 
101 


486   )  C900       pounds. 

101  }  X  n}  =  ]m^ 

101    )  (187j', 


900  -:-  486  =  155  =:  ratio  of  increase  for  each  proportional  quantity. 

Explanation.— 1  a  all  problems  of  this  kind,  we  fust  find  the  average  or  mean  price  of  the 
articles  or  injjredient.s  given,  according  to  the  principles  of  Alligation  Medial.  See  Article  1376. 
Second,  %vo  tind  the  jiroportional  quantities  of  the  average  price  of  the  articles  given  and  of  the 
articles  required,  according  to  the  principles  of  case  1.  See  Article  1378.  Third,  we  find  the  ratio 
number  with  which  to  multiply  the  proportional  quantities  so  that  900  pounds  of  the  given  articles 
may  be  used. 

As  .shown  in  the  operation,  the  proportional  quantities  are  486  pounds  of  the  given  articles 
and  101  pounds  each  of  the  nngiven  quantities,  the  walnuts  and  the  almonds.  And  since  900  pounds 
of  the  given  articles  are  to  be  used,  we  divide  900  by  486  and  obtain  Ij}  as  the  ratio  number  with 
which  to  multiply  the  proportional  numbers. 


ALLIGATION. 


797 


2.  A  goldsmith  has  11  oz.  of  gold,  14  carats  fine,  12  oz.  of  15  carats,  and  17 
oz.  of  18  carats.  He  -wishes  to  mix  with  these,  gold  of  17  carats,  20  carats,  and  22 
carats,  so  as  to  make  the  compound  19  carats  fine.  How  much  of  each  of  the  three 
last  will  be  required  ?  Ans.  See  operations. 

OPERATIONS. 

FIRST  STEP. 

11  oz.  at  14  carats  =  154  carats. 

12  oz.  at  1.5   "   =  180   " 
17  oz.  at  18   "   =  306   " 

40  oz.  =  640   " 

640  -r  40  =  1  oz.,  average  16  carats. 

SECOND    STEP. 


19 


(-16 
J  17 
]  20 
[22 


1 

2 

3 

4 

i 

1 

i 

1 

1 

2 

i 

1 

Oz. 

Proportional    1  (^ 
quantities.      2 


!J 


X  40  = 


40  —  1  =:  40  =^  ratio  multiplier. 
Or  thus,  for  the  second  step  : 


Oz.    Carata.      Carate. 

(  40  at  16  =  640 
J  40  at  17  =  680 
]  80  at  20  =   1600 

t40at22=     880 

3800 


^^]  20— 

[22-1 


Oz. 

1  X  40  =  40 
3  X  40  =  120 
3  X  40  =  120 

2  X  40  ^    80 


Or 
thus: 


Carats. 
10— 

19 !  11] 


1  20-' 
[22- 


Os. 

3  X  13i  =  40 

1  X  13i  =  13i 

2  X  13i  =  26J 

3  X  13i  =  40 


40- 


1  =  40  ratio  number. 


40  . 


■  3  =  13i  ratio  number. 


3.  A  broker  has  60  shares  of  stock  that  cost  $40  per  share,  and  40  shares 
that  cost  $75  per  share.  How  many  more  shares  must  he  buy  at  $90  per  share,  so 
that  he  can  sell  the  whole  at  $50  per  share  and  lose  33J  per  cent  ? 

Ans.  140  shares  at  '• 


OPERATIONS. 


Statement  to  find  average 
cost  to  lose  33J°i,'. 


66S 


100 
50 

$75   cost. 


(40 

75  .;  75 
(HO 


By  the  1st  method. 

1 


7  X  20  =  140     Ans. 
60  —  3  =  20  ratio  multiplier. 


I>y  loss  and  gain  method 

40  35  gain  =  7  =  3 
70  <  75 

(  90   15  loss   =  3  =  7  X  20  =  140    Ans. 

60  —  3  =  20  ratio  multiplier. 


By  the  linking  method. 

I  40,  15  =  3 
75  I  75 

|-90-l  35  =  7  X  20  =  140     Ans, 

60  —  3  =  20  ratio  multiplier. 


798 


SOQLES    rillLOSOI'HIC    PRACTICAL    MATHEMATICS. 


Case— IV. 
138-t.     AVlien  the  quantity  and  rate  of  the  mixture  to  be  made  ivS  given,  to  find 
the  several  quantities  composing  it,  whose  rates  are  given. 

1.  A  manufacturer  lias  cotton  worth  5  cents,  jute  wortli  9  cents,  and  wool 
wortli  20  cents  per  pound.  He  desires  to  make  a  mixture  of  533  pounds  worth  12 
cents  per  i)ouud.     IIow  many  jiounds  of  each  must  he  use  ? 

Ans.  1G4;  each  of  cotton  and  jute,  and  L-'O")  of  wool. 

OPERATION. 

Explanation. — We  first  find  the  proportional 
quantities  according  to  Case  1,  Article  1378. 
Then  we  aild  tlie  ]irop<irtional  (juantities 
together  which  gives  26.  Then,  since  the  re- 
quired mixture  is  to  contain  533  pounds,  we 
divide  533  by  26  and  obtain  20i  as  the  ratio 
multiplier  for  all  the  ]iroporti<)ual  quantities,  - 
thus  producing  164  jiounds  each  of  cotton  and 
jute,  and  205  pounds  of  wool. 

2.  A  wine  merchant  has  an  order  for  a  cask  of  wine  containing  12G  gallons, 
to  cost  just  $1.20  per  gallon.  Not  having  any  wine  of  tliis  price,  he  makes  a 
mixture  of  other  wines  wortli  respectively  80  cents,  fl.OO,  $1.30,  $1.50,  $1.80,  and 
water  worth  nothing.    How  many  gallons  of  each  kind  will  be  required  ? 

Ans.  See  operation. 


^2( 


i 

« 

i 

s 

i 

i 

"+ 

3 

8  X  20}  =  164 

8  X  20i  =  164 

10  X  20i  =  205 


26 


533 


FIH.ST    OPERATION. 


120  <; 


1 

r  0 

tU 

80 

A 

3 

100 

■A, 

1 

130 

»> 

150 

.1. 

4 

180 

A 

2 

Proportional 
numbers. 


x9;',= 


gals. 

29A 

19  ,^ 

38fJ 

1 19,^ 


■water  at 


0 

80c. 

|;1.00 

1.30 

1.50 

1.80 


:  $000. 

23.26^!, 
9.69,^ 
25.20 
58.15,^, 
34.89  ft 


13  126  gals,  at  .$1.20  =        $151.20 

126  —  13  =  9,\  the  ratio  multiplier. 

NoTK. — Pifferent  .answers  may  be  produced  by  different  methods  forming  couplets,    or   by 
linking,  or  by  the  method  shown  in  Article  1379. 

Operation  and  proof  by  the  loss  and  gain  method  as  elucidated  in  Article  1379. 


Gal.  Ct3. 

'  1,  water  at     0 
1,  wine 

-     y  1.      " 
1.20"^  1, 


a 


at  80  =  + 
at  100  =  + 
at  130  =  — 
at  150  =  — 
at  180  =  — 


Katio. 
120) 
40  '■  =  180  G.  =  9 
21)  > 
10^ 

30  '>  =  100  1..  =  5 
GO^ 


Gals.       Cts. 

$ 

5  X  3  =  15  at     0  = 

0.00 

5  X  3  =  15  at    80  = 

12.00 

5x3  =  15  at  100  = 

15.00 

9  X  3  =  27  at  130  = 

35.10 

9X3  =  27  at  150  = 

40.50 

9  X  3  =  27  at  180  = 

48.60 

126  at  120  =  151.20 


126  -H  42  =  3  proportional  multiplier. 


Operation  and  proof  by  the  linking  method  as  elucidated  in  Article  1380. 


r    0- 

80- 
inn 


,20,100-, 


130 
I  150— 
1^180 1 


60 
30 
10 
20 
40 
120 


6x4}  = 
3  X  4i  = 
1  X  4}  = 
2X4}  = 
4x4}  = 
12  X  4}  = 


27  gals. 

13}  " 
4}  " 
9   " 

18   " 

54   " 


at  Oc.  = 
at  80c.  = 
at  100c.  = 
at  130c.  = 
at  150c.  = 
at  180c.  = 


.00 
10.80 
4.50 
11.70 
27.00 
97.20 


28)126(4^  126 


at  120c.  =  $151.20 


ALLIGATION. 


799 


3.     A  liquor  dealer  has  ■whiskey  of  100°  proof,  and  another  grade  of  05°  proof. 
How  many  gallons  of  each  must  he  take  to  make  a  barrel  of  46  gallons  of  90° 


proof? 


Ans.  3Uf  gallons  of  100°  proof. 


OPERATION. 


90 


i    65 
'100 


Gal. 

I  16 

7     9 


13|  gallons  of  65°  proof. 


Gal. 
I  46 
7    5 


2  =  f-  of  653 
5  =  f  of  100>-" 

I  m  I  m 

4.  A  farmer  carried  120  fowls  to  market,  consisting  of  turkeys,  geese,  ducks, 
and  chickens,  and  sold  the  turkeys  for  $2.25  a  piece,  the  geese  for  $1.50  a  piece,  the 
ducks  for  50  cents  a  piece,  and  the  chickens  for  25  cents  a  piece.  He  received  $120 
for  the  lot,  averaging  $1  a  piece  for  all.    How  many  of  each  kind  were  there? 

Ans.  See  operation. 
Operation  by  the  analytic  method. 
5 


1 

2 

3 
5 

4 

r  25 

yS 

100  J  ^0 

^""i  1.50 

1 
1 

[225 

Th 

3 

60  chickens. 

12  ducks. 

12  geese. 

36  turkeys. 

10  120 

120  —  10  =  12  =  ratio  multiplier. 

Answer  by  loss  and  gain  method.  Answer  by  the  linking  method. 


Chickens, 
Ducks 
Geese, 
Turkeys, 


35 
35 
25 
25 


Chickens,  50 
Ducks,  20 
Geese,  20 
Turkeys,    30 


20 
50 
30 
20 


5.  A  contractor  hired  5  men  to  work.  He  paid  per  day  to  each  as  follows, 
viz :  to  A.  $5,  to  B.  $4,  to  C.  $3,  to  D.  $2,  to  E.  $1.  Altogether  they  worked  120 
days,  for  which  lie  paid  them  $420.  How  many  days  did  each  work,  and  what 
amount  did  each  receive  ? 

Operation  and  answer,  by  the  analytic  method. 

7 


1 

2 

3 

4 

5 

6 

E   r  1 

t 

6 

D 

2 

t 

2 

C  3i<^ 

3 

2 

2 

B 

4 

2 

2 

6+ 

2 

A 

I  5 

i 

10 

6  1 
2 

2  ^  X  4^  =  ^ 

8 

0  J 

r25?  X  1  =$  254E. 

8?  X  2  =   171  D. 

8f  X  3  =   25^  C. 

34f  X  4  =  1371  B. 

,  42?  X  5  =  214t  A. 

8     Days, 

120       $420 
3i 

120  —  28  =  4?  multiplier. 
$420  -i-  120  =  $3i  per  day. 


$420 


Answer  by  the  loss  and  gain  method, 
worked  36  days  at  $5  per  day  : 


36 

11 

$4 

16 

11 

$3 

16 

1( 

$2 

16 

11 

$1 

=  144, 

=  48. 

=  32. 

=  16. 


lao 


$420. 


8oo 


SOULE  S    THILOSOPHIC    PRACTICAL    MATHEMATICS. 


G.     A  stock  raiser  sold  100  iuiimals  for  $100  ;  lie  sold  i>i{j;s  at  $^  each,  calves 
at  $1  J,  aud  sheep  at  $3^.     How  mauy  of  each  kind  did  he  sell  ? 

Aus.  -1:. 

OI'IORATION. 


l)igs,  54  calves,  i 


sheep. 


1 

2 

3 

Piss 
Calves 

$1       . 

f  * 
li 

i 

3 

15+ 

Sheep 

[si 

1 

4 

19 

1 

TO  — 

19  = 

81 

(3)  =  1 
(6)  =  -2 


81  -^  3  =  27  the   ratio   inultiplier   of  the   4th 
column  of  proportionals. 


Explanalinii. — By  tlin  nnalytio  iiicthoil  of  Case 
1,  we  find  tliat  tlio  lirNt  ('(niiiU't,  as  tiliown  in 
column  3,  rc(|uiros  15  pigs  to  4  sheep;  and  tliat 
second  couplet,  as  shown  in  colinim  4,  requires 
1  ]iig  to  2  calves.  Then  liy  inspection,  we  see 
that  the  sum  of  these  projiortioiial  columns,  3 
aud  4,  which  is  (19  +  3)  22,  will  not  divide 
100,  the  number  of  animals  sold,  without  a 
remainder,  and  as  any  remainder  would  give  a 
fraction  of  au  animal,  we  cannot,  therefore,  in 
a  problem  where  animals  are  the  articles  or 
ingredients  use  a  multiplier  containing  a  frac- 
tion. Heuce  we  must  determine  some  uumlier 
without  a  fraction  with  which  to  multiply  tlie 
pro])ortional  numbers  in  one  of  the  proportional 
This  we  do  by  trial  as  follows:    First  subtract  3,   the  sum  of 


columns  to  produce  the  100  animals. 

the  4th  column  of  proportionals,  from  100  =  97  remainder.  Now  as  this  remainder  cannot  be 
divided  by  19,  the  sum  of  the  3d  column  of  proportionals,  without  a  remainder,  our  first  trial  fails 
and  we  must  try  another  way.  We  now  subtract  19,  the  sum  of  the  3d  column  of  jiroportionals, 
from  100  =  81,  which  can  be  divided  by  3,  the  sum  of  the  4th  column  of  j)roi)ortio!ial8,  giving  a 
cjuotient  of  27.  This  27  is  the  ratio  multiplier  of  the  4th  column  of  jiroportionals.  Thus:  1  (I'ig) 
X  27  =:  27  pigs ;  to  which  we  add  15  pigs  shown  in  the  3d  cidunin  of  proportionals,  =  42  jiigs ;  then, 
2  (calves)  X  27  =  54  calves.     Then,  as  shown  in  the  3d  columu  of  proportionals,  there  are  4  sheep. 

Note  1. — Whenever  the  division  of  the  remainder  by  the  sum  of  the  unused  proportioual 
column,  does  not  give  an  integer  as  .a  multiplier,  we  must,  by  mnltij)licatiou  of  one  of  the  other  of 
the  proportioual  columns  and  a  trial  at  division,  try  to  find  some  iutegral  multijdier. 

Note  2. — It  should  be  remembered,  as  stated  in  note  on  page  791,  that  a  common  factor  may 
be  used  or  omitted  in  auy  couplet  or  column  of  proportional  quautities  without  changing  the  pro- 
portional parts. 

7.  A  man  purchased  100  killed  and  dressed  animals  consisting  of  cows, 
goats  and  pigs,  for  $100.  The  ])riee  of  each  was  as  follows :  Cows  $10  a  ])iece,  goats 
$3  a  piece,  and  pigs  50  cents  a  piece.     How  mauy  of  ench  kind  did  he  ))nrchase  ? 

Aus.  91 1  pigs,  4g  goats,  4g  cows. 


Average 
Cost 


f^ 


10 


1 

2 

3 

4 

! 

t 
i 

18+ 

4 

1 

1 
y 

1 

100- 


221 

1  I 


'91 J   pigs    at$i    =  $451 


l}Xii=  i    ih   goats  at  $3    =  };12i 
J  [   4J   cows  at  $10  =  $41J 


■24: 


24  100   animals 

4J  ratio  multiplier. 


JIlOO 


8.     Suppose  in  the  above  iiroblem  that  the  animals  were  alive,  how  many  of 


each  kind  did  he  jiurchase  ? 


OPERATION. 

1 

2 

3 

4 

(    i      2 

^10    4 

2 

i 

18+ 

1 

4 

1 

19 


18  X 
1  X 


Aus.  1)4  pigs,  1  goat,  aud  5  cows. 


100  —  5  =  95 

95  ^-  19  =  5  ratio  multiplier. 

5  =  90  +  4  =  94  pigs   at  S*    =  $47 

5  =  5  cows  at  ,$iO  =  $50 

1  goat  at  $3    =  $  3 


100  animals 


$100 


ALLIGATION. 


80 1 


9.  Supxiose  ill  the  above  problem  the  price  of  the  cows  had  been  $10,  the 
jn-ice  of  the  goats  had  been  $1,  and  the  pigs  12J  cents.  How  many  of  each  would 
he  have  purchased  ?  Aus.  7  cows,  21  goats,  and  72  jjigs. 

10.  A  person  bought  100  head  of  cattle  for  $100  to-wit :  Cows  $10 ;  yearlings 
$5;  calves  $2;  and  goats  at  50  cents  each.     How  many  were  there  of  each  kind? 

Answers  1  and  2  as  i>er  operations  below.  Answers  3,  i,  5,  G,  7,  8, 
9,  and  10  may  be  produced  according  to  the  manner 
of  forming  couplets  and  variations  in  finding  ratio 
multipliers. 


Cows, 
Yearlings, 
Calves, 
Goats, 

1 

1 

21 

7-t 

4 

1 

5 

90 

1111 

2        3        4        5 

21      18      15      12 

76      78     80      82 

1 

6 

9 

84 

1 

7 

6 

86 

1 

8 

3 

88 

4 

2 

2 

92 

100 

100 

100    100    100    100 

OPERATION. 

100 

100 

100 

108 

Co-n-s  f  10 

Yearlings        5 
Calves   $1<    2 


Goats 


i 


1 

2 

3 

4 

5 

J 

y 

1 

4 

1 

1 

1 

t 

T 

! 

18 
19 

8 
9 

X  1  =  1  cow  at  $10  =  $10 
X  1  =  1  yearling  at  5  =  5 
1  X  24  =       24  calves   at  2  =  48 


;         X     1  =  18) 
'2x1=    8V  = 
X24  =  48i      - 


74  goats        at    |    =    37 
3  100  animals  flOO 

100—  (19  +  9)  =  72 
72  -^  3  =  24  ratio  number  for  proportional  column  6.     The  ratio  number  for  columns  4  and  3  is  1. 

Or,  we  may  vary  the  work  as  follows  : 

5  times  3,  the  sum  of  column  6,  ^  15 
To  which  add  sum  of  column  5.         9 


Gives, 


24 


Then,   100  —  24  =  76.     76  -^  19  =  4   the   integer  ratio  multiplier  for  column  4.     1  is  the 
multiplier  fer  column  5,  and  5  is  the  multiplier  for  column  6.     See  the  following  partial  operation : 


4 

0 

1 

1 

18 

8 

4  = 
1 

5  = 


X  4  =  72  ) 
X  1  =  8  ■ 
X  5  =  10  ^ 


4  cows. 

1   yearling. 

5  calves. 
90   goats. 

100   animals. 


11.  A  drover  bought  100  head  of  sheep,  for  which  he  paid  $100,  paying  $3 
a  piece  for  the  wethers ;  $1.50  a  piece  for  the  ewes ;  and  50  cents  a  piece  for  the 
lambs  ;  how  many  of  each  did  he  buy  1 


8o: 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


OPKItATIOX. 


1 

2 

3 

4 

J 11 

2 

2 
2 

4 

2 
2 

i    3 

i 

1 

5 

i 

Explanation. — Having  found  tlie  proportional 

4x5  ^20.  numbers  shown  iu  columna  3  and  4,  according 

to   Case   ],    and   Method    1,    then,    in   trial    fur 

100  —  20  =  80.  integer  multipliers,    multiply    4,    the    Kuni    of 

column  4,  bv  5  =  20,  aud  sulttraot  this  product 

80  —  5  =  16.  from  100.     Then  divide  the  renuiimler  80  by  5, 

the  sum  of  column  3,  =  16  which  is  the   ratio 

iniilti])lier  for  column  3.     The   ratio   multiplier 

for  column  4  is  5,  the  sum  of  column  3. 


Proportional  columns. 
3        4 


=  1 


X  16  =  64  ) 
X    5  =  10  > 

=    74   lambs      at  $  i  =    $37 

X    5  = 

10   ewes         .at    li-  =      15 

X  16  = 

16   wethers  at    3    =      48 

100   animals                    $100 

Several  other  answers  may  be  produced  by  difl'erent  methods  and  by  variations  in  finding 
ratio  multipliers. 


pl^rtnershipc 


1385.  Partnership  is  a  contract  -whereby  two  or  more  persons  agree  to  combine 
their  property,  hibor,  knowledge,  credit,  good  will,  or  skill,  (one  or  more  of  these) 
for  the  purpose  of  a  common  undertaking  and  the  acquisition  of  a  common  profit. 

The  partnership  or  business  is  called  Firm,  Hoxxse,  Concern,  or  Company. 

The  money,  merchandise,  or  other  kind  of  property  that  is  contributed  or 
brought  into  the  partnership  by  the  members  thereof,  is  called  Capital,  or  Capital 
Stock,  or  Capital  in  Trade. 

Sharing  Gains  and  Losses. 

The  gains  and  losses  of  a  partnership  are  shared  most  generally  in  certain 
fractfonal  or  percentage  jiarts  agreed  npon  at  the  time  the  partnership  is  formed. 

In  determining  these  parts,  the  money  or  property  investment,  the  business 
experience,  the  skill,  the  commercial  standing  or  credit  of  each  partner,  and  the 
service  to  be  rendered  by  each,  are  all  considered^as  elements  in  fixing  the  respective 
shares  of  gains  and  losses. 

Sometimes  one  partner  furnishes  the  money  capital  and  the  other  supplies 
the  business  knowledge  and  credit,  and  each  shares  the  gains  and  losses  equally. 
In  other  cases  in  which  the  business  knowledge  and  experience  are  about  equal  and 
the  monetary  capital  invested  is  unequal,  an  Account  Current  and  Interest  Account 
is  kept  with  each  partner,  and  the  gains  and  losses  are  divided  equally. 

In  a  Commercial  or  a  General  Partnership  in  -which  no  agreement  has  been 
made  regarding  the  division  of  gains  and  losses,  they  "will  be  divided  equally,  by  law. 

NoTE.^Iu  the  practice  of  accouuts,  it  is  quite  common  to  keep  two  Ledger  Accounts  Tvith  each 
partner;  one  is  called  Capital  Account  or  Stock  Account  or  Investment  Account.  To  this  account, 
all  investments  are  credited.  The  other  account  is  called  Private  Account  or  Current  Account.  To 
this  account  all  withdrawals  are  charged  and  salary,  if  any,  is  credited. 

When  the  books  are  closed,  the  loss  or  gain  of  each  partner  is  carried  to  his  Private  or  Current 
Account,  and  then  this  account  is  closed  into  his  Capital  or  Stock  Account. 

1386.  Partnership  is  divided  into  different  classes,  as  shown  in  the  following 
synopsis : 

1387.  Commercial  or  General  Partners  are  such  as  conduct  any  regular  and 
lawful  business  for  the  purpose  of  gain. 

PARTNERSHIPS  i:S^  LOUISIANA. 

In  Louisiana,  the  only  Civil  Law  State  in  the  Union,  there  are  the  following 
classes  of  Partnerships  :  1.  Commercial  Partners.  2.  Ordinary  Partners.  3.  Part- 
ners In  Commendam.     4.  Limited  Partners. 

1388.  Commercial  Partnership  embraces  the  following  lines  of  business:  1. 
For  the  piu-chase  of  any  personal  property  and  the  sale  thereof,  either  in  the  same 
state  or  changed  by  manufacture.  2.  For  buying  or  selling  any  personal  property,  as 
factors  or  brokers.  3.  For  carrying  personal  property  for  hire  in  shijjs  or  other 
vessels.     Commercial  Partnerships  are  divided  into  General  and  Special. 

(803). 


8o4 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


13S9.  Every  commercial  partner  is  tbe  accredited  agent  of  his  lirm,  and  each 
member  of  a  firm  is  liable  to  the  creditors  thereof,  Avithout  limit, 

1390.  Community  of  Profit  is  the  critical  test  to  prove  the  existence  of  a  part- 
nersLii]). 

1391.  Ordinary  Partners  are  all  that  are  not  commercial,  as  mechanics, 
farmers,  lawyers,  hotel-keepers,  etc.  Ordinary  Partners  are  divided  into  TJnivei'- 
sal  and  Particular. 

1392.  Universal  Partnersliip  is  one  in  -which  the  parties  make  common  stock 
of  all  the  property  tliey  respectively  possess,  real  and  personal,  or  personal  oidy. 

1393.  Particular  Partnership  is  one  formed  for  a  special  business  not  of  a 
commercial  nature. 

1394.  Ordinary  Partners  are  liable  to  the  creditors  of  the  firm  pro  rata,  i.  e. 
in  proportion  to  the  number  of  iiartiiers,  without  reference  to  the  proportion  of  the 
stock  or  profits  to  which  each  is  entitled. 

1395.  Partners  In  Conimendani.  In  Louisiana,  a  Partner  In  Commeudam  is 
one  who  contracts  with  some  house  or  firm,  either  Commercial  or  Oi^linary,  which  is 
already  established,  to  invest  a  specified  amount  and  to  be  exemjit  from  any  liability 
exceeding  the  sum  or  amoujit  invested  or  agreed  to  be  invested.  A  Partner  In 
Commendam  can  take  no  part  in  the  business ;  and  the  business  must  not  be  carried 
on  in  the  name  of  such  partner,  though  it  may  be  used  after  the  firm  name,  with  the 
•words.  Partner  In  Commendam.  The  violation  of  either  of  these  conditions  will 
make  the  Partner  In  Commendam  a  Commercial  Partner,  liable  without  limit. 

1396,.  Limited  Partners  in  Louisiana.  By  an  Act  of  the  Legislature  of 
Louisiana,  passed  June  -9,  1888,  it  is  made  lawful  for  any  luimber  of  i>ersons,  not 
less  than  three,  who  shall  contribute  not  less  than  $5000,  to  organize  a  corjioration 
for  the  purpose  of  carrying  on  any  lawful  business  or  enterprise,  not  otherwise  spe- 
cially jirovided  for,  excepting  the  stock  jobbing  business  of  any  kind. 

All  Limited  Partners  of  this  kind  shall  place  the  word  "  Limited,"  as  the  last 
word  of  the  corporate  name  wherever  it  is  written  or  used.  Failure  to  so  use  the  word 
"Limited,"  renders  every  member  participant  in  such  omission,  liable  for  any 
indebtedness,  damage  or  liability  arising  therefrom. 

PAETNERSHIPS  IN  THE  COMMON  LAW  STATES. 

1397.  Partnerships,  under  the  Common  Law,  are  as  follows:  1.  Trading. 
2.  Non-Trading. 

1398.  Trading  or  Conuuercial  Partnerships  are  such  as  are  formed  for  buy- 
ing, selling,  or  manufacturing,  or  otherwise  procuring  for  sale  and  selling  articles 
for  a  profit.     All  such  partners  are  responsible  to  creditors,  without  limit. 

1399.  Non-Trading  Partners  are  those  engaged  in  the  prosecution  of  some 
occupation  or  calling,  not  of  a  trading  or  commercial  character,  as  lawj'ers,  mechan- 
ics, i)hysiciaus,  etc.  In  Non-Trading  Partnershijis,  a  partner  does  not  possess 
general  power  to  bind  the  firm,  and,  as  a  general  ride,  has  no  power  to  conti'act 
partnership  debts  and  issue  commercial  paper. 


*  PARTNERSHIP.  8o5 

1100.  Trading  Partnerships  are  sub-divitlert  into  the  following  classes: 
1.  Universal.     2.  General.     3.  Particular.     4.  Limited. 

l-tOl.  A  UniTers.aI  Partnership  is  one  in  which  all  property  of  every  uatiire 
owned  by  the  parties  is  contributed,  and  aU  jirofits,  however  made,  are  for  joint 
benefit. 

1402.  A  General  Partnership  is  one  in  which  tlie  parties  carry  on  all  their 
trade  and  business  for  the  joint  benefit  and  profit  of  all,  whether  the  investments 
are  equal  or  unequal. 

1-10:{.  A  Particular  Partnership  is  one  formed  for  a  special  branch  of  trade, 
as  distinguished  from  the  general  business  or  employ  of  the  partners,  or  one  of  them. 

1404?  A  Limited  Partnership  is  one  wherein  one  or  more  of  the  partners  are, 
by  the  terms  of  the  contract,  liable  to  creditors  for  sucli  an  amount  ouly  as  be  or 
they  shall  Lave  invested,  or  shall  have  agreed  to  invest  in  the  partnership.  The  fol- 
lowiug  points  must  be  observed  in  all  partnerships  of  this  class  :  1,  There  must  be  at 
least  one  Commercial,  General,  or  Unlimited  Partner.  2.  The  contract  of  partner- 
ship must  be  duly  recorded  with  the  Clerk  of  the  Coiuity  Coiut,  or  with  some  officer 
specified  by  law. 

Note.— In  some  States,  Limitt-d  Partners  are  DOt  permitted  to  take  an  active  part  in  the  busi- 
ness, or  to  have  their  names  apx>ear  in  the  tirm  name. 

SPECIAL  NAMES   GIVEN   TO    PAETNEES. 

1405.  The  following  special  appellations  aie  given  to  partners  in  their  differ- 
ent relations : 

140(J.  A  Nominal  or  Ostensible  Partner  is  one  who  holds  liimself  out  to  the 
world  as  such,  but  who  has  no  interest  in  the  business ;  he  merely  lends  his  name 
and  credit  to  the  firm.  All  such  partners  are  liable  for  tbe  debts  of  the  firm, 
although  they  receive  none  of  the  protits. 

1407.    A  Silent,  a  Secret,  a  Sleeping,  or  a  Dormant  Partner  is  one  who 

is  not  announced  or  known  to  the  public  as  a  partner,  but  who  j)articii)ates  in  the 
profits.     All  such  partners  are  liable,  without  liuut,  to  the  creditors  of  the  firm. 

140S„     A  Sub-Partner  is  one  who  agrees  with  one  of  the  partners  to  i)artici- 

l)ate  in  that  i)artner's  share  of  the  gain,  and  to  bear  a  part  of  his  losses. 

1409,  Partners  are  responsible  to  one  another  for  any  viohition  of  the  articles 
of  agreement  among  themselves. 

FORMATION    OF    PARTNEESHIPS. 

1410.  Partnerships  may  be  formed  by  three  diflerent  methods :  1.  By  Articles 
of  Agi'cement  formally  executed  and  delivered.  2.  By  a  Verbal  Agreement.  3. 
By  the  Acts  of  the  Parties. 

"VVheu  a  contract  of  partnership  is  e\idenced  by  written  articles,  which,  under 
all  circumstances,  should  be  done,  great  care  should  be  taken  to  express  clearly  all 
of  the  pro\isions,  stiimlatious  and  terms  that  the  parties  desire  to  have  enforced 
against  their  co-i)artners. 


So6  soule's  philosophic  practical  mathematics.  ■* 

Note. — The  Law  of  Partnership  is  a  suliject  of  general  interest.  With  tlie  duties,  ohlig.ations, 
and  responsibilities  of  this  peculiar  connection,  every  mercantile  man  should  bo  acquainted.  It  is  a 
species  of  connection  which  pervades  the  mercantile  world — which  has  existed  and  will  exist  in  all 
ages  and  nations.  It  is  a  word  of  most  extensive  usage ;  it  comprehends  equally  the  luiion  of  a  day, 
and  those  venerable  guilds  and  ancient  associations,  which,  in  the  country  whence  our  laws  have 
been  derived,  have  existed  for  centuries;  it  aiiplies  at  once  to  the  ephemeral  operations  of  advent- 
urers whose  projects  connnenee  and  terminate  iu  the  compass  of  a  few  hours,  aiid  to  the  mighty 
transactions  of  thatcompany  of  merchants,  which,  commencing  with  a  factory  on  the  shores  of  India, 
has  overthrown  kingdoms,  and  from  their  ruins  cemented  a  splendid  empire.  The  inducement  to 
its  formation  results  from  the  combination  of  qualities  and  the  concurrence  of  circumst.ances  neces- 
sary to  successful  conuuercial  adventure.  It  is  a  connection  than  which  none  more  close  and  intimate 
can  exist  among  incn.  It  places  in  the  power  of  him  with  whom  you  form  it,  your  property  and 
your  reputation.  His  virtues  and  his  skill,  in  unison  with  your  own,  may  raise  you  to  the  pinnacle 
of  prosperity.  His  folly,  or  his  crimes,  may  strip  you  of  every  flourishing  branch  and  leaf,  and 
leave  you  a  naked,  withered,  and  dishonored  trunk.  It  should  be  formed,  therefore,  with  little 
less  care  than  the  marriage  tie,  to  which  it  has  been,  by  a  late  chancellor,  not  inai)tly  compared. — 
Law  Journal,  Vol.  1. 

DISSOLUTION  OF  PAETNEESHIPS. 

1411.  Partnerships  maybe  dissolved  for  any  one  of  the  following  reasons: 
1.  By  the  expiration  of  the  time  for  which  the  partuershij)  was  formed.  2.  By  the 
niutnal  consent  of  all  the  partners.  3.  By  death,  or  the  i)hysical  or  mental  incai)ac- 
ity  of  a  partner  to  perform  his  duties  in  the  partnership.  4.  By  the  personal  bank- 
ruptcy of  a  partner.  5.  By  a  decree  of  the  court  for  intemperance,  immorality, 
violation  of  the  articles  of  agreement,  fi-audident  conduct,  etc.  C.  By  the  comple- 
tion of  the  work  or  objects  for  which  it  was  formed. 

1413.  The  following  is  one  of  the  most  common  forms  for  written  instruments 
of  Partnership  Agreement,  or  Articles  of  Partnership : 

AETICLES  OF  PARTNBESHIP,  OR  ARTICLES  OF  AGREEMENT. 

Articles  of  Partnership  made  and  entered  into  this  first  day  of  January,  A.  D.  1893,  between 
Battle  Bell,  N.  L.  Baui'Ield  and  K.  M.  Clement,  all  of  the  City  of  New  Orleans  and  State  of 
'Louisiana,  Wituesseth  :  That  the  said  parties  hereto,  having  mutual  confidence  in  each  other,  do 
hereby  form  with  each  other  a  Partnerahip  Agreement  on  the  terms  and  conditions  following,  that 
is  to  say : 

1st.  The  partnership  shall  be  for  carrying  on  the  Grocery,  Western  Produce  and  General 
Commission  Business,  from  the  first  day  of  Jauuary,  A.  D.  1893,  for  the  term  of  three  years,  theuce 
ensuing  with  the  condition  that  any  party  may  be  at  liberty  to  dissolve  the  partnership  at  the 
expiration  of  each  and  every  six  months,  from  the  period  of  its  commencement,  by  serving  upon  the 
others,  at  least  three  months  previous  to  such  exi)iratiou,  a  written  notice  specifying  that  dissolu- 
tion will  take  place  at  such  time. 

2d.  The  said  partnership  shall  bo  conducted  and  carried  on  under  the  partnership  name, 
style  and  firm  of  Bell,  Baiifield  &  Co.,  and  the  place  of  business  shall  be  in  the  City  of  New 
Orleans,  State  of  Louisiana. 

3d.  That  the  said  Battle  Bell  shall  contribute  to  the  capita]  of  said  partnership  twenty 
thousand  dollars  legal  currency.  That  N.  L.  Bareield  shall  contribute  to  the  capital  of  the 
partnership  fifteen  thousand  dollars  legal  currency.  That  R.  M.  Clement  shall  contribute  to  the 
capital  of  said  partnership  ten  thousand  dollars  legal  currency. 

These  investments  by  the  respective  parties  to  the  said  partnership,  shall  constitute  tho  capi- 
tal of  said  partnership,  for  the  prosecution  of  the  said  business.  But  each  partner  is  entitled  to 
draw  out  of  tho  profits  or  capital  of  said  business,  for  his  own  separate  use,  a  sum  not  exceeding  one 
hundred  and  fifty  dollars  per  mouth,  while  the  said  iiartnership  continues. 

4th.  It  is  agreed  that  the  gains  and  losses  shall  be  shared  equally,  i.  e.  one-third  each  by 
each  partner;  and  that  an  interest  account  shall  be  kept  with  each  jiartner. 


PARTNERSHIP. 


807 


5th.  That  each  of  tho  parties  hereto  shall  diligeutly  employ  his  time  and  capacity  in  the 
business  of  said  partnership,  and  be  faithful  to  the  others  in  all  transactions  relating  to  the  same; 
and  that  none  will,  without  the  written  consent  of  the  others,  employ  either  the  cajiital  or  credit  of 
the  partnership  in  any  other  than  partnership  business. 

6th.  That  in  all  matters  respecting  the  transactions  of  the  partnership  and  the  management 
of  tho  business,  the  expressed  opinion  of  the  majority  of  the  parties  to  this  agreement  shall  govern 
and  be  binding  on  all  of  said  parties;  and  in  cases  of  difficulty  they  shall  have  power  to  settle  ui> 
or  sell  the  concern. 

7th.  That  the  books  of  account  shall  be  kept  according  to  the  principles  of  Double  Entry 
by  the  said  firm,  and  correct  entries  made  therein  of  all  moneys,  goods,  efi'ects,  debts,  sales,  pur- 
chases, receipts,  payments,  and  all  other  transactions  of  the  said  partnership. 

8th.  That  none  of  the  partners,  during  the  continuance  of  this  partnership,  shall  assume  any 
liability  for  another  or  others,  by  means  of  indorsing  or  of  becoming  guarantor  or  surety,  without 
first  obtaining  the  written  consent  of  the  other  parties  thereto,  nor  shall  any  partner  lend  any  of 
the  funds  of  the  partnership  without  such  consent  of  each  of  the  other  partners. 

9th.  That  at  the  expiration  of  each  and  every  year,  from  the  commencement  of  this  partner- 
ship, an  account  of  stock,  effects,  credits,  debits,  and  all  partnership  transactions  shall  be  taken, 
the  books  balanced,  and  the  net  gain  or  loss  carried  to  the  respective  partners'  accounts,  according 
to  article  four  of  this  agreement. 

10th.  That  at  the  close  of  the  partnership  from  whatever  cause,  the  parties  hereto  agree  to 
make  a  true,  just  and  final  account  of  all  things,  truly  adjust  the  same,  and  after  all  the  partnership 
business  is  adjusted,  that  they  will  equitably  settle  with  each  other,  according  to  the  conditions  of 
this  agreement. 

In  Testimony  Whereof,  the  parties  to  these  presents  have  hereunto  set  their  hands  and  seal, 
the  day  and  year  first  above  written. 


Witness  : 


J.  N.  Bradford. 
Jajies  Hancock. 

W.    B.    CUNYUS. 


BATTLE  BELL,  l^'^'l 

N.  L.  BARFIELD,  \  l-  s.  \ 

E.  M.  CLEMENT.  ^  l.  s.  ^ 


141lj.  By  article  four  of  the  above  Articles  of  Partnersliip  Agreement,  we 
see  that  although  the  investments  by  the  diflerent  partners  are  not  the  same, 
the  gains  and  losses  are  to  be  shared  equally,  and  that  an  interest  account  is  to  be 
kept.  The  keeping  of  an  interest  account  between  the  partners,  at  the  same  rate 
per  cent.,  is  in  efl'ect  equalizing  the  capital  of  the  diflerent  partners;  i.  e.,  it  remu- 
nerates the  partners  having  the  excess  of  capital  out  of  the  private  funds,  or  the 
indi\idual  gain  of  the  partners  who  have  the  lesser  capital. 

14:14.  The  money  invested  by  each  partner  is  in  efl'ect  a  loan  to  the  firm,  and 
which  the  firm  holds  in  trust  for  the  protection  of  the  creditor,  and  at  the  same 
time  uses  to  carry  on  the  business. 

Any  other  articles  or  clauses  that  the  parties  may  desire,  and  which  are  not 
inconsistent  with  established  law,  may  be  inserted  in  Articles  of  Partnership. 

In  all  partnership  contracts,  each  partner  should  have,  and  should  carefidly 
preserve,  a  copy  of  the  written  evidence  of  contract. 


liartnership  Settlements. 

— >  <>>  < — 

1115.  What  Quadratics  is  to  Algebra,  wbat  Dififerential  Calculus  is  to  Matlie- 
matics,  what  the  Supreme  Court  is  to  our  System  of  Jurisprudence,  so  is  Partner- 
ship Settlements  to  the  Science  of  Accounts. 

By  Partnershij)  Settlements  we  mean  :  1.  The  finding  and  the  adjustment  of 
the  flnaiu'ial  standing  ot  partners  at  the  close  of  the  business,  i.  e.,  finding  what  is 
the  monetary  interest  of  net  capital,  or  the  net  iusolveucy  of  each  partner  in  the 
business  at  the  time  the  statement  is  made  and  the  adjustment  thereof  on  the  books 
or  among  the  partners.  2.  The  finding  and  the  adjustment  of  the  unsettled  finan- 
cial affairs  of  business  men  who  are  not  i)artners,  but  who  have  had  business  trans- 
actions with  each  other,  which  involve  monetary  debits  aud  credits. 

The  adjustment  of  these  relationships  under  all  the  multitudinous  conditions 
which  aiise  among  business  men,  requires  on  the  i)art  of  the  accountant  a  clear 
mind,  skill  in  accounts,  thoroughness  in  practical  mathematics,  and  an  intimate 
acquaintance  with  the  principles  of  commercial  law,  business  customs,  and  com- 
mercial ethics.  His  attention  should  be  given  first,  to  the  account  current,  or  the 
monetary  debit  and  credit  standing  of  each  partner,  or  of  each  party  in  interest; 
and  secondly,  to  the  law  governing  the  business  of  the  parties  modified  by  the  spe- 
cial contracts  entered  into  by  them.  Where  no  specific  contracts  were  made,  and 
where  law  or  business  custom  does  not  cover  the  case,  equity  must  be  applied,  and 
strict  justice  between  man  and  man  must  be  rendered. 

The  imperative  necessity  for  partnership  settlements  is  of  frequent  occur- 
rence. In  the  regidar  course  of  business,  they  should  be  made  as  often  as  once  a  year. 
Whenever  a  new  partner  is  admitted,  or  an  old  partner  retires,  and  whenever  a 
change  of  interest  or  share  in  the  gains  and  losses  of  the  business  takes  place 
among  the  partners,  and  at  the  final  dissolution  of  the  firm,  partnership  settlements 
must  injustice  to  all  i^arties  be  made. 

GAESfS  AND  LOSSES. 

1416.  The  most  important  question,  and  the  first  to  be  ascertained,  in  settle- 
ments between  pai'tners,  is  the  determination  of  the  gain  or  loss  of  the  business. 

TERMS   AND  DEFINITIONS. 

1417.  The  following  terms  and  definitions  are  necessary  to  be  known  in  work- 
ing Partnership  Settlements : 

141S.  Kesources,  Assets  or  Effects  are  the  property  of  all  kinds  possessing 
value,  aud  belonging  to  a  firm,  corporation,  or  individual. 

KOTE. — The  amoniit  of  witbclrawiils  of  cash  or  property  liy  tbe  proprietor  or  jiartiiers,  is  to 
the  Ijusiuesa  or  tirin,  a  resouree. 

1419.  Liabilities  are  the  debts  or  obligations  of  all  kinds  owed  by  a  firm, 
corporation,  or  individual. 

NoTK. — The  amouut  of  iuvestmeut  hy  the  proprietor  or  jiartuers,  is  to  the  busiuoss  or  firm,  a 
liability. 

(808J. 


*  PARTNERSHIP    SETTLEMENTS.  S09 

14'J0.  luTestiueut  is  the  amount  of  money  and  property  contributed  to  tbe 
cajiital  of  a  firm,  or  api>ropriated  for  tbe  transaction  of  business  on  private  account. 

1421.  Net  Iiivestinent  is  tbe  excess  of  investments  over  tbe  witbdrawals;  or 
in  case  of  no  witbdrawals,  it  is  tbe  total  investment. 

1422.  Average  Investment  is  a  siun  wbicb,  invested  for  tbe  wbolc  term  of  tbe 
business,  will  produce  an  amount  of  interest  equal  to  tbe  interest  on  tbe  difterent 
actual  investments  for  tbe  time  tbat  tbey  were  respectively  in  tbe  business. 

1423.  Capital  is  tbe  total  value  of  tbe  resources  of  a  firm  or  iudividual. 

1424.  Net  Capital,  or  Present  Worth  is  tbe  excess  of  tbe  resoiuces  over 

the  liabilities  at  tbe  close  of  business,  or  at  any  time  during  business. 

NoTK. — The  net  investment  of  the  proprietor  or  partners,  is  not  to  be  included  in  the  liabilities 
■wbeu  tbe  net  capital  is  to  be  found. 

1425.  Insolvency  is  tbe  indebtedness  of  a  firm  or  indi^^dual  in  excess  of  tbe 

resources. 

IsoTK. — The  withdrawals  by  the  proprietor  or  partners  are  not  to  be  inclvuled  in  tbe  resources 
when  tbe  insolvency  is  to  be  found. 

1426.  Net  Gain  is  tbe  excess  of  tbe  resources  over  the  liabilities,  including  the 
withdrawals  and  investments  by  tbe  proprietor  or  partners. 

1427.  Net  Loss  is  the  excess  of  tbe  liabilities  over  the  resources  including 
the  -withdrawals  and  investments  by  tbe  xiroprietor  or  partners. 

142.S.  The  Net  Gain  or  Net  Loss  is  also  tbe  difference  between  tbe  total  gain 
and  total  loss. 

TWO  METHODS  OF  DETERMINraO  THE  GAIN  OR  LOSS  OF  A  BUSINESS. 

1429.  There  are  two  methods  of  determining  tbe  gain  or  loss  at  the  close  of 
a  business,  or  at  any  time  during  the  business.    Tbey  are  as  follows : 

1st.  To  collect  and  classify  all  tbe  various  items  of  gaiu  and  loss,  and  then 
find  the  difl'erence  betweeu  their  respective  sums.  This  method  can  only  be  used 
■when  the  books  have  been  correctly  kept  by  Double  Entry. 

2d.  To  collect  and  classify  all  the  various  items  of  resources  and  liabilities 
of  tbe  firm,  and  then  find  tbe  difference  between  their  respective  sums  or  aggregates. 
Tbis  difference  is  a  gain,  when  tbe  resources  exceed  the  liabilities,  and  a  loss,  wbeu 
tbe  liabilities  exceed  tbe  resoiu'ces. 

Tbis  fact  is  clear,  for  the  reason  tbat  when  the  business  was  commenced,  no 
matter  what  was  the  net  capital  or  net  insolvency,  the  resources  and  liabilities  were 
in  balance,  and  it  is  obvious  to  the  accountant  that  nothing  but  a  gaiu  or  a  loss  could 
destroy  that  balance. 

N.  B.  Wbeu  classifying  the  items  of  the  resources  aud  liabilities,  to  find  the 
gain  or  loss  of  a  business  as  above  directed,  include  as  resources  the  witbdrawals  of 
the  proprietor  or  partners,  and  include  as  liabilities,  tbe  investments  of  the  proprie- 
tor or  partners. 

NoTK.— Instead  of  including  both  tbe  investments  and  witbdrawals  of  the  partners  in  tho 
list  of  resoiaces  and  liabilities,  the  net  investment  or  net  withdrawal  may  be  used,  thus  abridging 
the  work. 

14;»().    The  principles  of  Accounts  upon  xchich  this  is  based,  are  as  foliates  : 
1st.    Tbat  tbe  firm  owes  each  partner  for  what  be  invests,  the  same  as  it  woidd 
owe  any  other  person  for  money  or  ijroperty  deposited  with  the  firm 


Sio  soule's  philosophic  practical  mathematics.  * 

2d.  That  eacli  partner  owes  the  firm  of  wliich  he  is  a  member,  for  all  with- 
drawals, the  same  as  any  other  person  would  owe  the  fii-m  who  received  money  or 
goods  from  it. 

Note  1. — To  the  critical  acconntaut,  tliero  is  a  clear  distinction  between  the  firm  itself  and 
the  meniliors  roniposiug  it. 

NoTK  2. — 111  tlio  case  of  a  single  proprietorship  hnsiness,  it  is  considered  that  the  business 
owes  the  proprietor  for  liis  investments,  and  that  lie  owes  the  business  for  his  withdrawals.  Thns  a 
single  proprietor  of  a  business  possesses  two  individualities,  one  as  an  individual,  and  one  as  a 
business  man. 

The  second  method  of  determining  the  gain  or  loss,  as  above  described,  can 
be  nsed  under  all  circnmstances  where  it  is  possible  to  find  the  resources  and  lia- 
bilities, and  it  is  the  only  method  that  can  be  used  when  the  books  have  been  kept 
by  Single  Entry,  or  have  been  incorrectly  kept  by  Double  Entry,  or  when  no  books 
have  been  kept — only  meraoraudnms  have  been  made. 

1431.    Items  of  Gain  and  Loss  not  to  be  included  in  the  state- 
ment  OF  RESOURCES  AND  LIABILITIES. 

WTien  making  the  statement  of  resources  and  liabilities,  care  must  be  taken 
not  to  include  any  items  of  gain  or  loss. 

All  of  the  various  items  of  gain  and  loss  are  represented  in  the  list  of  resources 
and  liabilities.  This  is  evident  from  the  fact  1st,  that  a  loss  is  the  result  of  a 
resource  disposed  of,  without  receiving  jiroperty  value  therefor,  and  hence  is  absent 
from  the  list  of  resources,  and  thus  the  amount  of  resources  is  decreased  and 
the  loss  is  correspondingly  increased.  2d,  that  a  gain  is  the  result  of  a  resource 
received,  without  giving  jiroperty  value  therefor,  and  hence  is  present  to  increase  the 
list  of  resources,  and  thereby  augments  the  gain. 

1432,  ORAL    EXERCISES. 

A.  commenced  business  with  a  capital  of  $8000,  and  closed  with  an  insolvency 
of  $1500.     What  was  his  loss  ?  Ans.  $9500. 

B.  commenced  business  with  an  insolvency  of  $3000,  and  closed  with  a  capital 
of  $2000.     \Miat  was  his  gain  ?  Ans.  $5000. 

0.  commenced  business  with  a  capital  of  $2000,  and  closed  with  a  capital  of 
What  was  his  loss  !  Ans.  $1000. 

D.  commenced  business  with  a  capital  of  $4000,  and  gained  $1400  dui'ing  the 
business.    What  was  his  capital  at  closing  f  Ans.  $5400. 

E.  commenced  business  with  an  insolvency  of  $3500,  and  gained  $1500  during 
the  year.    WTiat  was  his  indebtedness  at  the  close  of  the  year  ?  Ans.  $2000. 

F.  commenced  business  with  a  capital  of  $0200.  He  lost  during  business 
$2100.    \Miat  was  his  capital  at  closing?  Ans.  $4100. 

G.  closed  business  with  an  insolvency  of  $400.  He  gained  $1800  during  busi- 
ness.    What  was  his  insolvency  at  commencing?  Ans.  $2200. 

H.  commenced  business  with  an  insolvency  of  $1700.  Gained  during  the  year 
$2500.     What  was  his  capital  at  closing  ?  Ans.  $800. 

1433.  WRITTEN    PROBLEMS. 

1.  A.  and  B.  are  partners  equal  in  gains  and  losses.  At  the  close  of  business 
A.  has  a  net  credit  of  $0500,  and  B.  a  net  credit  of  $3500.     There  is  no  property  on 

.  band,  and  no  other  liabilities.    What  settlement  should  they  make? 

Ans.  B.  must  pay  A.  $1500  out  of  his  private  funds. 


PARTNERSHIP    SETTLEMENTS. 


8ii 


Besonrces. 


i) 


Liabilities. 
I  6500 
3500 

$10000  loss. 


OPEEATION. 

A's  credit  $6500— $5000  loss= 


DEBIT.         CREDIT. 

.  -  -  -  -  $1500 


B's  loss  $5000— $3500  credit=  -  -  $1500 


$  5000=A's  i  loss. 
5000=B's  I  loss. 

Eiplanaiion. — Considering  that  all  investments  of  partners  are  liabilities  of  the  firm,  as  shown 
above,  and  f  luit  these  investments  amount  to  $10000,  vs-ith  no  resources  to  oft'set  them,  it  is  there- 
fore clear  that  there  has  been  a  loss  of  this  amount,  which  must  be  shared  by  the  members  of  the 
firm  as  agreed  in  their  articles  of  partnership.     See  Articles  45  and  46. 

2.  A.  and  B.  are  partners,  equal  in  gains  and  losses.  At  tlie  close  of  the  busi- 
ness, A.  has  a  net  debit  of  $6500,  and  B.  a  net  debit  of  $3500.  There  are  no  other 
resources  and  no  liabilities.    What  settlement  should  they  make  ? 

Ans.  A.  must  pay  to  B.  $1500  out 
of  his  ijrivate  funds. 
OPERATION. 
Resources.    Liabilities.  debit.        credit. 


$6500 
3500 

2)$10000 


A's  debit  $6500  —  $5000  gain  =      -    $1500 
B's  gain  15000  — $3500  debit  =    -    -     - 


$1500 


$5000  ==  A's  net  gain. 

$5000  =  B's  net  gain. 

Explanation. — Since,  as  above  shown,  the  withdrawals  of  the  partners  are  to  the  firm  a  resource, 
and  since  the  resources  of  the  firm  are  $10000,  with  no  liabilities,  it  is  therefore  clear  that  there  has 
been  a  gain  to  the  firm  of  $10000,  which  is  to  be  shared  by  the  partners  according  to  the  agreement 
in  the  articles  of  partnership. 

3.  A.  and  B.  are  partners,  equal  in  gains  and  losses.  A.  has  a  net  credit  of 
$6500  and  B.  a  net  debit  of  $3500.  There  are  no  other  resources  or  liabilities. 
What  settlement  should  they  make  ?  Ans.  B.  must  pay  to  A.  $5000  out 

of  his  xjrivate  funds. 
OPERATION. 
Resources. 
$3500 


Liabilities. 
$6500 
3500 

A's  credit  $6500  —  $1500  loss  = 
B's  debit  $3500  -f  $1500  loss  = 

debit. 

credit 
*^nnf 

-      $5000 

i)$3000  loss. 

$1500  =  A's  loss. 
$1500  =  B's  loss. 

Explanation. — In  this  problem,  one  partner  owes  his  firm,  and  the  firm  owes  one  of  its  partners. 
The  excess  of  the  liabilities  over  the  resources  is  $3000,  which  is  the  loss  of  the  firm,  and  which  nuist 
bo  shared  according  to  agreement,  equally.  The  operation  shows  clearly,  that  B.  must  pay  to  his 
partner  $5000. 

4.  (3.  A.  Hines  and  J.  L.  Hall  are  partners.  Hines  f  and  Hall  J  in  gains  and 
losses.  During  the  business  year  G.  A.  Hines  invested  $24000  and  drew  out  $1700. 
J.  L.  Hall  invested  $6000  and  drew  out  $900.  They  have  Mdse.  $32000,  Cash  $4000, 
Bills  Keceivable  $5000,  Personal  Accounts  $11000,  Bank  Stock  $2000.  They  owe 
on  Bills  Payable  $3000,  and  on  Personal  Accounts  $6000.  They  gained  on  Mdse. 
$23308,  on  Interest  $162,  and  on  Bank  Stock  $320.  Expense  shows  a  loss  of  $5800, 
Insurance  a  loss  of  $360,  and  Exchange  a  loss  of  $30.  What  is  the  net  gain  or 
loss,  and  what  is  the  net  capital  of  each  partner  f  Ans.  $17600  net  gain. 

$35500  0.  A.  Hines'  net  capital.     $9500  J.  L.  Hall's  net  capitaL 


8i: 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Operation  to  fiiiil  gain 

by   tlin   Loss    and 

Gain  Jlothod. 


Operation  to  find  gain  l)y  the  Resource 
and  Liability  Mctbod. 


Loss 
$5800 
360 
30 

IG190 
Losses 


Gain 

$i23308 
102 
320 

$23790 
6190 


Resources. 

Liabilities. 

$32000 

C.  A.  Hines  $22300 

4000 

J.  L.  Hall         51110 

5000 

3000 

11000 

6000 

2000 

Resources   $54000 
Liabilities    3t)400 


$36400 


Net  gain    $17600 


$17600  =  net  gain. 


C.  A.  Hiues'  i  net  gain  ($17600)  is 
do  net  investment     -    • 


partners'  net  capital. 
J.  L 


-  $13200 

-  22300 


do 


Hall's  i  net  gain  ($17600)  is  - 
do  net  investment     -     - 


$4400 
5100 

net  capital  at  closing   -    -    $35500  do  net  capital  at  closing    -    -    $9500 

Explanation. — The  operations  of  this  problem  and  the  general  statements  and  directions  above 
given,  regarding  the  tTvo  methods  of  finding  the  gain  or  loss  of  a  business,  make  it  clear  without 
repeating  general  principles. 

5.  A.,  B.  and  G.  are  partner.^,  A.  J,  and  B.  and  C.  each  ^,  in  gains  and  losses. 
At  the  close  of  business,  their  accounts  stand  as  follows :  A.  is  debited  $3000  and 
credited  with  $15000;  B.  is  debited  with  $0000  and  0.  is  credited  with  $10000. 
They  have  cash  in  bank  $28000.  There  are  no  other  resoiu-ces  or  liabilities.  What 
is  the  financial  standing  of  each  partner,  aud  what  settlement  should  be  made? 

Ans.  B.  will  pay  $3000  from  his  private  funds. 

A.  will  receive  $18000  and  C.  will  receive  $13000. 


B's  net  withdrawal 
Cash 

Total  resources 
Total  liabilities  - 


OPERATION. 
Resources. 

-  $  6000 

-  28000 


Net  gain 

As  ^  net  gam  is 
A's  net  investment  is 

A's  net  capital  is 


$34000 
22000 

$12000 


A's  net  investment 
C's  net  investment 


Liabilities. 

$12000 

10000 


$  6000    B's  \  net  gain  is  -     -     - 
12000    B's  net  withdrawal  is - 


$3000    C's  J  net  gain  is   -    - 
6000    C's  net  investment  is 


-  $18000    B's  net  insolvency  is    - 


$3000 


C's  net  capital  is 


$22000 


$  3000 
10000 

$13000 


Explanation. — From  these  figures,  it  is  clear  that  B.  must  pay  $3000  out  of  his  private  funds, 
which,  with  the  $28000  cash  in  bank,  will  make  $31000 ;  of  which  A.  will  receive  $18000  and  C.  $13000. 


6.  Suppose,  in  the  preceding  i)roblem,  that  B.  has  no  private  property  aud 
proposes  to  give  A,  and  0.  each,  his  note.  What  should  be  the  face  of  the  note 
given  to  each  ?  Ans.  A.  $1741.94.     C.  $1258.06. 

OPERATION. 
First  divide  the  $28000  cash  between  A.  aud  C.  in  proportion  to  their  respective  net  capitals, 
thus : 

A's  Share. 

$28000  00 
$31000        18000  $31000 


A's  capital $18000 

C's  capital    -         -     -    -      13000 


A's  and  C's  Capital    - 

A's  net  capital     - 
A's  proportion  of  cash 

A's  face  of  note  - 


$31000 


$18000  00 
16258  06 

$  1741  94 


18000 
$16258  06 


C's  net  capital 

C's  i>roportiou  of  cash 

C's  face  of  note 


C's  Sh.are. 

$28000  00 
13000 


$11741  94 

$13000  00 
11741  94 

$  1258  06 


PARTNERSHIP    SETTLEMENTS. 


813 


7.  A.  and  B.  were  commercial  partners,  equal  in  gains  and  losses;  no  interest 
to  be  allowed  on  investments.  October,  1892,  A.  invested  $8000,  and  B.  $5000.  May 
1st,  1893,  the  tirm  was  dissolved,  and  fi'om  an  incorrectly  kept  set  of  books,  the  fol- 
lowing facts  and  figures  were  taken  by  the  partners  and  submitted  for  settlement: 

A.  has  a  net  debit  of  $2100;  B.  a  net  credit  of  $3500;  cash  on  hand,  $3400; 
10  shares  of  Bank  Stock,  market  value  $1100;  Expense  Account  is  debited  $5100; 
Profit  and  Loss  is  debited  $3000  and  credited  $500;  the  firm  owe  on  a  note  $2000,  and 
interest  on  same  $60.  Approved,  A. 

Make,  in  practical  accounting  form,  the  statement  showing  the  correct  settle- 
ment of  the  partnership  affairs  of  this  firm. 

STATEMENT  SETTLEMENT. 

STATEMENT  SHOWING  THE  RESOURCES  AND  LIABILITIES,  THE  NET  GAIN,  THE  CON- 
DITION OF  THE  PARTNERS'  ACCOUNTS  AND  THE  FINAL  SETTLEMENT  OF  THE 
COMMERCIAL  FIRM  OF  A.  &  B.,  BASED  UPON  A  STATEMENT  OF  FACTS  APPROVED 
BY  THEM  MAY  1,  1893. 


A's  i  net  gain 
B's  i  net  f;aiu 


RESOURCES  : 


A's  net  Tvithdrawal 
Cash  on  band     -        .         - 
Bank  Stock,  market  value 


Total  amount  of  resources 

LIABILITIES  : 

B's  net  investment  .        .        .        -        - 

Bills  Payable 

Int   due 


face  $2000 ) 
60( 


Total  amount  of  liabilities    - 
Net  gain  of  the  firm        -        .        . 

DIVISION  OF  galn: 


A's  NET  INSOLVENCY: 

A's  net  withdrawal  -        -        -        .        - 

A's  i  net  gain,  deducted  ----- 

A's  present  net  insolvency       -        -        -        - 


B's  NET  capital: 


B's  net  investment 
B's  J  net  gain  added 

B's  present  net  capital 

Net  capital  of  firm 


FINAL  settlement: 


In  final  settlement,  A.  must  pay  to  B.,  out  of  his  private  funds,  or 

give  his  individual  note  for 

B.  must  receive  the  cash  ------- 

and  the  bank  stock    --------- 


making  a  total  of 
B.  must  then  pay  the  note 
and  interest  on  same 


B.  will  then  have,  as  above,  a  net  capital  of 


2100 
3400 
1100 


3500 
2060 


520 
520 


2100 
520 


3500 
520 


1580 
3400 
1100 


2000 
60 


6600 


5560 


1040 


00 


00 
00 


1040 


00 


1580 


4020 


2440 


6080 
2060 


4020 


00 


00 
00 


00 
00 
00 


New  Orleans,  May  1,  1893. 


Geo.  Soul£,  Accountant. 


8i4 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


8.    The  following  is  a  Trial   Balance  of  a  Double  Entry  Ledger,  and  the 
inventory  of  stock  at  the  close  of  tlie  fiscal  year,  July  1, 1893 : 

TRIAL  BALANCE. 


E.  P.  Heald,  (partuer> 

A.  Harris,  (partner) 

Cash  -        -        - 

Mdso. 

J.  M.  Swoop 

\V.  Voges 

Expense     -         -        - 

Profit  and  Loss 

City  Railroad  Stock 

Commission 


2050 

00 

15500 

2550 

00 

4800 

18567 

00 

16892 

6972 

00 

5852 

400 

00 

900 

967 

50 

520 

00 

6743 

00 

5500 

00 

125 

50 

200 

44269 

44269 

00 
00 
00 
50 
00 


00 
00 


50 


Inventory  of  property  unsold :  Merchandise  $680.40,  Railroad  Stock  IS^OO. 

Note. — The  gains  and  losBes  are  divided  as  follows :  E.  P.  Heald  two-thirds,  and  A.  Harris 
one-third. 

9. ,  Required,  the  financial  standing  of  the  partners  by  the  two  methods  of 
determining  gains  and  losses. 


FIRST  OPERATION  BY  THE  RESOURCE  AND  LIABILITY  METHOD. 

STATEMENT  OR  BALANCE  SHEET  SHOWING  THE  RESOURCES  AND  LIABILITIES,  THE 
NET  LOSS  OF  THE  FIRM,  THE  NET  CAPITAL  OF  E.  P.  HEALD,  AND  THE  NET 
INSOLVENCY  OP  A.  HARRIS,  JULY  1,  1893. 


RESOURCES  : 

Mdse.  as  per  inventory    ------ 

Railroad  Stock,  as  per  inventory    -        -        -        - 

Cash  on  hand     -------- 

Personal  Accounts,  Dr. 

Total  amount  of  resources    -        -        -        - 

LIABILITIES : 

E.  P.  Heald's  net  investment  .        -        .        . 

Archie  Harris'  net  investment  -        -        -        - 

Personal  Accounts,  Cr.     ------ 

Total  amount  of  liabilities    -        -        -        . 

Net  loss  of  business        ----- 
partners'  present  net  capital: 
E.  P.  Heald's  net  investment  .        -        .        - 

E.  P.  Heald's  two-thirds  net  loss  -        -        -        . 

E.  P.  Heald's  net  capital  at  closing       -        -        . 

Archie  Harris'  net  investment         -        -        -        - 
Archie  Harris'  one-third  loss    -        -        -        -        - 

Archie  Harris'  net  insolvency  at  closing 


680 
5450 
1675 

967 


13450 

2250 

500 


13450 
4951 


2250 
2475 


8772 


16200 


7427 


8498 


225  r70 


90 


60 


I 


PARTNERSHIP    SETTLEMENTS. 


8l5 


SECOND    OPERATION    BY   THE    LOSS   AND 


MERCHAXDISK. 


ItAII.liOAD    STOCK. 


Cost  $6972 


$(3972 


Salos    $5852.50 
Invent.    680.40 


P.  &L.   439.10 


$6972.00 


Cost      $5500 
P.  &  L.       75 


$.5575 


Dividend 
Invent. 


$  125 
5450 


Net  Debit 


GAIN    METHOD. 

PROFIT  AND   LOSS. 

$6743.00  $  75  gainE.R.  Stock 


$5575    LossonMdse.   439.10 
Loss  by  Exp.     520.00 


Total  loss 


$7702.10 
275.00 


200  gain  on  Com. 


$275  total  gain. 


E.  P.  He.ald's  two-tbirds  net  loss  is    $  4951  40 
do  net  investment  is      -    -     13450  00 

do  net  capital  is       ...    $8498  60 


$7427.10  net  loss. 
Harris'  oue-tbird  net  loss  is      -    -    $2475  70 
do        net  investment  is     -    -    -      2250  00 


do       net  insolvency  is 


$  225  70 


10.  Two  physicians,  Jones  and  Smith,  formed  a  partnershii)  for  one  year, 
eqnal  in  gains  and  losses.  They  opened  two  offices,  one  in  the  upper  and  one  in  the 
lower  iiart  of  their  city.  Jones  took  charge  of  the  iipjier  office  and  had  a  practice 
aniouiitiug  to  $4510 ;  of  this  amount  he  collected  $3000,  leading  $910  outstanding. 
The  expenses  of  his  oflBce  were  $1100,  which  he  paid  out  of  the  money  collected. 
He  also  paid  $300  from  the  money  collected  on  account  of  the  expenses  contracted 
by  his  partner  for  the  lower  office. 

Smith  took  charge  of  the  lower  oflQce  and  had  a  practice  to  the  amount  of 
$3812 ;  of  this  sum  he  collected  and  used  x^ersonally,  $3050.  He  also  collected  and 
paid  on  account  of  exjjenses  $-i00,  leaving  $302  outstanding,  of  which  $200  is  worth- 
less. He  contracted  expenses  amounting  to  $1041,  and  gave  the  firm's  note  on  account 
of  expenses  for  $350  which  is  still  unpaid. 

At  the  close  of  the  year  Smith  retires  and  Jones  continues  the  practice. 
They  engaged  an  accountant  to  make  a  correct  statement  showing  how  they  stood 
financially  with  their  business,  and  with  each  other.  How  should  that  statement  be 
made  ?  Answer,  as  follows : 

FIRST    STATEMENT. 
STATEMENT    OP   SETTLEMENT  OF  THE  PARTNERSHIP  AFFAIRS  OF  JONES   &   SMITH. 


Amount  earned  by  the  practice  of  Jones        -        -        -        -        - 
Amount  earned  by  the  practice  of  Smith        .        -        -      $3812  00 
Less  worthless  accounts         -----  200  00 

Total  amount  earned      ---         -...- 

Expenses  of  Jones'  office  -.....-- 

"        "    Smith's  office        -         ------- 

Total  expenses        --------- 

Net  gain  or  revenue        -------- 

Jones'  one-half  net  gain   --------- 

Smith's  one-half  net  gain         -------- 


JONES'   ACCOUNT. 
By  one-half  net  gain         .         -        .        -        - 
To  collections  ------ 

Less  amount  paid  for  expenses  $1100  -{-  300  = 


Balance  due  Jones    ----- 

smith's  account. 
By  one-half  net  gain        -        -        .        - 
To  collections  used  personally 


Balance  due  by  Smith  to  Jones 


4510 
3612 


1100 
1641 


2690 
2690 


5381 


3600 
1400 


00 


8122 


2741 


5381 


5381 


2690 
2200 


490 


00 


00 


00 


00 


50 
00 


50 


2690 
3050 


359 


50 
00 


50 


CCotitinued.) 


8i6 


souLE  s  rHiLosonnc  practical  mathematics. 

(Continued  from  ike  2>receding  page.) 


liKSOUUCES  : 
Personal  accounts  dne  on  Jones'  practice 
Personal  aeeonuts  <liie  on  Smith's  practice 
Less  worthless  accounts  ... 


$362 
200 


Smith's  indebtedness  as  above         ---..-- 
Total  resources  of  the  firm    --.---- 

LIABILITIES   OF   KIUM; 

As  shown  above  the  total  expenses  were          -        -        -  -        . 

Of  this  amount  there  was  paid  by  Jones  $1100  +  300  =  $1400 

By  Smith  $400  cash,  $350  note  = 750 

Total  expenses  paid       ...        ..... 

Balance  due  for  expenses  ........ 

Face  of  note  due       .......... 

Total  amount  dne  by  the  firm       -.-... 

Net  capital  of  firm  (due  to  Jones)         ..... 

FINAL   SETTLEMENT. 

In  settlement,  Smith  must  jiay  to  Jones  from  his  private  funds,  his 

indebtedness  as  above 

Jones  will  collect  the  resources  due,  which  are      ... 

thus  placing  in  his  hands 

From  this  sum  we  will  pay  the  balance  due  on  expenses- 

and  the  note  outstanding  ....... 

Total  amount  to  be  paid  by  Jones         .        .        .        .        - 

Balance  in  his  possession  is      --...-.        - 
which  is  his  net  capital  as  above. 


910 
IGi 


1072 
359 


2741 


2150 


591 
350 


591 

350 


00 


1431 


941 


490 


359 
1072 


50 


00 


50 


1431 


941 


50 
00 


50 


OO 


490  50 


SECOND    OPERATION    TO    FIND    THE    GAIN. 


RESOURCES  : 

■personal  accounts  due  Jones'  office 
Personal  accounts  due  Smith     $362  00 
Less  worthless  accounts    -    -    200  00 


$    910  00 


162  00 


Jones'  withdrawals 2200  00 

Smith's         "  .------.-    3050  00 


Total  resources 
Total  liabilities 


Net  gain 


$6322  00 
941  00 

$5381  00 


LIAItlLITlES: 

Due  for  expenses       ....    $  591  OO 
Due  for  note    ---..-.     350  00 


Total  liabilities 


$  941  OO 


Having  the  gain,  tbe  statement  settlement  would  be  made  as  sliown  above. 


.a 


PARTNERSHIP  SETTLEMENTS. 


817 


11.  The  following  facts  and  figures  are  submitted  by  the  members  of  a  com- 
mercial fii'm  for  adjustment : 

Jones  and  Smith  are  commercial  partners.  January  1,  1892,  Jones  invested 
$20000,  and  Smith  $5000.  The  business  was  continued  until  January  1,  1893. 
Diu'ing  tlie  business,  each  partner  devoted  his  services  to  the  business,  and  neither 
drew  out,  nor  added  to,  his  capital.  There  was  no  agreement  regarding  their 
respective  shares  of  gain  or  loss.  Jaiuiary  1,  1893,  a  dissolution  took  place,  and 
after  the  goods  were  sold,  the  liabilities  to  third  jiarties  paid,  and  the  resources  col- 
lected, there  was  cash  in  bank  $35000. 

According  to  law,  custom,  and  equity,  what  is  the  amount  due  each  j)artner  ? 

STATEMENT  OF  SETTLEMENT. 

STATE>tENT  OF  THE  RESOURCES  AND  LIABILITIES,  NET  GAIN,  INTEREST  ON  INVEST- 
SIENTS,  AND  THE  PARTNERS'  NET  CAPITAL,  OF  THE  COMMERCIAl,  FIRM  OF 
JONES  &  SMITH,  JANUARY  1,  1893: 


RESOURCES : 

Casli  on  hand    ------ 

LIABILITIES  : 

Jones'  net  investment        -        -        -        - 
Smith's  net  investment    -        -        -        - 

Total  liabilities  .... 

Net  gain  of  the  firm  .  -  -  - 
Jones'  one-half  net  gain  -  -  -  - 
Smith's  one-half  net  gain         -        -        - 


partners'  interest  account: 
Jones'  investment  $20000.     1  yr.  -3)  8%"  -        - 

Smith's  investment  $5000.     1  yr.  'S>  8% 

Balance  of  interest  favor  of  Jones 
Smith's  one-half  debit  of  interest  is        -        -        - 
Jones'  one-half  credit  of  interest  is         .        .        . 


JONES'   net   capital  AT  CLOSING: 

Jones'  net  investment  as  above  -  .  -  -  . 
"  one-half  net  gain  as  above  -  -  -  -  . 
"    one-half  credit  of  interest    -        -        -        .        . 


Jones'  present  net  capital         ----- 

smith's   net   CAPITAL  AT   CLOSING: 

Smith's  net  investment  as  above      -        .        -        - 
"       one-half  gain  as  above        -        -        -        - 


total  investment  and  gain 
one-half  debit  of  interest  deducted 


"       present  net  capital 
Total  net  capital  of  firm  at  closing 


20000 
5000 


5000 
5000 


10000 


00 


1600 
400 


600 
600 


1200 


20000 

5000 

600 


5000 
5000 


10000 
600 


00 


35000 


00 
00 


10000 


00 


1200 


00 


1200 


25600 


9400 


3500000 


00 


00 


00 


00 


New  Orleans,  January  1,  1893.  .  Geo.  Soul£,  Accountant. 

Note. — According  to  law  and  custom,  commercial  partners,  in  the  absence  of  any  agreement 
pertaining  to  the  division  of  the  gains  and  losses,  share  the  same  equally.  And  in  accordance  with 
equity,  if  no  agreement  is  made  to  the  contrary,  interest  is  allowed  on  the  investments  and  with- 
drawals of  each  partner. 


8iS 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 
SECOND    FORMULA    OF    STATEMENT. 


RESOURCES : 
Cash  on  hand     ------..... 

I-IAHILITIES  : 
Jones'  net  investment       -------       $20000 

Interest  on  same,  one  year  at  8"^     .        -        .        -        .  i(jOO 


Jones'  total  investment    ----... 

Smith's  net  investment     [--.---- 

"       interest  on  same,  one  year  at  8^i^         -        .        - 

"      total  investment  .        -        .        .        . 

Net  gain  of  the  firm,  allowing  interest  on  investments 
.Tones'  one-half  net  gain  ------ 

Smith's  one-half  net  gain  -...-- 


$5000 
400 


partners'  net  capital  at  closing  : 
Jones'  net  investment  as  above        -----       $21600 
"     one-half  net  gain  ------  4000 

"      present  net  capital        -------- 

Smith's  net  investment  as  above      -----        $5400 

"     one-half  net  gain  ------  4000 

"     present  net  capital        -  ---'--. 

Total  net  capital  of  firm  at  closing  -        -        -        .        - 


21600 


5400 


4000 
4000 


8000 


25600 


9400 


00 


00 


00 


00 


00 


35000 


27000 


8000 


00 

00 


8000 


$35000  00 


00 


00 


12.  The  following  statemeut  was  received  from  a  book-keejier : 

"Adams  and  Lee  were  equal  partners  in  business,  the  aujount  of  capital 
was  $2500  each.  By  bad  management  the  business  was  unsuccessful,  and  on  closing 
their  books,  they  found  that  the  money  and  goods  on  hand  would  just  pay  cer- 
tain debts  which  they  owed  in  New  Orleans.  This  being  done,  the  flrm  had  no 
debts  due  them  nor  did  they  owe  any.  But  each  partner  had  a  special  cash  account 
on  the  books,  as  each  had  had,  from  time  to  time,  small  amounts  placed  to  his 
credit,  which  amounts  were  used  for  the  benefit  of  the  firm,  and  on  closing 
the  books,  Adams  had  a  cash  balance  due  him  of  $800,  and  Lee  a  balance  of  $200. 
They  now  desire  to  settle  this  between  themselves  by  a  note,  and  asked  my  advice, 
and  I  iiroiiosed  this  settlement :  that  Adams  is  to  give  his  note  to  Lee  for  $200,  and 
Lee  to  give  his  note  to  Adams  for  $800,  or  to  make  one  note  from  Lee  to  Adams  for 
$600.  I  am  not  positive  that  I  am  right,  and  so  refer  the  matter  to  higher  authority 
and  request  your  decision  whether  or  not  I  am  correct." 

Is  the  solution  of  the  book-keeper  correct  ? 

Ans.  No,    Lee  should  give  Adams  his  note  for  $300. 

See  problem  1, 

13.  The  following  statement  was  received  from  a  book-keeper : 

"  Will  you  be  kind  enough  to  explain  to  me  how  to  close  my  books,  the 
Trial  Balance  of  which  I  hand  you  herewith  ? 

"  The  merchandise  on  hand  as  per  Stock  Book  is  $38116.23. 
statemeut  showing  gain  of  business." 

(^Continued.) 


Please  also  make 


I 


PARTNERSHIP    SETTLEMENTS. 


819 


The  followiuff  is  the 


TEIAL   BALANCE: 


Cash  oil  hand     -     - 
Mdse.  as  per  Ledger 
Profit  and  Loss 
Sundry  accounts     - 


$  443  49 
14232  01 
22187  87 
10520  03 

$47383  40 


Stock        ....  $24012  52 

Bills  Payable    -    - 4728  24 

Sundry  accounts   ..-.--    18642  64 


$47383  40 


SOLUTION. 

Statement  of  the  resources  aud  liabilities  aud  the  net  gain  of  the  commercial 
firm  of ,  Galveston,  Tex.,  August  1,  1893 : 


liESOURCES  : 


Casli  on  hand     ... 
Mdse.  as  jier  account  stock 
Sundry  porsoual  accounts 

Total  resources  of  firm 

Stock — net  investment 
Bills  payable      ... 
Sundry  personal  accounts 

Total  liabilities  of  firm 

Net  gain  of  firm 


LIABILITIKS: 


443 

49 

38110 

23 

10520 

03 

49079 

75 

24012 

52 

4728 

24 

18642 

61 

47383 

40 

1696 

35 

To  close  the  books,  first  credit  Mdse.  account  with  the  stock  oti  hand,  and  then 
close  the  account  to  profit  aud  loss ;  then,  there  being,  as  per  statement  submitted, 
no  other  loss  and  gain  account,  close  profit  and  loss  to  stock  account.  Then  close 
stock  account  and  all  other  open  accounts  to  or  by  balance ;  then  ride  and  add  the 
accounts,  bring  the  balances  down  and  take  a  trial  balance  of  the  balances. 

14.  The  following  statement  and  question  were  presented  by  a  book-keeper : 
"  A.,  B.,  C.  and  D.  are  partners,  equal  in  gains  aud  losses.    During  the  time 

business  was  continued,  A.  drew  $260;  E,  drew  $155;  C.  drew  $180  ;  and  D.  drew 
$10.  There  is  due  A.  for  rent  $150 ;  to  B.  for  ser\ices  $150 ;  to  O.  for  services  $240, 
and  to  D.  for  services  $120. 

"  The  net  gain  of  the  business  independent  of  salaries  due  was  $1860.  "What 
is  the  amount  due  each  partner?"  Ans.  A.  $190.        C.  $300. 

B.     295.         D.     410. 

OPERATION. 

Net  gain,  per  statement,  is--------  $1860  00 

Less  .amount  due  for  rent  and  salaries            .....  660  00 

Amount  to  be  ei|ually  divided 1200  00 

Each  partner's  share  of  gain           -        - 300  00 

GAIXS.  WITHDRAWALS.  BALANCE.      KENTS  A  SALARIES.  AMT.    DUE. 

A.  $300.00   —   $260.00    =    $  40.00    -f    $150.00    =    $190.00  A. 

B.  300.00    —    155.00    =     145.00    -f-     150.00    =    295.00  B. 

C.  300.00   —    180.00    =    120.00    -f-    240.00    =    360.00  C. 
I).   300.00   —     10.00    =    290.00    -|-    120.00    =    410.00  D. 

15.  The  following  statement  of  facts  occurred  with  two  joint  owners  of  a 
steamboat  and  was  submitted  for  settlement  as  per    conditions    named   below. 

"J.  Jones  owns  five-eighths  of  the  steamboat  Star  Light,  and  S.  Smith 
owns  three-eighths  of  said  steamer.     The  boat  is  valued  at  $16000.     The  boat 


820 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


owes  liabilities  amounting  to  $G324.28;  $3490.13  of  tlie  liabilities  are  owing  to 
J.  Jones,  one  of  the  joint  owners.  The  remainder  of  the  liabilities  are  due  to  various 
liarties  and  amount  to  $2834.15. 

"  J.  Jones  holds  a  note  of  S.  Smith  for  $1000,  now  due. 

"  It  is  agreed  that  one  of  the  joint  owners  shall  buy  the  other's  interest,  and 

they  therefore  desire  a  statement  showing  how  much  Jones  shoiUd  pay  Smith  should 

he  purchase  his  interest,  and  how  much  Smith  shoidd  pay  Jones  should  he  piurchase 

his  interest  in  the  boat."    What  is  the  statement  to  be  made  ? 

NoTK. — Joint  owners  of  steamboats  are  respousihle  as  to  themselves  for  the  liabilities  of  the 
boat  iu  proportion  to  their  respective  interests  iu  the  boat. 

STATEMENT  BASED  UPON  THE  ABOVE  FACTS  SHOWING  AMOUNT  DUE  TO  EITHER 
OWNER  OF  THE  STEAMBOAT  STAR  LIGHT  BY  '.THE  OTHER  IN  CASE  EITHER 
SHOULD  PURCHASE  OF  THE  OTHER  HIS  INTEREST  IN  THE  SAID  BOAT  AND 
ASSUME  ALL  LIABILITIES  TO  THIRD  PARTIES. 


1. 
Steamboat  Star  Light,  valued  at    - 
J.  Jones'  live-eighths  interest  is       -        - 
S.  Smith's  three-eighths  interest  is 

2. 
Liabilities  of  the  boat  are        ... 
J.  Jones'  tive-eighths  of  liabilities  are 
S.  Smith's  three-eighths  of  liabilities  are 


3. 

Should  J.  Jones  buy  S.  Smith's  interest  in  the  boat  he  would  pay  as 

follows : 
For  S.  Smith's  three-eighths  interest       ------ 

For  his,  (J.  Jones')  live-eighths  of  liabilities         -        -        -        - 

Total  amount  to  be  paid  ------- 

Less  the  amount  advanced  by  him  and  included  in  the  liabilities 
Less  S.  Smith's  note  held  by  him  as  above     -        -        -        -        - 

Total  amount  of  deductions      ------- 

Net  amount  to  be  paid  by  J.  Jones  -..--. 

Liabilities  to  third  parties  assumed  by  him    -         -         -        -        - 

Net  amount  to  be  paid  to  S.  Smith         --...- 

4. 
Should  S.  Smith  buy  J.  Jones'  interest  in  the  boat  he  would  pay  as 

follows: 
For  J.  Jones' five-eighths  interest        -        -        -         ^        .        - 
For  his,  (S.  Smith's)  three-eighths  of  liabilities         -        .        . 
For  his,  (S.  Smith's)  note  held  by  J.  Jones  -         -         .         . 

Total  amount  to  be  paid  by  S.  Smith     .        -         ,        .         . 
Less  Liabilities  due  third  parties         .-..-. 

Net  amount  to  be  paid  to  J.  Jones        ....-- 


10000 
6000 


16000 


3952 
2371 


6324 


6000 
3952 


3490 
1000 


10000 
2371 
1000 


16000 


16000 


6324 


6324 


9952 


4490 

5462 
2834 


2628 


13371 

2834 


00 


00 


28 


28 


68 

13 

55 
15 


40 


60 
15 


10537  45 


16.    The  following  question  was  received  from  a  book-keeper : 
"Messrs.  A.  and  B.  entered  into  partnership,  share  and  share  alike  in  the 
gains  and  losses.    They  agreed  to  adjust  their  current  accounts  at  the  end  of  each 
year,  so  that  each  shall  draw  equally.   Now,  at  the  close  of  the  first  year,  the  debit  of 


*  PARTNERSHIP    SETTLEMENTS.  82 1 

A's  ctirrent  account  is  $627.24  in  excess  of  B's.    Wliat  entry  should  be  made  to 
close  their  current  accounts  ?  "    Answer  as  follows : 

B.  iiinst  (lra\y  from  the  firm  $627.24,  or  A.  must  pay  to  B.,  out  of  bis  private  funds,  one-half  of 
the  $627.24  which  he  has  drawn  in  excess.  In  case  B.  tlraws  the  §627.24,  the  entry  -svoukl  be,  B.  to 
Cash  for  the  amount. 

In  case  A.  paid  B.  one-half  of  his  excess  of  withdrawals,  the  entry  would  be  made  thus: 

B. $313  62 

To  A.        --------        -       S313  62 

In  case  it  is  desired  that  A.  shall  pay  B.,  and  in  case  A.  has  no  private  funds  and  is  obliged  to 
draw  the  necessary  amount  from  the  firm  to  equalize  the  current  accounts,  the  entries  would  be  as 
follows : 

FIRST. 

A. -        .        .        |;627  24 

To  Cash  ..-..---       $627  24 

SECOND. 

B. .a;G27  24 

To  A. $627  24 

In  place  of  the  two  entries  above,  one  resultant  entry  may  be  mad«  thus : 
B.  .....--.        $627  24 

To  Cash -        $627  24 

With  proper  explanations. 

17.     Smith  aiul  Brown  have  42000  logs  to  drive  down  a  certain  river.    Of 

this  number,  15000  have  been  taken  for  other  parties  at  the  agreed  rate  of  $0  jjer 

100.    Whatever  the  balance  costs  to  drive,  they  are  to  pay  in  proportion  to  the 

amount  of  logs  each  of  them  owns.    Smith  has  12000  logs  and  Brown  15000.    Smith 

pays  in  wages  and  supplies  $900 ;  Brown  pays  in  wages  and  supplies  $800 ;  Smith 

coUected  $600  for  10000  logs  at  $6  per  100 ;  Brown  collects  $300  for  5000  logs  at  $6 

per  100 ;  Smith's  books  show  the  accoiuit  of  Smith  and  Brown  charged  with  $900 

and  credited  with  $600.    The  question  is :  What  further  Journal  Entry  or  entries 

has  Smith  to  make  to  close  this  account  and  show  his  cost  or  expense  of  driving  his 

12000  logs  ?    What  has  he  to  pay  Brown,  or  what  to  receive  from  Brown,  as  the 

case  may  be  ? 

Note. — The  above  problem  was  submitted  by  a  correspondent  to  the  readers  of  "Business."    The 
following  is  our  solution. 

OPEEATION : 

1.  The  expense  of  driving  the  42000  logs  was  $1700,  of  which  Smith  paid 
$900  and  Brown  $800.  We  therefore  cretlit  Smith  and  Brown  respectively,  with  the 
amount  paid. 

Dr.        Smith.        Cr.  'Dr.        Brown.        Cr. 

$600  00  I  .$900  00  $300  00  I  $800  00 

355  56  I  444  44  | 

2.  Of  the  $900  collected  for  driving  the  logs  of  third  parties.  Smith  collected 
$600  and  Brown  $300.    We  debit  each  for  the  amoimt  he  collected. 

3.  By  deducting  the  $900  received  for  driving  the  logs  belonging  to  tliii-d 
parties  from  the  $1700,  the  cost  of  shipping  42000  logs,  there  is  a  balance  of  $800, 
which,  by  the  terms  of  the  problem,  is  to  be  borne  by  Smith  and  Bro^\Ti,  in  proportion 
to  the  logs  o^vned  by  them  respectively. 

OPERATION   TO   FIND   THE   AMOUNT   EACH   OWES   OF   THE   EXPENSE. 

Smith  12000  logs.  Smith.  Brown. 

Brown  1.5000     "  I  |    800  00  I  I    800  00 

27000  I    12000  00  27000  |    15000  00 

^'^^^  $    355  56  $    444  44 


82  2  SOULES   PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

4.  We  noTV  charge  each  in  the  above  accouuts  with  the  amouut  he  owes  for 
expense.  Tlic  acoounts  then  show  that  Smith  owes  Brown  $55.56.  When  this  is 
jmid,  Brown  will  be  debited  and  Smith  credited. 

Smith  had  debited  in  his  own  books,  the  account  of  Smith  &  Brown  with 
the  $900  paid  by  hiui,  and  credited  the  same  account  with  the  $000  collected.  The 
entries  for  these  transactions  were  as  follows : 

1.  Smith  it  Brown $900  00 

To  .Siiiith  (represeuting  the  cash  or  property  paid  out)  -        $900  00 

2.  Smith  (representiuj;  the  cash  received)  ...        $600  00 

To  Smith  &  Browu $600  00 

These  entries  produce  the  following  accounts : 

Dr.     Smith  A  Brown.     Cr.  Dr.        Smith.        Cr. 

$900  00  I  $600  00  $600  00  I  .$900  00 

Note. — To  close  these  accounts,  Smith  is  to  credit  Smith  &  Brown  with  |355.36,  which  is  the 
amonnt  it  cost  to  ship  his  logs,  and  to  debit  Smith  &  Brown  with  $55.56,  the  amount  he  paid  to 
Brown. 

The  following  journal  entries  should  be  made  to  effect  the  object : 

1.  Log  Expense  (or  Smith)       ......        $355  56 

To  Smith  &,  Brown  --------      $355  56 

2.  Smith  &  Brown  -        -        -        -        -        -        -        $  55  56 

To  Smith  (representing  cash)  -        -        -        -        -        -$55  56 

When  these  entries  are  posted,  the  accounts  will  stand  as  follows : 

Dr.     Smith  &  Brown.     Cr.  Dr.     Smith.     Cr.  Dr.     Log  Expense.     Cr. 


$900  00 
55  56 


$  600  00  1600  00 

355  56 


$900  00  1355  56 

55  56 


1955  56  $  955  56 

By  closing  Log  Expense  to  Smith,  will  i^roduce  a  balance  of  both  accounts. 

Note  1. — It  will  be  observed  that  the  account  of  Brown  in  the  first  part  of  the  statement 
does  not  Itelong  to  the  books  of  Smith. 

Note  2. — If  it  is  preferred,  the  debit  item  of  Log  Expense  may  be  charged  directly  to  Smith. 

18.  Jones,  a  planter,  agreed  with  his  neighbor.  Smith,  to  manufacture  shin- 
gles from  timber  furnished  by  Smitb,  and  to  receive  for  his  services  and  the  use  of 
machinery,  one-fourth  of  the  number  manufactured.  He  manufactured  and  delivered 
to  Smith  300,000  shingles.  How  many  must  be  manufactured  for  himself  to  pay 
him  according  to  contract  ?  Ans.  100,000  shingles. 

FIEST   OPERATION. 

1  — i  =  f.  Then  300,000  is  three-fourths  of  the  number  to  be  manufactured  ;  and  since  three- 
fourths  of  the  number  is  300,000,  one-fourth  of  the  number  would  be  the  third  part,  which  is  100,000. 

SECOND   OPEEATION. 

According  to  the  terms  of  the  contract  1  =  the  whole  number  manufactured. 

and  i  =    "  number  paid  for  manufacturing. 

and  the  difference,  f  =    "  number  manufactured  for  Smith. 

Hence  the  300,000  ^  J  of  the  number  to  be  manufactured  and  conversely  J  =  300,000;  then 
since  f  =  300,000  i  =  the  third  part  which  is  100,000  and  J  or  the  whole  number  =  4  times  as  much 
which  is  400,000,  J  of  which  is  100,000,  the  number  to  be  manufactured  to  pay  for  manufacturing 
300,000. 

19.  A  merchant  received  from  a  planter  500,000  pounds  of  seed  cotton  to  gin 
and  bale.    He  is  to  receive  for  the  service  -^  of  the  seed  cotton  ginned  and  baled. 


t 


PARTNERSHIP    SETTLEMENTS. 


8: 


How  many  of  the  500,000  pounds  of  seed  cotton  received  must  the  merchant  gin  for 
the  planter?  Ans.  454545 1^  pounds  seed  cotton  to  gin  for  the  planter. 

OPERATION. 
1      or  100  =  assumed  pounils  to  gin. 
-j'l;    or    10  =  poimds  due  for  giuuiug. 


ly^j  or  110  =  pouiids  of  seed  cottou  required  to  gin  1  or  100  pounds  of  seed  cotton. 


Then :  lA  =  1  :  =  500,000  :  x  or,  110  :  100  :  :  500,000 :   X  ;   or  thus : 


11 


1 

10 
500,000 


110 


100 
500,000 


I  454545t^  Ifes. 
454545i^  fts.  Ans. 

20.    A  father  wishes  to  divide  $14  between  his  son  and  daughter,  so  that  his 
daughter  will  receive  one-third  more  thau  his  son.     How  much  will  he  give  to  each? 

Ans.  $6  to  sou.     $S  to  daughter. 

OPERATION. 

$1    =  son's  share.  Son  Daugh^ter. 

IJ  =  daughter's  share.  $  $ 


24^  =  5  =  the  sum  of  the  proportions 

according  to  the  condition. 


14 
3 


S6 


3 


14 
3 
4 


21.  A  father  willed  his  estate  valued  at  $40000  to  his  three  children,  in  pro- 
portion, as  follows:  John  one-third,  Henry  one-fomth,  and  Katie  one-fifth.  Before 
the  settlement  was  made,  Henry  died.  What  sum  should  John  and  Katie  each 
receive  ?  Ans.  John  $25000.    Katie  $15000. 

OPERATION. 


the  sum  of  the  proportional 
interests  of  John  and  Katie. 


John. 

40000 
15 


Katie. 
40000 
15 


$25000  §15000 

Note.' Henry  having  died,  his  one-fourth  proportional  interest  is  therefore  to  he  divided 

bet^veen  John  and  Katie  in  proportion  to  their  proportional  interests,  as  is  also  the  remainder  of  the 
estate. 

EXAMPLES. 

22.  B.  A.  Root  and  J.  H.  Knox  have  been  doing  a  partnership  business 
under  the  firm  and  style  of  Root  &  Knox.  They  now  wish  to  settle  their  business 
and  dissolve  co-partnershii).  The  following  is  a  list  of  their  resources  and  liabilities, 
obtained  by  taking  an  account  of  stock,  and  from  their  memorandum  books:  cash 
on  hand  and  in  bank,  $2000;  merchandise  in  store,  $7400;  bank  stock,  market 
value,  $1400;  bills  receivable,  $2100;  personal  accounts  due  the  firm,  $3000.  They 
owe  on  notes  $800,  and  on  open  accounts  $500.  B.  A.  Eoot  invested  $5000  and 
-withdrew  $700.  J.  H.  Knox  invested  $4500  and  withdrew  $500.  \Vliat  was  the 
gain  or  loss  of  the  business  ?  Ans.  $0900  gain. 

OPERATION. 

resources: 
Cash        -  -        -        - 

Merchandise  -  -  -  - 
Bank  stock  -  -  -  - 
Bills  receival)le  -  -  .  - 
Personal  accounts 


LIABILITIES  : 

|;2600 

B.  A.  Root's  net  iu\ 

estment 

$4300 

7400 

J.  H.  Knox's     " 

" 

4000 

1400 

Bills  payable 

- 

800 

2100 

Personal  accounts 

- 

500 

3000 

816500 

9000 

Total  liabilities 

- 

-        $9600 

Total  resources 
"     liabilities 

Net  gain 

Explanation. - 
liabilities  of  the  firm,  and  then  find  the  difference  between  their  sums,  which,  because  the  resources 
are  the  greater,  is  a  gain. 


$0900  Ans. 
-In  the  operation  of  this  question,  we  simply  classify  and  add  the  resources  and 


824 


souLE  s  riiiLosoriiic  practical  mathematics. 


23.     Suppose,   in   tlie   foregoing  example,  that  tlie  partners  liad  sbarecl  the 
gains  ecpially,  what  would  have  been  the  net  capital  of  eai'li  jtartner  on  elosiiig? 

Ans.  $7750  Hoot's  capital.     $7450  Knox's  eax)ital. 


OPERATION. 


Net  Riiiii  as  slio'n'u  above,  fROOO. 
B.  A.  Koot's  oiie-lialf  net  gain    - 
"  invcstineiit 

Less  his  withtlrawals 


$5000 
700 


B.  A.  Root's  net  investment 
"  net  capital 


J.  H.  Knox's  one-half  net  gain 
"  investment 

Less  his  withdrawals 

J.  H.  Kuox's  net  investment 

"  net  capital 


$4500 
500 


$3450 


4300 


$7750 
$3450 


400O 


$7450 


24.  M.  W.  Conoly,  Cbas.  Block,  and  C.  Eiggs  have  been  doing  business  iinder 
the  commercial  firm  and  name  of  Oonoly,  Block  &  Kiggs.  The  gains  and  losses  are 
to  be  shared  as  follows :  Couoly  one-half,  Block  one-fourth,  and  Eiggs  one-fourth. 
During  the  business,  Gonoly  invested  $12000  and  drew  out  $1200 ;  Block  invested 
$5000  and  drew  out  $1600 ;  and  Eiggs  invested  $0000  and  drew  out  $700. 

The  foUowiug  is  a  list  of  the  resources  and  liabilities  of  the  firm  at  the  time  of 
closing:  cash  on  hand  and  in  bank,  $3000;  merchandise  in  store,  per  inventory,  $10850; 
house  and  lot,  No.  1428  St.  Charles  street,  valued  at  $4800 ;  bank  stock  of  A^arious 
banks,  market  value,  $2250 ;  office  and  store  fixtures,  $1300 ;  horse  and  wagon,  $540; 
bills  receivable,  $3200 ;  iiersonal  accounts  due  the  firm,  $2500,  on  wliich  it  is  agreed 
to  allow  10%  for  doubtful  debts.  The  liabilities  are:  bills  payable,  $5000;  mort- 
gage on  tlie  real  estate  for  $2000 ;  personal  accounts,  $4500.  WTiat  is  each  partner's 
net  capital  and  the  present  worth  of  the  firm  on  the  closing  of  the  business  ? 

Ans.  Conoly's  net  capital,  $9695.00. 
Block's  "  2847.50 


Eiggs' 


4747.50 


Present  worth  of  firm,  $17290.00, 


OPERATION   TO   FIND   THE   GAIN. 


RESOURCES : 

Cash        .       .        -        -       . 
Merchandise  -        -        - 
Real  estate     -        -        - 
Bank  stock     -        -        - 
Office  and  store  fixtures 
Horse  and  wagon 
Bills  receivable 
Personal  accounts 
Less  mi' 


$  3600 
10850 
4800 
2250 
1300 
540 
3200 


$2500 
250 


2250 


Total  resources 


$28790 


liabilities: 


M.  W.  Conoly  invested   -    $12000 
"  withdrew    -    1200 


M.  W.  Conoly's  net  invest. 
Chas.  Block  invested 

"  withdrew 

Chas.  Block's  net  invest. 
I  invested    - 
withdrew  - 

C.  Riggs'  net  investment 
Bills  payable 
Jlortgage  .        .        - 

Personal  accounts     - 

Total  liabilities 
"    resources 

Net  loss      ... 


$5000 
1(500 

$10800 

$6000 
700 

3400 

5300 
5000 
2000 
4500 

$31000 
28790 

$2210 


a 


PARTNERSHIP    SETTLEMENTS. 


825 


EACir  partner's  share  of  loss. 

M.  "W.  Couoly's  i  of  w'liicU  is  $1105  00 

Chas.  Block's      j  of  which  is  552  50 

C.  Riggs'             i  of  which  is  552  50 


c.  block's  net  capital. 


Chas.  Block's  net  investment 
"  i  net  loss      -    - 


$3400  00 
-  552  50 


net  capital  at  closing 


$2847  50 


M.    W.    CONOLY'S  NET  CAPITAL. 

M.  "W.  Conoly's  net  investment  $10800  00 
"  i  net  loss    -    -      1105  00 


"  net  cap'l  at  closing  $9695  00 

C.    riggs'   net  CAPITAL. 

C.  Riggs'  net  investment     -    -    $5300  00 
"         i  net  loss      -     -     .     -        552  50 


net  capital  at  closing 


747  50 


The  net  capital  of  the  firm  at  closing  is  $9695.00  +  $2847.50  +  $4747.50  =  $17290.00. 


Note. — In  the  following  questions,  we  have  the  resonrces,  liabilities  fexcept  the  partners' 
investments)  and  the  net  gain  or  loss  given  to  find  the  net  capital  at  commencing. 

25.  J.  A.  Y.  Barton  lias  just  closed  his  business  with  a  gain  of  $5280.  Ilia 
present  resources  and  liabilities  are  as  follows :  cash,  $9000 ;  merchandise  unsold, 
$1500 ;  bills  receivable,  face  value,  $0000 ;  personal  accounts,  $4500.  He  owes  on 
notes  $5000,  and  to  sundry  persons  $3500.  What  was  his  net  capital  at  com- 
mencing 1!  Ans.  $7220. 

OPERATION. 


RESOURCES : 


Cash         .         -         - 
Merchandise    - 
Bills  receivable 
Personal  accounts  - 

Total  resources 


1500 
6000 
4500 

$21000 


Bills  payable 
Personal  accounts 

Total  liabilities 


LIABILITIES  . 


Resources 
Liabilities 


Pre-sent  net  capital       -        .         - 
From  which  we  deduct  the  net  gain 


$5000 
3500 

$8500 


$21000 
8500 

$12500 
5280 


$7220 


And  obtain    - 

which  is  the  amount  of  net  capital  at  commencing. 

Explanation. — In  the  solution  of  this  example,  we  first  find  the  total  amount  of  resources  and 
liabilities;  second,  we  deduct  the  liabilities  from  the  resources  and  obtain  $12500  as  the  present  net 
capital.  Then,  since  the  present  net  capital  includes  the  gain,  it  is  clear  that  by  deducting  the  gain, 
$5280,  from  the  present  net  capital,  we  have  in  the  remainder  the  net  capital  at  commencing,  which 
is  as  above  $7220. 

26.  A.  and  Z.  lost,  dm-ing  business,  $8400.  Their  resources  at  closing  are: 
cash,  $10000 ;  merchandise,  $8000 ;  bUls  receivable,  $6000 ;  personal  accounts,  $3500. 
They  owe  on  notes  $12000,  and  personal  accounts,  $7500.  A.  invested  three- 
fourths  and  Z.  one-fourth  of  the  capital.  \Miat  was  the  amount  invested  by  each 
at  the  commencement  of  the  business  ?  Ans.  $12300  by  A.    $4100  by  Z, 

OPERATION. 


RESOURCES : 


Cash      - 
Merchandise 
Bills  receivable 
Personal  accounts 

Total  resources 


$10000 
8000 
6000 
3500 

$27500 


Bills  p.ayable 
Personal  accounts 

Total  liabilities 


LIABILITIES  : 


$12000 
7500 

$19500 


826 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Resources 
Liabilities 


Present  net  capital  or  worth 
To  which  we  add  the  loss     - 


$27500 
19500 

$8000 
8400 

$16400 


And  ohtain  ............        a        ^ 

which  is  the  net  capital  of  the  firm  at  commencement. 

Then,  as  A.  invested  three-fourths  of  the  capital  and  Z.  one-fourth,  it  is  clear  that  three-fourths 
of  $16400  is  A's  capital,  and  one-fourth  of  |16400  is  Z's  capital. 

J  of  $16400  =  $12300,  A's  capital.  i  of  $16400  =  $4100,  Z's  capital. 


27.  A.  aud  Z.  were  partners.  They  -were  both  insolvent  when  they  com- 
menced business,  and  the  firm  assumed  the  liabilities  of  each.  The  jiroportion  of 
insolvency  of  the  partners  was,  A.  §  and  Z.  J.  They  were  equal  in  gains  and  losses. 
During  the  business  their  profits  were  $7100,  and  at  the  close  of  the  business  their 
resources  were :  cash,  $2800 ;  merchandise,  $11400  ;  and  personal  accounts,  $1800. 
Their  liabilities  at  closing  were:  bills  payable,  $12800;  and  personal  accounts,  $5400. 
What  was  the  net  insolvency  of  the  flrni  and  of  each  partner  at  commenciug? 
What  was  the  capital  or  insolvency  of  each  partner  at  closing  1 

Ans.  $9300,  insolvency  of  firm  at  commencing. 
$6200,  "  of  A.      "  " 

$3100,  "  of  Z.      "  " 

$2050,  "  of  A. 

$450,    net  capital  of  Z. 

OPEEATION. 


"  closing. 

u         u 


Cash 
Merchandise 
Personal  accounts 

Total  resources 

Resources 
Liabilities 


resources: 


$  2800 

11400 

1800 

$16000 


Bills  payable 
Personal  accounts 

Total  liabilities 


LIABILITIES  : 


$12800 
5400 

$18200 

$16000 
18200 


Net  insolvency  at  closing 
Gain  during  business  added 


$2200 
7100 


Net  insolvency  of  firm  at  commencing        -----..--.  $9300 

I  of  $9300=  $6200,  A's  insolvency  at  commencing.     J  of  $9300=  $3100,  Z's  insolvency  at  commencing. 
Total  gain,  $7100. 

A's  one-half  of  which  is         -        $3550        Z's  one-half  of  which  is        -        $3550. 


A's  insolvency  at  commencing 
"  i  net  gain  ... 

"  present  net  insolvency 


$0200 
3550 

$2650 


Z's  insolvency  at  commencing 
"  i  net  gain  .... 

"  present  net  capital 


$3100 
3550 

$450 


Explanation. — The  difference  between  the  resources  and  liabilities  at  closing,  not  including  the 
partners'  accounts,  gives  us  the  insolvency  of  the  firm  at  closing,  $2200.  To  this  amount,  we  add 
the  gain  .and  obtain  the  insolvency  at  commencing,  $9300.  Had  the  difference  betAveen  the  resources 
and  liabilities  at  closing  been  a  net  capital,  not  exceeding  the  gain,  which  by  the  terms  of  the  ques- 
tion it  could  not  h.ave  been,  we  would,  for  obvious  reasons,  have  deducted  the  same  from  the  gain  in 
order  to  have  found  the  insolvency  of  the  firm  at  commencing. 

28.  W.,  X.  and  Y.  have  dissolved  partnership,  and  from  their  irregularly 
kept  books  and  memoranda,  they  submit  the  following  statement  of  facts  and  figures 
for  a  final  settlement.    They  were  equal  in  gains  and  losses.    They  have  cash  on 


PARTNERSHIP   SETTLEMENTS. 


827 


hand,  $14500;  merclianclise  in  store,  per  inventory,  $24200 ;  bank  stock,  market 
value,  $2100.  They  own  the  store  that  they  occupy,  which  is  valued  at  $31000.  The 
store  and  office  furniture  is  valued  at  $3800.  The  personal  accounts  due  them 
amount  to  $9500,  on  which  it  is  agreed  to  allow  20%  for  doubtful  accounts. 

They  hold  bills  receivable  amounting,  face  value,  to  $8000,  on  which  there  is 
$540  interest  due.  They  owe  on  notes,  $5000,  on  which  there  is  $150  interest  due, 
and  on  open  accounts  they  owe  $4500. 

During  the  business  W.  invested  $25000  and  withdrew  $3250. 
"  "  X.  '  22500  "  2810. 

"  "  T.  "  21000  "  4100. 

They  have  kept  a  private  book  which  shows  the  expense  to  have  been  $5768. 
W.  desires  to  withdraw  from  the  business,  and  hence  wishes  to  know  his  present 
net  capital.     How  much  is  it  ?  Ans.  $29CG6.67. 


OPERATION. 


RESOURCES  : 

Cash         -        .        .        -        . 

Merchandise 

Bauk  stock      -        .        -        - 

Real  estate      -        -        -        - 

Store  aud  office  furniture 

Porsoual  accounts 


$9500 
1900 


$14500 
24200 

2100 
31000 

3800 


Bills  receivable 
Interest  on  same 


Total  resources 


7600 

8000 

540 

$91740 


LIABILITLKS  : 

W's  net  investment 

X's     "            "  -  - 

Y's     "            "  .  - 

Bills  payable           .  -  . 

Interest  on  same     -  -  - 

Personal  accounts  -  -  . 

Total  liabilities 


Resources 
Liabilities 


Total  net  gain 


$21750 

19690 

16900 

5000 

150 

4500 

$67990 


$91740  00 
67990  00 

i  )$23750  00 


W's  one- third  net  gain  is 
"    net  investment 


$  7916  67 
21750  00 


"    present  net  capital 


$29666  67 


29.  A.,  B.  and  C.  enter  into  co-partnership  and  agree  to  share  the  gains 
and  losses  equally.  It  is  also  agreed  that  A.  shall  keep  the  books  and  receive  there- 
for a  salary  of  $1800  per  year;  that  B.  shall  act  as  chief  salesman  and  receive  there- 
for a  salary  of  $1500 ;  and  that  O.  shall  act  as  salesman  and  receive  a  salary  therefor 
of  $1200.  At  the  close  of  the  year,  there  was  a  net  gain,  exclusive  of  the  $4500  sala- 
ries, of  $3000.     How  much  of  the  gain  is  due  to  each  partner  1 

Ans.  A.  $1500.     B.  $1200.     C.  $900. 


A's  salary 
B's      " 
C's     " 

OPERATION. 

Cr. 

$1800    — 
1500    — 
1200     — 

Dr.              Cr. 

$300    =     $1500 
300     =      1200 
300    =        900 

Total  liabilities    - 
Less  gain 

Net  loss 

$4500 
3600 

-     i  )  $900 

£aeh  partner's  cue-third  loss         $309 


828 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


30.  W.  S.  Meek,  doing  business  in  Lewicville,  Ark.,  had  liis  store  and  goods 
burned.  His  books  were  also  destroyed,  and  from  a  trial  balance,  taken  the  day 
that  the  fire  occurred,  and  ])reserved  by  being  at  Ms  residence,  the  face  value  of  the 
ledger  is  shown  to  be  as  follows : 


W.  S.  Meek 

Merchandise 

Cash       -        -        .        . 

Bills  receivable 

Bills  payable 

Interest 

M.  &  T.  R.  R.  stock      - 

Expense  ... 

Personal  accounts  tine 

Personal  accounts  oTving 


Dr. 

Cr. 

1  1200 

114400 

32fi00 

24800 

50524 

47215 

1600 

600 

1100 

3100 

120 

50 

1000 

780 

4500 

3259  _ 

$93424 

$93424 

His  goods  were  fidly  insured,  and  in  settling  with  the  insurance  company,  it 
was  agreed,  after  thorough  investigation  and  close  figuring,  to  allow  35%  gain  on 
the  sales  of  merchandise.  The  R.  K.  stock  is  valued  at  cost,  making  an  allowance 
of  10%  on  the  personal  accounts  due  for  worthless  accounts.  What  was  the  net  gain 
of  W.  S.  Meek  during  business,  and  what  was  his  present  net  capital  ? 

Ans.  $5129.63  gain.    $18329.03  capital. 


RESOURCES  : 
Insurance  Co.  for  Mdse.  burned, 

lowing  35%  gain  on  sales 
Cash  on  hand 
Bills  receivable 
M.  &  T.  R.  R.  stock 
Personal  accounts 
Less  10%      - 


Total  resources 


OPERATION. 

LIABILITIES 

,  al. 

W.  S.  Meek's  net  investment 

. 

$14229  63 

Bills  pavable             .        .         . 

- 

3309  on 

Personal  accounts 

- 

1000  00 

. 

1000  00 

Total  liabilities 

$4500 

450 

4050  00 

. 

$23588  63 

Resources 
Liabilities 


Net  gain  during  business 
Net  investment  added 

Net  capital  at  closing 


$13200  00 
2000  00 
3259  00 

$18459  00 


$23588  63 
18459  00 


$  5129  63 
13200  00 


$18329  63 


Explanaiion. — It  Tvill  be  observed  that  in  the  statement  of  resources  and  liabilities,  we  did  not 
include  the  expense  or  interest  accounts.  They  were  omitted  for  the  reason  that  they  do  not  here 
show  any  available  resource  or  real  li.ibility,  and  hence  could  not  be  embraced  in  a  stiitcment  of  this 
character  to  obtain  gain  or  loss.  It  is  true  that  the  gains  and  losses  of  the  business  were  affected 
by  these  accounts,  but  it  was  done  through  some  property  account  at  the  time  the  debits  and  credits 
were  created.  This  is  the  case  with  all  accounts  of  this  character,  such  as:  Premium,  Discount, 
Commission,  Profit  and  Loss,  etc.,  whirh  appear  on  the  face  of  a  double  entry  Ledger,  and  which 
must  not  therefore  be  classed  as  resources  or  liabilities.  In  problem  No.  28,  we  included  th© 
interest  due  on  the  matured  bills  receivable  and  bills  payable  because  the  amounts  there  represented 
actual  resources  and  liabilities. 

To  find  the  amount  of  merchandise  on  hand,  wo  first  find  the  cost  of  the  amount  sold,  and 
then  by  deducting  the  cost  of  the  sales  from  the  total  cost,  we  obtain  the  amount  lost  by  fire. 


PARTNERSHIP   SETTLEMENTS. 


829 


31. 


The  following  is  a  trial  balance  of  W.  B.  Taylor's  Ledger: 

Dr.  Ch. 


W.  B.  Taylor 

Cash 

Real  estate 

Merchandise    - 

Expeuse 

Uauk  stock 

E.  R.  stock 

Interest 

Commission     - 

Personal  accounts,  Dr. 

Personal  accounts,  Cr. 


$  ii;(X) 

$15000 

60(X)0 

55000 

4000 

5000 

21000 

27500 

4,500 

800 

1050 

1000 

350 

100 

1000 

13900 

2500 

$107150 

$107150 

Tlie  net  gain  of  the  business  lias  been  $921-1.60.    The  only  property  on  hand, 
unsold,  is  merchandise  and  R.  E.  stock.     What  is  the  value  of  the  merchandise  ? 

Ans.  $5214.60. 

OPEEATION. 


RESOURCES : 

Cash $  5000 

K.  R.  stock  ------  1000 

Personal  accounts          -        -        -        -  13900 


Kesonrces,  not  including  merchan- 
dise -        -        .        -        . 


LIABILITIES  : 

"W.  B.  Taylor's  net  investment 
Personal  accounts 


Liabilities 
Net  gain  added 
$19900 

Total  liabilities  and  gain    - 
Less  known  resources 

Value  of  merchapdise 

Or, 


$13400  00 
2500  00 

$15900  00 
9214  60 

$25114  60 
19900  00 

$  5214  60 


Known  resources,  as  above 
Liabilities,  as  above 


Gain       - 

Which  deducted  from  the  whole  gain 

Gives  a  gain  balance  of       -        -        - 


$19900  00 
15900  00 

$4000  00 
9214  60 

$5214  60 


which  must  be  the  value  of  the  merchandise  on  hand  in  order  to  produce  the  whole  gain. 

32.  A.  and  Z.  are  partners  in  business  -with  equal  investments  and  equal  in 
gains  and  losses.  During  the  business  their  total  net  gain  is  84410.  A.  i)roi)oses  to 
draw  out  25%  of  his  share  of  the  gain,  allowing  the  balance  to  remain  as  an  addi- 
tional investment.  Z.  proposes  to  draw  out  20%  of  his  gain,  leaving  the  balance  as 
an  additional  investment.  The  total  net  gain  is  equal  to  12^%  of  the  net  invest- 
ment. \Miat  is  the  net  investment  of  each  after  the  gains  have  been  adjusted 
according  to  the  above  basis  ?  Ans.  A's,  $19293.75. 

Z's,  $19404.00. 

OPEKATION, 


12i 


I  100  =  assumed  net  investment  of  A.  and  Z. 
4410 


$35280  total  original  investment  by  A.  and  Z. 

$17640  A's  one-half  of  original  investment. 
17640  Z's         "  "  " 


830 


soule's  philosophic  practical  mathematics. 


A's  one-half  of  net  gain  is 
''  25;^  withdnuvals 

"  adtlitioiial  iiivestmeiit 

"  original  net  investment     - 

"  present  net  investment    - 


$2205  00 
551  25 

Z's  one-half  of  net  Rain  is 
"  20?„'  withdrawals 

"  additional  investment 

"  original  net  investment     - 

"  present  net  investment     - 

$2305  00 
441  00 

$  11)53  75 
17(540  00 

$  1764  00 
17640  00 

$19293  75 

$19404  00 

33.  A  dishonest  book-keeper  closed  a  luercliaut's  books  and  found  the  net 
gain  to  be  $1341i!.45.  After  he  had  closed  the  books,  he  destroyed  the  Cash  Book 
and  the  cash  account  in  the  Ledger,  and  ran  away  with  the  cash  on  hand.  The 
resources  of  the  house,  not  including  the  cash,  amount  to  $85418.20,  and  the  liabili- 
ties amount  to  $91300.15.    What  amount  of  cash  did  the  book-keeper  steal  ? 

Ans.  $19294.40. 

OPERATION. 


Resonrces  ...... 

Liabilities         ...  .        - 

Loss,  by  reason  of  not  including  cash    - 
Net  gain  ...... 


Amount  of  cash  stolen 


$85418  20 
91300  15 

$  5881  95 
13412  45 

$19294  40 


34.  A.  J.  Gibson,  D.  Kinsella,  and  J.  W.  Sandidge  associate  themselves 
in  business  with  equal  investments  and  equal  interests  in  the  gains  and  losses.  It 
is  agreed  that  each  jiartner  shall  be  charged  $3  per  day  for  absence  from  duty. 
During  the  year  A.  J.  Gibson  was  absent  41  days;  D.  luusella,  65  days;  and  J.  W. 
Sandidge,  34  days.  How  shall  the  partners  adjust  the  matter  between  themselves 
without  making  any  entries  in  their  books  ?  Ans.  D.  Kinsella  must  pay  to  A.  J. 

Gibson  $17,  and  to  J.  W.  Sandidge  $38. 


OPERATION. 

Pi!. 

A.  J.  Gibson        owes  for  41  days  at  $3    =    $123 
D.  Kinsella  "      "   65     "      "      3    =      195 

J.  "W.  Sandidge       "      "    34     "      "     3     =      102 

Total  amount  duo  the  partners        -        $420 

One-third  of  which  due  to  each  is  $140 


Cu. 

$140 
140 
140 


Dr. 

Cr. 

$55 

$17 
38 

$55 

$55 

35.  X.,  Y.  and  Z.  are  partners.  X.  one-half,  Y.  one-fourth,  and  Z.  one-fourth 
interest  in  gains  and  losses.  On  (^losing  business,  after  all  the  property  has  been 
di\nded,  it  is  found  that  X's  account  has  $500  excess  of  debit,  and  that  Y's  account 
has  $150  excess  of  debit.  What  settlement  is  necessary  to  properly  adjust  the 
matter  between  the  partners  ?  Ans.  X.  must  pay  Y.  $12.50,  and  Z.  $102.50. 


OPERATION. 


RESOURCES  : 


X's  debit 
y's  debit 


Total  resources 


$500 
150 

$650 


X's  one-half  of  resources  is 
Y's  one-fourth  "  " 

Z's  one-fourth  "  " 


$325  00 
162  50 
162  50 


X's 
Y's 
Z's 


Dr. 

$500 
150 


Or. 

$325.00 
162.50 
162.50 


Dr. 

$175 


Or. 

$  12.50 
162.50 


I 


PARTNERSHIP   SETTLEMENTS. 


8.V 


36,  Geo.  J.  Siegel  and  Chas.  Herzog  contracted  with  B.  W.  Brown  &  Co.  to 
erect  a  cotton  factory  for  the  sum  of  $18000.  Not  wishing  to  incur  the  expense  of 
a  bookkeeper,  they  agreed  that  each  should  keep  a  correct  account  of  all  receipts 
and  disbursements  on  account  of  the  contract,  and  when  the  work  was  completed 
they  woidd  have  a  general  settlement. 

On  the  completion  of  the  factory,  their  accounts-  aud  business  affairs  pertain- 
ing thereto  stand  thus : 

G.  J.  Siegel  has  paid  for  material  and  wages  $7124,  aud  received  from  B.  W. 
Brown  &  Co.,  at  various  times,  on  account  of  the  contract,  $5250. 

Chas.  Herzog  has  paid  for  material  and  wages  $5418,  and  received  from  B. 
W.  Brown  &  Co.,  on  account  of  the  contract,  $4870.  They  owe  to  various  parties 
for  material  fiiruished  aud  labor  performed,  $1985.  What  has  beeu  the  profit?  How 
much  is  due  from  B.  W.  Brown  &  Co.,  and  how  much  of  it  is  due  to  Siegel  and  how 
much  to  Herzog,  after  paying  the  amount  due  for  the  material  and  labor  ? 

Ans.  $3473  profit ;  $3610.50  due  to  Siegel ; 
$2284.50  due  to  Herzog;  $7880  due  by  B.  W.  Brown  &  Co, 


OPERATION. 


Contract  price  to  build  tbe  factory 

Contractors'  costs  to  build  the  same  are: 
Amounts  paid  by  G.  J.  Siegel 

"  "         Chas.  Herzog,   - 

"        unpaid  ... 


Total  contractors'  cost 
Net  profit 


$7124 
5418 
1985 


$18000 


14527 
$5473 


G.  J.  Siegel's  one-half  net  gain  is 
"  credit  (am't  paid)  is 


11736.50 
7124.00 


total  credit  is       -    -    -    $8860.50 
debit  (am't  received)  is      5250.00 


balance  of  credit  is       -    $3610.50 


Chas.  Herzog's  one-half  net  gain  is    -    $1736.50 
"  credit  (am't  paid)  is  -      5418.00 

"  total  credit  is    -    -    -    $7154.50 

"  debit  (am't  received)  is    4870.00 


balance  of  credit  is   -    $2284  50 


B.  AV.  Brown  &  Co.  contracted  to  pay      -        -        -        -  $18000 

"  "  paid  to  Siegel $5250 

"  "  paid  to  Herzog 4870 

10120 

"  "  owe  a  balance  of $7880 

RECAPITULATION   OF   BALANCES   DUE. 

Balance  due  to  G.  J.  Siegel $3610.50 

"  "     Chas.  Herzog 2284.50 

"  "    for  material  and  labor      ...        -  1985.00 

Which  added  gives  $7880.00 

■which  is  the  balance  due  by  B.  W.  Brown  &  Co. 

37.  T.  R.  Winn,  of  New  Orleans,  and  C.  P.  Johnson,  of  Texas,  enter  into  a 
contract  to  buy  aud  sell  cattle,  and  share  the  gains  and  losses  equally.  To  com- 
mence the  business,  T.  R.  Winn  remitted  to  C.  P.  Johnson  a  sight  check  for  $20000. 
Johnson  i)urchased  at  different  times  to  the  amount  of  $35421.50,  and  shipped  to 


^r- 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Wiiiu  at  clifiFerent  times,  from  which  he  lias  sold  to  the  amount  of  $28740.    Johnson 
has  sold  to  the  amount  of  $7190. 

It  is  now  proposed  and  agreed  to  dissolve  the  partnership.  In  making  ui> 
their  memorandum  accounts  for  final  settlement,  it  is  found  that  Winn  has  paid  for 
expenses  of  the  business  $785.30,  and  Johnson  $1014.05.  That  Winn  has  cattle  on 
hand,  valued  at  New  Orleans  market  price,  amounting  to  $2110,  and  that  Johnson 
has  on  hand  cattle  that  cost  $1871.  It  is  agreed  in  making  the  settlement  that  each, 
shall  retain  the  cattle  in  his  possession  at  the  valuation  above  given  and  pay  cash 
for  the  balance  due  the  other.  What  was  the  gam  or  loss?  WTiat  is  the  cash 
balance  due  by  cue  partner  to  the  other,  and  by  which  jiartner  is  it  due  1 

Ans.  $2701  55,  gain. 

$8719.93,  bal.  due  by  Winn  to  Johnson, 


OPEKATION. 


T.    R.    WINN'S   account: 

Dk. 
By  cash,  original  investment 

"        paid  for  expenses 
To  cash,  rec'd  from  sale  of  cattle  $28740 
"  cattle,  value  ou  haud         -         2116 


Cii. 

120000.00 
7«5.30 


$30856  $20785.30 


RESOURCES : 


T.  R.  Wiun  withdrew 
C.  P.  Johiison     '• 

Total  resources 
Net  gain 


c.  p.  Johnson's  account: 

Dk. 
By  cash  investment 

"       ]>iiid  for  expenses 
To  cash  rec'd  from  sale  of  cattle  $7196 
"  cattle,  value  on  hand         -        1*^71 


RECAPITULATION. 


$30856.00 
9067.00 

$39923.00 
37221.45 


$2701.55 


liabilities: 


T.  R.  Winn's  investment   - 
C.  P.  Johnson's      " 

Total  liahilities 


Cr. 
$15421. .50 
1014.  eS' 


$9067    $16436.15 


$20785.30 
16436.15 


57221.45 


T.  R.  AViun's  investment 

one-half  net  gain 

"  total  interest 

"  withdrawals 

"  balance  of  indebtedness 

which  he  owes  to  his  partner. 


$20783.30 
1350.77 

$22136.07 
30856.00 

$8719.93 


C.  P.  Johnson's  investment 

"  one-half  net  gain    - 

"  total  interest 

"  withdrawals 

"  balance  of  capital 

which  is  dne  to  him  by  his  partner. 


$16436.15 
1350.78 

$17786.93 
9067.0(> 

$8719.93 


38.  "A  co-partnership  existed  between  C,  D.,  E.  &  F.,  of  New  York;  they 
agreed  to  put  into  the  business  $25000  each.  0.  was  to  have  one-eighth,  D.  one-sixth, 
E.  one-fourth,  and  F.  one-third;  and  at  the  expiration  of  the  year,  the  balance  of 
gain  in  excess  of  such  proportions,  was  to  be  placed  to  jirofit  and  loss  account  for 
the  next  year,  as  well  as  to  cover  any  loss  that  might  arise  after  balancing  their  books. 
On  adjusting  their  accounts,  they  found  a  gain  of  $95070.15,  but  that,  instead  of  the 
partners  fulfilling  their  contract,  C.  and  D.  had  put  in  on  account  of  their  capital 
only  $15000  each.  In  this  case,  what  sum  or  proportion  of  the  gain  must  be  jilaced 
to  their  respective  credits?" 

The  foregoing  problem  we  take  from  a  recent  treatise  on  Commercial  Calcu- 
lation, wherein  the  author  made,  by  an  approximation  solution  only,  1290  figures. 


PARTNERSHIP  SETTLEMENTS. 


833 


In  our  solution,  ^vorking  all  our  statements  and  counting  all  the  figui'es,  ■sve  make 
but  269.    The  following  is 

OUR   SOLUTION.  • 

By  the  first  coiiditinn  of  tlio  co-partnership,  each  partner,  C,  D.,  E.  and  F.,  Tvas  required  to 
invest  $25000;  and  by  the  second  condition,  they  were  to  share  the  gains,  resi)ectively  |,  I,  i,  J, 
(which  added,  equals  |,)  leaving  |  undivided,  which  was  to  be  placed  to  the  credit  of  their  profit 
and  loss  account. 

Now,  since  the  first  condition  of  the  co-partnership  was  not  complied  with  by  C.  and  D., 
they  having  invested  less  than  they  were  required  to  invest  by  the  first  condition,  their  respective 
shares  of  the  gain,  i  and  J,  must  be  reduced  in  the  same  proportion  as  tlie  amounts  invested  by 
them  were  reduced  from  the  amounts  required  to  be  invested  by  the  first  condition  of  the  co-partner- 
ship. And  as  C.  invested  but  $15000,  instead  of  $25000  as  required  by  the  first  condition,  and  since 
$15000  is  f  of  $25000,  C.  is  therefore  to  receive  J  of  |  of  the  gain,  according  to  the  second  condition, 
which  is  -^  for  his  proportional  share. 

D.  having  invested  only  $15000,  which  is  l)ut  ?  of  the  amount  required  by  the  first  condition, 
he  is  therefore  to  receive  but  3  of  the  5  of  the  gain  specified  in  the  second  condition,  which  is  1^  for 
his  proportional  share. 

E.  and  F.  having  complied  with  the  first  condition  of  the  co-partnership,  their  respective 
shares,  and  also  the  i  of  the  gain  to  be  credited  to  profit  and  los.s,  according  to  the  second  condition, 
remain  unchanged. 

From  the  foregoing  deductions,  we  have  the  following  proportional  interests  of  the  gain: 

C.  A,     D-  Vn,     E.  i,     F.  i,     and    P.  and  L.  i. 

Having  ascertained  the  respective  proportional  interests  of  the  gain  at  the  close  of  the  year, 
we  have  but  to  divide  the  whole  gain  in  accordance  therewith. 

By  adding  C's  4',;,  D's  i^i,  E's  i,  F's  i,  and  P.  and  L's  i,  we  find  they  sum  up  but  |§S.  The 
deficit  of  -^i^,  of  unity  is  occasioned  by  reason  of  the  decrease  of  C's  and  D's  shares  of  gain  and  will 
result  in  augmenting  the  share  of  gain  of  each  partner  and  of  profit  and  loss,  iu  proportion  to  their 
respective  shares. 

These  increased  proportions  are  respectively  : 


C's 

•A 

of 

120 

= 

9 

= 

TnE 

D's 

1^ 

of 

120 

z= 

12 

^ 

T^<?ff 

E's 

i 

of 

120 

^= 

30 

— 

t=.."s 

F's 

i 

of 

120 

:= 

40 

^ 

^?e 

P.  and  L's 

i 

of 

120 

= 

15 

:^ 

1  ■, 
lirff 

Note.— The  106  is  the  snm 
of  the  respective  shares  as 
shown  above. 


The  whole  gain  of  the  business  being  $95676.15,  yj^  of  the  gain  is  : 

106  )  95676.15  (902.6052 


$902.6052 


C's 

Share  of  the  gain  is    9  times 

D's 

,.       "  ..         ly     .. 

E's 

"          "        30     " 

F's 

.<          "        40     " 

f  = 


P.  and  L's 


15 


$  8123.45 
10831.26 
27078.15 
36104.21 
13539.08 

$95676.15 


39.  A.  contracted  with  Z.  to  serve  him  iu  the  capacity  of  agent  in  the  pur- 
chase and  sale  of  goods.  July  1,  Z.  delivered  to  A.  merchandise  amounting  to 
$820.70,  and  cash  SGOO.  A.  bought  at  various  times  merchandise  valued  at  $4039.20 
and  sold  at  various  times  merchandise  amounting  to  $3615.55.  At  the  end  of  the 
year,  Z.  wishes  to  close  the  agency  and  settle  with  A.  A.  is  allowed  $300  for  services. 
There  is  $1508.'10  worth  of  merchandise  on  hand,  which  A.  returns  to  Z.     What 


S34 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


has  been  the  gain  or  loss  on  the  sales  of  merchandise,  the  gain  or  loss  of  the  busi- 
ness, and  how  nuicli  does  Z.  owe  A.  or  A.  owe  Z.  ? 

Ans.  $2G4.05,  gain  on  sales. 

$3.j. !),■»,  loss  of  the,  business. 

Z.  owes  A.  $123.05. 


OPERATION 

To  fiutl  bow  A.  &  Z.  stand  iu  a,  Dr.  and  Cr.  sense. 

A.  IN  Account  with  Z. 


Dr. 

Cr. 

To  cash  paid  to  him     - 

$  600.00 

' '      lec'd  from  sales  of  Mdse 

3615.55 

"  Mdse  on  baud  the  day  of  set- 

tlement       -        .        - 

1508.40 

By   cash  paid  for  Mdse. 

])urchased     - 

$4039.20 

"    Mdse.  returned    - 

1508.40 

"     salary  for  services 

300.00 

To  Balayice  (^amount  due  to  A.) 

123.65 

$5847.60  $5847.60 


OPERATION 

To  find  the  Gain  on  the  Sales  of  Merchandise. 

Merchandise  Account. 

Dr.  Cr. 

To  value  of  Milse.  invested  in 

the  business  -        -       .$  820.70 

"  value   of    Mdse.    purchased 

during  business    -        -        4039.20 
By  amount  received  from 

Mdse.  sold         -  $3615.55 

"   value    of    Mdse.     on 

hand  -        -  1508.40 

To  Profit  and  Loss  (for  gain  on 

sales)  ....        S64.0S 


$5123.95  $5123.95 


OPERATION 
To  iind  the  Loss  of  the  Business. 

Expense  incurred  to  conduct  the  business  (A's  salary)  .......        $300.00 

Gain  as  above  ._....---  .....  264.05 

Net  loss  of  the  business $35.95 


SECOND   OPERATION   TO   FIND   BALANCE  DUE   TO   A. 
A.  iu  Account  with  Z. 


Dr. 
To  cash,  paid  to  him    - 

"  Mdse.  delivered  to  him 
"  Gain  on  sales 
To  Jialance  due  to  A. 


$600.00 
820.70 
264.05 
123.65 


$1808.40 


Cr. 


By  Mdse.  returned 
"  salary 


$1508.40 
300.00 


$1808.40 


40.  J.  Jones,  a  merchant  in  New  Orleans,  has  a  branch  store  in  the  country. 
He  places  in  charge  of  said  store,  H.  Smith,  at  a  salary  of  $600.  At  the  time  Smith 
took  charge,  the  books  showed  the  following  accounts:  Goods  on  hand,  $.2100; 
cash,  $50;  bills  receivable,  $130  ;  personal  accounts  receivable,  $340;  bills  payable, 
$C20.     The  rent  of  the  store  was  $20  per  month. 

At  the  end  of  the  year,  Jones  examines  the  business  and  finds  goods  in  stock, 
$1800;  goods  bought,  $4500 ;  goods  sold,  $6300 ;  expenses  paid,  $72;  rent  paid,  $50 ; 
salary  i)aid,  $250;  bills  receivable  in  safe,  $300;  bills  i^ayable,  outstanding,  $750; 
personal  accounts  receivable,  $840 ;  personal  accounts  payable,  $120.  Eeceived  for 
interest,  $36 ;  for  running  a  post  ofiice,  $180 ;  for  other  services,  $24,  Paid  the 
proprietor,  cash,  $300. 


PARTNERSHIP    SETTLEMENTS. 


835 


Eeqiiired,  1.  The  net  gain,  allowing  10%  of  the  bills  receivable  and  personal 
accounts  i-cceivable  for  bad  debts.  2.  The  present  worth  of  the  country  house. 
3.  The  correct  cash  balance.  Ans.  $714  net  gain. 

$2414  present  worth. 
$998  cash  on  hand. 
The  above  iirobletn  was  presented  to  us  with  a  request  to  supply  the  required 
answers. 

The  following  are  our  solutions : 

FIRST   SOLUTION. 
1.     OPERATION    TO    FIND    NET    INVESTMENT    OF    J.    JONES. 

J.   JONES. 


To  bills  payable 
"  cash  drawn 
To  balance,  7iet  inve$tment 


$  620 

300 

2700 


12620 


By  Mdse.  ... 

"  cash  ..... 
"  bills  receivable  ... 
"  personal  accounts  receivable 


$2100 

50 

130 

340 

$2620 


OPERATION   TO    FIND    CASH    BALANCE. 
J.  Jones,  Branch  Store. 


To  Mdse.  at  opening        ....  $2100 

"       "      purchased         ....  4500 

"  cash,  for  expenses       ....  72 

"       "      "    rent      ...        -        -  50 

"       "      "    salary 250 

"  personal  accounts  receivable  at  closing  840 

"  bills  receivable  at  closing  .        -        -  300 

Total  debits    ' $8112 


By  net  investment            ....  $1700 

•  '■'  Mdse.  sales 6300 

"  bills  payable  at  closing     ...  750 

"  personal  accounts  at  closiug      -        -  120 

"  interest  received         ....  36 

"  post  ofiBce  salary         ....  igQ 

"  service,  cash        .....  24 

Total  credits $9110 

"    debits 8112 

Cash  balance    -----        .  $  998 


OPERATION  TO  FIND  THE  NET  GAIN  AND  THE  NET  WORTH  OF  THE  BUSINESS. 

Resources.  Liabilities. 


Cash  on  hand    - 
Mdse.  "     " 
Bills  receivable 
Less  10"^ 

Personal  accounts     - 
Less  10% 

Total  resources 

"       liabilities 


$300  > 

30  i 

840* 

84  f 


Net 


$  998 
1800 

270 
756 

$3824 
3110 

$  714 


Bills  payable    -        -        . 
Personal  accounts     - 
Due  for  rent 

'•'      salary 
J.  Jones'  net  investment 


Total  liabilities 


$  750 

120 

190 

350 

1700 


$3110 


J.  Jones'  net  investment 
J.  Jones'  net  gain 


$1700 
714 


J.  Jones'  net  Tvorth        -         - $2414 


Explanation. — In  problems  where  the  conditions  of  the  business  are  given  at  the  opening  and 
at  the  closing,  the  correct  cash  balance  maybe  determined  as  f(dlo\vs :  1.  Open  an  account  with 
the  proprietor  and  find  his  net  investment  by  crediting  his  account  with  the  cash  and  all  other  pro- 


8^,6 


soule's  philosophic  practical  mathematics. 


perty  advancod  on  commencing  or  during  flic  husines.s;  and  then  dehiiing  the  same  with  all  liabili- 
ties at  commencing,  and  for  all  ■withdrawals  during  the  business,  as  shown  in  the  first  operation 
above. 

2.  Open  an  account  with  thu  Ilranch  Store  and  debit  the  same  with  the  balance  of  all  real 
accounts  except  cash  which  is  udt  known,  and  also  with  the  total  of  the  representative  or  loss  and 
gain  accounts,  such  as  mercllan(Us(^,  real  estate,  ex]ieuse,  rent,  salary,  etc.  Then  credit  the  Branch 
Ktore  account  as  follows:  Ist. — With  the  pr(i]irietiir's  net  investment  as  shown  in  thl^  account  tirst 
o])Cued  with  him.  2d. — Credit  it  with  the  credit  balances  of  all  real  accounts,  and  with  the  totals 
of  all  representative  accounts,  such  as  merchaniliso,  interest,  etc.  When  this  lias  been  done,  take  the 
dill'ereneo  between  the  aggregate  debits  and  the  aggregate  credits,  and  the  same  will  be  the  correct 
cash  balance,  as  shown  in  the  operation.  Now,  having  the  cash  balance,  to  find  the  net  gain  and 
the  net  capital  make  the  usual  statement  of  resources  and  liabilities  as  shown  in  the  third  opera- 
tion. 

Note. — In  case  of  partnerships,  first  open  accounts  with  each  partner  and  find  the  net  invest- 
ment, as  above. 

SECOND   SOLUTION. 

J.    JONES. 


To  bills  payable 
"  cash  drawn 
To  balance  net  capital 


$  620 

300 

S414 


$3334 


By  Mdse.  .        .        .        . 

"   cash  -        -        -        - 

"   bills  receivable 
"   personal  accounts  receivable 
"  net  gain  from  F.  and  L. 


12100 

.50 

130 

340 

714 

$3334 


Merchandise. 


I 


To  inventory  at  commencing 
"  pnrch.ases     -        -         -        . 
To  profit  and  loss 


$2100 
4.500 
1500 

$8100 


By  siiles    -        -        - 
"  inventory  at  closing 


1800 


$8100 


CASH. 


To  amount  on  hand  at  commencing 
"  interest        .        .        .        -        - 
"  P.  O.  receipts      -        -        -        - 
"  sundries        -        -         -        - 
"  Mdse.  sales  .        -        - 

Less  increase  P.  acct's  recv.  $500  ) 
"  '•        on  bills      "        170) 


$6300) 
670  f 


f    50 

36 

180 

24 

5630 


15920 


By  expenses      -        -        - 
"  rent,    paid  on  account 
"  salary      "     "         " 
"  J.  Jones"     "         " 
"  Mdse.  purchases    - 

Less  increase  on  bills  pay. 
"         "        P.  aect's  pay. 


By  balance  on  hand 


$    72 

50 

250 

300 

$4500) 

^120 1    ^^\     ^^^*' 


$4922 
99S 

$5920 


33ills    Receivable. 


Dr.  at  commencing 
"    at  closing 


increase  during  business 


$130 
300 

$170 


PARTNERSHIP    SETTLEMENTS. 

Personal   Accounts    Receivable. 


837 


Dr.  at  commencing 
"   atclosiDK 


"  increase  duriufr  business 


$500 


Bills    Payable. 


Cr.  at  commencing    - 
"  at  closing      -        -        - 

"    increase  iluriug  business 


750 


OPERATION  TO  FIND  THE  NET  GAIN  BY  THE  RESOURCE  AND  LIABILITY  METHOD. 
Resources    and.    Liabilities. 


Mdse.  inventory  at  closing 
Cash,  at  closing 
Bills  receivable  at  closing  - 
Less  10,"„'  for  bad  debts 
Personal  accounts  at  closing 
i,ess  10%  lor  bad  debts 

Total  resources 

"     liabilities 


Net  gain 

J.  Jones'  net  investment 

"  "  capital 


$300) 

30  i 

840; 

84^ 


^1800 
998 

270 
756 


$3824 
3110 

$  714 
1700 

$2414 


Due  for  rent 

"      salary 
Bills  payable 
Personal  accounts 
J.  Jones'  net  investment 

Total  liabilities 


$  190 

350 

750 

120 

1700 

$3110 


OPERATION  TO  FIND  THE   NET   GAIN  BY  THE  LOSS  AND  GAIN  METHOD. 
Profit    and    Loss. 


To  rent      -------$  240 

"  salary  of  clerk     -----  600 

"  expenses       ------  72 

"  10J'„'  loss  on  personal  accounts  -  84 

"  W'a  loss  on  bills  receivable       -        -  30 


Total  loss 


$1026 


By  Mdse.  - 
"  interest 
"  P.  O.  salary 
"  sundry  services 


Total  gain 
"      loss 


$1500 

36 

180 

24 


$1740 
1026 

$  714 


Net  gain     ----- 

Note. — This  form  of  statement  can  be  made  only  in  cases  where  all  the  items  of  gain  and  loss 
are  given,  or  where  they  can  be  found  from  the  facts  presented. 

Explanation. — In  this  solution  we  have  opened  accounts  with  the  business  man,  and  with  the 
resources  and  liabilities,  and  thus  show  the  details  of  the  business  more  fully  thau  were  shown  by 
the  first  solution.  In  the  solution,  we  have  shown  the  two  methods  of  determining  the  gains  of  a 
business,  and  two  methods  of  showing  the  pn)2)rietor's  net  capital  or  net  worth. 

THIRD   METHOD   OF   SOLUTION. 

The  problem  may  be  solved  by  first  finding  the  net  gain  of  the  business  as  shown  in  the  second 
solution  by  the  loss  and  gain  method.  Then,  having  the  net  gain,  make  the  resource  and  liability 
statement  as  shown  in  the  second  solution,  excej)!  the  item  of  cash,  which  is  yet  unknown.  This 
statement  will  give  resources  $2826  and  liabilities  $3110.  Then  since  there  was'a  gain  of  $714,  it  is 
evident,  from  the  principles  given  in  article  1429  that  the  resources  must  exceed  the  liabili- 
ties by  that  sum,  $714,  which  would  make  the  aggregate  of  the  resources  $3110  -|-  $714  =  $3824. 
Then,  since  the  resources  with  the  item  of  cash  omitte<l,  amount  only  to  $2826,  therefore  the  differ- 
ence between  $3824  and  $2826,  which  is  $998,  must  be  the  correct  cash  balance.  Having  the  cash 
balance  and  the  net  gain,  it  remains  only  to  add  the  net  gain  to  the  proprietor's  net  investment 
and  thus  produce  his  net  worth  at  closing. 


838 


SOULE'S   PHILOSOPHIC   PRACTICAL   MATHEMATICS. 


41.  S.  Hicks  established  a  branch  store  in  a  country  town  and  placed  W. 
Wood  in  charge,  at  a  salary  of  $40  per  mouth,  and  a  commission  of  10%  of  the 
profits.  Rent  of  store  was  $300  jier  year  and  salary  of  assistant  clerk,  $25  per 
month.  S.  Hicks  placed  iu  the  store  merchandise  amounting  to  $2210,  and  handed 
W.  Wood  $100  for  account  of  the  business.  At  the  expiration  of  sixteen  months,  Mr. 
Wood  having  become  dissipated  and  untrustworthy,  a  settlement  was  made.  From 
the  irregularly  kept  books  and  the  inventory  of  goods,  the  followiiig  facts  were 

learned :    Merchandise    on    hand,    $3440 ;    bills    receivable    on    hand,    $3100 ; 

personal  accounts  due  the  honse,  $1050;  three  shares  of  leather  factory 
stock,  $270  5  bills  payable  outstanding,  $4820;  personal  accounts  payable,  $2300, 
not  including  the  clerks'  salaries.  Cash  on  hand,  $1280 ;  paid  on  account  of 
rent,  $400 ;  paid  assistant  clerk  on  account  of  salary,  $350 ;  paid  S.  Hicks,  $1200  j 
W.  Wood  drew  $900.  What  has  been  the  gain  of  the  business?  What  is  the 
net  capital  of  S.  Hicks  ?    Wliat  is  the  condition  of  W.  Wood's  account  ? 

Ans.  1.  Net  gain,  $1040.    2.  S.  Hicks'  net  capital,  $2046, 
3.  W.  Wood  owes  S.  Hicks,  $156. 

SOLUTION. 

OPEEATION  TO  FIND  S.  HICKS'  NET  INVESTMENT. , 

S.    HICKS. 


To  cash,  drawn  out 

To  balance  net  investment 


$1200 
1110 


$2310 


By  Mdse.  invested 

"  cash  " 


$2210 
100 

$231» 


OPERATION  TO  FIND  THE  NET  GAIN  OF  THE  BUSINESS. 
Resources.  Liabilities. 


Mdse.  on  hand $3440 

Bills  receivable 3100 

Personal  accounts  receivable           -        -  1050 

Leather  factory  stock       -        -        -        -  270 

Cash  on  hand            .        .        .        .        .  logo 

W.  Wood        -        -           -        -        -     900 )  9gQ 
Less  salary,  16  months     -        -        -     640  \ 

Total  resources           -        .        -        .  $9400 

Total  liabilities          -        .        -        .  8360 

Net  gain $1040 


S.  Hicks'  net  investment 
Bills  payable    - 

Personal  accounts  - 
Due  for  rent 

Less  amount  paid  - 

Assistant's  salary  - 

Less  amount  paid  - 


Total  liabilities 


$480) 
400  ( 

$400) 
350^ 


$1110 
4820 
2300 

80 


50 


$8360 


OPERATION  TO  FIND  THE  BALANCE  OF 
W.  wood's  ACCOUNT. 


W.  Wood,  salarv  16  months  ®  $;40 
"        10%  of  SilOlO  profit      - 
Total  amount  due 
W.  Wood  has  drawn      -        -        - 

"        ovres  S.  Hicks 


$640) 
104$ 


$744 
900 


$156 


OPERATION    TO    FIND    S.    HICKS' 
CAPITAL. 


S.  Hicks'  net  investment 
"        90%  of  11040  profit 


SlllO 
936 


.S.  Hicks'  net  capital 


42.    Jones  of  New  Orleans,  employs  A.  to  conduct  his  branch  Louse  at  New 
Iberia,  which  bad  been  in  operation  for  a  year,  and  the  books  showed  at  the  last 


PARTNERSHIP    SETTLEMENTS. 


839 


closing  the  following  results,  viz :  Cash  on  hand,  $500 ;  Mdse.,  $2000 ;  bills  receiv- 
able, $1(500;  personal  accounts  receivable,  $900,  and  bills  payable,  $700;  and  per- 
sonal accounts  payable,  $1300, 

The  agent  manages  the  business  for  another  year,  and  presents  the  following 
statement  of  the  Ledger  accounts  at  the  closing  of  the  current  year,  except  the 
cash  account  not  given.  Mdse.  Dr.,  $15000  ;  Cr.,  $14000;  balance  on  hand,  $2400; 
real  estate  on  hand,  $1500;  bills  receivable,  $1200;  personal  accounts  receivable, 
$1000;  interest  account  Dr.,  $80 ;  bills  payable  account  Cr.,  $2000  ;  personal  accounts 
jjayable,  $1000;  commission  account  Cr.,  $50.  What  was  the  cash  balance,  net 
gain,  and  present  worth  ?  Ans.  cash,  $1870.    Net  gain,  $1370. 

Proprietor's  net  worth,  $4370. 
OPERATION  TO  FIND  THE  NET  INVESTMENT. 
Jones'    Account. 


Bills  payable       -         -        . 
Personal  accounts  payaljle 


Total  debit 


$700 
1300 


$2000 


Cash S  500 

M<lse. 2000 

Bills  receivable 1600 

Personal  accounts  receivable  -        -        .  900 

Total  Cr. $5000 

"Dr. 2000 

Net  investment         -        .        .        -        .  $3000 


OPERATION  TO  FIND  THE  CASH  BALANCE. 
Branch   Account. 


Mdse.  Dr.  -  -  -  - 
Real  estate  -  -  -  - 
Bills  receivable  -  -  - 
Personal  accounts  receivable 
Interest  and  discount 


$15000 

1500 

1200 

1000 

iSO 


Total 
To  cash  balance 


$18780 
1S70 


Mdse.  Cr.        -        -        - 
Bills  payable 
Per,sonal  accounts  payable 
Commissions   -         -         - 
Proprietor's  net  investment 


$20650 

OPERATION  TO  FIND  THE  NET  GAIN. 
Resources.  Liabilities. 


$14000 

2000 

1600 

50 

300O 


$20650 


Cash  as  above  -  -  - 
Real  estate  -  -  -  - 
Bills  receivable  -  -  - 
Personal  accounts  receivable 
Mdse.  on  hand         .        -        - 


$1870 
1500 
1200 
1000 
2400 


Total  resources 


Proprietor's  net  investment 

Bills  jiayable 

Personal  accounts  payable 

Total  liabilities 
"     resources 


7970  Net  gain 

OPERATION  TO  FIND  THE  NET  'WORTH. 


$3000 
2000 
1600 

$6600 
7970 

$1370 


Jones'  net  investment 
Jones'  net  gain 

Jones'  net  worth 


$3000 
1370 

$4370 


Explanation.— T\i\s  problem  may  be  solved  in  the  same  manner  as  shown  in  the  several  solutions 
of  problem  40,  and  of  problem  41.     But  to  abridge  aud  simjjlify  the  work,  wo  have  solved  it  by 


840 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


another  method  based  011  the  ultimate  results  of  the  accounts,  which  method  is  applicable  to  all 
similar  problems. 

The  I'dllowiug  directions  render  this  method  clear: 

1st.  Open  an  account  for  the  proprietor,  and  credit  the  same  with  the  cash,  merchandise,  per- 
sonal accounts  receivable,  bills  receivable  and  all  other  property,  furnished  at  couniuMicinj;,  and 
with  any  other  advances  made  durinj;  the  business.  Then  debit  the  account  with  the  bills  payable, 
personal  accounts  payabh^,  and  any  other  oblip;ations  due  on  couuuenciug,  and  for  any  withdrawals 
during  the  business.     The  balance  of  his  account  will  be  Ins  net  investment. 

2d.  Open  an  accoinit  with  the  branch  house  and  debit  the  s^amv,  first,  with  the  debit  biilances 
of  all  the  real  accounts,  except  cash,  which  is  not  known  ;  second,  witli  the  totals  of  all  the  ]iroperty 
and  represi^ntative  aeeouuts.  Then  credit  tl)e  liranch  house  account  first,  with  the  i)roprietor's  uet 
investment;  second,  with  the  credit  balances  of  all  real  accounts;  aud  ;Ai>rf,  for  the  totals  of  all 
projjerty  and  representative  accounts. 

The  sum  of  all  tue  debits,  subtracted  from  the  sum  of  all  the  credits,  should  be  the  b.alauce  of 
cash  on  hand. 

3d.  Make  a  statement  of  all  the  given  net  resources  including  cash,  and  of  all  the  given  net 
liabilities,  including  the  projjrietor's  net  investment.  Their  difi'erence  will  be  the  net  gain  of  the 
business. 

Note. — If  allowiinees  are  to  be  made  for  doubtful  debts,  unpaid  expenses,  rents,  salaries,  etc., 
for  which  no  entries  have  been  made,  and  for  licenses,  taxes,  insurance,  etc.,  that  have  been  paid, 
but  whose  time  has  not  expired,  make  the  necessary  adjusting  entries  in  the  list  of  resources  and 
liabilities 


43.  A.  M.  Perdue,  a  merchant  in  Pine  Bluff,  Arkansas,  engages  M.  J.  Mul- 
ler  to  take  charge  of  his  branch  store  at  Linuwood.  The  following  assets  and  liabil- 
ities are  shown  by  a  statement  made  the  day  M.  J.  Miiller  assumed  charge  :  Cash, 
$275;  merchandise  in  stock,  $2990;  bills  receivable,  $180;  personal  accounts  Or., 
$340:  bills  payable,  $210.  For  conducting  the  business,  M.  J.  Muller  is  to  receive 
$40  per  month ;  rent  of  store  is  to  be  $20  per  month.  At  the  end  of  fifteen  months, 
Perdue  wishes  to  sell  out  his  business,  and  oti  investigating  his  afl'airs,  finds  the 
following  state  of  things:  Insurance  p.aid,  $102.50;  sundry  expenses  paid,  $135; 
one  Jefferson  Co.  bond  ($100),  worth  $90 ;  merchandise  bought,  $3450 ;  merchandise 
sold,  $5905;  merchandise  on  hand,  $3245;  rent  paid,  $300;  salary  jiaid,  $500;  per- 
sonal accounts  Dr.,  $475 ;  personal  accounts  Or.,  $375 ;  bills  receivable,  $325 ;  bills 
payable,  $290 ;  paid  interest  and  discount,  $12.50 ;  received  interest  and  discount, 
$10.  Perdue  has  drawn  out  $200.  He  cancels  his  policy  of  insurance  and  is 
allowed  $55,  return  premiums.  Find  gain  of  business,  i)resent  net  capital  of  A.  M. 
Perdue,  and  cash  on  hand  ?  Aus.  Cash  on  hand,  $890.     Net  gain,  $1565. 

Net  capital,  $4200. 


First  Operation   by  the   Method   shown    in   Problem  42. 
Perdue's    Account. 


Person.al  accounts  payable 
Bills  payable 
Withdrawal      -        -        - 


Total  debit 


340 
210 
200 


$750 


Cash $  275 

Merchandise 2990 

Bills  receivable        -----  180 

Total  credit 13445 

"      debit         -----  750 

Net  investment          -        -        -        -  $2695 

Net  gain  as  below       -         -         .        -  1565 

Net  capital  at  closing        -                -  $4260 


PARTNERSHIP  SETTLEMENTS. 

Branch  Account. 


841 


Insurauco  paid         .        -        - 

-     $  162.50 

Insurance  returned 

-    $    55.00 

Suinlry  expenses  paid     - 

135.00 

Merchandise             .         .        - 

-      5905.00 

Jeft'ersou  Co.  bond 

90.00 

Personal  ar<'ounts  Cr.     - 

375.00 

Mdse.  bought           -        -        . 

-      3450.00 

Bills  payable 

290.00 

Kent  paid         .        -        -        . 

300.00 

Interest    -         -         -        .         - 

10.00 

Salary  paid      -        .         -        - 

500.00 

Perdue's  net  investment 

-        -      2695.00 

Mdso.  at  commenciug 

-      2990.00 

Personal  acconnts  receivable 

475.00 

Bills  receivable        -        -        - 

325.00 

Interest  and  discount 

12.50 

Total           ...        - 

-    $8440.00 

To  cash  balance         ... 

800.00 

$9330.00 

$9330.00 

Resources. 


Liabilities. 


Ca.sh  as  above 
Mdse.  on  hand 
Personal  accounts  receii 
Jetferson  Co.  bond   . 
Bills  receivable 

'able 

$  890 

3245 

475 

90 

325 

Perdue's  net  investment 
Bills  payal)le              .         .         - 
Personal  accounts  payable     - 
Salary  due                 ... 

Total  liabilities 

- 

$2695 
290 
375 
100 

Total  resources 
Less  liabilities 

$5025 
3460 

Net  gain    - 

$1565 

$3460 

Second  Operation  by  the  Detailed  Account  Method. 

Perdue's  Capital  Account.  Opeisatiox  to  yi.xD  Cash   Received   and 

Paid  ox  Accolnt  of  Merchandise. 

Merchandise. 


"Withdrawal    -     -     $  200 
Net  capital       -     -       4360 

Investment 
Gain    ... 

.    $2895 
-     .     1565 

$4460 

$4460 

Merclia 

idise. 

At  commencing   .     $2990 
Bought     ...     -     3450 
Profit  and  loss  -    -    S710 

Sold  -    .     -    - 
Inventory  .    - 

.    $5905 
-       3245 

$9150 

$9150 

Cash. 


On  hand      -    .    .    |  275 
Insurance  returned      55 
Interest  received     -       10 
Bec'd    from    Mdse. 
Bales       ....  5195 


$5545 


Expenses  paid     $ 
Rent  paid       -     - 
Perdue  withdrew 
Insurance  paid 
Salary  paid    -     ■ 
Mdse.  paid    - 
Interest  paid 
Balance  on  hand 


135.00 
300.00 
200.00 
162.50 
500.00 
3345.00 
12.50 

soo.oo 


$.5545.00 


Mdse.  bought  $3450 

Pd.  by  bills  pay.  $80 
Due  on  ])ersonal 
accounts     -       35 

—     115 


Mdse.  sold       -    -    $5905 
Rec'd  bills  rec.  $145 

"     bond        -     90 
Personal  acct's 

due     ...    475 

710 


Paid  in  cash    -        $3345  Received  cash  $5195 

Bills  Payable. 


On  couuuencing         $210 
"  closing    -     -  290 


Increase 


$  80 


Int.  and  Dis(Pt. 


Paid  $12.50 


Rec'd     $10 


Rent. 


Insurance. 


Pd.  $162.50 


Ret'd      $55 


$300 
Salary. 


Paid     $500 
Due         100 


Bond. 


Expense. 


$135  \ 


84.2  soule's  philosophic  practical  mathematics.  * 

Bills  Receivable.  Personal  Accounts  Cr.  Personal  Accounts  Dr. 


On  coninicuc'g  $180 
At  closini;     -       325 


Increase 


$145 


On  coninienc'g  $340 
"  I'losiui;     -       375 


Increase 


$  45 


On  closing     -     $475 


Profit  and  Loss. 


Rent  -        -  $300.00 

Salary      -        -     600.00 
Interest    -        -       12.50 

$162.50  I 
Insurance  55.50 )  107.50 
Expenses       -         135.00 
Perdue' s  i/ain        1505.00 


$2720.00 


Milse.  gain 
Interest 


$2710.00 
10.00 


$2720.00 


Pesources  and  Liabilities. 


Cash         -        -      $  890 
M.lse.         -        -        3245 
Hills  receivable          325 
Hond         -        -            90 
Personal  accounts     475 

Perdue'suet  inv.     $269.5 
Bills  ])ayable               290 
I'ersonai  accts.  Cr.    375 
Salary  due        -          100 

$3460 
Net  gain        -            1565 

$5025 

$5025 

44.  A  Temperance  town  placed  iu  the  hands  of  a  druggist,  who  is  to  act  as 
agent,  liquor  amounting  to  $415  and  cash  $50.  The  agency  continued  fifteeu 
nioTiths.  On  settlement,  it  was  found  that  the  agent  had  purchased  goods  amount- 
ing to  $4600 ;  sold  goods  $5810,  and  had  on  hand,  which  he  turned  over  to  the  town, 
authorities,  goods  amounting  to  $1005.50.  The  agent  was  to  receive  a  salary  of  $20 
per  month  aiul  10 f^  of  the  gross  profits.  What  does  the  agent  owe  the  town? 
What  was  the  net  gain  of  the  town  ?  Ans.  1.  Agent  owes  the  town  $779.05. 

2.  Net  gain  of  the  town  Is  $1320.45, 

First  Operation  to  find  the  Balance  Due  the  Town. 


Dr. 

Druggist. 

Cr. 

To  merchandise     - 

"  cash           .        .        .        . 
"  gain  on  sales     -        -  ■     - 

-      $  415.00 

50.00 

1800.50 

$2265.50 

By  merchandise  returned    - 
"  15  months'  salary    - 
"  lOj"^  of  profits 

By  balance  due  totvn 

- 

$1005.50 
300.00 
180.05 

$1485.55 
779.95 

$2265.50 

Second  Operation  to  find  the  Balance  Due  the  Town. 

Druggist. 


To  ca.sh,  at  commencing 
"      "      merchandise  sales 


$     .50.00 
5810.00 


$5860.00 


By  cash,  merchandise  bought 
"  15  months' salary    - 
"  10%of  profits 

By  balance  due  the  town 


$4600.00 
300.00 
180.05 

779.95 

$5860.00 


OPERATION  TO  FIND  THE  GAIN  ON  MERCHANDISE. 
Mercliaiadise. 


Merchandise  on  commencing    - 
"           purchased     - 

-  $  415 

4600 

-  $5015 

Merchandise  sales        .        -        - 
"           inventory 

Sales  and  inventory    - 

Total  cost      -        -        -        - 

Gain  on  merchandise    - 

$5810.00 
1005.50 

Total  cost             .        .        .        . 

$6815.50 
5015.00 

$1800.50 

PARTNERSHIP    SETTLEMENTS. 
OPERATION  TO  SHOW  TOAVN'S  NET  GAIN. 

Profit   and.   Loss. 


843 


Salary     .        -        .        -        - 
lOJo  commission  ou  gross  gaiu 
Town's  net  (jain       .        -        - 


$300.00 

180.05 

1330.45 

$1800.50 


Gaiu  ou  merchandise    - 


$1800.50 


$1800.50 


Explanation. — The  above  problem,  lil<e  problem  39,  merits  the  critical  attention  of  the 
accountant.  The  operations  are  made  so  explicit  that  we  deem  an  extended  explanation  unnecessary, 
for  tliose  familiar  with  accouuts.  Parties  who  are  not  familiar  with  accounts  are  not  expected  to 
comprehend  the  work. 


45.  A.,  B.  and  C.  bought  a  lot  of  merchandise  on  speculation  for  .$3000.  A. 
paid  SIjOO,  B.  jiaid  .$900  and  C.  paid  .$G00.  It  was  agreed  that  each  one's  interest  in 
the  goods  should  be  in  proportion  to  the  amount  of  money  invested.  They  then  sold 
to  D.  oue-fourth  interest  in  the  goods  for  $1800,  and  A.,  B.  and  C  agreed  to  .so  adjust 
the  matter  that  each  of  them  should  be  one-third  owner  in  the  remainder  of  the 
goods.     What  was  the  financial  adjustment  between  A.,  B.  and  C.  ? 

Ans.  C.  paid  $360.    A.  received  $1800. 
B.  received  $3G0. 

SOLUTION. 


OPERATION  TO   PINT)    THE    INTEREST   SOLD  TO   D., 
THE    INTEREST    THAT    A.    AND    B.    SOLD    RE- 
SPECTIVELY AND  THE  ADDITIONAL  INTER- 
EST  BOUGHT   BY   C. 


Amount 
paid. 

A.    $1500 

Interest 
owned. 

H    - 

Interest 
retained. 

iDtereat 
sold  by 

Interest 

bought 

byC. 

B.        900 

3TT         

^%    = 

TZU 

C.        600 

^ 

n 

m    = 

$3000 

U  H-    4 

=  m  = 

the  interest  sold  to  D. 

an  _  ^s^j!^  _  ^>^n^  _  i„t.  retained  by  A.,  B.  &  C- 
TiS  -r    3     =  -i^u  =  int  of  A., B.  &C.  respectively 


OPERATION  TO  FIND  WHAT  C.  MUST  PAY  FOR  -rfff 

AT  THE   RATE   OF  $1800    FOR    J    INTEREST  IN 

$3000,    IN  ORDER  TO  OWN  -pi^  OR  i  INT. 

IN  THE  REMAINING   GOODS. 


120 


$3G0 
or  thus : 

•    T2il   • 


OPERATION  TO  FIND  THE  AMOUNT  OF  MONEY  DUE   A.   AND   B.   RESPECTIVELY    FOR 
THE  INTERESTS   SOLD  TO  D.  AND  C. 


Amount  received  from  D. 
Amount  received  from  C. 

Total  amount  received  is 


$1800 
360 

$2160 


which  is  to  be  divided  between  A.  and  B.  in  i)ro- 
portiou  to  the  interest  each  sold. 


A.  sold  as  above 
U.  sold  as  above  - 

Total  interest  sold 


A. 


36 
120 


$2160 
120 
30 

$1800 


B. 


36 
120 


$2160 
120 
6 


-rftr 
1% 


844  soule's  philosophic  practical  mathematics.  * 

4G.  A  nuinufactm-iiif,'  corporatiou  lia.s  an  agency  in  a  distant  city,  for  the 
sale  of  its  goods.  At  the  close  of  the  year,  December  31st,  1891,  the  agency  showed 
the  following  inventoiy  or  statement : 

Mcrcliandise  in  stock          .        .        -        -        -                 -        -  $256,897.00 

Accounts  receivable    ---------  102,10.5.00 

Cash  on  hand -        -         -         .  2,220.40 

Store  tixtures       -        - 4,500.00 

1365,722.10 
December  31st,  1892,  the  agency  presented  the  following  resource  statement : 

Mercliaudiso  in  store           ._.-----  $238,305.00 

Accoiints   receivable 110,300.00 

Cask              -        - 3,515.15 

Store  fixtures      ----- 4,200.00 

$356,320.15 

During  the  fiscal  year,  the  agency  received  from  the  manufacturing  com^iany 
for  expenses  $30000.  and  remitted  to  the  company  $285,427.  The  sales  of  goods  for 
the  year  were  $452,512.  Allowing  the  expense  of  the  business  and  the  loss  oji  sales 
for  bad  debts  to  be  equal  to  183%  of  the  sales,  what  has  been  the  net  gain  of  the 
agency  ?  ^"S.  $iCl,178.75. 

OPERATION 


Agency  at-... 


To  Mdse  Dec.  31,  1891  -        -        $256,897.00 

"   accounts  receivable,  Dec.  31, 1891      102,105.00 


cash 

store  fixtures 


"   cash  recei\'ed  from  Mfg.  Co. 


2,220.40 
4,500.00 


$365,722.40 
30,000.00 

$395,722.40 


By  Mdse.  Deo.  31,  1892            -        -  $238,305.00 

"  accounts  receivable,  Dec.  31,  1892  110,300.00 

"  cash             -----  3,515.15 

"  store  fixtures      -        -        -        -  4,200.00 


"  cash  remitted  to  Mfc.  Co. 


Expenses  and  loss,  18i5'o  on  |;452,512  ; 
Net  gain  of  Agency 


Gain 


$3.56,320.  l.-j 
28,5,427.00 

$641,747.15 
395,722.40 

$246,024.75 
84,846.00 

$161,178.75 


Manufacturing  Company  Account  at  the  Agency. 


To  net  inventory,  Dec.  31,  1892 
"  cash  remitted    -        -        - 

Total  debit 

"      credit      -        -        - 

Net  debit  -        -        -         - 

Less  expense  as  above     - 


Net  debit 


$356,320.15 

285,427.00 

$641,747.15 
395,722.40 

$246,024.75 
84,846.00 

$161,178.75 


By  net  inventory,  Dec.  31,  1891    -        $36.5,722.40 
•"'  cash  receiveii      -        -        -        -  30,000.00 

Total  credit       -        -        .        .        $395,722.40 


PARTNERSHIP    SETTLEMENTS. 


845 


47.  A  firm  sold  goods  at  70%  discount  on  list  price,  the  net  sales  amounting 
to  $200,(100.  The  net  gain  on  the  sales  was  $00000.  They  now  wish  to  allow  75% 
discount  on  list  price  and  increase  the  sales  so  as  to  produce  the  same  amount  of 
gain.    What  will  be  the  amount  of  net  sales,  the  amount  of  list  price,  and  the  cost  ? 

Ans.  $375,000  net  sales.     $1,500,000  list  price.     $315,000  cost. 

OPERATION. 

1.  Find  the  list  price  of  $200,000  uet  sales  at  70%  discount.  30  :  100  :  :  200,000  :  S666,666J  = 
the  list  price. 

2.  |666,666|  list  price— TO^o'  =  $200,000  net  sales,  giving  a  gain  of  S60000,  and  a  cost  of 
$140,000. 

3.  $666,666f  list  price— 75,^   =  {il66,666f  net  sales,  —  $140,000  cost  =  $26666|  gain. 
Having  now  produced  the  gain,  S26666j,  that  would  have  been  realized  on  the  net  sales  of 

$166,666f  at  a  discount  of  735'o  on  list  price,  we  divide  the  gain  that  is  required,  whatever  it  may  V)e, 
in  this  case  $60000,  by  the  amount  of  gain  realized,  in  this  case  S26666j,  and  in  the  quotient  we  have 
the  ratio  between  the  gain  realized  and  the  gain  required.  Thus,  to  further  elucidate,  $60000  -i- 
$26666f  =  2i;  which  shows  that  the  gain  realized  at  75%  discount  ($26666j)  is  but  2J  part  of  the 
gain  required  ($60000).  It  also  shows  that  the  net  sales  which  produced  the  S26666f  gain,  are  but 
2i  part  of  the  net  sales  required  to  give  2J  times  as  much  gain.     Hence  we  multiply  the  $166,666j- 

NET  SALES  BY  2^  AND  PRODUCE  $375,000,  THE  NET  SALES  REQUIRED. 

Now,  since  the  net  sales  are  only  25%  of  the  list  price,  therefore  4  times  the  net  sales  will  be 
the  list  price,  which  is  $375,000  X  4  =  $1,500,000  or,  thus :  100  —  75%  =  25,  and  then  25  :  100  :  : 
$375,000  :  $1,500,000. 

PROOF. 
$1,500,000  list  price  —  75%  =  $375,000  net  sales  —  $60000  gain  =  $315,000  cost. 

SECOND  PROOF. 

$140,000  cost  gained  $60000  =  426%. 
$315,000  cost  gained  $60000  =  19/,%. 

42S%  -1-  192^%  =  -i  which  is  the  ratio  between  the  gain  when  70%  discount  was  allowed  and 
the  gain  when  75%  discount  was  allowed,  as  shown  above. 

ADJUSTMEXT  OF  INTEREST  OX   PARTNERS'  ACCOUNTS. 

Many  of  the  following  questions  involve  in  their  solution  the  principles 
of  the  science  of  Double  Entry  Book-keeping  as  well  as  the  principles  of  the  science 
of  numbers,  and  are  therefore  of  special  importance  to  young  accountants  and  mer- 
chants who  aspire  to  eminence  and  success  in  the  line  of  their  professions. 

48.  J.  M.  Bracey,  T.  A.  Hammons,  and  T.  G.  Mackie  are  partners.  They  share 
gains  and  losses  equally,  and  by  the  conditions  of  the  partnership,  an  account  cur- 
rent and  interest  account  is  to  be  kept  with  each  partner,  at  8%  ;  i.  e.,  interest  at  8% 
is  to  be  allowed  on  the  investments,  and  charged  on  the  withdrawals  of  each  part- 
ner, by  which  means  the  investments  of  each  partner  are  equalized. 

At  the  end  of  the  year,  when  they  wish  to  close  their  books,  it  is  found  that 
there  is  interest  on  the  debits  and  credits  of  the  partners'  accounts,  as  follows : 


Dr. 

Cr.                Net  Cr. 

J.  M. 

Bracey 

$750.00 

$2330.00    =    $1600.00 

T.  A. 

Hammons     - 

515.50 

1740.00     =       1224.50 

T.  G. 

Mackie 

160.75 

641.20     =         480.45 

Total  credit  of  interest $3304.95 


846  soule's  PHiLosorinc  practical  mathematics.  ♦ 

"VTliat  is  the  correct  Journal  Entry  to  be  made  iu  the  books  of  the  firm,  in 
order  to  proiierly  adjust  the  interest  between  the  partners  ? 

OPERATION. 
Amount  of  credit  interest   -----        $3304.95 


Each  partners'  one-third  of  which  is  -        -        $1101.65 

Dr.  Cr.  Dr.  Or. 

J.  M.    Bracey        -        -        -         $1101.65  $1600.00  =  $498.35 

T.  A.  Hamm'ous    -        -         -           1101.65  1224.,50  =  122.85 

T.  G.   Mackio        -        -        -          1101.65  480.45  =  $621.20 


$621.20  ,      $621.20 


From  these  figures,  the  following  entry  is  made : 


T.  G.  Mackie  To  Sundries  $621.20 

To  .J.  M.  Bracey  -        -        $498.35 

"  T.  A.  Hamm'ons 122.85 

The  above  entry  is  the  only  correct  and  proper  one  that  can  be  made  to 

adjust  the  interest  that  accrues  on  the  partners'  investments.     But  the  general  way 

of  adjusting  i)artiiership  interest  of  this  kind  is  by  passing  it  through  profit  and 

loss  account.     This  is  done  by  the  following  entry : 

Interest  (or  Profit  and  Loss)  To  Sundries      -        -        $3304.95 

To  J.  M.  Bracey     -        - $1600.00 

"  T.  A.  Hamiiions -        -  1224.50 

"  T.  G.  Mackib    -----------  480.45 

This  entry  produces  the  same  result,  so  far  as  the  partners  are  concerned,  as 
the  correct  one,  but  by  it  a  false  loss  is  represented,  and  hence  it  is  not  a  prosier 
entry  to  make. 

It  is  clear  to  any  accountant  that  no  loss  can  be  sustained  by  a  firm  unless 
tne  capital  of  that  firm  is  thereby  decreased,  and  it  is  also  clear  that  while  partner- 
ship interest  efl'ects  changes  in  the  investments  of  the  different  members  of  the 
firm,  it  does  not  decrease  the  capital  of  the  firm,  and  therefore  no  loss  has  occurred 
by  reason  of  i)artnership  interest,  and  consequently  the  books  should  not  represent 
a  loss. 

It  is  a  sorrowful  fact,  not  generally  known,  that  many  merchants  of  reputed 
respectability,  for  the  purpose  of  showing  small  profits  in  their  business,  charge  their 
interest  accounts  -with  12,  15,  or  20%  interest  on  capital,  and  their  expense  account 
■with  partners'  salaries  sufiiciently  large  to  absorb  nearly  all  of  the  gain. 

This  practice  has  been  indulged  in  to  such  an  extent  by  the  non-observers  of 
the  in'inciples  of  ethics,  as  to  bring  disgrace  upon  honorable  merchants  iu  aU 
sections  of  the  country.    The  practice  should  be  discontinued. 

49.    The  following  jjroblem  was  presented  by  a  teacher  for  solution : 

A.,  B.  and  C.  form  a  co-partnership  under  the  following  conditions  :  A.  is  to 
manage  the  business,  and  to  receive  therefor  $2400  per  annum,  which  amount  is  to 
be  credited  on  July  1st.  He  is  to  receive  interest  on  his  salary  and  to  pay  interest 
on  sums  withdrawn  at  the  rate  of  G%  i^er  annum.  B.  and  (J.  are  to  furnish  the  cap- 
ital, and  to  receive  interest  therefor  at  the  rate  of  G%  per  annum ;  the  net  gain  or 
loss  to  be  divided  equally.  B.  invests,  January  1st,  $10000;  April  1st,  $.5000.  G. 
invests,  January  1st,  $10000;  July  l.st,  $5000,  and  draws  out,  September  IGth,  $500. 
A.  draws  out,  February  1st,  $200 ;  March  1st,  $100 ;  July  11th,  $500 ;  October  1st, 


PARTNERSHIP   SETTLEMENTS. 


84; 


I 


$200 ;  November  21st,  $100.  At  the  end  of  the  year  the  gain — without  taking  into 
account  either  the  sahxry  to  be  paid  to  A.  or  the  interest  on  the  partners'  accounts — 
is  $84:37.10.  What  will  be  the  balance  of  each  jiartner's  account  when  all  the  items 
have  been  jn'operly  entered  ? 

Note. — Interest  to  bo  computed  iu  moutlis  of  30  days. 

Ans.  A's  credit  balance,  $2505.75.    B's  credit  balance,  $17307.58. 


C's  credit  balance,  $16723.83 


OPEEATION. 


accoujStt  cueeext  and  interest  account  with  paetnees. 

Dr.  a.  Cr. 


Fel)ru.ary  1,  11  months  $11.00  interest 


$48.84 


Dr. 


March  1, 

10 

it 

20.00 

n 

400 

July  11, 

5't 

tl 

14.17 

li 

500 

October  1, 

3 

it 

3.00 

ii 

200 

Nov.  21, 

IJ 

i( 

67 

a 

100 

$1400 


July  1,  6  months,  -        $72.00  int. 

By  balance  of  interest    48.84 


"  one-third  net  gain 


12400.00 
23.16 

1482.59 


5.75 
1400.00 

$2505.75 

Cr. 


Dr. 


January  1,  12  months  $600  interest 
Aprill,  9      "  225      " 

By  interest      .        -        -        - 
"  net  gain     -        -        .        - 


c. 


$10000.00 
5000.00 

825.00 
1482.58 

$17307.58 

Cr. 


September  16,  3i  months  $8.75  interest       $500.00 


.Tanuary  1,  12  months,  $600  interest 
.Tuly  1,  6        "         150 

$750 
By  balance  of  int.        8.75 


"  one-third  net  gain 


$10000.00 
5000.00 


741.25 

1482.58 

$17223.83 
500.00 

$16723.83 

OPERATION  TO  FIND  THE  NET  GAIN. 

Total  gain  of  the  business  is --        $8437.16 

From  this  deduct  as  follows : 

Balance  of  interest  due  A..         -     $  23.16 
"  "  "         B.        -       825.00 

"  "  '        C.        -       741.25  $1589.41 

Salary  ^aid  to  A.       ------        -        2400.00  3989.41 

2Iet  gain  of  business  ---..-....        $4447.75 

A's  one-third  net  gain  is $1482.59 

B's  one-third  of  net  gain  is-----.-..  1482.58 

C's  one-third  of  net  gain  is-----.--.  1482.58 


848 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Second    Operation   to   adjust  the    Interest   and    Gain. 


A. 
B. 
C. 


Int.  Pr. 


8.75 
Total  interest  Cr. 


Int.  t'r. 

8L'5.()0 
750.00 


Net  Int.  Cr. 
$  2:ilt;        529.80 
825.00        529.80 
741.25        529.81 


Dr. 

$500.64 


Cr. 

$295.20 
211.44 


Eaeli  partner's  one-tlilrd  int. 


J. $1589.41 
$529. 80i 


$506.64 


$506.64 


Total  gain        -        -        -        - 
From  which  deduct  salary  of  A. 


$8437.16 
2400.00 


$6037.16 
2012.381- 


Ket  gain,  exclusive  of  interest  on  partners'  accounts 
Each  partner's  one-third  of  net  gain       .        .         . 

By  this  method  of  adjusting  the  matter,  the  partners'  accounts  will  stand  as  follows  : 
Pr.  A.  Cr.         Dr.  B.  Cr.        Dr.  C. 


Cr. 


$200 
400 

500 
200 
100 
506.64 


$2400 
2012.39 


4412.39 
1906.64 


$2505.75  Cr. 


$10000 
5000 
295.20 
2012.38 

$17307.58  Cr. 


$500 


$1906.64 

50.    A.  and  B.  form  a  partnership,  each  investing  $15,000. 


$10000 
5000 

211.44 
2012.39 

$17223.83 
500.00 

$16723.83  Cr. 


A.  is  to  manage 
ehe  business,  and  B.,  who  is  to  be  absent,  agrees  to  pay  a  book-keeper  to  take  his 
place  in  the  house  at  $1200  per  aunimi.  The  gains  and  losses  are  to  be  divided 
equally.  The  business  was  conducted  for  fifteen  months,  diu'ing  which  time  the  book- 
keeper drew  his  salary  and  charged  the  same  to  expense.  The  books  were  closed 
at  the  expiration  of  the  fifteen  mouths.  What  entry  should  be  made  to  adjust  the 
salary  of  the  book-keeper  ?  Ans. 


FIRST  FORM   OP  ENTRY. 


To  A. 


$750 


$750 


For  one-half  of  $1500  salary  of  book-keeper, 
charged  to  expense  instead  of  to  B. 


SECOND  FORM   OF   ENTRY. 

B.        -----        -        ,$1500 

To  Profit  and  Loss,  (or  Expense) 


$1500 


For  salary  of  hook-keeper,  the  same  having 
been  charged  to  expense  when  it  should  have 
been  charged  to  B. 


51.  C.  and  D.  are  partners,  equal  in  gains  and  losses.  C.  is  to  keep  the  books 
and  manage  the  financial  affairs,  for  which  he  is  to  receive  $200  i^er  month.  D.  is 
to  superintend  the  purchasing  and  selling  of  goods,  for  which  he  is  to  receive  $150 
per  month.  At  the  close  of  two  years,  there  was  a  cash  gain  of  $9000,  exclusive 
of  the  amount  paid  on  account  of  salaries.  How  much  is  due  to  each  partner,  C. 
having  drawn  $2000  salary,  and  D.  having  drawn  $2500  salary? 

Ans.   Due  to  C.  $5350.    Due  to  D. 

OPERATION. 


Gain  as  above  .        .        .        . 

C's  salary  24  months  ®  $200        = 

Less  amount  received  on  account 


$4800 
2000 


C's  balance  due  on  salary  ... 

D's  salary  24  months  ®  $150  = 

Less  amount  received  on  account 


2500 


D's  balance  due  on  salary  .... 

Total  liabilitv  of  the  firm  for  salaries 


1100 


$3900' 


Net  gain  to  lie  equally  divided  between  C.  and  D. 


$5100 


PARTNERSHIP  SETTLEMENTS. 


849 


C's  ono-half  net  gaiu  is 
"  l)alauce  of  salary  is 

"  interest  in  the 


,f;2550 
2800 

$5350 


D's  one-half  net  gaiu  is 
"  balance  of  salary  is 

"  interest  in  the  $9000  is 


C's  interest  $5350  +  D's  interest  $3650  =  $9000. 


$2550 
1100 

$3650 


52.  W.,  X.,  Y.  and  Z.  are  partners.  An  interest  account  is  to  be  kept  witli 
each  partner,  for  bis  investments  and  withdrawals.  W.  is  to  sliare  one-half  of  the 
gains  and  losses,  and  X.,  Y.  and  Z.  are  to  share  eqnally  the  remaining  one-half. 

At  the  close  of  the  tiscal  year,  tlie  account  current  and  interest  accounts  of 
the  partners  show  the  following  interest  balances  :  W.  $1000  credit ;  X.  $500  debit ; 
Y.  $1700  credit ;  .and  Z.  $400  debit.  "VYhat  is  the  proper  Journal  entry  on  the  books 
of  the  firm  to  adjust  the  matter  ?  Ans. 


FIRST  FORM  OP  ENTRY. 

Sundries  to  Sundries 


X. 

z. 


$800 
700 


To  W. 
"  Y. 


$  100 
1400 


Interest  to  Sundries 

To  W.     - 

"  Z.       - 


SECOND  FORM  OF  ENTRY. 

$2700 


Sundries  to  Interest 
X.  ... 

z.  .       .       . 


500 
400 


$1000 
1700 

900 


RESOURCES : 


Due  by  X.       - 
•'      Z.      - 

Total  interest  resources 


$500 
400 


OPERATION. 
LIABILITIES  : 


Due  to  "W. 
"     Y. 


$1000 
1700 


Total  int.  liabilities         $2700 
"       "      resources  900 


Net 

a 

liabilities 

Ws 

^ 

int. 

liabilities 

X's 

1 

" 

it 

Y's 

IC 

It 

Z's 

i 

It 

It 

$900 
300 
300 
300 


partners'  int.  accounts: 


Ws  int.  credit 
"      "     liability 

"  net  credit  of  int. 

X's  int.  debit 
"       "    liability 

"  total  debit  of  int. 

Y's  int.  credit 
•'       "    liability 

"  net  credit  of  iut. 

Z's  int.  debit 
"       "    liability 


$1000 
900 

$100 

$500 
300 


$1700 
300 

$1400 

$400 
300 


'    debit  of  int.  -  $700 

Note. — We  regard  the  first  form  of  entry  as  the  better  method  of  adjusting  the  interest  on 
partners'  accounts.     See  problem  48. 

53.  Jones  and  Smith  are  partners,  Jones  two-thirds  interest  and  Smith  one- 
third  interest  in  the  gains  and  losses.  An  account  current  and  interest  account  is 
to  l>e  kept  with  each  partner.  At  the  close  of  the  fiscal  year,  Jones  has  $800  net 
debit  of  interest,  and  Smith  has  $100  net  debit  of  interest.  1  What  entry  is 
required  ou  the  books  to  adjust  the  matter  between  the  partners  f  2.  In  case  the 
firm  had  dissolved  and  settled  all  matters  except  the  interest,  what  settlement 
should  they  make  between  themselves  1 

Ans.  1.  Jones  to  Smith  $200.     $200. 

2.  Jones  should  jiay  Smith,  out  of  his  private  funds,  $200. 

OPERATION. 

Dr.  Cr.  Dr.  Cr. 

Jones  owes        ....        $800        —        $600        =        $200 
Smith  owes       ....  100        —  300        =  $200 

Total  resources        -        -        $900 

Jones'  two-thirds  interest  is  $600 

Smith's  one-third  interest  is  300 


85o 


SOULES    PHILOSOPHIC    PRACTICAL    iM.\TlIE.NL\TlCS. 


54.  Folu-  mei'liaiiics,  W.,  X.,  Y.  and  Z.,  form  a  partuership,  for  carryiug  ou 
their  trade.  Tbey  are  to  share  gaiius  and  losses  as  fallows :  W.  g,  X.  |,  Y.  |,  Z.  J. 
They  have  agreed  that  for  all  lost  time  they  shall  be  cliarged  respectively  as  follows : 
W.  $4  iier  day,  X.  and  Y.  $3.50  each  per  day,  aud  Z.  $3  per  day.  At  the  end  of  the 
first  year,  it  is  found  that  they  have  lost  time  respectively  as  follows:  W.  42  days, 
X.  31  days,  Y.  53  days,  Z.  27  days.  What  is  the  correct  entry  on  the  books  of  the 
firm  to  equitably  adjust  the  matter  among  the  partners  ?  Aus. 

Sundries  To  Sundries, 

y.         -        - $49.75 

Z. -        .  13.12 


To 

\\.            ... 
X.             ... 

. 

- 

- 

27.25 

OPEEATION. 

w. 

X. 

Y. 
Z. 

owes  for  42  days  at  $4             = 
"       "    31        "        3i-             = 
"      "    53        "        Si             = 
"      "    27        "        3               = 

Total  amount  due  tlio  iiartnei 

1168.00 

108..")0 

185..50 

81.00 

Ws  three.eightbs  of 

X's  two-eigliths 

Y's 

Z's  one-eighth 

amount  il 

U                       it 

It               tl 

ue  is 

- 

$203.62 
135.75 
135.75 

67.88 

s,         $543.00 

$543.00 

W.        -        -        -        - 
X.         .... 
Y.         .... 
Z.          .... 

Dr. 

1168.00 

108.50 

185.50 

81.00 

Cr. 

$203.62        = 

135.75        = 

135.75        = 

67.88        = 

Dr. 

$49.75 
13.12 

Or. 

$35.62 
27.25 

From  these  figures,  we  make  the  foregoing  correct  adjusting  entry  as  shown  iu 
the  answer. 

55.  A.  and  Z.  are  partners,  equal  in  gains  and  losses.  After  business  had 
been  conducted  for  four  months,  it  was  agreed  to  admit  W.,  and  not  wishing  to  close 
the  books  at  that  time  it  was  agreed,  after  careful  investigation  and  calculation,  to 
allow  A.  and  Z.  $3000,  as  the  net  gain  of  the  business  up  to  that  time;  and  that  the 
gains  and  losses  of  the  new  partnership  shall  be  shared  equally  (j^)  by  each  partner. 
What  entry  is  necessary  to  adjust  the  $3000  gain  between  A.  and  Z.  ? 

Ans. 
Profit  and  Loss  To  Sundries        .        -        .        $3000 

To  A.  ....  $1500 

To  Z.  ....  1500 

(With  explanation.) 

56.  A.,  B.  and  C.  are  partners,  At  the  close  of  the  year,  A.  retires  from  the 
business.  Prei^aratory  to  closing  the  books  and  making  a  final  settlement  with  A., 
it  is  agreed  to  allow  10%  discount  ou  all  personal  accounts  due  the  firm,  and  5% 
discount  on  all  the  bills  receivable,  as  doubtful  or  bad  debts.  The  personal  accounts 
due  amount  to  $21281.  The  bills  receivalile  amount  to  $14500.  What  entry  is 
necessary  to  adjust  this  anticipated  loss  ?  Ans. 


Profit  and  Loss,  old  account         ...---.- 

To  Profit  atid  Loss,  new  account      ....        - 

For  10%  allowed  as  worthless,  ou  $21281  personal  accounts  $2128.10 

"    5%    allowed  as  worthless  on  $14500  bills  receivable  725.00 

Total  amount  $2853.10 

From  these  figures,  we  make  the  adjusting  entry  as  above. 


$2853.10 


$2853.10 


*  PARTNERSHIP    SETTLEMENTS.  85 1 

57.  X.,  T.  and  Z.  are  partners  iu  the  wholesale  grocery  aud  commissiou 
business.  At  the  end  of  their  fiscal  year,  when  they  desire  to  close  their  books  and 
ascertain  their  actual  gain  or  loss,  they  find  that  "  Com.  Sales  "  account  has  a  credit 
of  84S."iO(>,  resulting-  from  the  jiartial  sales  of  consignments  which  are  unclosed  and 
for  which  no  account  sales  have  been  rendered.  They  also  find  that  "  Charo-es 
Account"  has  been  debited  with  .$4050  for  various  charges  paid  on  open  consign- 
ments unsold  and  partially  sold.  They  also  find  that  there  is  $1812  accrued  interest 
due  on  matured  bills  receivable,  and  $280  due  on  bills  payable. 

Allowing  lY/p  commission  on  the  $48,500,  what  are  the  necessary  entries  to 
liropei'ly  adjust  the  commission,  charges  and  interest '?  Ans. 

ENTRY  FOR  ADJUSTING  THE   COMMISSION. 

Commission  »e«i  account        ---.._.        fl212.50 

To  Commission  oW  occomhJ         -----..      $1212.50 
Eor  2y^  on  $48500  sales  of  nnclosed  consignments,  per  Com.  Sales  Ledger. 

ENTRY  FOR   ADJUSTING   THE   CHARGES. 

Charges  neio  account       ---------        |l4050 

To  Charges  oW  flccoKni       ----.-..  $4050 

For  .anionnt  paid  on  unclosed  consignments,  per  Com.  Sales  Ledger. 

ENTRY  FOR  ADJUSTING   THE   INTEREST. 

Interest  neic  account       ---------        $1532 

To  Interest  old  account      ------..  $1532 

For  Interest  on  hills  receivahle  $1812 

Less      "  "      "    payahle  280 


Interest  in  favor  of  iirm  $1532 

KOTE. — In  all  entries  of  this  Icind,  whether  new  acconnt  is  debtor  to  old  acconnt,  or  old 
.acconut  to  new  acconnt,  the  new  nccouut  part  of  the  entry  is  not  to  be  jiosted  until  the  old  account 
side  is  posted  aud  the  old  account  closed  and  ruled  up.  See  Higher  AVork  of  Closing  Ledgers  in 
Soul^'s  New  Science  aud  Practice  of  Accounts. 

58.  The  following  question  was  sent  to  us  by  the  cashier  of  a  banking  insti- 
tution in  Texas,  where  accounts  are  kept  in  both  coin  and  currency : 

"  Our  Bank  receives  deposits  in  currency  and  gold.  In  the  regular  course  of  business  we  have 
sold  $15000  or  $20000  gold  belonging  to  depositors,  and  credited  currency  exchange  account  with  the 
premium  realized.  Wo  now  wish  to  ascertain  our  not  gain  for  the  purpose  of  declaring  a  dividend 
andaswe  willh.ave  to  buy  gold  to  pay  our  gold  depositors  when  they  demand  their  funds,  it  is 
clear  that  to  close  our  books  as  the  accounts  now  stand,  we  would  show  a  false  gain  equal  to  the 
premium  paid  for  gold.  Under  these  circumstances,  what  entry  must  we  make  to  adjust  the  matter 
aud  keep  our  books  in  correct  balance  ?     Gold  is  now  ruling  at  lli"j^  premium." 

Ans. 
Exchange  old  account 

To  Exchange  new  account. 

The  amount  of  debit  and  credit  of  this  entiy  would  be  the  premium  on  the 
gold  sold,  belonging  to  depositors,  at  the  current  rate  per  cent.  The  credit  side  of 
this  entry  would  not  be  posted  until  the  old  account  was  closed  and  ruled  up. 

59.  The  following  questions  were  sent  to  us  by  a  Texas  merchant : 

"  I  am  doing  a  general  inerchanilising  business  and  keep  accounts  in  both  coin  and  currency. 
It  often  happens  that  a  customer  who  owes  a  balance  in  coin  wishes  to  pay  it  in  currency,  and  that  a 


852 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS, 


customer  who  owes  a  balance  in  currency  wishes  to  pay  it  in  coin.  Now,  suppose  the  first  customer 
to  l)e  Smith  and  the  second  Jones,  anil  the  balanci!  in  each  case  to  he  $550,  and  gold  10^^  premium, 
what  would  bo  the  correct  entries  to  reduce  the  coin  balance  to  currency  and  the  currency  balance  to 
coin,  and  to  settle  both  accounts  ?  " 

Aus. 

Journal  E>fTRY  to  reduce  the  Coin  balance  to  Currency. 

Smith,  currency  account  To  Sundries,     -----        $603 

To  Smith,  coin  account      -----  $550 

To  Exchange  (IO^q'  premium  on  gold)  53 

Cash  Entry  to  settle  with  Smith. 

To  Smith,  (enter  in  currency  cash  column,)       .-...-  605 

Journal  Entry  to  reduce  the  Currency  balance  to  Coin. 

Sundries  To  Jones,  currency  account        .        -        -        .  550 

Jones,  coin  account        -----------        $500 

Exchange  (10%  premium  on  gold),     --------  50 

Cash  Entry  to  settle  with  Jones. 

To  Jones,  (enter  in  gold  cash  column),       -------  500 


A  DIFFICTILT  SETTLEMENT  AOT)  JOTJEXAL  EXTET. 


60.  Smith,  Joues  and  Brown  are  commercial  partners,  Smitli  one-lialf,  and 
Jones  and  Brown  one-fourth  each  in  gains  and  losses.  In  the  regular  course  of  busi- 
ness, Smith,  without  consulting  Jones  and  Brown,  sells  $1800  worth  of  merchandise, 
and  receives  a  note  for  the  same  at  four  months,  which  the  book-keeper  entered  in 
the  books  in  the  regular  manner.  Jones  and  Brown,  on  learjiing  of  the  transaction, 
disapproved  of  it  on  the  ground  that  the  maker  of  the  note  is  an  irresponsible  person, 
and  proposed  to  Smith  to  sell  to  him  their  respective  interests  at  25%  discount,  to 
which  proposition  Smith  agreed.  Jones  and  Brown  then  indorsed  the  note  over  to 
Smith,  who  leaves  it  in  the  business,  and  the  book-keeper  is  instructed  to  make  the 
necessary  entry.    What  is  the  correct  resultant  entry  ?  Ans. 

Sundries  To  Smith        ..------  $225 

Jones ------         $112.50 

Brown       -         - -        .         .         .  112.50 

Statement  of  the  several  entries  which,  when  the  debits  and  credits  are  can- 
celled to  the  lowest  amounts,  give  the  resultant  entry. 

First  entry  on  receiving  the  note : 

Bills  receivable -         $1800 

To  Merchandise  --------  $1800 

Second  entry,  when  Jones  and  Brown  each  take  from  the  business  $450  wortii 
of  i)roperty  to  sell  to  Smith  at  25%  discount : 

Sundries  to  Bills  receivable 

Jones 1450 

Brown 450 


I 


*  PARTNERSHIP    SETTLEMENTS.  853 

Third  entrj',  when  Jones  and  Brown  sell  to  Smith  at  a  discount  of  25<fc  the 
property,  bills  receivable,  taken  out  of  the  business  : 

Smith  to  Sundries       .-......-.        $675 

"  Jones  -        -  $337.50 

"  Brown  -        - 337.50 

Fourth  entry,  when  Smith  invests  and  guarantees  the  $900  worth  of  property 
bought  for  $675. 

Bills  receivable  .        -        - -        $900 

To  Smith         -        -        -        - $900 

Note. — If  it  is  preferred,  the  -whole  note  may  be  taken  out  of  the  firm  by  the  partners,  before 
making  the  sale  to  Smith,  and  the  other  entries  made  to  correspond. 

Without  making  the  above  entries,  the  proper  debits  and  credits  may  be 
determined  by  the  following  process  of  reasoning : 

Jones  and  Bro-wn,  fearing  they  may  lose  $-150  each  at  the  maturity  of  the  note, 
now  agi-ee  with  their  partner,  Smith,  to  allow  him  $112.50  each,  (825%  of  $450)  if 
he  will  secure  them  against  this  anticipated  loss  of  $450  each.  In  consideration  of 
the  $112.50  from  each.  Smith,  agrees  or  guarantees  to  protect  them  against  the  antic 
ipated  loss,  and  hence  each  owes  him  $112.50. 

IjoTE. — In  case  the  note  proves  worthless,  Smith  will  be  debited  and  Bills  Receivable  credited. 

61.  W.,  X.,  Y.  and  Z.  were  partners,  each  one-fourth  in  gains  and  losses. 
They  dissolved  partnership,  paid  all  their  outstanding  liabilities,  and  divided  all 
their  resources.  They  then  examine  their  books  and  find  the  following  balances  to 
the  credit  of  each  partner:  W.  $1100;  X.  $240;  Y.  $810;  Z.  $680.  What  settle- 
ment nuist  be  made  between  the  partners  to  adjust  the  matter,  and  what  entry  must 
be  made  on  the  books  when  the  final  settlement  is  made  ?  Ans. 

Sundries  To  Sundries, 

■\y $392.50 

Y  " -        .        -  102.50 

ToX. $m.50 

ToZ. 27.50 

OPERATION. 

Cr.  Dr.                           Dr.                Cr. 

Amount  due  W.            -        -        -          $1100.00  $707.50  =                              $392.50 

<i        "    X.              ...             240.00  707.50  =          $467.50 

ti        i<    Y.              ...             810.00  707.50  =                                102.50 

"        "    Z.              .        .        -             680.00  707.50  =              27..50 

Total  liabilities  -        -  i  )  $2830.00  $495.00        $495.00 

Each  partner's  i  of  liabilities  $  707.50 

By  this  operation  we  see  that  there  is  due  to  W.  $392.50,  and  to  Y.  $102.5&, 
of  which  X.  owes  8467.50  and  Z.  $27.50. 

AMien  X.  and  Z.  pay  to  W.  and  Y.  the  respective  amounts  due  by  them,  it 
is  clear  that  W.  and  Y.  will  then  owe  them,  and  hence  the  above  adjusting  entry. 


854 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


OPEISTTN'G   ENTEY  TO   CtlANGE    SINGLE    ENTRY  BOOKS   TO   DOUBLE 
ENTRY,  AND  USE  THE  SLSTGLE   ENTRY  LEDGER. 

62.  A.  S.  Blafifer  and  P.  Birba  are  commercial  partners,  equal  iu  gains  and 
losses.  Tlu'v  are  keeping  tlicir  books  by  tlie  Single  Entry  System,  but  desire  to 
oliiinge  tbem  to  tbe  Double  Entry  System,  and  use  tbe  jn-esent  Single  Entry  Ledger. 
The  following  formula  of  work  accompUshes  the  desired  object : 

New  Orleans,  May  1, 1893. 

For  the  purpose  of  changing  our  books  from  single  entry  to  double  entry,  we 
make  the  following  abstract  of  our  commercial  affairs  this  date  : 


ITEMS. 


Merchandise  per  stock  book 

Cash  ou  hand  and  in  bank       ------- 

Bills  receivable,  Nos.  1,  2,  3  and  4  .        -        -        -         - 

Canal  Bank  stock    --------- 

Bills  payable,  Nos.  1  and  2 -- 

Our  resources  and  liabilities  shown  by  our  Single  Entry  Ledger, 

are  as  follows : 
A.  S.  BlafFer's  account    --..-.-- 
P.  Birba's  account  -        -        -         ...... 

231  i)ersonal  accounts       -----..  Dr. 

67  personal  accounts       ---....  Cr. 

Total  amount  of  resources    --..... 

Total  amount  of  lialiilitics  ..... 

Net  Rain  to  date     ------.. 

A.  S.  Blaiier's  one-half  net  gain  is        -        -         -         -        - 

P.  Birba's  one-half  net  gain  is        -----        - 

Approved  : 

A.   S.  Blaffer, 
P.  Biisba. 


RESOURCES. 


17000 

7800 
2500 
1000 


1150 
1800 
9100 

40350 

26750 


13600 


00 


LIABILITIES. 


"1850 


00 


8500 
12000 

4400 


26750 


6800 


00 

00 

00 


00 


00 
OO 


Erom  the  foregoing  statement,  the  following  Journal  entry  is  made,  which 
converts  the  Single  Entry  Ledger  into  a  Double  Entry  Ledger: 


FIRST  FORM    OF   ENTRY. 

Sundries  to  Sundries 
•V/A.  S.  Blaffer.  (amount  posted)  $  1150 
{/P.  Birba,  (amount  posted)     -         1800 
"1/231  ])ersonal  acciinnts,  (posted)      9100 
Mdse.  ]icr  inventory       -     -     -       1701)0 
Cash  ou  hand         -----         7S(I0 
Bills  receiv;iliIo     -----        2.")(I0 
Canal  Bank  stock      -     -     -     -         1000 
V     To  A.  .S.  Blaffer,  (am't  posted)  $8,500 

x/      "  P.  Birba,  (amount  ])<isti'd)  12000 

■/      "  67  personal  acc'ts,  (posted)  4 11)0 

"  P.  and  L.,  net  gain  to  date  ISdlMl 

"  Bills  payable lS,-)() 


SECOND   FOR.M    OF   E.NTRV. 

Sundries  to  Suudries 

t/A.  S.  Blaffer,  (amount  posted)   $  1150 

■l/P.  Birba,  (amount  ])osted)      -         1800 

{/231  personal  accounts,  (pcisted)      9100 

Mdsr.  iicr  inveutorv       -     -     -       17000 

Cash  ou  hand    -     -'   -     -     -     -         7800 

Hills  receivable     -----         2500 

Canal  Bank  stock      -     -     -     -         1000 

x/     To  A.  S.  Blaffer,  (am't  jiosted) 

"      "  "        +  net  gain 

x/      "  P.  Birba,  (am't  posted)  - 

"         "  i  net  gain 

i/       "  G7  personal  acc'ts,  (posted) 
"  Bills  jiayalile     -     -     -     -     - 


.$8500 
6800 

12000 
680O 
4400 
1850 


Note. — The  debits  and  credits,  in  either  of  the  forms  of  entry,  that  are  checked  and  marked 
"posted,"  must  not  be  rejimsted. 


PARTNERSHIP    SETTLEMENTS. 


855 


TO    OPEN    NEW    BOOKS    FROM    AN    OLD    SET    WHICH    HAS    BEEN 

INCOERECTLY  KEPT. 

G3.  A.  find  Z.  are  partners,  doing  a  mercantile  and  manufacturing  business, 
and  their  boolcs  are,  in  their  own  language,  "  in  an  aAvf'ully  mixed  condition." 
They  therefore  desire  a  ne\y  set  opened,  and  accordingly  employ  an  accountant  to 
perform  the  duty.  How  should  he  proceed  ?  Ans.  He  should  first,  on  loose  paper, 
from  the  old  books,  memorandum  records  of  all  kinds,  and  from  the  partners, 
ascertain  the  present  resources  and  liabilities  of  the  lirm.  He  should  then  arrange 
them  as  follows : 

List  of  the  Kcsources  and  Liabilities  of  A.  and  Z.,  January  1st,  1803,  as  shown  by 
their  old  irregularly  kept  books,  the  inventory  of  stock  on  hand,  and  memoran- 
dum accounts;  the  same  being  agreed  to  by  the  members  of  the  firm,  and 
arranged  as  per  subjoined  statement  to  facilitate  the  opening  of  a  new  set 
of  books  according  to  the  principles  of  double  entry. 

LIABILITIES  : 

Z's  net  investmt'ut  -----  §26843 
9  persouiil  accounts  due  bj-  tbe  firm  as 

per  old  Ledger  -----  1750 
24  personal  accounts  due  by  tbe  firm  as 

per  memorandums       -         -         -         .  925 

Bills  payable  as  per  B.  B.           -         -         .  1280 


l!i:soui!CES : 
Mdse.  per  account  of  stock       -        -        - 
Casb  on  hand      ------ 

10  shares  Germania  Bank  stock,  valued 

at  $130  .--... 

5  shares  N.  O  and  T.  R.  R.  stock,  valued 

at  $90 
Mill,  machinery,  tools,  etc.       -        -        - 
Bills  receivable  per  B.  B.  -         -         . 

Personal  accounts  due  the  firm,  as  per  the 

following  Journal  entry,  the  names 

being  omitted  to  economize  space 
A's  withdrawals  in  excess  of  his  iuvest- 

ineuts 


$1421,S 
9210 

1300 


450 

112WI 

4210 


Total  amount  of  resources 


7460 
3100 
-  $51228 


Total  liabilities   ----- 

Which  <Ieducted  from  the  resources 

as.  above  -        .        .        .        . 


Gives  the  net  gain  to  date 


§30798 
51228 


$20430 


From  the  foregoing  statement,  the  following  Journal  entry  is  made,  which 
opens  the  new  set  of  books: 


A's  one-half  of  net  gain  is        -        -        - 
From  which  deduct  his  net  withdrawals 

Gives  his  present  net  capital    -        .        . 


Z's  one-half  net  gain  is      - 

To  which  add  his  net  investment 

Gives  his  present  net  capital    - 


$10215 
3100 

$  7115 


10215 
26843 

$37058 


Sundries  To  Sundries 

Mdse. $14218 

Cash    - 9210 

Germania  Bank  stock         -        -  1300 

N.  O.  &  T.  R.  R.  stock       -        -  450 

Mill,  machinery,  tools,  etc.        -  11280 

Bills  receivable  -        -        -  4210 

81  personal  accounts,  7460 

(Insert  the  81  names) 

To  A.         -        -        -  $  7115 

"  Z.  ...  37058 

"  bills  payable        .  1280 

"  personal  accounts  2675 

(Insert  the  names  of  the 

33  persons  owed.) 

Note. — After  this  entry  is  made,  the  amount  of  cash  should  be  entered  in  the  Cash  Book  as 
amount  on  hand. 

64.    A.  and  B.  are  jiartners,  equal  in  gains  and  losses.    A.  owns  one-fourth 
and  B.  owns  three-fourths  of  the  net  capital. 

The  resources  and  liabilities   at  commencing,  not  including  the  partners' 
investments,  were  as  follows : 

Cash         -        -        $  5000  C  Bills  payable 

Mdse.       -        -  15000  Liabilities  •  J   Personal  accounts 

Resources  :  i  Bills  receivable         2000  '1  

Bank  stock      -  3000  I.      Total  liabilities         $21000 

Personal  accounts      8000  Total  resources  33000 


$15000 
6000 


Total  resources 


$33000 


(Continued. ) 


Net  capital  of  the  firm 


$12000 


856 


SOULE  S    I'HILOSOPIIIC    PRACTICAL    MATHEMATICS. 


The  resources  and  liabilities  at  the  close  of  the  fiscal  year,  not  iucluding  the 
partners'  investmeuts,  were  as  follows : 


Eesources  : 


Ciish     -        .        -  I  3000 

Mdse.    -        -         -  12000 

Bills  receivable   -  9000 

Personal  accounts  7000 

Bank  stock       -  -        5<IO0 

Real  estate      -  -        4000 

Total  resources  $40000 


Liabilities  : 


.1 


Bills  payable 
Personal  accounts 


$12000 
7000 


Total  liabilities        $19000 


What  has  been  the  net  gain,  and  what  is  each  partner's  net  capital  ? 

Ans.  Net  gain  $9000.    A's  net  capital  $7500. 
B's  net  capital  $13500. 


Kesources  at  commencing  as  above 
Liabilities  at  commencing  as  above 


OPERATION. 


$33000 
21000 


Net  capital  at  cominencing  as  above     |12000 


A's  one-fourth  capital  at  commencing  $3000 

B's  tbree-fourtbs  capital  at  commencing         900O 


Firm's  capital 


$12000 


A's  one-fourth  net  capital  at  commencing     $3000 
A's  one-half  net  gain  -        -        -        -        4500 

A's  net  capital  at  closing 


$7500 


Resources  at  closing  as  above         -        -      $40000 
Liabilities  at  closing  as  above        -        -        19000 

Net  capital  at  closing       .        .        .        21000 
Less  net  capital  at  commencing      -        12000 

• B's  three-fourths  net  capital  at  commencing  i 

Net  gain  during  business  -        -        $9000    B's  one-half  net  gain         ....        4500 

B's  net  capital  at  closing  ....    $13500 
A's  net  capital  $7500  -f  B's  net  capital  $13500  =  $21000  =  the  net  capital  of  the  firm  at  closing. 


G5.  J.  Jones,  J.  E.  SoiiM  and  B.  B.  Euston  are  partners,  doing  business 
under  the  name  and  firm  of  Jones,  SouM  &  Euston.  They  commenced  business 
January  1,  1893.  The  gains  and  losses  are  to  be  shared  equally,  each  one-third. 
July  1, 1893,  Jones  retires  from  the  concern,  and  Soule  and  Euston  continue  the  busi- 
ness and  liquidate  the  affairs  of  the  old  firm.  I71  order  to  settle  with  J.  Jones,  July 
1,  the  books  are  to  be  adjusted  so  that  his  account  may  exhibit  the  cash  balance  due 
to  him  on  that  date.  With  a  view  to  effect  this,  the  following  facts  have  been  ascer- 
tained and  agreed  to  by  all  of  the  partners : 

The  merchandise  on  hand  per  account  of  stock  is  valued  at  $19528. 

The  merchandise  on  hand  belonging  to  Mdse.  in  joint  account  with  H.  Davis, 
each  one-half,  is  valued  at  $1236.  Charges  incurred,  but  not  posted  to  said  account, 
amount  to  $210.  The  commission  on  sales  is  2J%.  The  charges  incurred,  but  not 
posted,  on  "  Commission  Sales  Account,"  amount  to  $087.65,  and  the  commission  on 
the  sales  effected,  but  not  posted,  amounts  to  $915.18.  It  is  agreed  to  allow  the  new 
firm  10%  on  all  personal  accounts  receivable  and  bills  receivable,  to  cover  doubtful 
debts.    Of  the  cash  on  hand,  $500  is  gold,  which  is  worth  10%  i^remium. 

The  balance  of  S.  Brown's  account  is  payable  in  San  Francisco,  and  exchange 
is  2%  discount.    Interest  is  to  be  charged  on  all  accounts  due  the  firm,  and  allowed 


PARTNERSHIP  SETTLEMENTS. 


857 


oil  all  accounts  tliat  tLe  firm  owe,  at  G%  ;  aud  also  on  the  investments  and  with- 
drawals of  tbe  partners,  at  the  same  rate. 

The  following  shows  the  face  of  the  Ledger  June  30 : 


Due  by 

Averai^e 

Dr. 

May 

4 

$1850 

00 

April 

20 

2408 

00 

June 

3 

1320 

00 

Aug. 

20 

50840 

142.50 

71410 

9810 

00 
60 

85 
00 

8621 

00 

7412 
810 

2130 

6418 
230 

5450 

40 
30 
00 
15 
00 
00 

May 
July 

28 
16 

13400 
17815 

00 
00 

Jan. 
Mar. 

1.5 
3 

5000 
9405 

00 
00 

1228581 

30 

J.  Jones'  capital  account  ----- 
"  current  account  -  -  -  -  - 
J.  E.  Soul^'s  capital  account        -        -        -        - 

"       current  account        .        -        .        - 
B.  B.  Euston's  capital  account      -        .        -        - 

"  current  account     -        -        -     *  - 

Bills  payable 

Bills  receivable      ------- 

Cash        --------- 

Merchandise    -------- 

Commission  sales,  ...-.- 

(The  balance  $32328,  due  by  average  August  19.) 
Merchandise  in  joint  account  with  H.Davis 
(The  debit  is  our  one-half  investment. ) 

Expense  - -        - 

Interest  -.-...-- 

Exchange        -.-..--- 
Charges  -------- 

Commission     -------- 

Profit  and  Loss       ---...- 

H.  Davis 

S.  Brown         -------- 

J.  Lewis  --. 

W.  Wood 

G.  B.  Brackett 

A.  Williams  

M.  Moses         .---.... 


Cr. 


130000 
30000 
30000 
10400 


31581 
42138 

14140 


1420 
3816 
4511 
7913 

800 
12000 


4450 


5409 


$228581 


Due  bv 

Avera^ro 


Jan. 
Jan. 
Jan. 
Nov. 


July 


Aug. 


April 


Sept. 


10 


17 


11 


What  are  the  Journal  entries  to  be  made,  in  accordance  with  the  foregoing 
agreements  aud  exhibit  of  facts,  to  effect  the  desired  settlement  with  the  retiring 
partner,  J.  Jones?  Ans. 

(Entry  to  credit  Merchandise  in  Joint  Account  with  H.  Davis). 

Mdse.  in  Joint  Account  with  H.  Davis,  new  accoMnf,         -         -         .         -        f2118 

To  Mdse.  in  Joint  Account  with  H.  Davis,  old  account,  |2118 

For  one-half  value  of  Mdse.  unsold. 


(Entry  to  close  the  account  of  Merchandise  in  Joint  Account  with  H.  Davis). 

To  Sundries,      -        -        -        $7637 


Mdse.  in  Joint  Account  with  H.  Davis 

To  Charges  -.-.-. 

"  Commission  at  2i!'o  on  smiles 
"  H.  Davis,  his  one-half  net  jiroceeds 
"  Profit  and  Loss,  our  one-half  net  gain  - 


$210.00 
353.50 

6788.25 
285.25 


<Entry  to  credit  Charges  with  the  amount  due, 
but  not  posted,  on  "Com.  Sales"). 


Charges,  new  account. 

To  Charges,  old  account, 


^7.65 


$687.65 


(Entry  to  credit  Com.  with  the  amount  due,  but 
not  posted,  on   "Com.  Sales"). 

Commission,  neic  account,  -      $915.18 

To  Commission,  old  account,  $915.18 


858 


souLE  s  PHILOSOPHIC  practical  mathematics. 


(Entry  to   allow   10?;^  on  all  Porsoual   Accounts 
Keceivablo  and  Bills  Receivable  due  tbo  Firm). 

Prolit  and  Loss,  old  account,       -        $!)64G 

To  Proljt  aud  Loss,  new  account,  $19G4G 


(Eutry  to  adjust  tbo  Preminui  on  Gold  aud  tbo 
Discount  on  S.  Brown's  account). 


Kxcbanjjfe,  old  account, 

To  Excbaugc,  new  account, 


$218 


For     2%  disc,  on  $13400  =  $208 

Less  10%  preni.  ou      500  gold  =      50 


$1218 


(Entry  to  adjust  tbe  Interest  ou   tbo  Partners' 
Investments  and  Witbdrawals). 


Sundries 
J.  .Jones 
J.  E.  SouI6 


To  B.  B.  Eustou 


J     .24 
ll.lli 


$11.40 


Or,  as  many  accountants  -nould  make  tbe  cntrv. 
tbus: 


Sundries 


To  Interest 


$.->-'.00 


.1.  .Tones,  57  ds.  ®  6^^  on  IllS.'iO  $17.57 
.1.  E.  S()ul6,  71  "  "  6%  "  2408  28.49 
B.  B.  Eustou,  27  "     "  6%  "     1320        5.94 


Or,  consideriug  tbe  interests  on  tbe  iiivestmeuts  for  181  days  at  6%,  aud  on 
tbe  -witbdrawals  as  in  tbe  i)recediug  eutry,  tbus : 

12663 

$887.43 


Interest 


To  Suudries 
"  .J.  .Jones 


J.  E.  S(ml6    - 
B.  B.  Eustou 


876.51 
899.06 


(Entry  to  adjust  tbe  Interest  or  Discount  due  tbe  Firm  on  Personal  Accounts  matured  and 
unmatured,  and  on  Bills  Payable  unmatured). 


Interest,  new  account,      .        .        -        -        . 

To  Interest,  old  account. 

For  aceruefl.  interest  on  tbe  following  credit  balances  : 


$1061.22 


$1061.22 


S.  Brown,  $13400.00  for    33  ds.  ®  G%  $  73.70 

G.  B.  Brackett,     5000.00  "    166   "    "  6;!„'    138.33 
A.  Williams,  9405.00  "    119    "    "  G^H    186.53 

H.  Davis,  12000.00  "      48   "    "  6»^      96.00 


H.  Davis, 
M.  Moses, 
Com.  Sales, 
Bills  Payable 


6788.25  for    10  ds. 

5409.80  "  73  " 
32328.00  "  50  " 
10400.00  "    127   " 


at  6%  11.31 

"  6J^  65.82 

"  6,V  269.40 

"  6%  220.13 


(Entry  to  adjust  tbe  Interest  or  Discount  due  by  tbe  Firm  ou  Personal  Accounts  aud  Bills 

Receivable). 


Interest,  old  account,         ------- 

To  Interest,  new  account,  -        -        - 

For  accrued  interest  ou  tbe  following  debit  balauces: 
.1.  Lews,  $17815  for  16  days  ®  6°^ 

Bills  r.'ceivable,       50840  "    51     "      "  6,^ 
AV.  Wood,  4450  "    90     "      "  6% 


$546.40 


$546.40 


$  47.51 

432.14 

66.75 


Note. — Wben  tbe  foregoing  entries  sball  have  been  posted,  except  the  new  account  debits  and 
credits,  tbe  gains  aud  losses  will  be  ascertained  in  tbe  usual  way,  and  tbe  books  closed.  Profit  aud 
Loss  should  be  closed  to  tbe  partners'  current  account,  and  their  current  account  should  be  closed 
to  their  capital  account. 

THE  ADMISSION   OP  A  NEW  PARTNER  INTO  A  COMMERCIAL   FIRM. 

G6.  There  are  many  reasons  for  admitting  uewpartuers  into  commercial  bouses. 
1.  Some  partners  are  admitted  I'or  tbeir  money.  2.  Some  for  tbeir  reinitatiou.  3, 
Otbers  for  tbeir  knowledge  or  skill. 

There  are  also  many  ways  of  admitting  partners.  1.  Some  are  admitted  to 
an  agreed  upon  interest  in  tbe  gains  and  losses,  by  making  a  specified  ca.sb  or  note 
investment.    2.  Some,  by  paying  to  the  old  members  of  tbe  firm,  in  cash  or  notes, 


'W 


*  PARTNERSHIP   SETTLEMENTS.  SSg 

a  stated  price  for  i,  ^,  ^,  or  J  interest,  of  the  gains  and  losses.  3.  By  paying  to  the 
members  of  the  firm,  a  specified  sum  for  a  part  or  share  of  their  respective  interests. 
4.  By  paying  to  the  firm  a  specified  sum  for  J,  }^,  J,  or  J  interest  or  ownership  iu  the 
entire  stock  or  cai>ital  of  the  firm,  aud  a  certain  share  of  the  gains  and  losses  of 
the  new  firm. 

There  are  also  other  ways  of  admittiug  uew  partners,  but  these  are  the  most 
general,  and  we  shall  briefly  consider  the  entries  to  be  made  under  the  foiu'  cases 
named. 

In  the  first  case,  where  a  specified  cash  investment  was  made  for  a  certain 
share  in  the  gains  and  losses,  cash  would  be  debited  aud  the  party  credited  for  the 
amount  invested. 

In  the  second  case,  where  a  specified  bonus  of  cash  or  uote  payment  was  made 
to  members  of  the  old  firm  for  a  certain  interest  in  the  gains  aud  losses,  no  entry 
would  be  made  in  the  books  for  the  cash  or  notes  received  by  the  old  members, 
uuless  they  shoidd  make  an  investment  of  the  same  in  the  firm.  They  sold  only  an 
interest  of  their  respective  interests  in  the  gain  aud  losses  of  the  business,  aud  the 
only  thing  necessary  to  be  doue  is  to  divide  the  amount  received  between  the  old 
partners,  or  invest  it  to  their  respective  credits  iu  proportion  to  the  interest  sold, 
aud  then  to  draw  new  articles  of  agreement  specifying  each  partner's  interest. 

To  illustrate,  let  us  suppose  that  Jones  and  Smith  are  partners,  Jones  §  aud 
Smith  i  iu  gains  aud  losses,  and  that  Brown  is  admitted  as  a  i)artuer  with  ^  interest 
iu  the  gains  and  losses  of  the  business  ou  the  payment  of  a  bonus  of  $10000  cash, 
aud  that  Jones  and  Smith  sell  to  Brown  the  J  interest  in  proportion  to  their 
respective  shares,  §  aud  ^,  and  invest  the  same  in  the  business.  What  is  the  entry, 
and  what  are  the  respective  interests  of  the  tliree  i>artuers  in  the  uew  firm  ? 

The  cash  entry  iu  Journal  form,  is  as  follows : 

Cash  to  Sundries         -        - $10000 

To  Jones,  Lis  two-thirds  of  $10000       -        -        .        -  $6666.67 

"  Smith,  his  oue-third  of  $10000       -        -        .        .  3333.33 

The  partners'  respective  interests  are  uow  as  follows : 

Joues  four-uinths,  Smith  two-ninths,  and  Brown  three-ninths  of  the 
gains  and  losses  of  the  new  business. 

In  the  third  case,  the  old  partners  who  sold,  should  be  debited  each  for  the 
l)roportiou  sold  aud  the  new  partner  credited  for  the  amount  bought. 

To  illustrate  this,  we  will  again  suppose  that  Jones  and  Smith  are  partners, 
Jones  §  and  Smith  J  in  gains  and  losses,  and  that  Brown  pays  to  Joues  aud  Smith 
$12000  for  a  partnership,  and  J  interest  in  the  gains  and  losses,  and  that  they  sell 
the  i  interest  in  proportion  to  their  present  shares.  What  is  the  entry,  aud  what 
is  the  respective  interest  of  each  partner  iu  the  gains  and  losses  ? 

The  entry  is  as  follows : 

Sundries  to  Brown $12000 

Joues,  his  two-thirds  of  $12000 $8000 

Smith,  his  one-third  of  $12000 4000 

The  partners'  interests  are,  Jones  four-ninths,  Smith  two-uiuths,  and  Brown 
tUree-ninths.  Should  Jones  aud  Smith  invest  this  money  in  the  firm,  Cash  would  be 
debited  aud  they  credited. 


86o  soule's  philosophic  practical  mathematics.  * 

In  the  fourth  case,  Tvhere  a  specified  amoimt  was  paid  for  an  interest  or  owner- 
ship of  a  certain  part  of  the  whole  capital  of  a  lirni,  the  amount  paid  for  the  interest 
should  be  divided  between  tlie  old  jiartners  in  i)roporti()n  to  the  net  capital  of  each ; 
and  then  the  amount  of  the  one-fourth  interest  sold  should  be  credited  to  the  pur- 
chaser, and  the  old  partners  debited  for  the  same  in  proportion  to  their  respective 
net  capital. 

To  illustrate  this,  we  will  suppose  Soul6  and  Weiss  to  compose  a  firm  that 
sells  to  Harris  J  of  the  net  capital  of  their  firm  for  $30000  cash,  and  that,  while  they 
are  now,  Soule  f  and  Weiss  J  in  gains  and  losses,  in  the  new  firm  the  interests  will 
be  Soul6  J,  Weiss  J,  and  Harris  J. 

We  will  further  suppose  that  the  books  have  been  closed  and  exhibit  the 
following  .statement  of  accounts : 

Soule.  Weiss.  Cash.  Mdse. 


$100000 


$10000  $9000 


$90000 


Bills  Recv.  Bills  Pay.  H.  Kent. 

$10000 


$5000  $6000 


This  exhibit  shows  the  net  resources  to  be  $110000,  ^  of  which  is  $27500. 
For  this  $27500  and  a  partnership  with  one-fourth  interest  in  gains  and  losses, 
Harris  pays  $30000.  This  $30000  divided  between  Soule  and  Weiss,  in  proportion 
to  their  net  capital,  gives  Soule  $27272.73,  and  Weiss  $2727.27. 

The  $27500  being  the  one-fourth  of  the  net  capital  purchased  by  Harris,  must 
be  credited  to  him,  and  Soul^  and  Weiss  debited  for  the  same  in  proportion  to  their 
respective  net  capital.    The  following  is  the  adjusting  entry : 

Sundries  to  Harris    - $27500 

Soul6,  his  proi)ortiou         -         -         -       $25000 
Weiss,  his  proportion        ...  2500 

Articles  of  Agreement  should  be  drawn  by  the  new  firm  specifying  the  invest- 
ments, interests  in  gains  and  losses,  whether  interest  is  to  be  allowed  on  the  part- 
ners' investments  and  charged  on  their  withdrawals,  etc.,  etc.  See  Articles  of  Agree- 
ment a  few  pages  fiu'ther  on. 

Shovdd  Soul6  and  Weiss  invest  the  money  received  for  the  sale  of  their  one- 
fourth  net  capital,  cash  would  be  debited  and  they  credited  in  the  usual  manner. 

A  party  desiring  to  purchase  an  interest  in  any  business  should  carefully 
examine  the  accounts  and  the  goods  in  stock.  He  should  give  close  attention  to  the 
loss  and  gain  accounts  and  to  the  personal  accounts  diie  the  firm.  He  should  have 
an  account  of  stock  taken  in  the  presence  of  some  person  who  is  thoroughly  informed 
on  the  value  of  the  line  of  goods  or  material  in  which  the  house  deals. 


partnership  and  Partnership  Average, 


For  a  definition  of  Partnership,  and  a  statement  of  the  different  classes 
of  Partnership,  the  Formation  of  Partnerships,  the  Dissohition  of  Partnerships, 
Articles  of  Partnerships,  Pai'tnership  Settlements,  etc.,  see  images  1  to  5. 

Partnership  Average  is  the  process  of  finding  what  proportional 
amoun'c  of  the  gain  and  loss  is  to  be  given  to  each  member  of  the^rm  on  final  set- 
tlement, or  at  any  specified  time. 

Sharing  Gains  and  Losses. 
The  amount  of  gain  or  loss  to  be  apportioned  to  each  partner,  depends  njion 
the  following  conditions :  1.  The  gain  or  loss  may  be  divided  in  proportion  to  the 
net  investment  or  cai^ital  of  each  partner,  or  some  other  agreed  upon  proportional 
basis.  2.  By  allowing  each  jiartuer  a  salary,  and  dividing  the  remainder  of  the 
profits  according  to  a  fixed  proportion.  3.  By  allowing  each  partner  interest  on  his 
net  investment,  and  dividing  the  remainder  of  the  profits  according  to  a  fixed  pro- 
portion. 4.  By  allowing  each  partner  interest  on  his  net  investment,  and  a  salary 
for  services  rendered,  for  skill  possessed,  or  for  credit  or  good-will,  and  then  divid- 
ing the  remainder  of  the  profits  according  to  some  fixed  proportion.  See  Partner- 
ship, page  1. 

CLASSIFICATION, 

The  subject  is  classified  into  simple  and  compound  partnership  average. 

Simple  Partnership  Average  is  the  process  of  finding  the  proportional 
amount  of  gain  or  loss  to  be  apportioned  to  each  member  of  the  firm  when  the  net 
capital  of  each  has  been  invested  for  the  same  length  of  time ;  or,  in  brief,  it  is  the 
proportional  division  of  the  gain  or  loss  between  the  partners,  according  to  the  capi- 
tal invested  by  each. 

Compound  Partnership  Average  is  the  process  of  finding  the  jiropor- 
tionai  ajnount  of  gain  or  loss  to  be  apportioned  to  each  member  of  the  firm  when  the 
net  capital  of  each  has  not  been  invested  for  the  same  length  of  time ;  or,  in  brief, 
it  is  the  iiroportional  division  of  the  gain  or  loss  between  the  partners,  according  to 
their  average  net  investments  and  the  periods  of  time  that  they  were  in  use  by  the 
firm. 

SIMPLE  PAETNERSHIP  AVERAGE  PROBLEMS. 

67.  Jones,  Smith  and  Brown  contracted  to  do  business  as  partners  for 
one  year.  Jones  invested  $5000,  Smith  $4000,  and  Brown  $1500.  The  gain  was 
$3150.     What  was  the  proportional  share  of  each  partner  ? 

Ans.  Jones  $1500.     Smith  $1200.     Brown  $450. 
(Continued.) 
(861). 


862 


s(juLiis  riiiLosoriiic  practical  mathematics. 


FIRST   OPERATION, 


Jonos'  capital 
SiiiitU's    " 
Bruwu's  " 


15000 
4000 
1500 


Statement   to   iiud 
Joues'  share. 

Statemeu 
Smith's 

t  to  liiul 
share. 

Statement  to   finil 
Browu's  share. 

10500 

$      GAIN. 

3150 
5000 

$1500 

10500 

$      GAIN. 

3150 
4000 

$      GAIN. 

3150 
10500      1500 

$1200 

l|  450 

Total  capital      $10500 

Explanation. — By  aililing  the  imlividual  capital  of  each  partner,  wo  obtain  a  total  capital  of 
$10500,  which  is  the  amount  that  gained  and  i.i  to  receive  tlie  .f 3150 ;  and  as  the  gain  is  to  be  divided  in 
})roportion  to  the  respective  capital  of  eacli  partner,  we  place  the  gain  on  our  statement  lino  and 
reason  tlius  :  If  $10500  capital  gains  $31.50,  $1  will  gain  the  10500th  part,  and  $5000  (Jones'  capital) 
will  gain  5000  times  as  much.  The  result  of  this  statement  gives  us  $1500  as  Jones'  share  of  the 
gain.  The  reasoning  for  the  statomcnt.s  to  obtain  Smith's  and  Brown's  respective  shares  is  the  same 
as  that  in  the  statemeiit  to  obtain  Jones'  share,  and  hence  i?^  omitted, 

SECOND   OPERATION. 


Jones'  capital     - 
Biuith's    '' 
Browu's   " 

$.^000     $10 
^000         8 

xm    3 

Statemen 
Jones' 

21 

t  to   find 
share. 

$ 

3150 
10 

Statement  to  tind 
Smith's  share. 

$ 

3150 
21      8 

Statement  to  find 
Brown's  share. 

$ 

3150 
21      3 

Total  capital 

$21 

$1500 

$1200 

$450 

ExpJanation. — In  this  solution,  we  cancel  the  investments  of  the  respective  partners  by  divid- 
ing eacli  amount  by  500,  which  lessens  the  figures  in  the  operation  without  changing  the  ratios. 
The  same  reasoning  is  applied  as  given  in  the  first  solution,  and  henco  is  omitted. 


THIRD   OPERATION. 


Statement  to  find  the 
rate  per  cent.  gain. 


10500 


$  GAIN. 

3150 
100 


30%  gain. 


Jones'     capital  - 
Smith's        " 
Brown's      " 

Total  capital 


$5000 
4000 
1500 

$10500 


30%  of  which  is  $1500 
30%        "        "        1200 


305 


450 


68.  A.  and  Z.  formed  a  partnership  for  the  purpose  of  siieculatiug  in 
various  kinds  of  business.  A.  invested  $8400,  and  Z.  $6000.  At  the  clo.se  of  the 
business,  the  net  loss  amounted  to  $1728.  What  was  each  partners'  proportional 
share  ?  Ans.  A's  $1008.    Z's  $720. 


A's  investment 

Z's  " 


$8400 
6000 

$144 


OPERATION 

Statement  to  find  A's  loss. 


144 


% 
1728 

84 


Statement  to  find  Z's  loss. 


144 


1728 
60 

$720 


69.  W.,  X.,  Y.  and  Z.  entered  into  i^artnership  for  the  i^urpose  of  conduct- 
ing a  general  merchandising  business.  W.  invested  $3500,  X.  $3000,  Y.  $2500,  and 
Z.  $1000.  At  the  expiration  of  the  partnership,  the  net  gain  was  $15000.  "VVTiat 
was  each  partner's  proportional  share  ?  Ans.  W.  $5250.    X.  $4500. 

Y.  $3750.     Z.  $1500. 

70.  Three  young  gentlemen,  A.,  O.  and  P.,  contracted  for  the  use  of  a  carriage 
for  thirty  days  for  $100,  which  they  agreed  to  pay  in  proportion  to  the  number  of 


PARTNERSHIP    SETTLEMENTS. 


863 


days  that  each  had  the  use  of  it.    A.  used  it  9  days;  O.  14  days;  and  P.  6  days, 
and  1  day  it  was  not  in  use.     What  is  the  amount  due  by  each  ? 

Ans.  A.  $31.03.     O.  $48.28.     P.  $20.69. 


71.  A  father  has  $2000  which  he  wishes  to  divide  among  his  three  children, 
in  i^roportiou  to  their  ages,  which  are  6,  8  and  11  years.  What  amount  will  each 
receive  1  Ans.  6  years,  $480.    8  years,  $640.    11  years,  $880. 

OPERATION. 

6  +  8  +  11  =  25  years. 


Statement  to  find  the  amount 
for  the  child  of  6  years. 


I    2000 
25      6 


Statement  to  find  the  amount 
for  the  child  of  8  years. 

$ 
2000 


$640 


Statement  to  find  the  amount 
for  the  child  of  11  years. 


25 


2000 
11 


CO]\IPOma>  PARTNERSHIP  AVERAGE  PROBLEMS. 

72.  Jones,  Smith  and  Brown  associated  themselves  together  in  business,  and 
agreed  to  share  the  gains  and  losses  in  proportion  to  their  respective  investments 
and  the  time  the  same  were  employed  by  the  firm. 

Jones  invested  $4000  for  12  mouths;  Smith  invested  $3500  for  9  months; 

Brown  invested  $2000  for  8  months.    The  gain  during  the  business  was  $9550. 

what  was  each  partner's  share  ?  Ans.  Jones  $4800 ;  Smith  $3150. 

Brown  $1600. 

OPERATION. 

Mos.     Investment  for  1  mo. 
Jones  invested       -        -        -        -        $4000     X     12     =    $48000 
Smith      "  .        -        .        .  3500     x       9     =      31500 

Brown     "  ....  2000     x       8     =      16000 


Statement  to  find  Jones'  share. 


95500 


9550 
48000 

$  4800 


Statement  to  find  Smith's  share. 


95500 


9550 
31500 

$  3150 


$95500 
Statement  to  find  Brown's  share. 


95500 


9550 
16000 

$  1600 


Explanation. — In  this  problem,  the  gain  being  shared  in  proportion  to  the  money  invested 
and  the  time  for  which  it  was  Invested,  and  the  amounts  and  jieriods  of  time  of  each  partner  being 
■dift'erent,  we  must,  therefore,  find  the  respective  equivalent  investment  of  each  partner  for  1  mouth. 
This  is  done,  as  shown  in  the  operation,  by  multiplying  the  amount  invested  by  the  time  that  it 
■was  invested. 

The  reasoning  for  the  work  is  as  follows :  $4000  invested  for  12  months  is  equivalent  to  12 
times  $4000  for  1  month,  which  is  $48000;  then  $3500  invested  for  9  months  is  equivalent  to  9  times 
as  many  dollars  for  1  month,  which  is  $31500 :  and  then  $2000  invested  for  8  months  is  equivalent  to 
S  times  as  many  dollars  for  1  month,  which  is  $16000.  By  this  work,  we  obtain  the  respective 
investments  of  each  partner  for  1  month.  We  then  add  the  same,  to  obtain  the  total  capital  invested 
for  1  month,  and  then  proceed  to  find  each  partner's  share  of  the  gain  by  our  line  statements,  in  the 
same  manner  and  with  the  same  reasoning  that  is  presented  in  the  first  problem  of  Simple  Part- 
nership Average. 

73.    A.  commenced  business  January  1,   1893,  with  a  capital  of  $2500;  on 
the  1st  of  May,  he  entered  into  partnershij)  with  B.  who  invested  $3000 ;  aud  on  the 


864 


SOULE  S    rillLOSOPIIIC    PRACTICAL    MATHEMATICS. 


1st  of  October,  O.  was  adiiiitti'd  as  a  third  partner  and  invested  $4000.     At  tlie  close 
of  the  year,  there  was  a  gain  of  $5500.     What  was  each  i)artner's  apportionment  of 


the  gain  1 


Ans.  A's  $2500. 
FIRST  OPERATION. 


B's  $2000.     C's  $1000. 


1893. 
January  1, 
May         1, 
October  1, 


A.  invested 

B.  " 
C. 


$2500 
3000 
4000 


Mo3.  Am't  for  1  mo. 

12  =  $30000 
8  =  24000 
3  =   12000 


Total  capital  fur  1  iiKiutli, 


Statement  to  find  A's  share. 


66000 


$      GAIN. 

5500 
30000 

$2500 


Statement  to  find  B's  share. 

$     GAIX. 
5500 
66000      24000 


$00000 
statement  to  find  C's  share 


66000 


I     GAIN. 

5500 
12000 


$1000 


66000 


$2000 
SECOND    OPERATION. 

To  find  the  respective  shares  of  each  XJartuer  by  first  finding  the  rate  per  cent, 
gain,  we  proceed  thus : 

tsoo'^'^"  ^'^  capital  for  1  month  is    $30000,     8L''o  of  which  is  -        $2500 

100  B's      "        "    1      "      "       24000,     8i5'^  of  which  is  -  2000 

C's      "        "    1      "       "        12000,     8i?„'  of  which  is  -  1000 

Si%  gain. 

J<:x2)lanation. — In  the  first  part  of  the  operation  of  this  problem,  we  conld  have  shortened  th& 
work  by  cancellation,  but  in  order  to  more  fully  elucidate  this  subject,  we  did  not  cancel. 

74.  W.,  X.,  Y.  and  Z.  engaged  the  services  of  a  teacher  for  10  months  for 
$600,  and  agreed  to  pay  the  same  in  proi^ortion  to  the  number  of  chiklreu  sent  and 
the  time  that  they  attend  school. 

W.  sent  3  for  200  days ;  X.  sent  2,  1  for  200  days  and  1  for  120  days ;  T.  sent 
5,  3  for  ISO  days,  1  for  150  days,  and  1  for  90  days  ;  Z.  sent  4,  1  for  200  days,  and  5 
for  160  days.    What  amount  was  due  from  each  ? 

Ans.  W.  $151.26.    X.  $80.07.     Y.  $196.64.     Z.  $171.43. 

OPERATION. 


No.  sent 

attendance. 

"W. 

. 

3 

200 

X. 

. 

1 
1 

200 
120 

Y. 

- 

3 

1 

1 

180 

150 

90 

Z. 

- 

1 
3 

200 
160 

Total  number  for  1  day 

Statement  to  find  the 
amount  due  by  W. 

Statement  to  find  the 
amount  due  by  X. 

Statemen 
amount 

2380 

$ 
600 
600 

2380 

$ 
600 
320 

2380 

$151.26 

$80.67 

Ko.  in  attendance 
lor  1  da}-. 

600 


=         200) 
=         120^ 


540) 

150  V 

90^ 

200 
480 


320 


780 


680 


2380 


600 
780 

$196.64 


Statement  to  find  th© 
amount  due  by  Z. 


2380 


600 
680 

$171.43 


Explanaiion. — In  problems  of  this  character,  where  there  are  several  parties  to  receive  the  ' 
gain  or  contribute  to  the  loss  or  expense,  it  would  be  better  first  to  find  the  rate  per  cent,  as  shown 
m  preceding  Droblems. 


PARTNERSHIP   SETTLEMENTS. 


865 


I 


75.  A.  and  B.  formed  a  jiartncrsliip  iu  which  A.  iuvested  $8000  for  10 
months,  B.  $5000  for  12  months.  They  lost,  during  the  partnership,  $700.  What 
was  the  loss  of  each  partner"^  Ans.  A.  $100.    B.  $300. 

70.  X.,  Y.  and  Z.  executed  aiiiece  of  work  for  which  they  received  $500.  X. 
worked  15  days  of  S  hours  each  ;  Y.  worked  15  days  of  10  hoiu's  each ;  and  Z.  worked 
12  days  of  CJ  hoiu's  each.     How  should  the  $500  be  divided  among  them  ? 

Ans.  X.  $175.41.     Y.  $219.30.     Z.  $105.20. 

77.  Jones,  Smith  and  Brown,  on  the  1st  of  January,  1893,  associated  them- 
selves together  as  partners  for  the  term  of  1  year. 

January  1,  Jones  invested  $10000  ;  May  1,  he  drew  out  $1000 ;  October  1,  he 
drew  out  $2000 ;  and  December  1,  he  invested  $1000. 

January  1,  Smith  invested  $6000;  July  1,  he  invested  $2000;  Xovember  1, 
he  drew  out  $1000. 

March  1,  Brown  invested  $4000 ;  April  1,  he  invested  $1000 ;  August  1,  he 
drew  out  $3000;  November  1,  he  drew  out  $2500. 

The  gain  at  the  close  of  the  year  amounted  to  $9700.  What  was  each  part- 
ner's share  of  the  gain,  aud  what  is  Brown's  indebtedness  to  the  firm  ? 

Ans.  Jones'  gain,  $4150.     Smith's  gain,  $4100 ; 

Brown's  gain,  $1450,  and  indebtedness,  $500. 

FIRST    OPERATION. 

„        Am't  Invested 
-°^'"'-     for  1  month. 

January      1,     Jones  invested        -        -        -        $10000        4        $40000 

May  1,         "      -vrithclrew      -        -        -  4000 


October       1, 

"      withdrew 
"     invested 

capital  for  1  month 
Smith  invested 

"     withdrew 

's  capital  for  1  month 
Brown  invested 

it             a 

"     withdrew    - 
"      withdrew    - 

flOOOO 
2000 

5 
2 
1 

Mos. 
6 

4 

2 

Mos. 

1 

4 
3 

30000 

December  1, 

14000 
1000 

8000 

$5000 

$6000 
2000 

$8000 
1000 

5000 

Jones' 

January      1, 
July          I, 

November  1, 

Am't  Invested 
for  1  month. 

$36000 
32000 

$7000 

$4000 
1000 

14000 

Smith 

March         1, 
April           1, 

Am't  Invested 
for  1  month. 
$  4000 

August       1, 

$5000 
3000 

20000 

November  1, 

$2000 
2500 

6000 

$83000 


82000 


Excess  of  withdrawals,  or  amount  due 
the  firm  by  Brown 


Excess  of  withdrawals  for  2  months  deducted 
Brown's  capital  for  1  month 
Total  capital  for  1  month 


$500 


$30000 
-    1000 


29000 

$194000 


866 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Having  the  total  capital  for  1  moutli,  to  fliul  each  partner's  share  of  the  gain, 
■we  have  but  to  proceed  as  shown  in  preceding  problems,  either  by  the  line  state- 
ment or  per  centum, 

SECOND   OPERATION. 

In  problems  of  this  character,  where  there  are  several  investments  and  with, 
drawals  by  the  difl'orent  members  of  the  firm,  it  is  best  first  to  average  the  account 
of  each  partner.    The  following  operation  will  elucidate  this  method : 


Dr. 


JONES. 


Or. 


1893.                             ^°'-                             irTZ. 
May         1,    }I4000           8        =        -        -    $32000 
October  1,      2000           3        =         -        -        6000 

1893. 
January      1,     flOOOO 
December  1,         1000 

Hos. 

12         = 
1         = 

Dollars 
for  i  mo. 

$120000 
1000 

$38000  1 

$121000 
38000 

Balance  of  dollars  for  1  month 

$83000 

Dr. 

SMITH. 

Cr. 

1893. 
November  1, 

Mo8.                                     Dollars 
for  1  mo. 

$1000        2        =        -        -       $2000 

1893. 
January  1,     $6000 
July         1,      2U00 

Mob. 

12         = 
6        = 

Dollars 
for  1  mo. 

-  $72000 

-  12000 

$84000 
2000 

Balance  of  dollars  for  1  mouth     - 

$82000 

Dr. 

BROWN. 

Cr. 

1893. 
August       1, 
November  1, 

M03.                                     Dollars 
for  1  mo. 

$3000          5          =          -          -     $15000 
2500        2        =        -        -        5000 

1893. 
March  1,     $4000 
April     1,       1000 

Mos. 

10         =,      - 
9        = 

Dollars 
fur  1  ino. 

-    $40000 
9000 

$5500                                          $20000 
5000 

$5000 

$49000 
20000 

Due  the  firm 

,     $500                       Balance  of  dollar 

s  for  1  month 

-    $29000 

EECAPIT 

Jones'  capital  for  1  month     - 
Smith's      "      for  1  month    - 
Brown's     "      for  1  month,    - 

ULATION. 

$83000 
82000 
29000 

Total  capital  for  1  month      -        -        -      $194000 
Having  thus  obtained  the  total  capital  of  the  firm  for  one  month,  we  would 
proceed  as  in  previous  examples  to  find  each  partner's  share. 

78.    A.  and  Z.  bought  a  house  for  $10000.    A.  paid  $6000  and  Z.  $4000.    The 
iouse  rents  for  $150  jjer  month.    To  how  much  of  the  monthly  rent  is  eatih  entitled  ? 

Ans.  A.  $90.     Z.  $60. 


PARTNERSHIP    SETTLEMENTS. 


867 


79.  X.,  Y.  aiul  Z.  hired  a  pasture  for  the  season  for  $90.  X.  pastured  9  head 
of  mules  for  150  days;  Y.  pastured  11  head  of  mules  for  110  days,  and  Z.  pastured 
24  head  of  mules  for  160  days.    How  much  is  each  to  pay  ? 

Ans.  X.  $18.98.    Y.  $17.02.    Z.  $54. 

80.  Four  persons,  A.,  B.,  C.  and  D.,  residing  contiguous  to  each  other,  agree 
to  build  a  schoolhouse,  and  contribute  to  the  payment  of  the  same  in  the  reci]3rocaI 
ratio  of  their  resi^ective  distances  from  the  schoolhouse.  The  cost  was  $1800.  The 
schoolhouse  was  located  J  of  a  mile  from  A's  residence ;  1  mile  from  B's  residence ; 
1 J  miles  from  C's  residence,  and  2  miles  from  D's  residence.  How  much  did  each 
contribute  ?  Ans.  A.  $837.20f f .    B.  $418.60f|. 

C.    334.SSi|.    I>. 
-The  reciprocal  of  a  number  is  tbe  quotient  of  1  divided  by  the  number. 

OPERATION. 


209.30H. 


NOTE.- 


A.  i    mile  distant.  The  reciprocal  of  J     =  2 

B.  i       "          "  "            "        "1     =  1 

C.  li     "          "  "            "        "    li  =  f 

D.  2       "          "  "            "        "    2     =  i 


Sum  of  reciprocals 


4A  =  n. 


statement  to  find  A's 
contribution. 

Statement  to  find  B's 
contribution. 

Statement  to  find  C's 
contribution. 

Statement  to  fin 
contribution 

43 

$ 

1800 
10 
2 

43 

1 
1800 
10 

1 

43 
5 

1800 
10 

4 

43 

2 

1800 
10 

$837.20}S 

$418.6013 

$334.88i§ 

$:209.30iJ 

AVERAGIXG  SALES  IN^  COMMISSIOX, 

81,  A  commission  merchant  received  from  three  different  correspondents, 
X.,  Y.  and  Z.,  consignments  as  follows :  from  X.,  of  Jefferson,  Texas,  he  received 
200  bbls.  beef;  from  Y.,  of  Shreveport,  La.,  he  received  120  bbls.  beef;  and 
from  Z.,  of  Galveston,  Tex.,  he  received  210  bbls.  beef.  A  sale  of  the  three  lots  was 
effected  at  $11.  But  on  inspection,  it  was  found  that  Y's  beef  was  25%  better  than 
Z's,  and  X's  was  10%  better  than  Y's.  What  amount  of  money  is  to  be  credited  to 
the  sales  account  of  each  correspondent? 

Ans.  X.  $2524.80.    Y.  $1377.17.    Z.  $1928.03. 

OPERATION. 


BUs. 

X. 

200 

Y. 

120 

Z. 

210 

530 

11 

+ 


Bbls.  of  tbe  quality  of  Z's. 


+    25%     = 


=    250    +    io;'o 


150 
210 

635  of  quality  of  Z's. 


$i5830 


Operation  to  find  X's  credit. 


635  I    5830 
275 


Operation  to  find  Y's  credit. 


635  I    5830 
150 


Operation  to  find  Z's  credit. 


$ 


635 


5830 
210 


$2524.80  I  $1377.17  $1928.03 

ExpJanaiion. — The  difference  in  the  qxiality  of  the  beef  being  given  in  per  cent.,  we,  there- 
fore, to  find  the  eiiuivaleut  number  of  barrels  of  X's  and  Y's  beef  according  to  the  standard  quality 


868 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


of  Z's,  add  to  Y's  SSJ^  of  the  numlier  of  barrels,  and  to  X's,  aa  hia  ia  -worth  10%  more  than  Y's,  we 
first  add  -5°,^  of  tlir  nmnber  of  barrels,  which  gives  250  of  tlio  quality  of  Y's,  and  then  to  this  we 
add  10",,  of  tlie  niiniiicr  of  l>arri'ls,  and  olitain  275  barrels  of  tlie  iiuality  of  Z's.  The  balance  of  the 
operation  is  too  simple  to  renuiro  explanation,  and  hence  we  omit  it. 

82.  A  commission  merchant  received  on  consignment  from  A.,  $300  barrels 
X  flonr;  from  B.,  2i0  barrels  XX  flour;  from  0., 50  barrels  XXX  flour;  and  from 
D.,  100  barrels  choice  flour.    The  current  market  price  for  X  is  $6.00 ;  for  XX,  $0.50 ; 


for  XXX,  $7.00  ;  and  for  choice  grade  $ 
lots  were  sold  at  the  round  figure,  $7.00. 
the  sales  account  of  each  consignor  ? 


i,  with  a  market  tending  upward.     The  four 
What  was  the  correct  amount  to  credit  to 
Ans.  A.  $1927.72.     B.  $1670.09, 
0.       374.83.    D,      856.76. 


OPERATION. 

Bbls. 

Market 
rale. 

Market 
value. 

A. 
B. 
C. 

300 

240 

50 

®          6          = 
&         6i       = 
®         7         = 

$1800 

1560 

350 

D. 


100 


@ 


690 
7 


800 
$4510    Total  market  value. 


$4830    Amount  of  sales. 
Statement  to  find  A's,  B's,  C's  and  D's  respective  credits. 


A's. 


4510 


4830 
1800 

$1927.72 


B's. 


4510 


4830 
1560 

$1670.69 


C's. 


4510 


4830 
350 

$374.83 


D's. 

$ 


4510 


4830 
800 

$856.76 


Explanation. — In  the  operation  of  this  problem,  wo  must  find  the  amount  of  sales,  then  the 
amount  of  the  market  value  of  each  grade,  and  then,  having  these  two  amounts,  we  divide  the 
sales  between  the  different  consignors  in  proportion  to  their  respective  amounts  at  the  market  rate. 
To  find  the  price  jjer  barrel,  we  wonld  divide  the  total  amount  due  each  by  the  number  of  barrels 
sold  for  each. 

COTTON  AVEEAGE  PEOBLEMS. 

83.    A  cotton  factor  has  four  bales  of  cotton,  weighing  as  follows :  450,  460, 
398  and  486  pounds.    What  is  the  average  weight  i)er  bale  ? 

Ans.  448J  pounds. 

OPERATION. 

450 
460 
398 
486 

4  )  1794 


448i  pounds.     Ans. 

84.  A  cotton  factor  has  four  lots  of  cotton  belonging  to  four  different  parties. 
The  first  lot  consists  of  3  bales,  weighing  1400  pounds,  and  is  classed  as  Ordinary, 
the  quoted  market  price  of  Avhich  is  16/.  The  second  lot  consists  of  8  bales,  weigh- 
ing 3500  pounds,  and  is  classed  as  Low  Middling,  the  quoted  market  price  of  which 
is  18J/.  The  third  lot  consists  of  4  bales,  weighing  1800  pounds,  and  is  classed  as 
Strict  Middling,  the  quoted  market  price  of  which  is  19^/.    The  foui-th  lot  consists 


J^MtK'f 


PARTNERSHIP    SETTLEMENTS. 


869 


of  5  bales,  weighing  2500  pounds,  and  is  classed  as  Good  Middling,  the  quoted 
market  priee  of  whieli  is  20/".     What  is  the  average  price  per  pound  ? 

Explanation. — This  method,  although  siifficiL'utly  correct  for 
practical  purposes,  is  not  quite  accurate,  for  the  reason  that  the 
weight  of  different  bales  of  cotton  is  not  the  same.  But  in  prac- 
tice the  weight  is  not  often  known  at  the  time  the  average  price  is 
required,  and  hence  the  necessity  in  those  cases  of  using  the  num- 
ber of  bales  which  really  represent  pounds.  To  have  added  the 
four  prices,  16'',  184<^,  19J<',  and  20)^  together,  and  divided  by  4, 
would  have  been  a  woeful  error. 


OrERATION 

By  the  practical  method. 

Bales.    Cts. 

3  ®  16  =     48,«> 
8  ®  18i  =  118^ 

4  ®   19i  =     77(^- 

5  ®  20  =  1000 


20 


)  $3.73 


l^W 


Ans. 

OPERATION   BY   THE   EXACT   METHOD 


Bales. 

Pouiida. 

3 

= 

1400 

® 

16  f' 

= 

$224.00 

8 

= 

3500 

® 

\m 

= 

647.50 

4 

= 

1800 

® 

19i)* 

= 

346..50 

5 

= 

2500 

-a 

20  f 

= 

500.00 

9200 


)  $1718.00  (  18J.W,     Ans. 


79800 
73600 


6200 


85.  Suppose,  iu  the  foregoing  problem,  that  a  sale  of  the  four  lots  of  cotton 
had  been  effected  at  19/,  what  would  have  been  the  correct  increased  price  for 
each  lot  and  grade  of  cotton  ? 


Pounds. 

u 

Market  Price 

'Amount, 

1400 

® 

16  f. 

= 

$224.00 

8500 

® 

ISiJ. 

= 

647.50 

1800 

-S 

190^. 

= 

346.50 

2500 

® 

20  /. 

^ 

500.00 

9200 

$1718.00 

19/. 

$1748.00 


State 
of 

1718 

ment  to  find  price 
the  Ordinary. 

16 

1748 

State 
of  tl 

0 
1718 

ment  to  fiiul  price 
0  Low  Middling. 

f- 

37 

1748 

State 
of  th 

4 
1718 

ment  to  find  price 
B  Strict  Middling. 

f 

77 

1748 

State 
of  th 

1718 

ment  to  find  price 
e  Good  Middling. 

20 
1748 

16.279B^<'.     Ans. 

18.823.^1-.     Ans. 

19.586ifyi'.     Ans. 

20.349i-g5f.     Ans 

Explattalinn. — In  the  operation  of  this  problem,  we  first  find  the  value  of  each  lot  of  cotton  at 
■the  quoted  market  price,  and  then  by  adding  the  same  we  obtain  the  total  market  value,  §171H 
Then,  by  the  use  of  our  reason,  wo  see  that  1(5,  18i,  19i,  and  20'-'  prices  gave  $1718  value,  and  by  trans- 
position that  $1718  value  required  16,  18^.  19},  and  20''  prices,  and  as  the  whole  cotton  was  sold  for 
$1748,  we  have  this  simple  proportional  question:  if  $1718  value  requires  16,  18i,  19},  and  20(» 
prices,  what  will  $1748  value  require  f  The  line  statements  give  the  exact  answers  to  this  question, 
and  complete  the  solution. 

Had  the  cotton  been  sold  at  a  price  less  than  the  average,  the  same  method 
of  work  would  be  used  to  find  the  decreased  i^rices. 


Sjo 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


It  lias  been  said  by  many  tliat  questions  lilce  tlie  preceding  one  could  not  be 
solved,  and  it  is  claimed  by  others,  in  wliose  business  similar  questions  are  of  daily 
occurrence,  that  because  the  correct  results  often  contain  such  i7icoiivenient  frac- 
tions, the  correct  soluticm  is  of  no  advantage.  The  method  in  general  use  is  the 
guessing  metliod ;  a  method  of  solving  arithmetical  questions  that  we  cannot  indorse. 
By  the  guessing  metliod  of  finding  the  increased  jirices,  injustice,  nnintentional  of 
course,  is  done  to  some  of  the  owners  of  the  different  lots  of  cotton,  for  the  reasoa 
that  the  guessing  cannot  be  made  exactly  in  the  correct  proportion;  and  hence,  even 
though  the  guessing  prices  absorb  the  exact  amount' received,  one  owner  will  have 
more  and  anothei'  less  than  his  correct  iiroi)ortional  increase. 

By  our  correct  solution  of  this  character  of  questions,  we  admit  that  the 
fractious  in  the  results  are  often  impracticable  according  to  custom,  but  by  these 
correct  results,  extending  the  figures  to  two  or  three  places  of  decimals,  we  cau 
readily  see  what  the  nearest  practical  fractions  are,  and  hence  do  the  nearest  possi- 
ble justice  to  all  the  owners. 

By  our  correct  method,  we  avoid  the  repeated  guessing  operations  that  are 
often  necessary  to  arrive  at  a  satisfactory  approximate  result,  and  never  have  any 
considerable  excess  or  deficit  of  money  by  reason  of  the  increase  or  decrease  in 
price. 

There  is  another  objection  that  is  sometimes  raised  ia  opposition  to  the  cor- 
rect method  of  work.  It  is,  that  because  there  is  more  cotton  of  a  high  grade  than 
there  is  of  a  low  grade,  or  vice  versa,  the  increase  or  decrease  of  price  should  nofc 
be  in  exact  proportion  to  the  quoted  market  rates.  If  this  circumstance  is  taken 
into  consideration  in  fixing  the  selling  price  of  the  different  lots,  then  we  would 
increase  or  decrease  the  quoted  market  rate  of  such  cotton  as  influenced  the  selling 
price,  and  proceed  by  the  exact  method  of  work. 

To  show  the  practical  prices  of  the  above  problem,  we  present  the  following : 


1400 

® 

16J-   =  $227.50 

Or,  1400 

® 

m 

=  $227.50 

Or,  1400 

-S 

16i 

=  $227.50 

3500 

® 

1813  =   658.43 

3500 

& 

m 

=   660.62 

3500 

® 

1811 

=   658.43 

1800 

® 

19,"e  =   352.12 

1800 

® 

19J 

=   351.00 

1800 

® 

m 

=   353.25 

2500 

® 

20|   =   509.37 

2500 

® 

20f 

=   509.37 

2500 

® 

20f 

=   509. 3T 

11747.42 


$1748.49 


$1748.55 


The  following  table  of  IGths  and  32nds  will  facilitate  the  changing  of  the 
decimals  of  the  cents  in  the  price  answers  to  practical  equivalent  common  fractions, 
in  all  problems  of  this  character : 

TABLE  OF  16THS  and  32nd8  AND  THEIR  EQUIVALENTS  IN  DECIMALS. 


sV 

= 

•= 

iV 

5^ 

= 

i 

Tftr 

= 

^ 

= 

l\ 

sV 

^= 

^  = 


.03125 

.0625 

.09375 

.125 

.15625 

.1875 

.21875 

.25 


=   T^   = 


2  2 

u  = 

1^  — 

a -J  — 

ii  = 

U  = 

M  = 


=  i 


28125 

i5 

= 

.53125 

3f 

3125 

K 

= 

0 

"1^  = 

.5625 

If 

34375 

i? 

= 

.59375 

u 

375 

n 

= 

i  = 

.625 

M 

40625 

u 

= 

.65625 

U 

4375 

it 

= 

u   = 

.6875 

M 

46875 

U 

= 

.71875 

a 

5 

u 

= 

i  = 

.75 

n 

=  ii  = 


.78125 
.8125 
.84375 
.875 
.90625 
.9375 
.96875 
1. 


PARTNERSHIP    SETTLEMENTS. 


S7r 


DlVISIOiSr  AND  PROPOETIOXAL  DIYISIOX. 

SC.    A  benevolent  man  lias  $000,  of  which  he  wishes  to  give  to  A.  J,  to  B.  ^, 
to  C.  I  and  to  I).  ^.     What  amount  will  each  receive  ? 

Ans.  A.  $200.     B.  $150.     C.  $120.     U.  $100. 

OPERATION. 

A's  one-tliinl   of  $600  =  $200 

B's  one-fourth  of     600  =  150 

C's  one-fifth      of     600  =  120 

D'a  one-sixth    of     600  =  100 

$570    total  amount  donated. 
30    remaining  on  hand. 

$600 

87.    Another  benevolent  man  has  $600  which  he  wishes  to  give  to  W.,  X.,  T, 
and  Z.,  in  jiroportion  to  ^,  ^,  A  and  ^.    What  amount  would  each  receive  ? 

Ans.  W.  $210.53.     X.  $157.89.     T.  $126.32.     Z.  $105.26. 

OPERATION. 


J  =:  Ws  proportion. 

i  =  X's  ■" 

i  -=  Y'8 

i  =  Z's 

iJ  =  the  sum  of  the  propor- 
tions which  are  to  re- 
ceive  the  amount   of 


Statement  to  find  Ws  propor- 
tional share. 


19 
3 


600 
20 


$210.53    Ans. 


Statement  to  find  X's  projior- 
tional  share. 


19 

4 


600 
20 


$157.89 


Statement   to  find  Y's 
proportional  share. 


19 


600 
20 


$126.32 


Statement  t^  find   Z's 
proportional  share. 


19 
6 


600 
20 


$105.26 


Explanation. — The  operation  of  this  proldem 
is  so  simple  that  special  exphiuation  is  deemed 
unnecessary.  The  preceding  exanijde,  No.  86,  is 
often  given,  when  the  answer  to  this  example  is 
required.  To  show  clearly  the  difference  between 
giving  certain  parts  of  a  sum  of  money  to  ditfer- 
eut  parties,  and  giving  the  whole  sum  of  money 
in  proportion  to  certain  parts,  we  have  presented 
the  two  preceding  questions. 


88.  A  planter  has  seventeen  horses  which  he  wishes  to  give  to  his  three  sons, 
Henry,  William  and  Robert,  in  proportion  to  i,  J  and  J.  How  many  horses  will 
each  receive  1  Ans.  Henry,  9.     William,  6,     Robert,  2. 


OPERATION. 

i  =  Henry's  proportion, 
i  =  William's          " 
i  =  Robert's            " 

■1^  ^  the  sum  of  the  pro- 
portional shares 

I 
B 

17 
o 

lENRY. 

orses. 
17 
18 

E 

17 
3 

'ILUAM. 

orses. 
17 

18 

E 

B 

17 
9 

OBERT 

[orses. 
17 
18 

which    are    to   re- 
ceive the  17  horses. 

9.     Ans. 

6.     Ans. 

2.     A 

89.    What  are  the  respective  amounts  resulting  from  the  division  of  $77  pro. 


portionally  to  J,  i,  i,  i  ? 


Ans.   $30,  $20,  $15,  $12. 


S7 


SOULE  S    PIIILOSUPHIC    PRACTICAL    MATHEMATICS. 


To  find 
n 

■ 
13 

the  amount  A. 
lUst  pay. 

$ 
10 
10 

To  tiud 
15 

the   amount  Z 
lust  pay. 

$ 
10 
5 

$6.66|.      Ans. 

$3.33i.      Ans. 

SETTLEMENTS  WITH   SPECIAL   CONDITIONS. 

90.  A.  and  Z.,  residents  of  New  Orleans,  liired  a  team  to  go  to  tlie  lake,  a 
distance  of  five  miles  and  return.  The  clnu  iLje  for  the  nse  of  the  horses  and  car- 
riage was  $10.  On  arriving  at  the  lake,  Z.  proposed  to  return  by  some  other  convey- 
ance, to  which  proposition  A.  agreed.  They  also  agreed  to  settle  the  charges  for  the 
use  of  the  horses  aud  carriage  in  pi-oportiou  to  the  distance  that  each  rode.  What 
must  each  pay  ?  Ans.  A.  f  6.G6§.    Z.  $3.33^. 

OPERATION. 


A.  rode  to  the  lake  and  returned  =  10  miles. 
Z.         "        "     lake  =    5  miles. 

Total  number  of  miles  ridden 

by  both        -        -        -        -    =  15  miles. 


91.  Suppose  in  the  above  problem,  that  they  had  agreed  to  settle  the  charge 
for  the  use  of  the  horses  and  carriage  justly,  what  would  each  have  paid  ? 

Ans.    A.  $7.50.     Z.  $2.50. 

SOLUTION. 

The  distance  to  the  lake  and  return  being  ten  miles,  and  the  charge  for  the  use  of  the  horses 
and  carriage  being  $10,  it  is  clear  that  ten  miles'  ride  costs  $10,  and  if  ten  miles'  ride  cost  JilO,  one 
mile  ride  will  cost  the  10th  part,  which  is  ($10  -H  10)  |1.  Then  A.  and  Z.  having  ridden  together  to 
the  lake,  a  distance  of  five  miles,  injustice,  they  are  required  to  settle  equally  the  expense  of  the 
ride  that  they  have  equally  enjoyed.  And  the  expense  for  the  five  miles  at  $1  per  mile  is  $5,  which, 
settled  equally,  is  ($5  ^  2)  $12.50  that  A.,  and  $2. .50  that  Z.  must  pay.  This  settles  the  expense  to  the 
lake,  where  Z.  retires  and  A.  becomes  solely  responsible.  The  expense  for  the  five  miles'  ride  by  A. 
from  the  lake  to  the  city,  at  $1  per  mile,  is  $5,  which  justice  requires  A.  to  settle,  .as  he  enjoyed  the 
ride  alone.  We  thus  obtain  $2.50  -|-  $5  =  $7.50  that  A.  would  have  paid,  and  $2.50  that  Z.  would 
have  paid,  had  they  settled  Jusf/y. 

Or,  thus : 

A.  and  Z.  having  ridden  together  to  the  lake,  which  is  -,%  of  the  whole  distance,  it  is  clear 
that  they  should,  equally,  pay  i^  of  the  whole  cost.  The  whole  cost  being  $10,  -j^,!  of  it  is  .$5,  and  + 
of  $5  is  $2.50,  the  amount  each  should  pay  for  riding  to  the  lake.  Then,  as  A.  rode  the  remaining 
^  of  the  whole  distance  alone,  it  is  clear  that  ho  should  alone  pay  the  remaining  -fi^  of  the  whole 
■expense  ;  -^  of  IjilO  is  |5,  which  added  to  the  $2.50  first  obtained  makes  $7.50  that  A.  is  to  pay.  This 
completes  the  solution,  and  gives  $7.50  as  A's  share  of  the  expense,  aud  $2.50  as  Z's  share  of  the 
•expense. 

92.  A.  and  Z.  bought  merchandise  to  the  amount  of  $800,  of  which  sum  A. 
paid  $500  and  Z.  paid  $300.  Soon  after  they  sold  to  G.  J  of  the  whole  stock  for 
$400.  How  much  of  the  $400  must  A.  and  Z.  receive  respectively,  in  order  to  con- 
stitute each,  A.,  Z.  aud  G.  J  owner  of  the  goods  ?  Aus.  A.  $350.     Z.  50. 


FIRST   OPERATION. 


Capital    Am't  of  invoice  Am't  of  iu- 
contrib-     each  retained,    voice  each 
uted.  sells. 


A. 
Z. 


300 


=     $233i 
266|     =        33i 


i  of  800 
$266J 


amount  of  invoice  sold. 


Statement  to  find 
amount  due  A. 


800 
3 


400 

3 

700 

$350.  Ans. 


Statement  to  find 
amount  due  Z. 


400 
800  I    3 
3      100 


$50.     Ans. 


PARTNERSHIP    SETTLEMENTS. 


S73 


SECOND   OPERATION. 


Amoimt  paid.     Tart  owned.    Part  retained. 
A.         -         $500         =         f         —         ^"4 

Z.        -  300        =        I        —        A        = 

4of  )g(A- 


Statement  to   find 

Statement  to    find 

sold 

amount  due  A. 

amount  due  Z. 

A- 
^ 

1 
400 

400 

— 

8 

24 

8 

24 

A 

24 

7 

24 

$350 

$50 

93.  Two  newsboys,  James  and  Henry,  on  taking  their  seats  to  eat  some 
cakes  that  they  had  inirchased,  were  requested  by  another  member  of  their  ft-ater- 
uity  named  "  Diek,"  to  allow  him  to  dine  with  them.  To  this  request,  they  cheer- 
fully assented,  and  the  three  ate  the  cakes.  When  they  had  finished,  Dick  laid  ou 
the  table  40/  for  his  share  of  the  dinner.  How  nuich  of  this  40/  must  James 
receive,  who  contributed  5  cakes  to  the  dinner,  aud  how  much  nuist  Henry  receive, 
■who  contributed  3  cakes  1  Aus.  James  35/.     Henry  5/. 


OPERATION. 


James     5 
Heury    3 

J  of  8 

2J 


=    2*     =     5 
=      i    =    i 


JAMES. 


40 
3 

7 

35>» 


40 
3 


5,i» 


94.    A  cotton  factor  insured  : 

200  bales  of  cottou  for  3  months  at  1J%  ; 

150     "       "         "       "    2       "         atli%; 
100     ''       "         "       "    1       <'         at    %%; 

What  was  the  average   rate  per   cent,  of  insiu'ance,  supposing   that   each   bale 


■was  of  the  same  value  ? 


Ans.  l,29i%. 


FIRST   OPERATION. 


Bales. 

Per  cent. 

on  1  bale. 

200 

® 

li"o 

=         300.00 

1.50 

® 

n% 

=        206 

25 

100 

® 

=          75 

00 

450 


Bales.      Assumed  value. 


)  581.25  (  1.29^%    Ans. 
SECOND   OPERATION. 


Insurance. 


200  ®  $100  =  $20000  ®  U%  =  1300.00 
150  -S  100  =  15000  ®  1|%  =  206.25 
100     ®       100     =       10000     ®       i<!<    =        75.00 


145000 


$581.25 


Operation  to  tind  the  rate  per  cent. 
45000 


581.25 
100 


Aus. 


95.     A.  contracted  with  Z.  to  deliver  to  him : 

78  bales,  38.531  pounds.  Good  Middling  cotton,  at  lo.J/, 
aud  72      "       36500        "        Middling  "       at  14/; 

(Continued.) 


874 


SOULE  S    nilLOSOPHIC    PRACTICAL    MATHEMATICS. 


but  A.  having  disposed  of  the  foregoing  cotton  to  other  parties,  by  subsequent 
agreement  A.  delivei'ed  and  Z.  received: 

2475  jjounds  of  Inferior  at    9/ ; 

6681         "       "   Ordinary  at  10/; 

12219         "       "   Good  Ordinary  at  12/; 
7218        "       "   Low  Middling     atl3|/; 

Z.  then  demanded  the  balance  due  him  in  Middling  and  Good  Middling  cotton. 
How  many  pounds  of  each  grade,  Middling  and  Good  Middling,  were  due  him  ? 

Ans.  25527 1-4  iiounds  Middling. 

26917  If      "        Good  Middling. 

OPERATION. 

Balea. 
78 
72 


= 

38531     pounds  Good  Middliug  at 

ISif? 

$5972.30 

= 

36500          "       Middling              at 
Total  value        -        -        -        - 

Wi 

5110.00 

$11082.30 

KECEIPTS. 

2475 

pounds  Inferior                 at      9? 

$222.75 

6681 

"       Ordinary              at    lO* 

668.10 

2219 

"        Good  Ordinary  at    12!< 

1466.28 

7218 

Low  Middling    at    13i^' 
Total  receipts 

974  43 

3331.56 

Balance  due  ... 

wliicL  is  to  lie  paid  in  Good  Middling  and  Middling  at  contract  price, 
in  proportion  to  $5972.30  and  $5110. 


$7750.74 


Statement  to  find  the  amount  to  be  paid  in 
Good  Middliug. 


11082.30 


7750,74 
5972.30 


$4176.91 


$4176.91  ■ 


■  15i<*  =  26947H  lbs.  of  Good 
Middliug  cotton. 


Statement  to  find  the  amount  to  be  paid  in 
Middliug. 


11082.30 


7750.74 
5110 

$3573.83 


$3573.83  —  14<?  =  25527,^  lbs.  of 
Middliug  cotton. 


SETTLING  ESTATES. 


9G.  A  man  had  two  sons  and  three  daughters;  to  the  youngest  son,  he  gave 
i  of  li  times  $3000,  which  was  f  of  the  elder  son's  share,  and  what  the  elder  son 
received  was  f  of  J  of  the  whole  estate ;  the  balance  was  divided  among  his  three 
daughters  in  reciprocal  proportion  to  their  ages,  8,  10, 12  years.  What  was  each 
daughter's  share  ?  Ans.  $7114.87  for  the  one  8  years  old. 

5091.89     "         "     10  " 

4743.24     "         "     12  " 

SOLUTION. 

i  of  n  =  f,  and  f  of  $3600  =  2700  =  the  younger  son's  share  ;  then  if  $2700  is  j  of  the  elder 
son's  share,  |  is  the  i  of  it, -which  is  $1350,  and  J  or  the  whole  share  is  5  times  as  much,  which  is 
$6750;  then  if  $6750  is  t  of  |  =  J  of  the  estate,  4  or  the  whole  estate  is  4  times  as  much,  which  is 


PARTNERSHIP    SETTLEMENTS. 


S75 


$27000.     Then  the  two  sons'  shares,  $2700  +  $6750  =  $9450,  deducted  from  the  whole  estate,  $27000, 
leaves  $17550,  helougmg  to  the  three  daughters  iu  reciprocal  proportion  to  their  ages,  H,  10  aud  12 


years. 


The  reciprocal  of  8  is     ^ 

"   10  is     tV 
"   12  is    tV 

i     +     To     +    TZ     =     "ft'cT 


Having  now  tbe  reciprocals,  the  following  statements  give  the  respective 
shares  of  each  daughter : 


37 


17550 
120 


$7114.87 


37 

10 


17550 
120 


$5691.89 


37 
12 


17550 
120 


1743.24 


To  perform  the  first  part  of  the  solution  without  the  reasoning,  we  would  make 
the  following  statements : 


Statement  to  find  the  younger 
sou's  share. 


3600 

3 

$2700 


Statement  to  find  the  elder 
son's  share. 


2700 
5 


$6750 


Statement  to  find  the  whole 
estate. 


6750 

7 


$27000 


97.  A.,  B.  and  C.  were  to  receive  $G000  in  proportion  to  ^,  J  and  i ;  but  B. 
"having  died,  it  is  required  to  divide  the  money  between  A.  and  C.  What  sum 
should  each  receive?  Ans.  A.  $4000.    C.  $L'000. 


OPERATION. 


i   +    i    = 


6000 


$4000    A's  share. 


6000 


$2000    C's  share. 


Explanafton. — B.  having  died,  his  J  pro- 
portional interest  is  therefore  divided  be- 
tween A.  and  C.  in  proportion  to  their  pro- 
portional interests,  \  and  ^ ;  we  therefore 
hare,  as  .shown  in  the  operation,  hut  to  di- 
vide the  whole  sum  in  projiortioutoland^. 


98.  A  father  willed  his  estate  valued  at  $40000  to  his  three  children  in  pro- 
portion as  follows :  John  J,  Henry  I  and  Katie  \.  Before  the  settlement  was  made, 
Henry  died.     What  sum  should  John  and  Katie  each  receive  ? 

Ans.  John  $25000.    Katie  $15000. 


OPERATION. 


*     +     * 


the  Slim  of  the  proportional 
interests  of  John  and  Katie. 


Then 


40000 
15 


$25000 


40000 
15 


$15000 


Note. — Henry  having  died,  his  J  proportional  interest  is  therefore  to  be  divided  between  John 
and  Katie  in  proportion  to  their  proportional  interests,  as  shown  in  the  operation. 


876  soule's  philosophic  practical  mathematics.  * 

99.  A  man  at  liis  death  U'f't  an  estate  amoniitins  to  $o5()00,  to  liis  ■wife  and 
two  eliiklren,  a  son,  and  daughter.  His  eliildieii  being  absent  in  Euro[)e,  hedirected 
by  will,  that  if  his  son  retni iied,  his  wife  shonld  ha\e  J  of  the  estate,  and  the  son 
the  remainder;  but  if  the  daughter  returned,  his  wife  should  have  §  and  the  daugh- 
ter the  remainder.  i?ow,  it  so  happened  that  they  both  returned.  What  sum 
should  each  receive  ? 

Aus.  $20000  sou.     $10000  wife.     $5000  daughter. 

SOLUTION. 

Bj'  the  first  condition  of  the  will,  the  son  was  to  receive  twice  as  much  as  the  wife ;  and  liy  the 
second  condition  the  wife  was  to  receive  twice  as  much  as  the  daughter.  Hence  to  comply  with  both 
conditions,  we  assume  1  to  represent  the  daughter's  share  ;  then  twice  1,  or  2  will  be  the  wife's  share, 
and  twice  2,  or  4  will  be  the  son's  share ;  and  l-(-2-|-4  =  7  =  the  sum  of  all  the  shares.  Then  4  of 
$35000  =  $5000,  the  daughter's  share;  t  of  $35000  =  $10000,  the  wife's  share;  and  f  of  $35000  = 
$20000,  the  son's  share. 

100.  A  father  leaves  a  number  of  children  and  a  certain  sum  of  money  to  be 
divided  among  them  as  follows:  The  first  is  to  receive  $100  and  yg  of  the  remain- 
der ;  the  second  is  to  receive  $200  and  -^  part  of  what  then  remains ;  again,  the 
third  is  to  receive  $300  and  -^^  of  the  residue,  and  so  on  ;  each  succeeding  child  is 
to  receive  $100  more  than  the  one  immediately  i)receding,  and  then  ^  part  of  what 
still  remains.  At  last  it  is  found  that  all  the  children  have  received  equal  sums. 
What  was  the  amount  left,  and  how  many  children  were  there  ? 

Ans.  The  sum  divided  was  $8100  and  the  number  of  children  9. 

SOLUTION. 

After  giving  the  first  $100,  there  remained  the  sum  minus  $100,  and  i^  of  this  is  i^  of  the  sum 
minus  $10;  heuco  the  first  received  $100  -f-  i^g  of  the  sum  minus  $10  or  $90  +  A  of  the  sum,  and 
there  remained  of  the  whole  -^  of  the  sum  minus  |90;  giving  the  second  $200,  there  remained  -1^^  of 
the  sum  minus  $290  and  -[\,-  of  this  is  tIj];  of  the  sum  minus  $29  ;  hence  the  second  received  $200  -f- 
V?n  of  the  sum  minus  $29  or  $171  +  jh  of  the  sum. 

Now  according  to  the  conditions  of  the  question,  the  first  received  the  same  as  the  second- 
hence  $90  -f  1-^,  of  the  sum  =  $171  +  rSis  of  the  sum  or  ,■!,;{,-  of  the  sum  =  ^ Sji  of  the  sum  +  $81,  or 
an-ain,  rhn  of  the  sum  =  $81 ;  hence  the  sum  =  $8100.  Now  the  first  received  $100  +  Ai  of  what 
remained  or  $100  -f  $800  =  $900. 

But  since  each  child  is  to  receive  an  equal  sum,  there  must  be  as  many  children  as  $8100  is 
equal  to  $900  which  is  9. 

PEOPOETIONAL  DIVISION. 

101.  Divide  IS  oranges  between  Katie  and  Sallie,  so  that  Katie  will  have  J 
jaore  than  Sallie.    What  number  will  each  receive  ? 

Ans.  Katie  10.     Sallie  8. 

FIKST  OPERATION. 

1    =  the  assumed  niunber  gi\en  to  Sallie.  Statements  to  find  the  proportional 

Ij  =  "  "  "        Katie.  share  of  each. 

2i  =  the  sum  of  the  ratio  numbers  by  which  SALLIE.  Katie. 


the  oranges  are  to  be  divided 


18  !  18 

4 
5 


10 


PARTNERSHIP    SETTLEMENTS. 


877 


SECOND   OPERATION. 


4  =  the  assumed  niimber  giveu  to  Sallie. 

5  =  "  "  "        Katie. 

9  =  the  sum  of  the  ratio  numbers. 


SALLIE. 

18 


KATIE. 

18 
5 

10 


102.  A  young  liusband  has  one  and  a  foiu'th  dozen  ai)j)les  which  he  Irishes  to 
divide  between  his  "wife  and  his  mother-in-hiw,  so  that  his  mother-iu-law  will  have 
g  less  than  his  wife.    How  many  will  he  give  to  each  ? 

Aus.  Wife  Hi.    Mother-iu-law  3|. 


Statements  to  find  the  share  of  each. 


4 

p-E. 
15 
3 

MOTHER-IX-LAW 
15 
4      3 

iii 

3 

3i 

OPERATION. 

1    ==  the  assumed  number  given  to  his  wife. 

J    =  "  "  "  mother-in-law. 

\\  =  the  sum  of  the  ratio  numbers,  by  which  the  a^j- 
ples  are  to  be  divided. 


Note. — If  preferred,  3  and  1  may  be  assumed  as  the  ratio  numbers,  in  place  of  1  and  J. 

103.  Divide  $2000  between  A.,  B.  and  C,  so  that  A's  part  will  be  to  B's  as  2 
to  3,  and  that  C.  will  have  as  much  as  A.  and  B.  together,  less  $200.  What  amount 
will  each  receive?  Ans.  A.  $440.    B.  $660.    C.  $900. 

SOLUTION. 

As  C.  is  to  have  as  much  as  A.  and  B.  together,  less  $200,  A.  and  B.  will  therefore  have  i  of 
$2000  +  +  of  $200,  which  is  $1100.  Then,  as  A.  and  B.  are  to  receive  this  in  the  proportion  of  2  to 
3,  A.  will  therefore  receive  |  and  E.  f  of  $1100.  j  of  $1100  =  $440  A's  share ;  f  of  §1100  =  $600  Ba 
share ;  and  lastly,  as  C  is  to  receive  as  much  as  A.  and  B.  together,  less  $200,  he  will  therefor©' 
receive  $1100  —  200  =  900. 

104.  A.  and  B.  made  an  investment  in  real  estate,  A.  invested  $26217.50  and 
B.  invested  $1707.40. 

From  the  proceeds  of  a  sale  of  a  portion  of  said  real  estate,  A.  drew  out 
$15289.25  and  B.  drew  out  $2908.95. 

B.  now  desires  to  be  an  equal  owner  with  A.  in  the  remainder.  How  much 
must  B.  pay  A.  to  be  au  equal  owner  ? 


OPERATION. 


A.  invested, 
A.  drew  out. 


A's  net  investment  of  which  B.  owes  A.  one-half, 

B.  invested, - 

B.  drew  out,  -        -        -        -        -        -   '     - 


$1707.40 
2908.95 


B's  net  withdrawals  of  which  he  owes  one-half  to  A. 


Total  amount  of  which  B.  owes  A.  one-half, 
Amount  li.  must  pay  to  A.  •        - 


$26217.50 
15289.25 

$10928.25 


1201.55 

i  )  $12129.80 
$()0G4.9O 


A's  total  investment, 
A's  withdrawals, 
A.  received  from  B. 


A's  i  net  investment. 


$26217.50 

21354.15 
$4863.35 


B.  invested, 
B.  paid  to  A. 

B's  total  investment, 
B.  drew  out, 


B's  net  investment, 


$1707.40 
6064.90 

$7772,30 
2908.95 

$4863.36- 


'8/8  soule's  philosophic  practical  mathematics.  * 

THE  LAND  AND   THE  DITCH  PEOBLEM. 

105.  A.  and  B.  paid  $C00  ($300  eacli)  for  300  acres  of  land.  In  dividing  tlie 
land  according  to  quality,  it  was  agreed  tliat  A.  should  have  the  best  quality,  and 
pay  therefor  75  cents  i)er  acre  more  than  B.  How  many  acres  did  each  receive,  and 
what  price  per  acre  did  each  pay  ? 

Ans.  A.  received  122.8  acres  and  paid  $2,443  +  per  acre. 
B.         "         177.2      "  "  1.693  +         " 

OPERATION. 

300  ®  75/  =  $225 ;  and  $000  —  225  =  $375  reserved  remainder.  Then  $225 
X  4  X  300  =  $270000  to  which  is  added  375^  =  140G25;  270000  +  140G25  = 
V410025  =  G40.8  + 

Then  G40.8  +  375  =  1015.8,  which  -  (300  x  2)  =  600  =  $1,093  +  the  price 
per  acre  paid  by  B.  Then  $1,693  +  75  cents  =  $2,443  +  the  price  iier  acre  paid 
by  A. 

Then  $300  -v-  1.693  =  177.2  acres  that  B.  receives;  and  $300  ^  $2,443  =  122.8 

acres  that  A  receives. 

Explanation. — This  and  the  following  proljlem  belong  strictly  to  algebra,  and  although  they 
may  be  solved  by  arithmetical  operations  deduced  from  algebraic  equations,  they  cannot  be  as 
analytically  explained  as  regular  arithmetical  problems.  The  solution  of  all  problems  of  this 
character  may  be  performed  by  observing  the  following  directions :  First  find  the  cost  of  the  whole 
number  of  acres  at  the  diflerence  between  the  prices  per  acre;  then  subtract  this  from  the  price  paid 
for  the  whole  landand  reservethe  remainder.  Then  multiply  the  value  produced  by  the  difference 
in  price  by  4  times  the  whole  amount  paid  by  him  who  paid  the  least  per  acre,  aud'to  the  product, 
add  the8c|uare  of  the  reserved  remainder;  then  extract  the  8(|uareroot  of  this  sum,  and  to  it  add 
file  resers'ed  remainder.  Then  divide  this  sum  by  twice  the  whole  number  of  acres,  and  the  quotient 
will  be  the  price  per  acre  paid  by  him  who  paid  the  least  Then  to  this  price  add  the  difference 
between  the  prices  paid  and  the  sum  will  be  the  price  paid  by  him  who  paid  the  greater  price. 
Then  divide  the  amount  jiaid  by  each  by  the  respective  price  paid  by  each,  and  the  quotients  will 
be  the  number  of  acres  that  each  is  to  receive. 

It  will  be  observed  that  there  is  in  the  problem  an  interminable  decimal  which, 
because  it  arises  in  extracting  a  root,  cannot  be  expressed  as  a  vulgar  fraction.  It 
may,  however,  be  ap])roximated  to  any  extent. 

Problems  of  this  character  are  often  misstated  by  giving  either  the  price  that 
each  is  to  pay  or  the  number  of  acres  that  each  is  to  receive.  To  give  either,  ren- 
ders the  problem  absurd  and  insolvable. 

106.  A.  and  B.,  together,  agree  to  dig  100  rods  of  ditch  for  $100.  The  part 
of  the  ditch  on  which  A.  was  employed  was  more  diflicult  of  excavation  than  the 
part  on  which  B.  was  employed ;  and  it  was  therefore  agreed  tliat  A.  should  receive 
for  each  rod  25  cents  more  than  B.  received  for  each  rod  that  he  dug.  How  many 
rods  must  each  dig,  and  at  what  price  so  that  each  may  receive  Just  $50  ? 

Ans.  B.  receives  $  .8903  +  per  rod,  and  digs  56.16  rods. 
A.         "        $].]403  per  rod,  and  digs  43.84  rods. 

OPEEATTCN. 

100  X  25/  =  $25;  and  $100  —  $25  =:  $75,  reserved  remainder.  Then  $25  x 
4  X  50  =  5000  to  which  we  add  75=  =  5625;  5000  +  5625  =  10625,  the  square  root 
of  which  is  103.07  + 

Then  103.07  +  75  =  178.07,  which  -^  (100  x  2)  =  200  =  $.8903  +  the  price  per 
rod  received  by  B.     Then  $.8903  +  25/  =  $1.1403  the  price  per  rod  received  by  A. 

Then  $50  -h  $.8903  =  56.16  rods  that  B.  digs;  and  $50  -4-  $1.1403  =  43.84 
rods  that  A.  digs. 


■*  PARTNERSHIP    SETTLEMENTS.  879 

107.  A.,  B.,  C.  and  D.  agree  to  do  a  piece  of  work  for  $5500.  A.,  B.  and  C. 
can  do  it  in  20  days ;  B.,  0.  and  D.  in  24  days;  C,  D.  and  A.  in  30  days ;  and  D., 
A.  and  B.  in  3G  days.  In  Low  many  days  can  all  do  it,  working  together;  in  liow 
many  days  can  each  do  it,  working  alone;  and  what  part  of  tbe  pay  ought  each  to 
receive  1  Ans.  See  the  solution. 

SOLUTION. 

As  A.,  B.  and  C.  can  do  the  work  in  20  days,  they  can  do  -^^  of  it  in  1  day ;  as  B.,  C,  and  D. 
can  do  it  in  24  days,  they  can  do  -^  of  it  in  1  day;  as  C.,  D.  and  A.  can  do  it  in  30  days,  they  can 
do  j^o  of  it  in  1  day  ;  and  as  D.,  A.  and  B.  can  do  it  in  36  days,  they  can  do  -/„■  of  it  in  1  day ;  there- 
fore, A.,  B.,  C.  and  D.,  by  working  3  days  eacli,  will  do  ,/^  +  -i,-  +  -jV  +  5^-  =  |^  of  tlie  work,  and 
in  1  day  they  will  do  j  of  4i  =  -sYb ',  'in'l  as  tliey  all  do  /,-V  of  tbe  work  in  1  day,  it  will  require  as 
many  days  to  do  the  whole  work  as  1  is  times  equal  to  ^r^,  which  is  (1  -h  ^,^  =)  IOiY  days. 

Now,  hy  deducting  successively  the  work  that  three  can  do  in  1  day,  when  they  work 
together,  from  what  the  four  can  do  in  1  day,  we  obtain  the  tvork  that  each  one  will  do  iu  1  day ; 
thus 

^iV  —  /u  =  ToVff   the  -svork  that  D.  can  do  iu  1  day. 

T^^  —  l\  =  ToT    the  vrork  that  A.  can  do  in  1  day. 

TTiS  —  'an  =  Ttihs  the  work  that  B.  can  do  in  1  day. 

■j'/ff  —  TiV  =  -E^b     the  work  that  C.  can  do  in  1  day. 
Having  now  the  work  that  each  can  do  in  1  day,  it  is  clear  that  it  will  require  each  as  many- 
days  to  do  the  work  alone  as  1  is  equal  to  the  jiart  of  the  work  performed  by  eachj  thus 

1  -:-  TifsTy  gives  1080  days  that  it  would  require  D. 

1  —  ToT  gives    108  days  that  it  would  require  A. 

1  -r-  tobtt  gives  56|f  days  that  it  would  require  B. 

1  H-  lis  gives  43^    days  that  it  would  require  C. 
In  dividing  the  money,  it  is  evident  that  each  should  receive  such   a  part  of  the   whole 
amount  paid  as  he  did  of  the  whole  work. 

This  is  shown  by  multiplying  the  amount  of  work  performed  hy  each  daily  by  the  number 
of  days  that  all  worked  j  thus 

^-     rhs   X  191^  =  -fr    A's  proportion. 

B.  -rihXlS-h^U     B's 

C.  jh  X  19A-  =  i\    C's 

D.  -nsVu  X  19A-  =  -bV    D's 
According  to  these  proportions  the  money  is  divided  thus : 

1^1-  of  $5500  is  $1000,  A's  share. 
^1  of    5500  is    1900,  B's      " 
~fr  of    5500  is    2500,  C's      " 
jV  of    5500  is      100,  D's      " 


Total  amount,  .....        $5500 

111.  A  young  hoodlum,  a  modern  evolvement  of  the  human  race,  stole  a  bas- 
ket of  peaches  and  divided  them  among  three  brother  hoodlums  and  himself  as 
follows:  To  the  first,  he  gave  i  of  the  whole  number  and  ^  of  a  peach  more;  to  the 
second,  he  gave  J  of  what  remained  and  ^  of  a  peach  more ;  to  the  third,  he  gave  J 
of  what  remained  and  J  of  a  peach  more.  The  stealer  retained  what  was  then  left, 
for  himself,  which  was  J  the  number  he  gave  to  the  first  hoodlum.  What  was  the 
number  of  peaches  stolen,  and  how  many  did  each  hoodlum  receive? 

Ans.  7  peaches  were  stolen.    1st  hoodlum  received  2. 

2d         "  "        2. 

3d         "  "        2. 

The  stealer  had  1. 

SOLUTION. 
In  all  problems  of  this  kind,  the  following  principle  governs:  That  when  the  fractional  parts 
of  the  successive  portions  are  consecutively  increasing,  each  of  the  parts  or  shares  of  the  several 


SSo 


soule's  niiLosoriiic  i'r.'\.ctical  m.\tiie.matics. 


porsons  are  equal,  except  the  last  one,  -which,  from  the  nature  of  thing.s,  must  be  one  less  than  tho 
averan-e.  Henoo  in  tlie  given  jirolilem,  the  fact  stated  that  the  last  one's  share  is  half  of  the  first 
(and  each  of  the  others  also)  and  from  the  above  principle  that  it  must  be  one  less,  it  nni.st  neces- 
tarily  be  OXE,  and  since  each  of  the  others  is  one  greater,  they  must  be  two  each,  and  since  there 
are  three  persons  having  equal  portions,  there  will  be  6  +  the  1  for  tho  last,  making  7  in  all. 
Hence,  to  solve  problems  of  this  kind :  add  1  to  the  last  number,  (ONE),  and  multiply  by  the  num- 
ber of  persons,  and  then  subtract  1  from  the  product. 

112.  Three  persons,  say  A.,  B.  and  C,  buy  a  piece  of  laiul  for  $45G9,  and  the 
parts  for  which  they  pay  bear  the  following  jiroportioiis  to  each  other,  viz  :  the  sum 
of  the  first  and  second,  the  sum  of  the  first  and  third,  and  the  sum  of  the  second 
and  third,  are  to  each  other  as  |,  f ,  and  xo.  How  much  did  each  pay,  and  what  part 
did  each  own?  Ans.  A.  paid  $1015^  and  owns  f. 

9* 
4. 
9* 


B. 

u 

1523 

a 

0. 

u 

2030| 

u 

OPERATION. 

A.  and  B. 

=    i    =    -f J    =    5  shares. 

A.  and  C. 

=     ?     =    ,s,     =     6       " 

B.  and  C. 

=    iV  =    /j    =    7       " 

18  shares  ;  but  as  each  party's  share  has  been 
counted  twice,  we  therefore  divide  the  same  by  2  and  obtain  9  shares. 

Then,  to  find  the  respective  shares   of  A.,  B.  and  C,  we  deduct  from  the  sum  of  the  share* 
the  number  of  shares  o%vned  by  each  two  of  the  parties,  thus  : 


A.,  B.  and  C.  own  9  shares. 
A.  and  B.  own        5      " 


A.,  B.  and  C.  own  9  shares. 
A.  and  C.  own         6      " 


A.,  B.  and  C.  own  9  shares. 
B.  and  C.  own         7      " 


C  owns 


B.  owns 


A.  owns 


Having  the  respective  shares  of  each  party,  we  make  the  followiag  state- 
ments and  produce  the  amount  due  to  each : 

A's  |.  B's  |.  C's  5. 

A.  2  shares 

4569 

2  9 


B.  3 

C.  4 


$1015i 


4569 
3 

•11523 


4569 

4 

$2030J 


113.  Three  men  take  an  interest  iu  a  mining  company.  A.  put  in  $480  for  6 
months ;  B.  a  sum  not  named  for  12  months ;  and  G.  $320  for  a  time  not  named : 
when  the  accounts  were  settled,  A.  received  $600  for  his  stock  and  i)roflts ;  B.  $1200 
for  his;  and  G.  $520  for  his.    What  was  B's  stock,  and  C's  time? 

Ans.  B's  stock  $800.     C's  time  15  mos. 


Investment. 

A.  $480. 

B.  ? 

Time  invested. 

6  mos. 

12  mos. 

Am*t  rec'd  in  settlement. 
$  600 
1200 

Gain. 
$120     = 

1 

m 

=    i  of  stock 

C.    320. 

l 

520 

200 

OPEEATION. 

Since  A's  investment  of  $480  gained  $120  in  6  months,  in  12  months,  the  time  B's  was  invested^ 
it  would  gain  twice  as  much  or  .$240,  which  is  jJJ  =  |  of  his  investment. 

Then,  since  the  gain  of  each  is  iu  like  proportion,  the  $1200  received  iu  settlement  by  B.  for 
l2'months  is  i  of  his  investment  of  stock;  and  since  $1200  is  3  of  B's  investment,  i  is  the  third  part 
or  $400,  and  two  halves  or  the  whole  investment  is  $800,  and  his  gain  is  $1200  —  $800  =  $400. 

Now  having  B's  investment  $800,  and  his  gain  for  12  months  $400,  and  also  C's  investment 


PARTNERSHIP    SETTLEMENTS. 


88i 


Moiitbs. 
12 

Investment. 
B's         $800 
C's          320 

Gain. 

$400 

200 

Months. 
12 
1 

320 
400 

800 
200 

15  nios. 

$320,  and  his  gain  $200,  to  find  C's  lime,  vre  make  the  following  classiacatiouaiul  proportioual  state- 
mout : 

CLASSIFICATION.  STATEMENT. 

"\fmitlis. 

Explanation. — In  making  the 
statement,  we  reason  as  fol- 
lows :  Since  an  |800  invest- 
ment requires  12  months  time 
to  produce  $400  gain,  $1  would 
require  800  times  as  many 
months,  and  §320  investment  the  320th  part.  Then,  since  $400  gain  requires  12  mouths  time  to  pro- 
duce it,  $1  gain  will  require  the  400th  part,  and  $200  gain  will  require  200  times  as  many  months. 

114.  Jones,  Smith,  Brown  and  Wood  formed  a  partnersliip.  It  was  agreed 
that  Jones  should  receive  ^  of  the  gains,  Smith  J,  Brown  1,  and  Wood  ^.  The 
remainder  of  the  gains  was  to  be  credited  to  Eeserve  Fund,  to  be  used  for  the  con- 
tingencies of  the  business.  It  was  agreed  that  each  partner  should  invest  $15000, 
on  condition  of  receiving  the  above  proportional  share  of  profits.  Jones  however 
invested  only  $12000  and  Smith  invested  only  $10000.  The  net  gain  amounted  to 
$16500.  Since  Jones  and  Smith  faUed  to  invest  the  required  amount,  how  should 
the  gain  be  divided  1  Ans.  Jones  $5225 ;  Smith  3265.62J ; 

Brown  $3918.75  ;  Wood  $32C5.62J. 
Eeserve  Fund  $825. 
SOLUTION. 

1.  By  the  terms  of  the  partnership,  each  partner  was  to  invest  $15000  which  sum  was  to 
entitle  each  to  the  following  shares  of  the  profits  :  Jones  i  -j-  Smith  i  -{■  Browu  -|  -(-  Wood  ^  =  ^|i ; 
thus  leaving  ^g  to  he  credited  to  Eeserve  Fund. 

2.  Since  Jones  and  Smith  foiled  to  invest  the  full  amount  of  $15000  each,  the  respective 
shares  of  their  gain  must  ho  reduced  in  the  same  proportion  as  their  respective  investments  were 
reduced  from  the  §15000  required  hy  the  terms  of  the  agreement.  The  following  statements  give 
their  reduced  shares : 

$15000  :  12000  :  :  J  :  i^-  =  Jones'  share  of  gain  for  his  $12000  investment. 
$15000  :  10000  :  :  i  :   ^  =  Smith's  share  of  gain  for  his  $10000  investment. 

3.  The  respective  shares  of  the  partners  in  the  ^S  of  the  distrihutahle  gain  are  now  as  fol- 
lows :  Jones  fV;  Smith  ^,  Brown  4  and  Wood  ^  =  i  which  is  the  sum  of  the  interests  or  shares  that 
are  to  receive  ^S  of  the  gain. 

4.  The  total  gain  is  $16500 
/„-  deducted  for  Reserve  Fund  is       825 


Amount  to  be  divided  is  $15675 

5.     The  following  statements  give  the  share  of  each  partner : 

JONES.  SMITH. 

$15675  $15675 


4 
15 


BROWN. 

$15675 
5 


$3265. 62i 


WOOD. 

$15675 
4      5 
6 


$3918.75 


$3265.62i 


Note. — By  the  terms  of  the  agreement,  the  partners  were  to  receive  in  different  proportions 
•JS  of  the  gain,  and  the  remainder,  ^V.  '"'iis  to  he  credited  to  Reserve  Fund.  The  respective  interest 
of  each  partner  was  conditioned  upon  an  investment  of  $15000 ;  hence  those  X'^rtners  who  invested 
less  than  $15000  are  to  have  their  original  shares  reduced,  as  shown  above,  and  then  the  ^g  of 
the  gaiu  is  to  be  divided  among  the  partners  in  proportion  to  the  revised  shares,  as  shown  in  the 
operation.  Reserve  Fund  thus  receives  ^,!  of  the  gain,  as  per  agreement.  By  this  method,  the  part- 
ners received  the  l',\  of  gain,  in  equitable  proportion  to  their  actual  investments,  based  upon  the 
terms  of  agreement.     This  we  believe  to  be  the  proper  method  of  adjusting  the  matter. 

In  Problem  38,  pg.  832,  which  involves  the  same  conditions  as  this  problem,  the  ^  interest, 
which  was  to  be  credited  to  Profit  and  Loss,  was  allowed  to  participate  in  the  gain  resulting  from 
the  decrease  in  the  respective  shares  of  C.  and  D.  because  they  failed  to  make  full  investments  as 
they  had  agreed.     By  thus  giving  Profit  and  Loss  a  part  of  this  gaiu,  more  than.  |  of  the  gain  of  the 


882  PARTNERSHIP    SETTLEMENTS.  * 

business  is  creditefl  to  this  aceonnt,  and  consequently  less  than  J  of  the  gain  is  distributed  among 
the  partners.  In  the  solution  of  Problem  38,  which  problem  was  selected  from  another  work,  we 
adopted  the  method  of  the  author  of  the  problem,  by  which  he  credited  Profit  and  Loss  with  more 
than  I  of  the  giiiu. 

We  now  present  the  above  problem  and  give  a  different  solutiou  and  submit  the  two  methods 
for  the  consideratiou  of  accoiiutauts. 

115.  Supi)ose,  in  the  above  problem,  that  the  partners  had  made  the  following 
investments:  Jones  $12()()(),  Smith  $25000,  Brown  $15000  and  Wood  $15000.  llow 
should  the  gain  have  been  divided  ? 

Ans,  Jones,  $3980.95  +,  Smith,  $6220.24  — . 
Brown,  2985.71+.  Wood,  2488.10—. 
Eeserve  Fund,  $825. 

OPERATION. 

$15000  :  12000  t  :  J  :  iV  =  Jones'  share  of  the  gain  for  his  .$12000  investment. 
15000  :  25000  :  :  J  :  -.Af  =  Smith's     "        "  "         "         25000        " 

Then  Jones'  -fi;  +  Smith's  fV  +  Brown's  J  +  Wood's  J  =  §3  which  is  the  sum  of  the  shares  or 
interests  that  are  to  receive  jo  of  the  gain. 

The  total  gain  is       -        -        $16500 
jig  deducted  for  reserve  fund         825 


Gain  to  be  divided  is        -        $15675 

The  following  statements  give  the  proportional  share  of  eacli  partner : 


Joi 

63 
15 

les'  A. 
$15675 
60 
4 

Smith 

63 
12 

$15675 
60 
5 

Brc 

63 

5 

wn's  i. 

115675 
60 

W( 

63 
6 

)od's  J. 
$15675 
60 

$3980.95  + 

$6220.24  — 

$2985.71  + 

12488.10  — 

H^6  A.,  B.,  and  C.  are  general  partners  in  business,  equal  in  gains  and  losses.  They  conclude  to 
dissolve  their  commercial  or  trading  partnership  and  form  a  stock  company.  Accordingly  they 
close  their  books  and  tiud  the  net  capital  of  each  to  be  as  follows:  A.  $32238,  B.  $28712,  and  C. 
$28644.     Total  $89594. 

The  stock  company  is  organized  with  a  capital  of  1000  shares  at  $100,  amounting  to  $100000, 
of  which  each  of  the  partners  of  the  former  business  is  to  subscribe  for  3o3J  shares. 

The  excess  of  the  capital  of  the  stock  company  over  the  total  capital  of  the   old   firm   is 
($100000  —  $89594)  $10406. 

It  is  agreed  that  to  offset  or  balance  this  excess,  they  will  value  some  patents  which  the  old 
firm  possess  at  $10406,  and  ))lace  the  same  in  the  list  of  assets. 

What  amount  of  money  must  13.  and  C.  p.iy  to  A.  respectively,  to  adjust  the  matter  equitably 
among  the  three  ?  Ans.  See  operation. 

OPERATION. — First  Step. 
The  valuation  of  the  patents  creates  a  new  asset,  which  belongs  to  the  members  of  the  old 
firm  in  proportion  as  they  sliared  the  gains  and  losses,  J  each.     Hence  $10406  equals  $3468.66f  due 
respectively  to  A.,  B.,  C. 

Second  Step. 
A's  credit  $32238  +  $3468.67  =  $35706,67  —  $33333.34  =  A's  Cr.  $2373.33 

B's  credit    28712  +    3468.67  =    32180.67  —    33333.33  =  B's  Dr.    1152.66 
C's  credit    28644+    3468.66=    32112.66—   33333.33  =  C's  Dr.    1220.67 

$89594  $100000.00  $2373.33 

10406  valuation  placed  on  patents. 

$100000  -f-  3  =  $33333. 33J  =  each  partner's  interest  in  the  stock  company. 
From  these  statements  it  is  clear  that  B.  and  C.  must  pay  to  A.  respectively,  $1152.66  and 
$1220.67. 


H  n\/olL4tion. 


InToIution  is  the  in-ocess  of  multiplying  .1  number  by  itself  a  certain 
number  of  times.     The  several  products  are  called  the  powers  of  the  number. 

The  number  used  as  a  factor  is  called  the  root  or  first  power. 

The  second  power  or  square  is  the  root  used  twice  as  a  factor. 

The  third  power  or  cube  is  the  root  taken  three  times  as  a  factor. 

The  fourth  power  or  biquadrate  is  the  root  taken  four  times  as  a  factor  ;  and 
all  liifrhcr  jiDwers  indicate  the  number  of  times  the  given  number  or  root  enters  as 
a  factor. 

The  number  of  times  the  root  is  used  as  a  factor  is  called  a  degree,  as  the 
1st,  2d,  3d,  and  4th  degrees;  and  the  number  denoting  tluit  degree  is  called  the 
index,  or  exponent  of  the  power.  Thus,  the  cube  of  5  is  represented  by  5^.  The 
exponent  should  be  a  small  figure,  and  written  to  the  right  of  the  ujiper  part  of  the 
root  or  number.  A  fraction  should  be  inclosed  in  a  parenthesis  before  writing  the 
exponent,  to  indicate  that  it  belongs  to  the  whole  expression,  otherwise  it  might  be 
considered  as  belonging  to  the  numerator  alone.     Thus  the  square  of  f  is  denoted 

The  following  arrangement  will  illustrate  the  formation  of  degrees  and 
powers  for  the  number  5 ;  thus : 

5"  =  5  -H  5  =      1  =  zero  power  of  5.  • 

5'  =  1X5  =      5  =  1st     power  of  5. 

6^  =  5  X  5  =    25  =  2il      power  or  square  of  5. 

5' =  5x5x5  =  125  =  3d      power  or  cube  of  5. 

5*  =5x5x5x5  =  625  =  4tli    power  or  biquadrate  of  5. 

We  here  see  that  there  are  three  things  to  be  noted  in  Involution,  viz. :  1st. 
The  root  or  number;  2d.  The  exponent;  and  3d.  The  power.  Considering  the  1st 
power  as  a  multix)licaiid,  we  see  the  number  of  multiplications  will  bo  one  less  than 
expressed  by  the  exponent.  Any  power  of  1  is  1 ;  any  power  of  a  number  greater 
than  1,  is  greater  than  the  given  number ;  and  any  power  of  a  number  less  than 
1,  is  less  than  the  number  itself. 

When  the  rcMjuired  power  is  of  a  high  degree,  instead  of  multiplying  by  the 
root  continually,  we  can  involve  any  of  the  subordinate  powers,  the  sum  of  whose 
degrees  amounts  to  the  degree  of  the  required  power;  always  observing  that  the 
product  of  the  poircrs  corresponds  to  the  snm  of  the  exponents.  Thus,  to  find  the 
7tli  power  of  the  number  3,  we  may  multiply  its  4tli  power  (81)  by  its  3d  power 
(27),  or  3"  =  3*  X  3^ 

When  the  exponent  of  the  required  power  is  a  composite  number,  we  may 
raise  the  root  to  a  power  indicated  by  one  of  the  factors  of  the  given  exponent,  and 
then  raise  this  result  to  a  power  indicated  by  the  other  factor.  Thus,  to  raise  5  to 
the  12th  power,  or  5  ^  %  we  first  find  the  4th  power  of  5  to  be  625 ;  we  then  raise  tlie 
625  to  the  3d  power,  because  5^^  =  (5^)''. 

(883) 


884  SOULe's    rillLOSOPHIC    PRACTICAL    MATHEMATICS.  * 

The  difference  between  tlie  last  and  tlie  preceding  proposition  sliould  be 
clearly  understood  to  prevent  confusion,  their  jjrinciples  being  in  nowise  analogous. 

To  raise  a  vulgar  fraction  to  any  power,  laise  both  numerator  and  denomina- 
tor to  that  power.  Mixed  luimbers  should  be  changed  to  imiirojier  fractions.  To 
find  any  power  of  a  decimal,  or  mixed  decimal,  multiply  as  in  whole  numbers,  and 
point  off  in  the  result  as  many  places  for  decimals  as  will  equal  the  number  of 
decimal  places  in  the  root,  multiplied  by  the  exponent  of  the  power. 

EXAMPLES. 

1.  Whatis  the  2d  power  of  3405?  Ans.  1200G225. 

2.  What  is  tlie  Dth  jtower  of  2.5G  1  Ans.  109951 1627776. 

3.  What  is  the  3d  power  of  i  'I  Ans.  f||-. 

4.  Wliatis  the6thpowerof  3|»  Ans.  2430 ,V<,-. 

5.  Whatis  the  4th  power  of  .125?  Ans.  .000244140625. 

6.  Wliat  is  tlie  cube  of  3.024?  Ans.  27.«r)3107824. 

7.  Wliatis  the  7th  power  of  .004?       •  Ans.  .0000000(10000000016384. 

8.  What  is  the  10th  power  of  7  ?  xVns.  282475249. 

The  2d  power  of  numbers  ending  in  5,  when  not  too  large,  may  be  very 
readily  found  by  taking  all  the  figures  except  the  5,  and  then  multiply  the  number 
thus  taken  by  itself  increased  by  1,  and  to  the  jiroduct  annex  25,  the  square  of  the 
6.  Thus,  to  find  the  square  of  25,  we  omit  the  5,  and  multiply  the  2  by  3  =  6,  then 
annexing  25,  we  obtain  625. 

Find  the  square  of  15,  of  35,  of  55,  of  75,  of  105,  of  225,  of  575. 

To  find  any  power  of  a  number  that  is  an  aliquot  part,  or  aliquot  multii)le  of 
a  hundred  or  thousand,  we  form  a  fraction  by  taking  the  given  number  for  a 
iiumerator,  and  100,  or  1000,  for  a  denominator,  and  reduce  to  its  lowest  terms;  then 
raise  the  numerator  of  this  fraction  to  the  required  power,  and  annex  as  many 
naughts  as  would  equal  the  number  of  naughts  in  the  100  or  1000  used,  multiijlied 
by  the  exijonent  of  the  jjower;  then  divide  by  the  denominator  of  the  reduced 
fraction,  as  many  times  as  there  are  units  in  the  exponent.  Thus,  to  find  the  cube 
of  375,  we  have 

aiA  _  i5 .  tlicn,  153  =  3375000000  -^  4  cubed,  =  Ans.  52734375. 

Or, 
^;  then,  3'  =  27000000000  -i-  8  cubed,  =  Ans.  52734375, 


1000   — 


EXAMPLES. 


o 


Find  the  square  of  875.  Ans.  765625. 

3.  Find  the  cube  of  625.  Ans.  244140625. 

4.  Find  the  4th  power  of  375.  Ans.  1077.5390625. 

5.  Find  the  5th  power  of  225.  Ans.  576650390625. 

6.  Find  the  6th  power  of  125.  Ans.  3814697265625. 

7.  Find  the  7th  power  of  75.  Ans.  13.348388671875. 

8.  Find  the  8th  power  of  25.  Ans.  152587890625. 

9.  Find  the  6th  power  of  16§.  Ans.  21433470f4t. 

10.  Find  the  5th  power  of  33|.  Ans.  41152263#J3-- 

11.  Find  the  4th  power  of  666§.  Ans.  197530864197||. 

12.  Find  the  product  of  the  cube  of  66§  by  the  square  of  3333J. 

Ans.  3292181069958|f|. 


^.\/olutiori. 


Evolution,  or  the  extraction  of  roots,  is  the  reverse  of  Involution,  and 
is  the  process  of  finding  tLe  eqnal  factors,  or  roots,  of  numbers  or  qiumtities. 

A  root  of  a  number,  as  illustrated  in  involution,  is  a  factor  which,  multiplied 
by  itself  a  certain  number  of  times,  will  produce  that  number. 

The  name  or  degree  of  the  root  is  correlative  with  that  of  the  power.  Tlras, 
the  square  root  is  the  correlative  term  of  the  square  or  second  power,  and  the  cube 
root  that  of  the  third  power  or  cube. 

The  root  to  be  extracted  is  indicated  by  placing  the  radical  sign  (V)  before 

the  number  or  power,  with  the  index  of  the  root  over  the  sign.  Thus,  v  125  denotes 
the  cube  root  of  125.  The  index  of  the  square  root  is  generally  omitted,  the  radical 
sign  itself  always  indicating  the  square  root.     The  root  is  also  sometimes  indicated 

by  means  of  a  fractional  exponent ;  thus,  36^  denotes  the  square  root  of  3G ;  and 

64*  indicates  the  cube  root  of  the  square  of  C4,  or  the  square  of  the  cube  root  of  64. 

A  number  that  is  a  perfect  power  is  called  a  rational  niimher,  and  its  root  a 
rational  root ;  thus  27,  being  a  perfect  cube,  is  rational.  A  number  that  is  not  a 
perfect  power  is  called  an  irrational  number  or  surd,  and  its  root  an  irrational  or 
surd  root;  thus  24,  being  an  imperfect  power,  is  a  surd,  and  also  its  root.  A  num- 
ber, however,  may  be  rational  to  one  degree  and  a  surd  to  another.  Thus  27  is  a 
rational  cube,  but  an  irrational  square. 

All  the  rational  roots  of  whole  numbers  are  whole  numbers. 

All  the  powers  of  fractional  numbers  are  fractional  numbers. 

All  prime  numbers,  except,  1,  have  no  rational  roots. 

All  coiniiosite  numbers  that  have  rational  roots,  must  have  all  the  exponents 

3  . 

of  their  prime  factors  divisible  by  the  exponent  of  the  required  root.  Thus,  v  216 
=  \^3^  X  2^  =   Voa"  =  6. 

The  number  of  rational  roots  is  comparatively  small. 

The  extraction  of  roots  higher  than  the  square  and  cube  is  not  usually  given 
in  arithmetics,  though  roots  of  any  degree  may  be  easily  found  by  an  extension  of 
the  method  given  in  this  work ;  but  all  higher  roots  are  more  readily  obtained  by 
the  use  of  logarithms. 

EXTEACTION  OF  THE  SQUAEE  ROOT. 

The  extraction  of  the  square  root  of  a  number  is  the  process  of  finding 
one  of  its  two  equal  factors. 

The  following  numbers  in  the  first  line  are  the  square  roots  of  those  in  the 
second. 

Eoots:  1,     2,     3,       4,       5,       G,       7,       8,       9,       10,  100. 

Squares:         1,     4,     9,     IG,     25,     36,     49,     64,     81,     100,     10.000. 

(885) 


886 

We  here  see  tliat  the  square  root  of  a  number  less  than  100,  is  less  than  10  j 
the  square  root  of  a  number  less  than  10000,  is  less  than  100,  etc.  Therefore  the 
square  of  any  single  figure  will  always  be  found  in  the  first  two  right  hand  jjlaces 
of  the  jjower,  and  the  square  of  a  number  of  two  figures,  in  the  firist  four  places  of 
the  power,  every  two  figures  in  the  square  giving  one  figure  in  the  root.  For  this 
reason,  in  the  extraction  of  the  square  root,  we  separate  the  numbers  into  periods  of 
two  figures  each,  from  the  right,  though  the  left  hand  period  will  sometimes  contain 
but  one  figure,  when  its  root  is  less  than  4. 

To  investigate  the  principles  for  the  extraction  of  the  square  root,  let  us  take 
a  number  of  two  places  of  figures  as  54,  and  represent  the  digit  in  the  tens'  i)lac© 
by  X  (5),  and  the  digit  in  the  units'  jjlace  by  u  (4) ;  then  the  number  54  or  (50  +  4) 
^nll  be  represented  by  lOj:-  +  «.  We  assume  this  binomial  for  the  basis  of  our 
operations,  instead  of  the  number  itself,  because  first,  principles  can  be  more  clearly 
demonstrated  by  generalizing  their  formula  tlian  by  using  ordinary  figures,  as  these 
special  figures  would  serve  to  obscure  the  exhibition  of  general  i)rinciples;  and  we 
prefer  the  above  form  of  the  binomial  (10,r  +  v)  to  the  one  generally  used,  viz.,  (izr 
-1-  m),  because  the  relative  position  of  these  factors,  x  and  u,  in  the  expansion,  is 
more  easily  perceived,  and  the  reason  for  certain  operations,  and  the  location  of 
their  results  in  the  extraction  of  roots,  can  be  more  readily  explained. 

BXPANSION. 

10a! -{- «  Explanalion. — We  here  square  lOx  +  «,   t.   e., 

lOx -j- «  multiply  it  by  itself.     Tlie  process  is  so  plain 

that  no  explanation  is  necessary.     We  find  the 


lOxiJ       u^  result  to  be  lOdx-  +  20x!t  -(-  «-',   or  iu   common 

IOOj;' +  lOj-ij  langnafje   would  he   read  tbns:    One    hundred 

•  times  tbe  sqnare  of  the  tens,  plus  twenty  times 

lOOx-  -\-  20j-«  +  ti^  the  tens  mnltii)lied  by  the  units,  plus  the  square 

of  the  units. 


Let  us  now  arrange  the  formula  in  a  vertical  column  on  tbe  Irft,  tbcn  substitute  the  dibits 

of  tbe  number  54  for  tbeir  representatives  x  and 

IdOar'  ^=  5  X  5  X  100  :=  2500  or  2.")  u,  and  perform  tlie  operations  indicated,  and  wo 

20xit  =  5  X  4  X  20    =    4U0  or    40  bave  for  tbe  several  terms  tbese  partial  results, 

tt2  =  4x4  =      16  or      16  viz.  :  Kor  IOO.p^  =  2.'00;   for  20.r«  =  400;   for  w^ 

=  16;    and  tbeir  Bum  =  2916  for  tbe  square   of 

lOOx^  +  20xtt  +  u'      =  2916  =  2916  54.     Also  observe  tbat  tlie  coefficient  of  x-  is 

100,  and  contains  two  cipliers  ;  tbe  coefficient  of 
Xtt  is  20,  and  contains  one  cipher  less  than  the  first,  nwiX  u'  lias  no  cipber,  or  one  less  tlian  tbe  last. 
By  virtue  of  tbis  c<mstant  airanfiement,  we  can  cancel  or  discaril  tbe  cipbcrs  altotrether,  and 
arrange  tbe  results  as  in  tlie  last  colunin,  wbere  we  see  tbat  after  setting  down  the  value  of  x-  ^ 
25,  we  place  tbe  next  term  2xit  =  40,  one  place  to  tbe  rigbt,  and  u'  =  16,  one  place  furtber  to  tb» 
right.  This  arrangement  gives  tbe  ordinary  binomial  S(|nare  of  a;  -f  u  =  jr'^  -|-  '2xu  +  u',  and  in 
this  form  we  will  use  it  for  convenience  after  tbe  fust  illustrations. 

Let  us  now  find  tlie  square  root  of  2310,  from  tbe  above  formula,  by  resolving  it  into  its 
Actors. 


EVOLUTION. 


887 


THEORETICAL  OPERATION. 


100^2  +  20xu  +  «2 

=  29' 16  (  5 
25  00 

+ 

4 

IOOj^ 

u 

(20x  = 
«  = 

100 
4 

)  4  16 

X 
«  =  4 

X  104 

4  16 

20x«  + 

w^     = 

0 

Explaiutiioii.- — Since  evolution  is  tbe  reverse 
of  involution,  we  will  coinnieuco  with  tlie  term 
on  the  left,  lOO.r-  ;  we  iirst  divide  by  the  100.  by 
pointing  oft"  from  the  right,  two  places;  now 
we  know  that  j,  the  square  root  of  x'^,  must  bo- 
found  wholly  in  the  figures  to  the  left  of  the 
point,  viz.,  29,  the  greatest  perfect  equari'  of 
whicli  is  25,  whose  square  root  is  5;  we  plnce 
this  to  the  right  like  a  quotient  in  division,  and 
designate  it  by  x;  we  then  subtract  its  square 
25  from  29,  leaving  a  remainder  4,  to  which  we 
annex  the  second  period  16.  This  new  dividend 
416,  we  see  is  what  remains  after  subtracting 
PRACTICAL  OPERATION'.  the  value  of  IOOj*  from  2916,  and  hence  the  41& 

is  equal  to  the  (20,r((  +  u')  ;  now.  as  we  do  not- 
29'16     (  54  yet  know  the  value  of   w,  we  will  eliniiuate    >i 

25  from  both  terms  of  2nj-n  -f  u',  which   gives  u» 

(20x  +  i()  w  =  416,  which  we  arrange  vertically 

104  )     4 16  as  in  tlie  operation;   and  as   we   do   know   the 

416  value  of  20x  =  20  x  5  =   100.    and   that    it    i» 

many  times  greater  than  u,  we  can   consider  it 

0  as  a  trial  divisor,   and  say.  20jr  or  100   divided 

into  416  gives  ajqiroximately  a.  quotient  4  ^  »(, 
%rhich  we  place  in  the  root.  Having  now  found  u  =  4,  we  add  it  to  100,  making  104  =  20j;  4-  «,  for 
the  true  divisor;  we  now  multiply  this  by  u  =  4,  and  obtain  20j;h  +  u'^  =  416;  subtracting  this 
from  the  dividend,  and  liuding  no  remainder,  completes  the  work,  and  gives  the  exact  root,  54  =  K 
-1-  u,  or  rather  lOj;  +  «. 

In  the  practi<al  operation  we  omit  the  two  naughts  of  the  100,  and  find  the  first  figure  of  the 
root  as  before,  subtract  its  square,  and  bring  down  the  nest  period  for  a  new  dividend.  We  therh 
omit  the  naught  of  the  20,  and  use  only  the  2  to  niultijily  into  the  tens'  figure  of  the  root  5,  thus.  5 
X  2  =  10,  for  a  trial  divisor;  to  compensate  for  the  elision  of  the  divi.sor,  we  cut  oil'  the  right 
hand  figure  of  the  dividend  before  diviiling  to  find  the  units'  figure  of  the  root,  or  u,  that  is,  we 
Bay  10  into  41  =  4;  we  then  OHnci  this  4  to  the  10,  making  104  for  the  true  divisor,  which  being 
multiplied  by  4,  and  the  product  subtracted  from  the  dividend,  leaves  no  remainder ;  we  then 
place  the  4  in  the  root,  which  completes  the  woru. 

The  philosophy  of  the  theoretical  process  slioukl  be  thorong;hly  understood, 
and  then  the  reason  for  the  contractions  in  the  ])raetical  method  will  bo  apparent. 

8ince  all  numbers  above  units  nmy  be  considered  as  composed  of  tens  and 
units,  thus  450  may  be  called  45  tens  anil  6  units,  therefore  when  there  are  several 
figures  in  the  required  root  we  may,  after  having  found  the  first  two  left  hand 
figures  of  the  root,  consider  them  as  tens,  and  the  next  figure  to  be  found  as  the 
units' figure,  and  use  the  same  formula  and  process  for  finding  the  third  figure  as 
"we  did  for  tlie  second;  and  the  same  principle  and  ai)])lication  may  be  extended 
indefinitely  to  any  inimber  of  jilaces,  so  that,  by  the  same  kind  of  operation,  the 
Bquare  root  of  any  number  may  be  found,  or  approximated,  as  illustrated  in  th© 
following  examjile : 

2.     Find  the  square  root  of  5()475225. 

OPERATION. 

Hoot.  Explanafion. — We  first  point  off  into  periods 

56'47'52'25  (  7515  of  two  figures  each  from  the  right,  then  find  tho 

4<j  greatest  perfect  square   in   the   first   lelt   hand 

jieriod.  56,  to  be  49,  the  S([uaro  root  of  whicli  is 

7       X  2  =  115       )  747  7,  which   we   place   to   the   right   for   the   first 

5           725  figure  of  the  root,  and  its  sqiuue,  49.  we  subtract 

. from  56  ;  to  tbe  remain<ler  7  we  bring  down  or 

75     X  2  =  1501       )  2252  annex  the  second  period,  47,  giving  us  747   for 

1           1501  the  first  dividend ;  we  then  multiply  the  figure 

ill  the  root  7,  by  2  =  14  for  the  first  trial  divisor, 

751  X  2  =  15025      )  75125  and  ascertain  how  often  it  is  contained  in  the 

75125  dividend  exclusive  of  the  right  hand  figure, 

■  i.  e.,    in   74.    (for   the   reason    explained   in   the 

0  previous  example,   having  multiplied   the  root 

bv  2  instead  of  20) ;  we  find  the  nearest  quotient 


8S8  soule's  niiLosopnic  practical  mathematics.  * 

to  he  5,  ■wbicli  we  place  in  llio  root,  and  also  at  the  right  of  tlie  trial  divisor  14,  giving  145  for  the 
true  divisor,  ■which  we  iimltiply  by  the  5,  and  snbtract  the  product  7-'5  from  the  dividend  747, 
leaving  a  remainder  (if  22,  to  wiiicli  we  annex  the  tliird  jieriod,  02,  giving  2252  for  the  second 
ilividend;  -we  then  mnltiidy  llie  (ignrcs  in  the  root,  75  l)y  2,  or  add  the  last  lignre  of  the  root,  5,  to 
llic  last  divisor,  145,  giving  150  for  tlie  second  trial  divisor,  which  -ne  fmd  is  contained  iu  the  new 
dividend  exclusive  of  tlie  riglit  hand  lignre  ("2251,  1  tune;  wo  place  this  (igiiro  1  in  the  root,  and 
also  to  the  right  of  the  trial  divisor  150,  giving  1501  for  a  true  divisor,  which  we  multiply  hy  tlie 
1,  and  snbtract  the  ))roduct  from  the  dividend ;  to  the  remainder,  751,  wp  annex  the  fourth  period, 
25,  giving  7512"  fi>r  anew  dividend,  then  donlile  the  figures  of  the  root,  751,  or  add  the  last  root 
lignre,  1,  to  the  last  divisor,  1501.  giving  us  1502  for  a  new  trial  divisor,  and  operating  as  before 
explained,  we  find  the  next  root  figure  to  be  5,  which  we  place  in  the  root  and  at  the  right  of  the 
divisor;  then  after  multiiilying  aud  subtracting,  as  there  is  uo  remainder,  we  have  the  exact  root 
7515. 

If  there  was  .a  remainder,  we  could  annex  periods  of  ciphers,  and  continue  the  operations  in 
the  same  manner  until  sufficiently  exact,  each  period  of  ciphers  annexed  giving  one  decimal  pl.ace 
in  the  root.  Whenever  the  signiticant  figures  of  the  given  number  do  not  give  an  exact  root,  the 
number  is  a  surd,  and  the  root  indeterminate,  and  cannot  be  ex.actly  expressed  by  any  combination 
of  fractious,  either  vulgar  or  decimal,  but  admits  of  an  infinitesimal  approximation  by  decimals. 

When  the  product  of  any  divisor  by  its  root  figure  is  greater  than  its  corresponding  dividend, 
we  must  diminish  the  last  figure  of  the  root  and  divisor. 

When  the  dividend,  exclusive  of  its  right  hand  figure,  is  less  than  the  trial  divisor,  a  cipher 
must  he  placed  in  the  root,  and  also  at  the  right  of  the  divisor,  then  bring  down  the  next  period  to 
the  right  of  the  dividend,  and  proceed  as  before. 

When  the  given  number  is  a  decimal,  or  a  whole  number  and  decimal,  we  point  off  the 
<lecimal  part  into  periods,  towards  the  right  from  the  decimal  point,  annexing  a  cipher  if  necessary 
to  make  the  number  of  places  even. 

AVhen  the  given  number  is  a  fraction,  reduce  it  to  its  lowest  terms ;  if  it  is  a  mixed  number, 
<?hange  it  to  an  improper  fraction;  then  if  both  terms  of  the  fraction  are  perfect  powers,  extract 
the  root  of  each;  if  one  or  both  terms  are  surds,  multiply  them  together  and  extract  the  root  of 
the  product  to  a  sufficient  approximation,  then  divide  the  result  by  the  denominator;  or  reduce 
the  fraction  to  a  decimal  and  extract  the  root. 

Tlius,  tlie  square  root  of  -j^,  or  Vj^  =  V-s.^  =  s^   Ans. 

And  the  square  root  of  12^  =  V^=  |  =  3J,   Aus. 

And  the  square  root  of  §  =   v'o  x  8  =   VW=  C.3245  -f  ,  divided  by  8  =  .7905 

-+,   Ans. 

Or,   Vf=  V'.G250  =  .7905  +,  Ans. 

EXAMPLES. 

1.  Wliat  isibe  square  root  of  1377894-4  7 

2.  ^Yllat  is  the  square  root  of  210C.S1 1 

3.  ^Vhat  is  the  square  root  of  .00007225 1 

4.  Vrhat  is  the  square  root  (  f  7  i\r  ? 

5.  What  is  the  square  root  of  4257-/6 ' 

G.     What  is  the  value  of  59049^  ? 

7.  What  is  the  value  of  "/sF? 

8.  AVliat  is  the  square  root  of  781260581179521?  Ans. 

APPLICATIOiSrS  OF  SQUARE  EOOT. 

1429.  1.  Wliat  is  the  length  of  the  side  of  a  square  field,  in  yards,  that 
contains  40  acres  ?  Ans.  440  yards. 

2.     Wliat  is  the  length  and  width  of  a  rectangular  field,  in  yards,  that  is  2J 
times  as  long  as  wide,  and  contains  25  acres  ?  Ans.  550  yards  long. 

220  yards  wide. 


Ans. 

3712. 

Ans. 

45.9. 

Ans.  . 

0085. 

An 

s.  2f. 

Ans 

651. 

Ans 

243. 

Ans 

729. 

5.  27951039. 

EVOLUTION. 


88g 


SOLUTION. 

25  acres  =  121000  square  yards;  then  121000  —  2.J  =  48-100  =  square  of  shortest  side;  then 
v/JsIOO  =  220  yards,  shortest  side. 

It  is  demonstrated  in  geometry  that  the  square  of  tlie  liypotenuse  of  a  right-angled  triangle 

is  equal  to  the  sum  of  the  squares  of  the  other 
C  two  sides.     In  the  annexed  diagram  we  see  that 

the  square  of  the  hypotenuse  «(',  is  equal  to  the 
sum  of  the  squares  of  the  base  ab  and  i>er]ieu- 
dicular  be,  and  hence  when  any  two  of  the  sides 
are  given,  the  third  can  be  found ;  thus,  when 
the  base  and  perpendicular  are  given,  we  lind 
the  square  root  of  the  siini  of  their  squares  for 
the  hypotenuse ;  and  when  the  hypotenuse  and 
Base.  B  one  sliie  are  given,   we   find   the   square   root   of 

the  difference   of  their  squares  for  the    other 
ac'  =  ai'  4-  6"'  side. 

1.  e.,       5-  =    42+32 
Or,      25    =  16    +   9 

We  also  observe  that  the  areas  of  all  similar  triangles,  -with  the  same  base,  have  the  same 
ratio  to  each  other  as  their  altitudes,  and 

All  similar  triangles  have  their  areas  proportional  to  the  squares  of  their  corresponding 
sides. 

3.  A  wall  40  feet  high  stands  on  the  side  of  a  uioat  30  feet  -wide;  find  the 
length  of  a  ladder  that  will  reach  from  the  opposite  side  of  the  uioat  to  the  top  of 
the  wall?  Ans.  50  feet. 

4,  In  the  center  of  a  garden  400  feet  square,  there  is  a  pole  100  feet  high. 
What  is  the  distance  from  each  corner  of  the  garden  to  the  foot  of  the  pole,  and 
also  to  the  top  of  the  pole  ? 

Ans.  Distance  from  each  corner  to  foot  of  pole,  282.84+  feet. 
Distance  from  each  corner  to  top  of  pole,  300  feet. 

SOLUTION. 

i  of  400  =  200 ;  then,  V20(F"x~2  =  1^80000  —  282.84+ 
And        VlOO^  +  80000  =  300. 

0.  A  square  plot  of  ground  contains  2J  acres;  what  will  be  the  area,  the 
length  of  sides,  and  the  diagonal,  in  yards,  of  another  square  field  whose  sides  are 
teu  times  as  great?  Ans.  Area,  2.^0  acres. 

Length  of  side,  1100  yards. 

Length  of  diagonal,  1555.63+,  yards. 


SOLUTION. 

2J  acres  X  10^  =  250  acres. 
Then,         V250  X  4840  =  1100  yards  on  each  side. 
And  ■/iTxTlOO^  =  1555.63+,  yards  diagonal. 

6.  A  room  is  IS  feet  wide,  24  feet  long,  and  1 2i  feet  high  ;  what  is  the  length 
of  the  diagonal  drawn  from  one  of  the  lower  cornei'S  to  the  upper  corner  farthest 
from  it  ?  Ans.  32J  feet. 

SOLUTION. 


1/18'  +  242  +  (12i)2  =  32i. 


S90  SOULe's    nilLOSOPHlC    PRACTICAL    MATHEMATICS.  * 

7.  Two  sliips  sail  from  the  same  port ;  one  due  east  at  the  rate  of  300  miles 
per  day,  the  other  due  south  at  the  rate  of  125  miles  i)er  day.  After  a  certain  time 
they  were  1625  miles  apart ;  allowing'  the  oeean  to  be  a  plane,  how  many  days  were 
they  out  ?  Aus.  5  days. 

,  OPKRATION    INDICATED. 

300'  =     90000 
1252  =    15625 

V  105825  =  325  miles  apart  the  1st  day. 
Then,         1625  -i-  325  =  5  clays. 

8.  A  rectangular  field  is  2f  times  longer  than  it  is  wide,  and  the  length  of 
its  diagonal  is  520  rods.     How  many  acres  does  it  contain  ?  Aus.  COO  acres. 

SOLUTION. 

Vl"  -|-  (21)"  =  2f,  proportional  diagonal. 

And  520  -f-  2f  =  200,  ratio  of  diagonals, 

Thi'n,         1  rod  X  200  =  200  rods  wide. 

And  2J  rods  X  200  =  480  rods  long. 

480  X  200 
Then  — jyg =  600  acres. 

9.  A  liidder  15  feet  long,  with  its  foot  a  certain  distance  from  a  wall,  just 
reaches  the  toj)  of  the  wall;  another  ladder  20  feet  long,  will  reach  the  top  of  the 
wall  when  its  foot  is  7  feet  farther  from  the  wall  than  the  first.  What  is  the  height 
of  the  wall,  and  the  distauce  of  the  foot  of  each  ladder  from  the  wall? 

Aus.  Shortest  base,  9  feet. 
Longest  base,  16  feet. 
Height  of  wall,  12  feet. 

SOLUTION. 

20^  —  (15^  +  72)  =  126  —  (2  X  7)  =  9  feet  shortest  base. 

The  other  conditions  are  now  easily  found. 

The  above  solution  depends  on  the  following  principle,  viz,:  The  difference 
between  the  squares  of  the  liypoteiuises  of  two  right-angled  triangles,  whose 
altitudes  are  the  same,  diminished  by  the  square  of  the  difference  of  their  bases,  is 
equal  to  twice  the  rectangle  (or  product)  of  the  difference  of  the  bases  by  the  shorter 
side. 

10.  A  pole  10  feet  from  a  wall  just  reaches  the  top ;  when  removed  10.8  feet 
farther  from  the  wall,  the  upper  end  descends  8.4  feet.  What  is  the  length  of  th© 
pole  and  height  of  the  wall ?  Aus.  Pole,  20  feet;  Wall,  24  feet. 

SOLUTION. 

(10  +  10.8)2  —  (10)2  _f_  (84)2  ^  (2  X  8.4)  =  15.6 

Then,         15.6  +  8.4  =  24  feet,  height  of  wall. 

Or  hy  principle  of  next  example,  thus: 

(20.8  4-  10)  X  10.8  =  332.64  -H  8.4  =  39.6 

39.6  +  8.4 
Then,        5 =  24  feet,  height  of  wall. 


*  CUBE    ROOT,  891 

11.  The  foot  of  a  ladder  placed  7  feet  from  the  middle  of  a  street,  tlie  upper 
end  will  reach  a  point  on  a  wall  on  the  nearer  side,  CO  feet  from  the  ground,  and  if 
turned  to  the  other  side,  will  leach  a  point  52  feet  from  the  ground.  What  is  the 
length  of  the  ladder,  the  width  of  the  street,  and  tlie  distance  of  the  foot  of  the 
ladder  from  each  side  of  the  street  ? 

Ans.  Length  of  ladder,  C5  feet. 
Width  of  street,  Cii  feet. 
Distance  of  ladder  from  nearer  wall,  25  feet. 
Distance  of  ladder  from  farther  wall,  39  feet. 

SOLUTION. 

60  4-  52  =  112  X  8  =  896  -^  (2  X  7)  or  14  =  64  feet,  -svidth  of  street ;  then,  64  —  2  =  32  feet,  half 
the  width  of  the  street ;  aud  32  —  7  =  25  feet,  distance  to  nearest  wall ;  and  32  -|-  7  =  39 
feet,  distance  to  the  farther  wall. 

Then,         VgO-  +  25^  or  V52'  +  31)-  =  65  feet,  length  of  ladder. 

The  above  solution  depends  on  the  following  principle,  Aiz.:  In  two  right- 
angled  triangles  with  the  same  hypotenuse,  the  difference  of  the  squares  of  two 
similar  sides,  is  equal  to  the  difference  of  the  squares  of  the  other  two  similar  sides, 
hence  CO^  —  52-  =  896  is  also  equal  to  the  difference  of  the  squares  of  the  two 
portions  of  the  width  of  the  street;  but  the  difference  of  the  squares  of  two  num- 
bers is  equal  to  the  sum  of  the  two  numbers  multiplied  by  the  difference,  hence  60 
+  52  =  112  X  8  =  890;  and  since  the  difference  of  the  two  portions  of  the  street 
is  equal  to  twice  the  distance  of  the  ladder  from  the  centre,  or  2  x  7  =  ]4,  and  as 
the  sum  of  the  two  portions  of  the  street  multiplied  by  their  difference  is,  by  the 
above  principle,  equal  to  890.  if  we  divide  890  by  14,  the  quotient,  04  feet,  will  be 
the  widtli  of  the  street;  the  remaining  part  of  the  oiieratiou  is  self-evident. 

CUBE    ROOT. 

I-ISI.  Since  the  cube  of  a  number  is  tlie  continued  product  of  the  luiniber 
taken  three  times  as  a  factor,  or  raised  to  the  third  power,  as  explained  in  involu- 
tion, the  cube  root  of  a  number  is  the  converse  of  the  cube,  or  one  of  the  three 
equal  factors  of  the  given  number. 

The  following  comparison  of  the  digits,  etc.,  and  their  cubes,  will  illustrate 
the  relation  between  numbers  in  general  and  their  third  powers,  or  cubes. 

The  cubes  of  the  nine  digits  should  be  committed  to  memory. 

Numbers,        1,     2,       3,       4,         5,         0,         7,         8,         9,         10,  99. 

Cubes,  1,     8,     27,     64,     125,     210,     343,     512,     729,     1000,     970299. 

By  comparing  the  above,  we  observe  that  the  cube  of  the  smallest  number  of 
a  single  figure  (1),  contains  one  figure  (1),  and  the  cube  of  the  largest  single  figure 
(9)  contains  three  tignres  (729),  or  two  more  than  the  smallest;  also  the  cube  of  10 
=  1000,  and  the  cube  of  99  =  970299,  or  two  places  more  than  the  10.  Therefore 
we  see  that  the  cube  of  a  single  figure  will  contain  from  one  to  three  places;  the 
cube  of  a  number  of  two  figures  will  contain  from  four  to  six  ]ilaccs;   and  in 


892  soule's  riiiLosopiiic  practical  mathematics.  * 

general,  the  number  of  x>liU'es  in  the  cube  will  be  at  most  three  times  the  given 
number  of  places,  and  at  least  three  times  the  given  number,  less  two ;  and  con- 
versely, we  may  deduce  the  geiieral  principle  that  a  number,  or  cube,  consisting  of 
one  to  three  figures,  will  give  one  figure  in  its  root;  wlien  of  four  to  six  figures,  its 
root  will  contain  two  figures;  from  seven  to  nine  figures,  gives  three  figures  in  the 
root,  etc.  This  is  the  reason  why,  in  extracting  the  cube  root,  we  point  off  the 
number  into  periods  of  three  figures  each,  from  the  right,  each  period  giving  one 
place  in  the  root.  The  left  hand  period  may  contain  one,  two,  or  three  figures,  and 
its  root  is  always  found  by  inspection,  or  from  the  tabular  cubes  of  the  nine  digits 
given  above. 

To  demonstrate  the  formation  of  a  cube,  let  us  take  a  number  of  two  places 
of  figures,  as  54,  and  represent  them,  as  in  the  elucidation  of  the  square  root,  by 
lOj;  +  M,  respectively.  Then,  raising  this  to  the  third  power  by  simjjle  multiplica- 
tion, we  obtain  the  following  exijansion,  i.  e.,  (lOj;  +  «)  ^  =  lOOOo;^  +  SOOx'ii  + 
SOxu"  +  u^.  From  this  quantity  we  see  that  the  third  power  of  x  is  taken  1000 
times,  and  hence  it  is  necessary,  before  finding  the  value  of  x,  to  divide  by  1000, 
because  the  third  power  of  lO.r  has  no  representation  in  the  first  three  right  hand 
figures  of  the  entire  cube  of  lOj;  +  u,  but  lies  wholly  to  the  left.  From  the  above 
formula  we  see  that  the  cube  of  any  number  consisting  of  tens  and  units,  is  equal 
to  1000  times  the  cirbe  of  the  tens,  plus  300  times  the  square  of  the  tens  multiplied 
by  the  units,  plus  30  times  the  tens  niultiiilied  by  the  square  of  the  units,  plus  the 
cube  of  the  units. 

We  will  now  expand  the  number  54  to  the  third  power,  by  the  formula,  and 
for  convenience  will  arrange  the  terms  in  columns,  to  correspond  with  the  numerical 
denominations, 

Thus,  (lOr +  »/)'  =  (50  +  4)3 

Or,  1000x3     =  5  X  5  X  5  X  1000  =  125000 

300x^(  =5x5x4X    300=    30000 

30x«2    =5x4x4x      30=      2400 

«'  =  4  X  4  X  4  =         64 


1000x3  ^  300x-'«  +  30x«'^  +  m3  =  157464 

We  here  make  a;  =  5  in  ten's  x)Iace,  and  m  =  4  in  unit's  place;  then  perform- 
ing  the  operations  indicated  in  the  formula,  we  obtain  the  several  results  corres- 
ponding to  the  equivalent  terms  of  the  formula.  W^e  see  from  these  partial  results 
that  the  third  power  of  the  tens  (5)  has  been  multiplied  by  1000,  giving  125000; 
therefore  in  extracting  the  cube  root,  we  must  first  divide  this  number  by  1000 
before  proceeding  to  resolve  the  125  into  its  three  equal  factors  ;  the  first  root  figure 
or  factor  must  always  be  found  by  inspection  or  trial,  and  in  this  case  we  find  it  to  be 
5  tens,  the  5  being  the  cube  root  of  125,  and  the  tens  the  cube  of  the  1000,  which 
was  rejected  by  division,  or  pointing  off.  If  our  number  was  just  125000,  then  5 
tens  or  50  would  be  the  required  root,  but  as  the  whole  luimber  is  1574C4,  we  know 
there  must  be  a  unit  figure  to  produce  the  other  partial  results,  and  with  the  125000 
make  n\}  the  total  157404,  whose  cube  root  is  required. 


M 


CUBE    ROOT. 


895 


To  fiiui  the  unit's  figure,  let  us  resume  our  analysis  by   taking   tlie   entire 
number  to  operate  011  in  regular  order,  instead  of  the  partial  results,  tlius: 


1000j3     =  53 

JSOOx'^     =  52  X  300 
30x«     =5    X  4  X 
m"   =  4    X  4 
X 
u    or   4     X 


=  7500 

30=    600 

=      16 

8116 


157'4Gl  (  54 
125 

32464 


2464 


We  first  place  tlie  formula  on  tbe  left,  observing  to  eliminate  u  from  tlie  three 
last  terms,  as  indicated  by  tbe  brace,  and  place  tlie  u  at  the  bottom,  as  a  final 
multiplier  •when  found;  we  then  point  off  the  number  into  periods  of  three  figures, 
as  before  explained,  and  find  by  trial  the  largest  cube  in  the  left  hand  period  to  be 
125,  the  root  of  -which  is  5  tens ;  'we  place  this  root  figure  to  the  right,  as  a  quotient 
in  division,  and  its  cube  125  under  the  first  period,  and  after  subtracting,  annex  the 
second  period.  This  constitutes  the  first  dividend,  and  from  the  formula  ue  see 
that  this  amount  32464,  contains  SOOx'u  +  30xu'  +  w^,  but  as  we  do  not  yet  know 
the  unit  figure  (m),  we  will  eliminate,  or  take  out,  («)  from  each  term,  and  obtain 
(SOOa;^  +  30x21  +  u-)u  =  324C4.  Now  if  we  knew  the  value  of  the  quantities  con- 
tained in  the  parenthesis,  we  could  find  u  by  dividing  32404  by  that  value.  We  do- 
know,  however,  the  value  of  the  first  term  SOOj?^,  which  is  7500,  and  from  a  com- 
parison of  the  several  terms,  we  see  that  the  first  term  SOOj;-  is  many  times  greater 
than  all  the  other  terms ;  we  will  therefore  use  this  value,  7500,  for  a  trial  divisor,^ 
and  find  that  it  is  contained  in  the  dividend,  4  times,  with  a  remainder  which  we 
ignore;  we  place  the  4  in  the  root,  and  then  perform  the  operations  indicated  by 
the  remaining  terms  in  the  parenthesis  or  brace  (30xm  -f  u"),  that  is,  we  add  to  the 
trial  divisor,  7500,  30  times  the  ten's  figure  5  multiplied  by  the  unit's  figure  4,  or  5 
X  4  X  30  =  600,  and  also  the  square  of  the  unit's  figure  4,  or  4  x  4  =  IG ;  the  sum 
of  these,  8116,  is  the  complete  divisor,  and  this  multiplied  by  the  last  root  figure,  i, 
and  the  product  extended  and  subtracted  from  the  dividend,  leaves  no  remainder  f 
therefore  54  is  the  exact  cube  root. 

2.     Find  the  cube  root  of  279726264. 


lOOOx'      =6^                    = 

279'726'264     (  654 
216 

300x2     ^  6^  X  300        =  1  (  10800 
30xtt     =6    X5x30  =  *^     9001 
1*==   =52                    =^(       25  U 

X                         — r 

«    or  5    X                          11725  J 

)  63726           1st  dividend. 

=    58623 

0 

2 

a  trial  div.  65'  X  300      =        1267500 

65  X  4  X  30  =             7800 

4"          =                 16 

)    5101264     2d  dividend. 

(1  complete  divisor,                    1275316 

=      5101264 

0   remainder. 

After  pointing  off  into  periods,  we  find  the  greatest  cube  iu  the  first  period 
to  be  216,  the  cube  root  of  which  is  6 ;  we  place  this  to  the  right  for  the  first  figure 


894  soule's  niiLosoi'iirc  practical  mathematics.  * 

of  the  root,  and  subtract  its  cube  210,  from  the  first  period  279,  and  to  the  remainder 
oa  sinuex  tlie  second  jieriod,  malcinj^-  (!3726  for  tbe  first  dividend ;  tlien,  according  to 
tlie  formula,  we  take  300  times  tlie  square  of  the  root,  C,  or  3(3  x  300,  giving  10800 
for  the  first  trial  divisor,  and  tlien  find  that  this  is  contained  in  tlie  dividend  5 
times;  we  place  the  5  for  the  second  figure  of  the  root,  and  then  to  complete  the 
divisor,  we  add  to  the  trial  divisor  30  times  the  product  of  the  iirevions  root  figure 
l)y  the  last,  or  G  x  5  x  30  =  900,  and  also  the  square  of  the  last  root  figure  5  = 
25,  the  sum  of  all  giving  11725  for  the  first  complete  divisor,  which  we  multiply  by 
the  last  figure  of  the  root,  5,  and  subtract  the  product  58G25  from  the  first  dividend, 
and  to  the  remainder  bring  down  tlie  next  period  for  the  second  dividend. 

We  then  take  300  times  the  square  of  all  the  figures  in  the  root,  or  05^  x  300 
=  12C7500,  for  the  second  trial  divisor,  which  we  find  is  contained  in  the  dividend 
4  times;  we  place  the  4  in  the  root,  and  then  proceed  to  comi)lete  the  divisor  as 
before,  that  is,  we  add  to  the  trial  divisor  30  times  the  product  of  tlie  previous 
figures  of  the  root,  C5,  by  tlie  last,  4,  or  G5  x  4  x  30  =  7800,  and  the  square  of  4 
=  IG;  the  sum  of  these,  1275310,  gives  the  2d  complete  divisor,  which  we  multiply 
by  4  and  subtract  the  product  from  the  dividend,  and  as  there  is  110  remainder,  054 
is  the  exact  cube  root. 

If  there  was  a  remainder,  we  would  annex  periods  of  three  ciphers  each,  and 
continue  the  operation  in  the  same  manner,  until  sufficiently  exact,  each  period  of 
ciphers  giving  one  decimal  place  in  the  root. 

KoTK. — The  following  new  and  original  nictboils  of  finding  tbe  trial  divisors,  after  the  first, 
embrace  a  beautiful  princijjle,  ap]ilicable  to  roots  of  every  degree.  Thus,  multiply  tbe  numbera 
conned ed  by  tbe  brackets  on  tbe  left,  by  tbeir  respective  positions,  counting  downwards,  as  1st, 
2d  :-iil,  its  inilicated  by  tbe  small  figures  to  tbe  left,  and  their  sum  with  two  ciphers  annexed  gives 
tbe  next  trial  divisor."  which  is  written  below.  This  is  most  easily  done  by  multiplying,  tirst  tbe 
riglit  hand  figure  of  the  3d  line  by  3,  and  the  significant  /igures  if  there  be  any,  above  it,  by  2  and 
1  respectively,  and  set  tbcir  sum  below  the  last  complete  divisor,  then  multiply  and  add  tbe 
successive  loit  hand  columns  in  the  same  Tiuinner.  Or  we  may  find  tbe  trial  divisor  by  adding  the 
three  numbers  as  shown  connected  by  the  right  band  bracket,  observing  to  niulti|)ly  by  2,  or 
double  the  iigures  of  tbe  middle  line  or  squ.are  of  units,  in  adding,  and  annex  two  ciphers  to  tbe 
result,  in  either  case,  for  tbe  trial  divisor. 

3.    What  is  the  cube  root  of  58050510848  ? 

OrEKATIOX. 

Power  =  58  OSOolO  848     (  3872  root. 
3'  =27 


Ist  trial  divisor,   3'  X  3        =27  )  31050  =  1st  dividend. 

3  x  8  X  3  =    72  1 


«2         =      64  [ 


2 


1st  complete  divisor,  3484  J       =27872 

2d  trial  divisor,  4332  )  3178510         =  2d   dividend. 

38  X  7  X  3  =      798  1 
=         49  i, 


2d  complete  divisor,  441229]      = 


3088G03 


3d  trial  divisor,  449307  }  89907848  =  3d  dividend. 

387  X  2  X  3  =       2322  -| 
2^         =  4^2 

3d  complete  divisor,  44953924]      —      89907848 

0   =  remainder. 


.    *  CUBE    ROOT.  895 

In  tliis  Illustration  we  dispense  with  all  unnecessary  figures,  by  setting  the 
numbers  that  form  the  complete  divisors  one  i)lace  successively  to  the  right,  because 
we  omit  the  ciphers  in  the  multipliers  300  and  30,  and  in  dividing  by  the  trial 
divisors,  observing  to  exclude  the  two  right  hand  figures  of  the  dividend  to  find  the 
next  root  figure. 

To  find  the  trial  divisors,  after  the  first,  we  add  the  last  complete  divisor  and 
the  two  numbers  immediately  above,  as  shown  by  the  brace  on  the  right,  observing 
to  double  the  figures  of  the  middle  number  in  adding  ;  or  the  first  method  mentiMied 
in  the  note  to  last  example,  is  the  most  scientific. 

If  the  luimber  is  a  mixed  or  pure  decimal,  we  point  off  as  in  square  root,  only 
in  periods  of  three  figures  each,  to  the  left  of  the  decimal  point  for  the  integers,  and 
to  the  right  for  the  decimals,  annexing  ciphers  if  necessary  to  complete  the  last 
period. 

When  we  wish  to  limit  the  decimal  places  of  approximation  in  the  root  to  a 
certain  number,  we  may  extract  half,  or  one  more  than  half,  the  required  number  of 
root  figures  by  the  regular  method,  and  then  continue  to  divide  the  last  remainder 
by  the  last  complete  divisor,  until  the  required  number  of  root  figures  is  obtained. 
This  last  process  may  be  further  abbreviated  by  cutting  ofl',  or  canceling  a  figure 
from  the  right  of  the  divisor  at  each  consecutive  division,  no  ciphers  being  annexed 
to  the  remainders ;  the  last  decimal  root  figure,  by  this  process,  may  be  in  error. 
When  the  divisor  is  very  large,  we  need  use  of  it  only  one  more  than  the  number  of 
root  figures  yet  to  be  found,  and  also  cut  off  from  the  dividend  one  less  than  from 
the  divisor,  observing  in  multiplying  by  the  root  figure  to  carry  for  the  figure  cut 
off. 

In  the  following  example,  we  illustrate  the  above  processes,  and  also  present 
another  modified  method  of  operating  for  finding  the  trial  and  complete  divisors. 
Thus,  after  finding  the  first  trial  divisor  by  taking  3  times  the  square  of  the  first 
figure  of  the  root,  and  obtaining  the  second  figure  of  the  root,  we  then  take  3  times 
the  first  root  figure,  and  to  the  product  annex  the  second  root  figure,  multiply  the 
result  by  the  second  root  figure  and  set  the  product  two  places  to  the  right  of  the 
trial  divisor,  and  by  adding  we  obtain  the  first  complete  divisor. 

Then  to  find  the  2d  trial  divisor,  we  add  together,  as  shown  by  the  braces  on 
the  right,  the  last  complete  divisor,  the  number  just  above,  and  the  square  of  the 
last  root  figure,  which  may  be  written  below  or  added  mentally;  then,  after  finding 
the  3d  root  figure,  we  take  3  times  the  preceding  figures  of  the  root  and  annex  the 
last  root  figure,  and  multii)ly  the  result  by  the  last  root  figure,  and  extend  the  prod- 
uct two  places  to  the  right  of  the  trial  divisor  and  add  for  the  second  complete 
divisor,  and  proceed  in  the  same  manner  as  far  as  requisite.  This  is  i)erhaps  the 
most  concise  method  of  operations,  when  the  general  theory  is  first  well  understood. 


896  SOULe's    rillLOSOPHIC    I'RACTICAL    MATHEMATICS. 

4.    Fiud  the  cube  root  of  100.9oi5  to  eiglit  places  of  decimals. 

OPERATION. 

'1(56.054  500     (  5.50637828+ 
5'  =  125 


S'^  X  3  =  75  )  41954 

155  X  5  =    775  ~] 

1st  complete  divisor,  8275  ( 

b''  =      25  J 


2(1  trial  divisor,  907500  579500.000 

16506  X  6=        990361 

2d  complete  divisor,  90849036  \  ~  '>^->^^^  ^16 

6'  =  36  J 

3d  trial  divisor,  90948108 

165183  X  3  =  495549 

90953|06349  = 


3440578  4000 

2728.591  9047 

711986 

4953 

636671 

75315 

72762 

2553 

1819 

734 

727 

7 

=  remainder. 

When  any  product  is  greater  tlian  the  dividend  above  it,  diminish  the  last 

figure  of  the  root. 

When  any  trial  divisor  is  greater  than  its  corresponding  dividend,  place  a 

cipher  in  the  root,  and  two  ciphers  to  the  right  of  the  trial  divisor,  then  bring  down 

the  next  ])eriod  to  the  right  of  the  dividend  and  continue  as  before. 
To  find  the  cube  root  of  common  fractions,  we  may 
1st,  Extract  the  cube  root  of  each  term,  if  both  are  rational ;  or, 
2d,  If  surds,  multii>ly  the  numerator  by  the  square  of  the  denominator,  then 

extract  the  cube  root  of  this  product  to  any  approximation,  and  divide  the  result  by 

the  given  denominator;  or, 

3d,  Eeduce  the  fraction  to  a  decimal  and  extract  the  cube  root. 

EXAMPLES   AND   APPLICATIONS. 

1.  What  is  the  cube  root  of  C1S470208?  Ans.  852. 

2.  What  is  the  cube  root  of  34S415G090?  Ans.  loK!. 

3.  What  is  the  cube  root  of  19.G174G2873?  Ans.  2.097. 

4.  Find  the  value  of  Y 71!  to  nine  decimal  places.  Ans.  4.1C01G7C4C. 

5.  Find  the  value  of   a/27442.  Ans.  19(5.    ' 

6.  What  is  tiie  cube  root  of  2-7 1  •  Ans.  §. 


*  CUBE    ROOT.  897 

7.  What  is  the  cube  root  of  J  ?  Ans.  i. 

8.  What  is  the  cube  root  of  |  to  five  decimal  places?  Ans.  .87358. 

9.  What  is  the  cube  root  of  ^  to  six  decimal  places?  Ans.  .G299C1+. 

10.  How  many  quarter-inch  solid  blocks  are  equal  iu  volume  to   one   solid 
inch;  also  to  one  solid  foot ?  Ans.  64,  and  110592. 

11.  A  box  is  12  feet  long,  10§  feet  wide,  and  4  feet  deep;  what  is  the  side  of 
a  cubical  box  of  equal  capacity  ?  Ans.  8  feet. 

12.  A  vat  holds  C77G  hogsheads;  if  it  is  twice  as  wide  as  deep  and  twice  as 
long  as  wide,  what  are  its  dimensions  ? 

Ans.  19  ft.  3  in.  deep;  38  ft.  6  in.  wide;  and  77  ft.  long. 

SOLUTION. 

Extract  the  cnbe  root  of  (6776  X  63  x  231)  divided  T)y  (1x2x4);  and  multiply  the  result 
by  the  iiroportioual  parts,  1,  2,  and  4,  for  the  depth,  width  and  length.     Thus, 

3  , (  X  1  =  231    in.  deep. 

V  6776  X  63  X  -.'SI     ^     231    ?  X  2  =  462   in.  wide. 
1^-=^*  ^  X  4  =  924   in.  long. 

Geometry  demonstrates  that, 

Similar  solids  have  the  same  ratio  to  each  other  as  the  cubes  of  their  like 
dimensions,  or  as  the  cubes  of  the  ratio  of  their  like  dimensions;  and 

The  like  dimensions  of  similar  solids  have  the  same  ratio  as  the  cube  roots  of 
their  solidities,  or  as  the  cube  root  of  the  ratio  of  tlieir  solidities. 

13.  A  block  of  marble  is  12  inches  long,  and  contains  576  cubic  inches ;  what 
are  the  contents  of  a  similar  block  18  inches  long?  Ans.  1944  cubic  inches. 

14.  Tlie  solidities  of  two  similar  blocks  are  162  and  2058  cubic  inches;  the 
length  of  the  first  is  9  inches,  what  is  the  length  of  the  second  ?  Also,  what  are 
the  other  dimensions  of  each,  whose  ratios  are  as  1,  2,  3  ? 

Ans.  21  in.,  length  of  second  block. 

First  block,  6  in.  wide  and  3  in.  thick. 
Second  block,  14  in.  wide  and  7  in.  thick. 

SOLUTION   TO  FIRST   QUESTION. 

2,<>,A^  =  12^8,  ratio  of  solidities ;  then. 
V'l2i?  =  •/ W  =  3  =  2J,  ratio  of  lengths ;  then,  9  x  2J  =  21  in. 

Tlie  solution  of  the  second  question  is  the  same  as  in  example  12  or  by  direct 
proportion  from  the  lengths. 

15.  A  rectangular  reservoir  is  25  feet  deep ;  the  ratio  of  the  other  two  sides 
is  as  3  to  4,  and  the  longest  transverse  diagonal,  or  distance  from  one  lower  corner 
to  the  opposite  upper  corner,  is  65  feet;  what  are  tlie  otlier  dimensions  of  the 
reservoir;  how  many  gallons  does  it  contain,  and  what  would  be  the  side  and 
contents  of  a  cubical  reservoir  of  5  times  the  capacity  ? 

Ans.  Longest  side,  48  feet. 
Shortest  side,  36  feet. 
Contents,  3231584*-  gallons. 
Side  of  cubical  reservoir,  60  feet. 
Contents  of  cubical  reservoir,  1615792^f  gals. 


^ligS  soule's  philosophic  practical  mathematics.  * 

SOLUTION. 

V  65'  —  25'  =  60  feet,  diagonal  of  sides,  or  bottom. 

And,  V  3'  -f"  4*  =  5,  diagonal  ratio;  then,  60  -:-  5  ^  12,  multiple  of  ratio  of  sides. 

Then,  3  x  12  =  36,  width  ;  and  4  x  12  =  48,  length. 

The  other  conditions  are  now  easily  fonnd,  thus: 

V  48  X  36  X  25  X  5  =  60  feet,  side  of  cubical  reservoir. 

And,  (48  X  36  X  25  X  1728)  —  231  =  323158H  gals,  in  first  reservoir, 

IG.  A  man  has  two  rectaugular  plots  of  grouud  of  equal  extent ;  the  first  is 
twice  as  long  as  the  second,  and  the  second  is  twice  as  wide  as  the  first,  and  the 
squares  of  their  diagonals  are  S9189  yards  and  17141  yards  respectively.  lie 
exchanges  the  two  plots  for  a  square  garden  of  equal  area  to  the  two  plots.  He 
then  sells  the  garden  for  as  many  mills  i>er  square  yard  as  the  garden  is  yards 
square,  and  receives  $1331,  thereby  gaining  10  per  cent  on  original  valuation.  What 
were  the  dimensions  and  contents  of  each  piece  of  ground;  what  did  the  garden 
sell  for  per  square  yard,  and  what  was  the  original  value  of  each  piece? 

Ans.  First  plot,  242  yards  long  and  25  yards  wide,  =  1^  acres,  cost  $605. 
Second  plot,  121  yards  long  and  50  yards  wide,  =  IJ  acres,  cost  $605. 
Garden,  110  yards  square,  =  2J  acres,  cost  $1210,   or   10   cents   per 
square  yard,  and  sold  for  11  cents  jier  square  yard. 

SOLUTION. 

V'$1331.0(J0  =  110  mills ;  then  side  of  garden  =  110  yards,  and  area  of  garden  =  12100  square 
yards,  and  12100  -^  2  =  6050  square  yards,  the  area  of  each  plot.  110  mills  selling  price  at  10% 
gain,  gives  100  mills  or  10  cents  for  cost  per  square  yard. 

To  find  the  sidefs  of  the  two  plots,  we  first  find  the  sum  and  the  difference  of 
the  square  of  the  diagonals,  and  operate  as  follows : 

OPERATION. 

59189  square  of  1st  diagonal  59189 

17141  square  of  2d  diiigonal  17141 

2'  -f  1«  =  5)  76330  sum  of  squares  2*  —  1'  =  3)  42048  difference  of  squares. 

15266  =  sum  of  squares  of  shorter     14016   =  difference  of  squares  of  shorter  length 
length  and  width.  [and  width. 

Then  to    15266  And  from      15266 

add  14016  subtract        14016 


2)  29282  2  )     1250 

Square  of  shorter  length  V  14641  V'625  square  of  shorter  width. 

Extract  square  root,  =  121  shorter  length;  25  shorter  width. 

2  2 

242  greater  length ;         50  greater  width. 


*  CUBE    ROOT.  899 

MISCELLANEOUS  EXAMPLES  IN  SQUAEE  AND  CUBE  BOOT. 

1.  Two  men  formed  a  copartnership;  tbe  first  invested  $3000  for  864 
days,  and  received  f  of  the  gains ;  tlie  second  invested  as  many  dollars  as  he 
continued  days  in  business.     What  was  his  capital  ?  Ans.  $2160. 

OPERATION    INDICATED. 

$3600  X  8G4  days  =  3110100  =  |  gains  and  §  gains  =  4665600. 
Then  extract  square  root,  for  second  capital  and  number  of  days  =  V36OO  x  Sbl  X  i  =  2160. 

2.  If  it  take  8  planks  to  build  one  panel  of  fence,  and  there  are  just  25 
panels  on  one  side  of  a  square  acre,  how  many  miles  square  must  the  tract  be,  that 
will  contain  as  many  acres  as  it  will  take  planks  to  fence  it?  Also,  if  it  cost  $1 
per  acre  for  the  work,  how  many  men  must  be  employed  at  $1J  per  day,  to  do  the 
work  in  as  many  days  as  there  are  men  ? 

Aus.  Side  of  the  square  tract,  31.622776+  miles. 
Number  of  men  required,  600. 

SOLUTION. 

25  X  8  X  4  =  800  planks  for  one  acre ;  then,  800*  =  640000  acres. 
And  640000  —  640  =  1000  square  miles. 

Then,        1^1000  =  31.622776+  miles  square. 

SOLUTION   TO   SECOND   QirESTION. 

$640000  —  IJ  =  360000  days  of  work. 
And  V/36OOUO  =  600,  number  of  men. 

3.  A  farmer  has. a  rectangular  field,  containing  30  acres,  whose  sides  are  as 
3  to  4 ;  what  will  it  cost  to  dig  a  ditch  6  feet  wide  and  4-;-,-  feet  deep  diagonally 
through  it,  at  40  cents  per  cubic  yard  ?  Also,  how  many  men  will  it  take  to  do  the 
work,  each  man  working  twice  as  many  days  as  there  are  men,  and  receiving  twice 
as  many  cents  per  day  as  each  works  days? 

Ans.  $640,  cost.     20  men.    40  days  each.     SO  cents  per  day. 


SOLUTION. 


30  acres  =  1306800  square  feet.^ 
3  X  4  =  12  ratio  divisor. 
Gives  quotient  =  108900. 


Extract  square  root  =  330  feet,  multiple  of  ratios. 
Then,        3  x  330  =    990  feet  width  of  field. 

4  X  330  =  1320  feet  length  of  field. 

5  x  330  =  1650  feet  length  of  diagonal  of  field. 
Then,        1650  X  6  x  41,-  =  43200  cubic  feet  of  ditch. 

Divided  by  27  gives  1600  cubic  yards  of  ditch. 

1600  X  40  cents  =  $640.00  cost  of  work. 
Divide  by  1x2x4  =  8  =  8000. 

Extract  cube  root  ^  20  multiple  of  2d  ratios, 
Man      1  X  20  =  20  men. 
Days    2  X  20  =  40  days. 
Cents  4  X  20  =  80  cents  per  day. 


900 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


4.  A  man  bong^lit  a  square  piece  of  land,  paying  as  many  cents  per  square 
rod  as  the  piece  was  rods  square  ;  he  then  sokl  it  at  an  advauc'e  of  17  ,\-  per  cent, 
and  gained  $880.  How  many  acres  in  the  piece,  and  how  many  rods  square  ?  What 
was  the  total  cost  and  selling  in'ice,  and  the  cost  and  selling  price  i)er  square  rod? 

Ans.  Area,  40  acres, 

Or,  SO  rods  square. 

Total  cost,  $5120. 

Total  selling  i)rice,  $6000. 

Cost  per  square  rod,  80  cents. 

Selling  price  per  square  rod,  93^  cents. 


OPKRATION    INDICATED. 


To 

275 

in<l  cost. 

100 
16 

880 

|;5120   cost. 
880  gain. 

$6000   Sales. 

Since  the  land  was  a  certain  number  of  rods 
square  and  the  jirice  paid  was  as  many  cents 
per  square  rod  as  the  price  was  rods  square,  the 
cost  of  the  -whole  in  cents  was  the  cube  of  the 
price  per  square  rod. 


Hence  1^512000  =  80  cents. 

=  80  rods  square. 


:i 


jithmetical  Progression. 


Arithmetical  Progression,  sometimes  called  Frogression  hy  Difference, 
or  Equidifferent  Series,  is  a  series,  or  succession  of  iiuinbers  or  quantities  that 
continually  increases  or  decreases  from  term  to  term,  by  a  common  difference. 

The  numbers  constituting  a  Progression  are  called  the  terms  of  the  series,  of 
■which  the  first  and  last  are  denominated  the  Extremes,  and  all  the  intermediate 
terms,  the  Means. 

An  Arithmetical  Progression  is  an  ascending  series,  Trhen  each  snccessive 
term  is  formed  by  adding  the  common  difference  to  the  preceding  term ;  and  a 
descending  series,  when  the  common  difference  is  subtracted  consecutively  from  the 
first  to  the  last  term  ;  thus : 

1,       3,     5,     7,     9,  is  an  ascending  series,  common  difference  +  2 
13,     10,     7,     4,     1,  ia  a  descending  series,  common  diflei'ence  —  3 

In  an  arithmetical  i)rogression,  there  are  five  things  or  parts  to  be  conside/  ed, 
any  three  of  which  being  given,  the  others  can  be  determined. 

These  parts  are  named  and  represented  as  follows : 


1st. 

The  First  Term, 

represented  by 

a. 

2d. 

The  Last  Term, 

ii 

a 

I. 

3d. 

The  Common  Difference, 

u 

ti 

d. 

4th. 

The  Number  of  Terms, 

11 

u 

n. 

5th. 

The  Sum  of  the  Series, 

li 

a 

s. 

Since  any  one  of  these  five  parts  may  be  found  from  any  three  of  the  four 
remaining  parts,  there  will  arise  just  5  x  4  =  20  cases  of  problems,  prodncing  as 
many  formulas  and  resulting  solutions. 

Only  5  or  6  of  these  20  cases  are  usually  treated  in  arithmetics,  the  other 
cases  requiring  more  elaborate  algebraic  demonstrations  and  formulas,  and  are  of 
less  practical  application. 

Problem  I. 

To  find  the  last  term,  when   the   first   term,   common   difference,   and 
number  of  terms  are  given. 

1.  The  first  term  is  2,  the  common  difference  3,  and  the  number  of  terms 
5 ;  what  is  the  last  term  ? 

Illustration.     Thus,         2,  (2  +  3),  (5  +  3),  (8  +  3),  (11  +  3). 
Gives        2,        5,  8,  11,  14. 

(901) 


902  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  *■ 

Here  we  see  that  the  common  difference,  3,  is  added  as  many  times,  less  one, 
as  there  are  terms  j  that  is,  5  —  1  r=  4  times,  which  gives  12  to  be  added  to  the  first 
term  2,  =  14  for  the  last  term,  or  thus:  2  +  (5  —  1)  x  3  =  14. 

Hence  the  general  formula,  I  =  a  +  {n  —  l)d. 

Or,  expressed  in  common  language,  the  last  term  is  equal  to  the  first  term, 
plus  or  minus  the  product  of  the  number  of  terms,  less  one,  by  the  common  difier- 
ence. 

The  sign  plus  or  minus,  ±,  is  used  to  show  that  the  quantity  to  the  right,  (» 
—  l)d,  is  to  be  added  to  the  first  term  when  the  series  is  ascending,  but  to  be  sub- 
tracted from  the  first  term  when  the  series  is  descending. 

2.  The  first  term  of  an  increasing  arithmetical  progression  is  5,  the  common 
difference  is  6,  and  the  number  of  terms  16 ;  what  is  the  last  term  ? 

Ans.  6  +  (15  X  G)  =  95,  last  term. 

3.  The  first  term  of  a  decreasing  arithmetical  progression  is  450,  and  the 
common  difference  is  10 ;  what  is  the  33d  term  1 

Ans,  450  —  (32  x  10)  =  130  for  the  33d  term. 

4.  If  a  man's  salary  is  $20  the  first  month,  $30  the  second  month,  and  so 
on,  increasing  $10  each  month,  what  will  it  be  the  20th  month  ?  Ans.  $210. 

5.  What  will  be  the  amount  of  $1000  in  25  years,  at  6  per  cent  per  annum, 
simple  interest?  Ans,  $2500. 

SOLUTION. 

$106  +  (24  X  6)  =  $250,  amount  of  $100 ;  then  $1000  will  give  $2500. 

6.  The  first  term  of  an  ascending  series  is  ^,  and  the  common  difference  i* 
J;  what  is  the  1000th  term?  Ans.  600^, 

7.  The  first  term  of  a  descending  series  is  1000,  the  common  difference  -jV, 
and  the  number  of  terms  9000 ;  what  is  the  last  term  ?  Ans,  100.1. 


Problem  II. 

To  find  the  common  difference,  the  extremes  and  number  of  terms  being 
given. 

Since  the  number  of  common  difference  is  always  one  less  than  the  number  of 
terms,  and  the  sum  of  all  the  common  differences  is  equal  to  the  difference  between 
the  first  and  last  terms,  if  we  divide  the  difference  of  the  extremes  by  the  number 
of  terms  less  one,  the  quotient  will  be  the  common  difference;  hence  the 

Formula,  d  =  ?^=-^j  for  ascending,  and  d  =  ^^'  for  descending. 

1,  The  first  term  of  an  arithmetical  series  is  2,  the  last  term  74,  and  the 
number  of  terms  25 ;  what  is  the  common  difference  ? 

Ans.  ^^  =  |i  =  3,  common  difference. 


■^  ARITHMETICAL    PROGRESSION,  9O3 

2.  A  man  travels  15  days,  going  10  miles  the  first  day;  after  ■whicli  he 
increases  tlie  distance  an  equal  number  of  miles  each,  day ;  tlie  last  day's  journey 
was  80  miles;  what  was  the  daily  increase?  Ans.  5  miles. 

3.  In  a  company  of  101  persons,  whose  ages  are  in  arithmetical  progression? 
the  youngest  is  1  year  old,  and  the  oldest  is  51  years ;  what  is  the  common  difference 
of  their  ages  ?  Ans.  6  months. 

4.  The  first  term  of  an  arithmetical  series  is  21,  the  last  J,  and  the  number 
of  terms  83 ;  what  is  the  common  diflereuce?  Ans.  ^. 

To  insert  any  number  of  arithmetical  means  between  two  given  numbers,  is 
simply  to  fiud  the  common  difference  by  the  above  formula,  and  then  form  the 
series,  observing  that  the  whole  number  of  terms  is  two  more  than  the  given  number 
of  means,  viz.,  the  two  extremes. 

5.  Insert  5  arithmetical  means  between  10  and  40. 

SOLUTION. 
d  =  1^^zl9  =  5,  common  difference ;  then  10,  15,  20,  2.5,  30,  35,  40  =  series. 

6.  The  extremes  are  2^  and  52,  and  the  whole  number  of  terms  is  100;  what 
is  the  second  term,  and  the  next  to  the  last  term  f 

Ans.  Second  term,  3 ;  next  to  the  last  term,  olj. 

Pkoblem  hi. 

To  find  the  number  of  terms,  when  the  extremes  and  common  difference 
are  given. 

Since  the  difference  of  the  extremes,  divided  by  the  common  difference,  ia 
equal  to  the  number  of  common  differences,  then  by  adding  one  we  obtain  the  num- 
ber of  terms ;  hence  the 

Formula,  n  =  ?-=-?  +  1 

That  is,  the  number  of  terms  is  equal  to  the  difference  of  the  extremes 
divided  by  the  common  difference,  with  the  quotient  increased  by  1. 

1.  The  extremes  of  an  arithmetical  series  are  7  and  127,  and  the  common 
difference  is  5 ;  what  is  the  number  of  terms  ? 

SOLUTION. 
127  —7 


-f  1  =  24  +  1  =  25,  number  of  terms,  =  Ans. 

2.  How  long  will  it  take  a  man  to  make  a  journey,  if  he  travels  12  miles  the 
first  day  and  increases  the  distance  3  miles  each  succeeding  day,  the  last  day's 
journey  being  72  miles  ?  Ans.  21  days. 

3.  A  person  settles  his  account  by  paying  25  cents  the  first  day,  with  10 
cents  additional  each  succeeding  day;  the  last  day's  payment  was  $36.65;  how 
many  days  was  he  in  paying  ?  Ans.  365  days. 


904  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

Problem  IV. 

To  find  the  sum  of  all  the  terms,  or  series,  when  the  extremes  and  num- 
ber of  terms  are  given. 

If  we  take  any  series  and  write  nnder  it  the  same  series  in  an  inverse  order, 
the  snra  of  eacli  couplet  of  terms  must  be  equal,  and  the  summation  of  all  the 
couplets  will  be  twice  the  sum  of  the  given  series. 

Thus,  3+    5+    7+    9  +  11  =  3r),   buiu  of  given  series. 

And  11  +    i'  +    7  +    f)  -|-    3  =^  35,    sum  of  given  series  inverted. 


Then,         14  -|-  11  -(-  14  -f-  14  -f  11  =  70,    =  twice  the  sum  of  the  given  series. 

Here  we  see  that  the  sum  of  the  extremes,  14,  is  equal  to  the  sum  of  any  two 
terms  that  are  equally  distant  from  the  extremes;  therefore,  if  we  multiply  the  suui 
of  the  extremes,  l-i,  by  the  number  of  terms,  5,  the  product,  TO,  will  be  twice  the 
sum  of  the  series ;  then,  this  divided  by  2  will  give  35  for  the  sum  of  the  given 
series.  But  we  may  divide  either  factor,  14  or  5,  by  2,  with  the  same  result;  hence 
the 

Formula,  s  =  "  J"     x  n,   or,  7,  X  a  -\-  I 

That  is,  the  sum  of  the  series  is  equal  to  half  the  sum  of  the  extremes  multi- 
plied by  the  number  of  terms ;  or  half  the  number  of  terms  multiplied  into  the  sum 
of  the  extremes. 

1.  The  extremes  of  an  arithmetical  ]>rogression  are  5  and  1G5,  and  the  num- 
ber of  terms  33 ;  what  is  the  sum  of  the  series  ? 

Aus.  s  =  '1+^  X  33  =  85  X  33  =  2805,  sum. 

2.  The  extremes  are  0  and  560,  and  the  common  difference  is  7  ;  what  is  the 
sum  of  the  series  ?  Aus.  22080. 

We  first  find  the  number  of  terms  by  Problem  III,  and  then  the  stim  of  the 
series. 

3.  If  a  person  pay  $50  the  first  month,  and  $50  additional  each  subsequent 
mouth,  what  amount  of  debt  will  be  discharged  in  one  year?  Ans.  $3900. 

OPERATION    INDICATED. 

1  =  50  +  (11  X  50)  =  600. 

s  =  •'"  +  '^"»  X  12  =  3900. 
2 

4.  If  1000  oranges  be  laid  in  a  line  five  feet  from  each  other,  the  first  being 
five  feet  from  a  basket,  how  far  will  a  person  travel  who  gathers  them  up  singly, 
returning  with  them  one  by  one  to  the  basket?  Ans.  947  miles,  1613^  yards. 

SOLl'TION. 
5  feet  X  2  =        10       feet  =  1st  term. 
5000  feet  X  2  =  10000       feet  =  last  term. 

10010       feet  =  sum  of  extremes. 
500     =  i  number  of  terms. 

Divided  by  5280)  1005000  feet  =  -whole  distance  =  947  miles,  1613i  yards. 


*  ARITHMETICAL    PROGRESSION.  "  9o5 

We  append  a  lew  additional  formulas,  without  explauatiou.s,  for  some  of  the 
other  simjiler  cases. 

Peoblem  V. 

To  find  the  last  term,  when  the  first  term,  number  of  terras,  and  sum  of 
the  series  are  given. 

Formula,  I  ^  ~ a 

ji 

That  is,  the  last  term  is  equal  to  twice  the  sum  of  the  series  divided  by  the 
number  of  terms,  and  the  quotient  diminished  by  the  first  term. 

Problem  VI. 

To  find  the  first  term,  when  the  last  term,  number  of  terms,  and  sum  of 
the  series  are  given. 

Formula,  a  =  "^ I 

'  n 

That  is,  the  first  term  is  equal  to  twice  the  sura  of  the  series  divided  by  the 
number  of  terms,  and  the  quotient  diuunished  by  the  last  term. 

Problem  VII. 

To  find  the  number  of  terms,  when  the  extremes  and  sum  of  the  series 
are  given. 

Formula,  ?t  = 


a  +  I 

That  is,  the  number  of  terms  is  equal  to  twice  the  sum  of  the  series  divided 
by  the  sum  of  the  extremes. 

Problem  VIII. 

To  find  the  sum  of  the  series,  when  the  first  term,  common  difference, 
and  number  of  terms  are  given. 

Formula,  s  = '^^  \2a  +  (»  —  1)  X  d] 

That  is,  the  sum  of  the  series  is  equal  to  the  number  of  terms  less  one, 
multiplied  by  the  common  difference,  this  product  added  to  twice  the  first  term,  and 
the  sum  multiplied  by  half  the  number  of  terms. 

If  the  last  terra  were  given  in  Problem  VIII,  instead  of  the  first,  the  follow- 
ing would  be  the 

Formula,  s  =  ^  [21  —  (n  —  1)  x  d] 

Here  we  subtract  the  product  of  the  number  of  terms  less  one  by  the  common 
difference,  from  twice  the  last  term,  and  multiply  the  remainder  by  half  the  number 
of  terms,  for  the  sum  of  the  series. 


<jo6  soule's  philosophic  practical  mathematics.  • 

MISCELLANEOUS  EXAMPLES. 

NoTE^ — Questions  in  simple  interest  arc  solved  1)7  arithmetical  progression. 

1,     What  will  $10000  amount  to  in   50   years,   at  8  per  cent  simple 
interest  t  Ans.  $50000. 

SOLUTION. 

The  amount  of  $100  at  8  per  cent  for  one  year,  $108,  is  the  first  term,  anil  the  amount  for  50 
years  Is  the  last  term;  the  interest  for  1  year,  $8,  is  the  common  diflerence,  aiul  50  years  the  num- 
ber of  terms;  hence,  to  lind  the  last  term,  wo  have  $108  +  (49  X  $8)  =  $500,  amount  of  $100  for 
50  years;  then  l)y  projiortion  $10000  will  amount  to  $50000. 

2.  What  sum  must  be  loaned  at  6  per  cent  simple  interest  for  20  years,  to 
amount  to  $4400 1  Ans.  $2000. 

SOLUTION. 
$106  4-  (19  X  $6)  —  $220,  amount  of  $100;  then  by  proportion  $4400  require  $2000. 

3.  At  what  rate  per  cent  simple  interest  must  $5000  be  loaned  to  amount  to 
$20000  in  30  years  ?  Ans.  10  per  cent. 

SOLUTION. 

$5050  +  (29  X  $50)  =  6500  —  5000  =  1500,  amount  of  interest  at  1  per  cent;  then  $15000  ~ 
1500  =  10  per  ceut. 

4.  How  long  a  time  must  $7000  be  loaned  at  7  per  cent  simple  interest,  to 
amount  to  $10430  ?  Ans.  7  years. 

SOLUTION. 

$7490  +  (0  X  $490)  =  $7490  —  7000  =  $490  interest  for  1  year ;  then  $3430  -^  490  =  7  years. 

5.  A  heavy  body  falling  unobstructed,  moves  through  a  distance  of  IG^Vfeet 
the  first  second,  with  an  accelerating  velocity  of  32^feetfor  each  additional  second; 
what  distance  will  it  fall  in  31  seconds  ?  Ans.  15456ii-2  feet. 

SOLUTION. 

This  question  may  be  solved  by  Problems  I  and  IV,  or  directly  by  Problem  VIII,  thus; 
s  =  V-  [2  X  16(V  +  (31  —  1)  X  32J]  =  15456tV 

6.  A  man  owes  $4510  to  41  different  persons,  in  arithmetical  j)rogression ;  to 
the  first  he  owes  $10,  how  much  does  he  owe  the  last,  and  what  is  the  common 
difference  of  the  debts  ?  Ans.  Amount  due  the  last,  $210. 

Common  difference  of  debts,  $5. 
Solution  of  first  question  by  Problem  V, 

I  =  ?  ^  ^^^"  —  10  =  210,  last  term. 
41 

7.  Two  persons  start  at  the  same  time  from  two  cities,  207  miles  apart,  to 
meet  each  other  ;  the  first  travels  10  miles  the  first  day,  and  20  nnles  the  last  day; 
the  second  travels  12  miles  the  first  day,  and  27  miles  the  last  day.  In  how  many 
days  will  they  meet ;  and  how  far  will  each  travel;  and  what  is  the  daily  increase 
of  each?  Ans.  6  days. 

First  travels  90  miles ;  with  a  daily  increase  of  2  miles. 
Second  travels  117  miles ;  with  a  daily  increase  of  3  miles. 


ARITHMETICAL    PROGRESSION. 


907 


Solution  of  first  question  liy  Problem  YII, 

First  day's  travel  of  l)otli,  10  +  12  =  22,  and  last  day's,  20  +  27  =  47   n  =  ^  ^  ^^^  =  6  days. 

'  22  +  47 

8.  A  man  earns  10  cents  the  first  day,  and  10  cents  additional  each  succeed- 
ing day,  for  30  days ;  his  expenses  are  $1  a  day;  when  will  he  be  most  in  debt; 
when  will  the  sum  of  his  earnings  and  expenses  be  equal ;  and  what  will  he  make 
clear  at  the  end  of  the  month  ? 

Ans.  Most  in  debt,  $4.50,  on  the  9th  and  10th. 

Sum  of  earnings  and  expenses  equal  on  the  19th,  $19. 
Clear  earnings  for  30  days,  $10.50. 

9.  Two  persons,  A.  and  B.  start  from  the  same  place  at  the  same  time ;  A. 
goes  due  north,  traveling  3f  miles  the  first  day,  with  a  regular  daily  increase 
afterward;  the  last  day's  journey  being  2Gf  miles.  B.  goes  due  west,  traveling  4| 
miles  the  first  day,  and  35^  miles  the  last  day.  lu  a  certain  time  they  are  600  miles 
apart,  on  a  straight  line;  how  many  days  have  they  traveled;  how  many  miles 
each ;  and  what  was  their  respective  daily  increase  ? 

Ans.  20  days. 

A.  traveled  300  miles  ;  daily  increase,  Ij  miles. 

B.  traveled  400  miles;  daily  increase,  If  miles. 


V(3§)'  +  (it)'  = 


SOLUTION 

Vl8«  +  24 «  _  ,  5  _ 


=  ^1  =  6  miles  apart  the  1st  day. 


Then  Tjy  proportion, 


6  :  500  :  :  3i  :  300  miles,  distance  A.  traveled. 
6  :  500  :  :  4i  :  400      "  "         B.         " 


Then  find  the  number  of  days  by  Problem  VII,  thus  : 


2X300     „^     2X400   ^ooday^ 


+  26j 


ii  4-  35^ 


eometrical  Progression. 


^ 


A  Geometrical  Progression  is  a  series  of  numbers  or  quantities, 
wlioso  formation  results  from  the  l;nv  of  a  constant  multiplier,  either  integral  or 
fractional,  called  the  Common  Ratio,  which  bears  the  same  relation  to  the  common 
difference  of  an  arithmetical  progression  that  multiplication  does  to  addition.  The 
series  of  a  geometrical  progression  is  ascending,  when  each  successive  term  increases 
by  a  constant  factor,  tliat  is  when  the  ratio  is  greater  than  one  ;  M'hen  the  ratio  is 
less  than  one,  or  a  proper  fraction,  the  terms  will  decrease  from  the  first,  and  the 
series  is  descending. 

Thus,  1,  3,  9,  27,  81,  etc.,  is  an  ascending  series,  whose  ratio  is  3, 
And     4,  2,  1,   i,    i,   etc.,  is  a  descending  series,  whose  ratio  is  J. 

A  descending  series  may  be  transformed  into  an  ascending  series  by  reversing 
the  order  of  its  terms,  substituting  the  last  for  the  first,  and  inverting  the  terms 
of  the  ft-actional  ratio. 

When  a  descending  series  is  continued  to  infinity,  that  is  when  the  number 
of  terms  is  infinite,  the  last  term  must  be  considered  etpial  to  zero  (0). 

In  geometrical  progression,  as  in  arithmetical,  tliere  are  fire  parts  or  quanti- 
ties, any  three  of  which  being  given,  the  others  can  be  determined;  they  are: 

1st.     The  First  Term,  rei^resented  by  A. 

2d.      The  Last  Term, 

3d.      The  Common  Eatio, 

4th.    The  Number  of  Terms, 

5th.     The  Sum  of  all  the  Terms, 

Since  any  one  of  these  five  pai'ts  may  be  required  from  any  three  of  the  four 
remaining  parts,  there  will  result  by  co>K&iMa<to«,, as  iu  arithmetipal  progression,  5 
X  4  =  20  cases  of  problems,  requiring  20  distinct  formulas  for  their  solution. 

Only  a  few  of  the  20  cases  in  geometrical  progression  admit  of  solution  by 
ordinary  arithmetical  operations,  because  the  common  ratio  (E)  enters  into  the 
series  by  continued  involution,  and  its  value  must  be  found  by  the  extraction  of  the 
higher  roots;  and  the  number  of  terms  (N)  is  represented  in  the  series  as  an 
exponent,  and  its  value,  iu  most  cases,  can  only  be  found  by  logarithmic  computa- 
tions. 

When  a  geometrical  progression  terminates,  or  is  limited  between  two  quan- 
tities, the  first  term  and  the  last  term  are  called  the  extremes,  and  the  intermediate 
terms,  the  geometrical  means. 

Problem  I. 

To  find  the  last,  or  any  proposed  term,  when  the  first  term,  common 
ratio,  and  number  of  terms  are  given. 

(90S) 


(( 

"     L. 

u 

"     R,  or  r  when  an  index. 

li 

"     N,  or  «  when  an  exponent. 

u 

"     S. 

*  GEOMETRICAL    PROGRESSION.  909 

1.  If  tlie  first  term  of  a  geometrical  progression  is  2,  the  ratio  3,  and  tlie 
number  of  terms  5,  what  is  the  hist  term  ? 

Illustration.     Tims,  3  (2  x  3)  (2  x  3=)  (2  x  3»)  (2  x  4^),  etc. 

Gives  2       C  IS  54  1C2         hist  term. 

We  here  see  that  the  second  term  is  equal  to  the  product  of  the  first  term,  by 
the  first  jioirer  of  the  ratio  ;  the  third  term  is  equal  to  the  product  of  the  first  term 
by  the  second  power  of  the  ratio,  and  in  general  any  term  is  equal  to  the  product  of 
the  first  term  by  that  power  of  the  ratio  whose  index  is  one  less  than  the  number 
of  the  required  term ;  hence  the 

Formula,  L  =  A  x  E'^' 

That  is,  the  last  term  is  equal  to  the  first  term  multiplied  by  the  ratio  raised 
to  a  power  one  less  than  the  number  of  terms. 

Problem  II. 

To  find  the  first  term,  when  the  last  term,  ratio  and  number  of  terms 
are  given. 

Since  the  last  term,  as  shown  in  Problem  I,  is  equal  to  the  product  of  the 
first  term  by  the  ratio  raised  to  a  known  power,  if  we  divide  the  last  term  by  that 
known  power  of  the  ratio,  the  quotient  will  necessarily  be  the  first  term;  hence  the 

Formula,  A  =  L  4-  R"-'     or    A  =  ~ 

That  is,  the  first  term  is  equal  to  the  last  term  divided  by  that  power  of  the 
ratio  whose  exponent  is  one  less  than  the  number  of  terms. 

2.  The  first  term  of  an  ascending  geometrical  series  is  7,  the  ratio  is  4,  and 
the  number  of  terms  is  6 ;  what  is  the  last  term. 

SOLUTION. 

L  =  7x4==7x  10-'  t  =  7168,  last  term. 

3.  The  first  term  of  a  descending  series  is  4,  the  ratio  ^  and  the  number  ot 
terras  5 ;  what  is  the  last  term  ? 

SOLUTION. 

L  =  4  X  (i)*  =  4  X  I'r  =  A,  last  term. 

4.  The  last  term  is  26244,  the  ratio  3,  and  the  number  of  terms  8 ;  what  is 
the  first  term  ? 

SOLUTION. 
A  =  26244  -f-  3^  =  26244  —  2187  =  12,  first  term. 

5.  The  last  term  is  -jfe,  the  ratio  J,  and  the  number  of '  terms  11 ;  what  is  the 
first  term  ? 

SOLUTION. 
A  -  5^5  ^  ay  =  sf^  -r  tbW  =  12,  first  term.   ■      • 


9IO  soule's  philosophic  practical  mathematics.  * 

6.     Bought  20  acres  of  land,  giving  5  cents  for  the  first  acre,  10  cents  for 
the  second,  20  cents  for  tlie  third,  and  so  on ;  what  did  tlie  last  acre  cost  ? 

Ans.  $26214.40. 

OPERATION    INDICATED. 

Z  =  5  X  2' »  =  $26214.40. 

Problem  III. 

To  Jind  the  sum  of  a  series,  when  tlie  extremes  and  ratio  are  given. 
If  we  take  .1  series,  as  2, 10,  SO,  250, 1250,  whose  ratio  is  5,  and  multiply  each 
terra  by  the  ratio,  we  will  obtain  a  new  series  5  times  greater  than  the  first,  as 
indicated  below. 

5,     10,     50,     250,     1250,     6250  =  5  times  given  series. 
2,    5,     10,     50,     250,     1250  =  1  time  given  series. 


—  2  6250 

2 


5  —  1  =  4  )  6248  =  4  times  the  sum  of  given  series.  ^ 
1562  =  sum  of  the  given  series. 
We  here  observe  that  all  the  terms  of  both  series  are  identical,  or  common  to 
each,  except  the  first  term  of  the  first  series  and  the  last  term  of  the  second  ;  hence, 
if  we  arrange  them  as  above,  so  that  the  terms  of  the  first  series  will  fall  under 
similar  terms  of  the  second,  and  subtract,we  see  that  the  like  terms  will  cancel  each 
other,  and  the  difl'erence  between  the  first  term  of  the  first  series  and  the  last  term 
of  the  second;  but  since  the  second  series  is  5  times  greater  than  the  first,  their 
difference  is  5  —  1,  or  4  times  greater  than  the  sum  of  the  first  series ;  then,  if  we 
divide  this  difference  by  4  (the  ratio  less  one)  the  quotient  will  be  the  sum  of  the 
given  series;  hence  we  deduce  the  following 

Formulas,  S  = _  ~     for  an  ascending  series. 

V  S  =     ~  _    — -  for  a  descending  series. 

That  is,  in  common  language  for  either  formula,  the  sum  of  the  series  is 
equal  to  the  difference  between  the  product  of  the  last  term  by  the  ratio,  and  the 
first  term,  divided  by  the  difference  between  the  ratio  and  (1). 

When  the  series  is  descending,  instead  of  using  the  second  formula,  we  may 
transform  it  into  an  ascending  series  by  reversing  the  order  of  its  terms,  and 
taking  the  reciprocal  of  the  ratio. 

1.  The  extremes  of  a  geometrical  progression  are  5  and  32805,  and  the  ratio 
is  3 ;  what  is  the  sum  of  the  series  ? 

SOLUTION. 

32805  X  3  =  98415  —  5  =  98410  —  2  =  49205,  sum  of  series. 

2.  The  extremes  of  a  descending  series  are  81  and  ^^-j,  and  the  ratio  is  Jj 
what  is  the  sum  of  the  series  ? 

SOLUTION. 

81  —  (jrb  X  i)  =  81  —  ris  =  80?  J|  ^  f  =  121Hi,  sum  of  series. 

Or, 
81  X  3  »=  243  —  rb  =  2i2iH  —  2  =  121ili,  sum  of  series. 


*  GEOMETRICAL    PROGRESSION.  9II 

3.  How  large  a  debt  can  be  discharged  in  one  year  by  paying  $5  the  first 
month,  and  twice  as  much  each  subsequent  month  as  the  preceding,  and  what  will 
be  the  last  payment?  Ans.  $10240,  last  month's  iiayment. 

$20475,  total  debt. 
First  find  the  last  term  by  Problem  I,  and  then  the  sum  of  the  series. 

4.  If  a  man  travel  5  miles  the  first  hour,  aiid  increases  his  speed  10  per  cent 
each  subsequeut  hour,  how  far  will  he  go  in  10  hours,  and  how  far  the  last  hour  ! 

Ans.  Last  hour,  11.789738455  miles. 

Whole  distance,  79.687123005  miles. 
Find  the  last  hour's  travel  by  Problem  I,  then  the  whole  distance  by  Problem 
III. 

5.  If  a  rifle  ball  moves  1000  feet  the  first  second,  and  diminishes  its  velocity 
10  per  cent  each  subsequent  second,  how  far  will  it  go  lu  10  seconds,  and  how  far  the 
last  second?  Ans.  387.420489  feet,  last  second. 

6513.215599  feet,  whole  distance. 

OPERATION     INDICATED. 

L  =  1000  X  (re)'  =  387.420489  =  last  term, 
g  ^  1000- (387.420489  X  A)  ^  6513.215599  =  sum  of  series. 

Problem  IV. 

To  find  the  sum  of  the  series,  when  the  first  term,  number  of  terms  and 
ratio  are  given,  without  finding  the  last  term. 

In  Problem  I  we  have  L  =  All"-',  and  in  Problem  III  we  have  S  =    ^  JJ     ; 

by  substituting  the  value  of  L,  in  Problem  I,  (AE«-'),  for  L  in  Problem  III,  we 

obtain,  S  =  — "'^  which  reduced  gives  the 

■c,  1        o         AR»  — A  a         A(R»  — 1) 

Formula,  S  =  — ;; — —  or  S  =  — ^-5 — ~ 

K  —  1  K  —  1 

That  is,  the  sum  of  the  series  is  equal  to  the  product  of  the  first  term  by  the 
ratio  raised  to  the  power  indicated  by  the  number  of  terms,  minus  the  first  term, 
and  the  result  divided  by  the  ratio  minus  1.  Or,  by  second  formula,  the  sum  is 
equal  to  the  known  (nth)  power  of  the  ratio,  less  one,  multiplied  by  the  first  term, 
and  the  product  divided  by  the  ratio  less  one. 

1.  The  first  term  is  10,  the  ratio  4,  and  the  number  of  terms  5;  what  is  the 
sum  of  the  series  ? 

SOLUTION. 

(^OXin-10    „r    12J<iiljzi2  =  3410,  sum  of  series. 
3  3 

2.  A  person  bought  10  acres  of  land,  agreeing  to  pay  5  cents  for  the  first 


91  2 

acre,  25  cents  for  tlie  second  ;icre,  iiiid  so  oii  in  a  quiutuple  ratio;    what  did  tlie 
whole  cost?  Aus.  Sl^liOTO.Sl. 

OPERATION   INDICATED. 

g  ^  5(o-"-l)  ^  5  (97r,5624)    ^  ^jo.OTO.Sl. 
4  4 

Note.— To  find  the  10th  power  of  5,  divide  the   10th   power  of  10,   or  1   wiih   10   ciphers 
annexed,  by  4  live  time.s. 

3.  If  a  man  travel  81  miles  the  first  day,  and  diminish  his  rate  of  travel  so 
that  each  succeeding  day's  journey  shall  be  |  of  the  previous  day,  how  far  will  he 
travel  in  7  days  ? 

SOLUTION. 

1-*  i  ' 

4.  If  a  man  earn  1  cent  the  first  day,  2  cents  the  second,  4  cents  the  third, 
and  so  on,  how  much  will  he  earn  in  30  days  ?  Ans.  $10737418.23. 

•             OPERATION     INDICATED. 
1     w  030  1 

S  =  tJLl i  =  $10737418.23. 

1 

To  find  the  30th  power  of  2,  we  may  cube  the  10th  power  of  2. 

Find  the  second  answers  to  examples  3  and  4,  in  Problem  III,  by  Problem 
IV. 

Problem  V. 

To  find  the  sum  of  a  decreasing  geometrical  series,  when  the  number  of 
terms  is  infinite,  the  first  term  and  ratio  being  given. 

When  a  descending  geometrical  progression  is  carried  to  infinity,  the  last 
term  must  be  considered  as  0 ;  hence  if  we  take  the  second  formula,  Problem  III, 
S  =  A  —  (L  X  R)^  ^jjjj  substitute  for  L,  its  value  0,  there  will  result  for  an  infinite 
series,  the 

Formula,  S  =  :; =7 

1  —  It 

That  is,  the  sum  of  an  infinite  series  is  equal  to  the  first  term,  divided  by 
one  minus  the  ratio. 

1.     What  is  the  sum  of  the  infinite  series,  9,  3,  1,  J,  ^,  etc  ? 

6OLUT10N. 

9  9 

S  =  =:  _  =  13J,  sum  of  the  series. 

1  —  i        J 


*  GEOMETRICAL    PROGRESSION.  913 

2.  What  is  the  sum  of  au  iufltiite  series,  whose  first  term  is  one  inilliou,  and 
ratio -jio?  Ans.  llllllli. 

If  the  ratio  was  j^,- 1  Ans.  10000000. 

If  tlic  ratio  was  J  ?  Ans.  11428574. 

If  the  ratio  was  j^o  ?  Ans.  lOlOlOlsjig. 

Tills  process  may  be  used  for  determining  the  vahie  of  circulating  decimals, 

because  their  repetends  are  in  geometrical  progression,  the  common  ratios  of  which 

^re  iJ-o,  -j^o,  T-oVo;  6tc.,  according  to  the  number  of  figures  in  the  repetend ;  thus,  the 

3  3 

Circulate  .333,  etc.,  gives  the  series  i^  +  jqi  +  jth,  etc. ;  and  .232323,  etc.,  gives  -^o-o 

23  '^S  •  -^ 

+  -— rj  +  -^,  etc.,  to  infinity ;  hence  the  value  of  the  circulate  .3  is  equal  to    ■'    , 

XUU  J.UU  X  —  "Jq" 

=  -j-g  -f-  -i%  =  J ;  and  the  value  of  .23  is  equal  to  -j-q^  -4-  1^0%  =  If  j  3,nd  as  the 
operation  is  uniform  for  the  summation  of  all  circulates  to  infinity,  we  have  the 
principle  demonstrated  that  the  value  of  any  circulating  decimal  is  equal  to  a  frac- 
tion with  the  given  repetend  for  a  numerator,  and  as  many  9's  for  a  denominator,  as 
there  are  places  in  the  repetend. 

3.  What  is  the  value  of  the  circidates  .'oi,  .003,  .6'98412,  .428571  f 

^Vll.>.      ,j,    333,    63,    7. 

4.  If  a  cannon  ball  move  2000  feet  the  first  second  of  time,  and  each 
succeeding  second  ^0  as  far  as  the  preceding,  until  it  comes  to  a  state  of  rest,  how 
far  will  it  move  ?  Ans.  20000  feet. 

OPERATION    INDICATED. 

S  =  2000  ^,,0000. 
10- 

5.  The  hour  and  minute  hands  of  a  clock  are  together  at  12  o'clock  or  noon, 
when  will  they  be  together  again  ?  Ans.  1  h.  5  m.  27 1\  sec.  P.  M. 

This  question  gives  the  infinite  series,  1  +  -jV  +  ytij  etc. 

Problem  YI. 

To  find  tlie  ratio,  when  the  extremes  and  number  of  terms  are  given. 
In  Problem  I  it  is  demonstrated  that  the  last  term  of  a  geometrical  progres- 
sion is  equal  to  the  first  term,  multiplied  by  the  ratio  raised  to  a  power  one  less 
than  the  number  of  terms ;  therefore  if  we  divide  the  last  term  by  the  first  term, 
the  quotient  will  be  equal  to  the  ratio  raised  to  the  n — 1th  power,  and  extracting 
the  11 — 1th  root  will  give  the  ratio ;  hence  the 

Formula,  E  =  ""^/-^ 

'  V    A 

1.  The  first  term  is  6,  the  last  term  750,  and  the  number  of  terms  4  ;  what  is 
the  ratio  I 

SOLUTION. 
R=     V^<i —Vvio —  ?>,TAi\o. 

2.  The  extremes  are  4  and  2916,  and  the  number  of  terms  7,  what  is  the 
ratio?  Ans.  3. 


914  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

To  fiud  the  Gtli  root  we  may  take  the  cube  root  of  the  square  root. 

3.     The  first  term  of  a  decreasing  geometrical  series  is  10,  the  last  term  /gj 
and  the  number  of  terms  10;  what  is  the  ratio?  Ans.  J. 

To  find  the  9th  root,  extract  the  cube  root  of  the  cube  root. 

NoTK. — Wlifn  llie,  iiuniber  nf  tornis  is  ]ai'{;e,  tlm  extractiiin  <if  llic  ]iiglicr  roots  should  be 
made  by  logarithms. 

The  above  Problem  is  used  to  insert  any  number  of  geometrical  means,  or 
mean  proportionals,  between  two  given  numbers. 

EXAMPLES. 

1.  Insert  three  geometrical  means  between  C  and  3750. 

Explanation. — Since  there  are  three  terms  to  be  inserted  between  two  extremes,  there  will  be 
fire  terms  in  the  series,  and  we  therefore  have  the  extremes  and  number  of  terms  given  to  liud  the 
ratio  by  Problem  VI;  that  is,  wo  divide  the  last  term,  3750,  by  the  iirst  term,  ti,  giving  625;  then, 
sinoe  the  number  of  terms  is  5,  we  extract  the  5—  1  or  4th  root  of  6J5,  which  gives  5  for  the  ratio: 
thf  n  to  complete  the  series,  wo  multiply  the  first  term  by  the  ratio  for  the  second,  and  the  second 
b'.  ihe  ratio  for  the  third,  and  so  on,  which  gives  for  this  example  the  series, 

6,    30,    150,    750,    3750. 

To  find  a  ^ngle  mean  jn-oportional  between  two   given    numbers,   we   may 
t»«tract  the  square  root  of  the  product  of  the  extremes  for  the  mean  term. 

2.  Find  and  insert  a  mean  proportional  between  4  and  324. 

Ans.  4,  36,  324,  series. 

3.  Fiud  and  insert  two  mean  proportionals  between  5  and  320. 

Ans.  5,  20,  80,  320,  series. 

4.  Find  and  insert  five  geometrical  means  between  4  and  2910. 

Ans.  4,  12,  30,  108,  324,  972,  2916,  series. 

5.  Find  eight  geometrical  means  between  3  and  1536. 

Ans.  0,  12,  24,  48,  96,  192,  384,  768,  means. 

To  find  the  9th  root  of  512,  extract  the  cube  root  of  the  cube  root. 

6.  Insert  seven  geometrical  means  between  9  and  f |f,  descending  series. 

Ans     Q    0    4    ^Z    1i    1  ^     SA    XS.S     25  e    oprioK 

.rt.ii.s.  J,  u,  t,  _)j,  ly,  i-jY,  B~f,  -M-js,  -fj9,  series. 
To  find  the  8th  root  of  eWr;  extract  the  square  root  three  times  consecutively. 

Problem  VII. 

To  find  the  ratio,  when  the  extremes  and  sum  of  the  series  are  given. 

Let  us  take  a  series,  as  4,  8, 16,  32,  64,  whose  ratio  is  2,  and  write  it  twice, 
omitting  the  first  term  the  first  time,  and  the  last  term  the  second  time,  thus : 

8,    16,    32,    64  =  120  =  sum  of  series,  less  the  first  term. 
4,      8,    16,    32  =    60  =  sum  of  series,  less  the  last  term. 

We  see  by  coupariug  the  two  series,  that  the  terms  and  sum  of  the  first  series 


*  GEOMETRICAL    PROGRESSION.  9l5 

axe  2  times  as  large  as  the  corresponding  terms  and  sum  of  the  second  series ;  hence 
2  must  be  the  common  ratio.    Or  we  may  demonstrate  it  thus:  Take  the  first 

formula,  Problem  III,  S  =  ^^Zi'  '"''^  ^^*^'i^'  ^*  ^f  fractions,  giving  ES  —  S  =  EL 
—  A ;  then  by  transposition  we  obtain,  E(S  —  L)  =  S  —  A,  and  to  find  E,  we  have 
by  division  the 

Formula,  E  =  S  — A 

S—  L 

That  is,  "the  ratio  is  equal  to  the  sum  of  the  series  minus  the  first  term,  divided 
by  the  sum  minus  the  last  term. 

1.  The  extremes  of  an  ascendiug  series  are  4  and  2500,  and  the  sum  of  the 
series  is  3124;  what  is  the  ratio?  Ans.  5,  ratio. 

SOLUTION. 

R  =     ^^^^  ~  ^     =  W4»  =  5,  ratio. 
3124  —  2500 

2.  The  extremes  of  a  descending  series  are  9  and  -gVj  and  the  sum  of  the 
series  is  13^^ ;  what  is  the  ratio  ?  Ans.  J,  ratio. 

'  SOLUTION. 

E=13iizzi=lii  =  il-J=i,  ratio. 
13i?-T^        13|f 

3.  The  first  term  of  an  infinite  series  is  16,  and  the  sum  of  the  series  is  64; 
what  is  the  ratio  ?  Ans.  ^,  ratio. 

SOLUTION. 

R  =-■  ^^f^  =  tS  =  i,  ratio. 

D-l  — •  0 

4.  The  distance  between  two  places  by  rail  is  260.SCS155  miles ;  if  a  train 
moves  oO  miles  the  first  hour,  and  20.57  a,V  miles  the  last  hour,  what  will  be  the  ratio 
of  the  decreasing  progression  of  the  rate  of  travel  ?  Aus.  -fg,  ratio. 

OPERATION  INDICATED. 

E  =  260.868155  —  50.      _  210.868155  ^  g  _  ^.^^..^ 
260.868155  —  26.570205    234.297950   ""^       • 

Problem  VIII. 

To  find  the  number  of  terms,  when  the  extremes  and  ratio  are  given. 
By  Problem  I  we  see  that  L  =  A  x  E""'  ;    dividing  both  numbers  by  the 
first  term  A,  we  obtain  the 

Formula,  t  =  K"-' 

'  A 

That  is,  the  quotient  of  the  last  term  by  the  first,  is  equal  to  a  power  of  the 
ratio  one  less  than  the  number  of  terms,  and  by  adding  one  to  that  power  we  have 
the  number  of  terms. 


o 
terms? 


916  SOULe's    nilLOSOPHIC    PKACTICAL    MATHEMATICS.  * 

Or  we  may  iiidiciito  it  thus : 

Foi inula,  li"  =  V  X  R 

A 

Fioni  tins  we  see,  that  the  number  of  terms  (n)  is  that  power  of  the  ratio 
which  is  equal  to  the  quotient  of  the  last  term  by  the  first,  multiplied  by  the  ratio. 

1.  The  extremes  are  3  and  375,  and  the  ratio  is  5;  what  is  the  number  of 
terms  ? 

SOLUTION. 

^^  =  125  =  5' ;  then,  3  +  1  =  4,  number  of  terms. 
Or,       ^5^  X  5  =  625  =  5*,  gives  4,  uumber  of  terms. 

The  first  term  is  125,  the  last  ji^,  and  the  ratio  3- ;  what  is  the  number  of 

SOLUTION. 

T^  -:-  125  =  TTri?7  =  ii)^  )  then,  6  -f-  1  =  7,  number  of  terras. 

3.  If  in  traveling  262f  miles,  a  locomotive  runs  32  miles  the  first  hour,  and 
78^  miles  the  last  hour,  what  was  the  ratio  of  the  increasing  rate  of  speed,  and  how 
many  hours  Mere  occupied  ?  Ans.  1^,  ratio.     5  hours. 

First  find  the  ratio  by  Problem  "VII,  and  then  the  luimber  of  terms. 

4.  If  I  pay  $5  the  first  month,  and  $98415  the  last  month,  in  geometrical 
progression,  the  common  ratio  being  3,  what  amount  will  bo  disbursed,  and  how 
many  months  will  it  require  ?  Ans.  Amount,  $147620.     Time,  10  months. 

OPERATION   INDICATED. 

g  _  (98415  X  3)  -  5  ^  ^i^-g,,Q  _  ^^^^^^^^ 

3n  =  SUlo  j^  3  ^  59049  ^  ^^jj  po^-ej.  „f  j-atio  3. 
5 

Find  the  amount  by  Problem  III,  and  the  time  by  Pi'oblem  VIII. 

Problem  IX. 

To  find  the  last  term,  when  the  first  term,  common  ratio,  and  sum  of  the 
series  are  given. 

If  we  take  the  formula  of  Problem  YII,  R  =     ~    ,  and  clear  it  of  fractions, 

we  obtain  ES  —  EL  =  S  —  A;    then  by  transposing  we  have,   EL  =  ES  —  S  + 
A,  and  reducing  gives  the 

Formula,  L  =  rr — ■ — 

That  is,  the  last  term  is  equal  to  the  jirodnct  of  the  sum  by  the  ratio  less 
one,  i)lus  the  first  term,  and  the  result  divided  l)y  tiie  ratio. 

1.    The  first  term  is  5,  the  ratio  4,  and  the  sum  6825 ;  what  is  the  last  term  t 

SOLUTION. 

6825  X  3  =  20475  +  5  =  20480  —  4  =  5120,  last  term. 


*  GEOMETRICAL    PROGRESSION.  91 7 

2.  If  a  mercliant  owes  $98410,  and  pays  $10  the  first  month,  and  triple  the 
amount  each  succeeding  month,  how  long  will  it  take  to  pay  the  debt,  and  what  will 
be  the  last  payment?  Ans.  $65610,  last  payment.     Time,  9  months. 

Find  the  last  jjayment  by  Problem  IX,  then  the  number  of  months  by  Prob- 
lem VIII. 


APPLICATIONS    OF    GEOMETEICAL    PROGEESSIOK    TO     COMPOUND 
INTEREST  AND  COMPOUND  DISCOUNT. 

Any  principal  at  compound  interest,  for  a  number  of  years,  will  produce 
for  the  successive  years,  or  other  intervals,  a  series  in  geometrical  progression. 

Thus  if  we  put  $1  at  compound  interest  at  6  per  cent  for  4  years,  we  will  have 
$1  for  the  first  term,  and  the  amount  of  $1  for  one  period  ($1.06)  for  the  ratio,  and 
4  years  for  the  number  of  terms,  which  gives  the  series  $1,  (1.06),  (1.06-),  (1.06^)> 
(IM*). 

The  last  term,  1.06*,  gives  the  amount  of  $1  for  4  years  at  6  per  cent. 

We  here  see  the  number  of  terms,  5,  is  one  greater  than  the  given  number  of 
years,  because  the  first  term,  $1,  existed  previous  to  any  interval  of  time,  and 
hence  if  we  raise  the  ratio  (the  amount  of  $1  for  a  specified  interval)  to  a  power 
indicated  by  the  number  of  years  or  periods,  we  will  have  the  compound  amount  of 
$1  for  that  time ;  aiid  since  all  amounts,  for  the  same  time  and  rate,  are  proportional 
to  their  principals,  if  we  multiply  the  amount  of  $1  by  any  principal,  we  will  obtain 
the  compound  amount  for  that  principal,  and  by  deducting  the  principal  from  the 
amount  we  have  the  compound  interest.  From  the  above,  and  principles  previoiisly 
explained  in  Problems  Yl  and  VIII,  Geometrical  Progression,  we  may  deduce  the 
following  conclusions,  viz. : 

That  in  compound  interest,  the  amount  and  interest  vary  directly,  or  in  the 
same  ratio  as  the  princijial,  Avhen  the  time  and  rate  are  the  same; 

Whereas  the  rate  varies  as  the  nth  root  of  the  quotient  of  the  amount  by  the 
principal ; 

And  the  time  varies  as  the  nth  i)ower  of  the  ratio ; 

That  is,  increasing  the  rate  or  time,  increases  the  interest  in  a  still  greater 
ratio. 

1.     What  is  the  amount  of  $i'500  for  3  years  at  8  per  cent  compound  interest? 

Ans.  $3149.28. 

SOLUTION. 

$2500  X  (1.08)'  =  $2500  x  1.259712  =  $3149.28 

Here  we  have  the  principal,  $2500,  for  the  first  term,  the  amount  of  $1  for 
one  year  at  8  i)er  cent  (■'SI. OS)  for  the  ratio,  and  3,  the  given  number  of  years  (one 
less  than  the  number  of  terms),  for  the  power  of  the  ratio;  which  corresponds  to 
Problem  I,  Geometrical  Progression. 


91 8  soule's  philosophic  practical,  mathematics.  * 

2.  What  is  the  amount  of  $10000  for  10  years  at  10  per  cent  compound 
interest!  Ans.  $25937.425. 

3.  What  is  the  compound  amount  and  interest  on  $5000  for  4  years  at  6  per 
cent,  compounded  semi-annually  ?  Ans.  Amount,  $6333.8504. 

Interest,  $1333.8504. 

SOLUTION. 
$5000  X  (1.03)« 

When  the  intervals  are  less  than  a  year,  we  take  such  a  part  of  the  rate  as 
the  given  interval  is  of  12  months,  -which  gives  (1.03)  for  the  ratio  in  the  last 
example,  to  be  raised  to  the  8th  power,  because  there  are  8  semi-annual  interyals  in 
4  years. 

4.  What  is  the  compound  interest  on  $50000,  at  8  per  cent  compounded 
quarterly,  for  4  years?  Ans.  Interest,  $18639.285. 

SOLUTION. 

From  Table,     (1.02)' «  =  1.3727857 

PRESENT  WORTH,  AND  COMPOUND  DISCOUNT. 

To  find  the  present  worth  at  compound  interest,  of  any  sum  due  at  a, 
future  time.  Since  diflerent  amounts  are  proportional  to  their  principals,  it  is  plain 
that  if  we  divide  any  given  amount  by  the  amount  of  $1,  for  the  given  rate  and 
time,  the  quotient  will  be  the  required  principal  or  present  worth  ;  and  the  difference 
between  the  given  amount  and  present  value,  will  be  the  compound  discount. 

The  above  principle  of  compound  discount  furnishes  the  only  correct  method 
of  finding  the  present  value  of  government  or  corporation  bonds,  due  at  a  future 
time,  or  of  estimating  the  value  of  investments  for  long  periods  of  time,  without 
gain  or  loss,  except  the  interest. 

1.  What  principal  will  amount  to  $11576.25  in  3  years,  at  5  per  cent  com- 
pound interest  ?  Ans.  $10000, 

SOLUTION. 

By  Problem  II,  Geometrical  Progression,     $11576.25  -f-  1.05'  =  $10000. 

Or  by  Table,  page  611.  Divide  the  given  amount  by  the  tabular  .amount  of  $1,  for  the  same 
time  and  rate,  because  the  amounts  in  the  table  are  the  different  powers  of  the  rates  at  top,  for 
the  number  of  years  opposite  the  amounts. 

2.  What  principal  will  produce  $724.50  compound  interest  at  7  per  cent  in  2 
years?  Ans.  $5000. 

SOLUTION. 

$724.50  —  (1.07'^)  —  1  =  $5000. 

Or  by  T.able.  Divide  the  given  compound  interest  by  the  compound  interest  of  $1,  for  the 
given  time  and  rate. 


*  GEOMETRICAL    PROGRESSION.  919 

3.  Ill  how  many  years  will  $12000  amouut  to  $17509.20  at  10  per  cent  per 
annum,  compound  interest  ?  Ans.  4  years. 

SOLUTION. 

By  Problem  VIII,  Geometrical  Progression,  17569.20  —  12000  =  1.4641  =  (1.10)* 
Or  liy  T<al>le.     In  the  colnnin  of  the  given  rate  per  cent,  find  the  quotient  of  the  amount  by 
the  principal,  and  then  find  the  year  opposite. 

4.  At  wliat  rate  in  3  years  will  $50000  amount  to  $62985.60? 

Ans.  S  per  cent. 

SOLUTION. 

By  Problem  VI,  $62985.60  H-  50000  =  \/l.259712  =  1.08  =  8  ^^ 

Or  by  Table.     Opposite  the  given  year,  find  the  rxuotient  of  the  amount  by  the  principal,  the 
column  at  top  will  shov  the  rate  per  cent 

5.  What  woukl  be  the  present  value  of  $25907.10  in  State  bonds  due  in  10 
years,  and  not  bearing  interest,  allowing  8  per  cent  compound  interest  on  money 
invested?  Ans.  $12000. 

Solution  by  Problem  II,  Geometrical  Progression. 

Or  by  Table.     Divide  amount  of  bonds,  $25907.10  by  the  tabular  .amount  of  $1,  at  8  per  cent 
for  10  years,  viz.,  by  2.158925. 

6.  What  would  be  the  present  value  of  $134550  in  government  bonds,  due  in 
20  years,  allowing  10  per  cent  compound  interest  on  investment  ?       Ans.  $20000. 

SOLUTION. 

Divide  $134550  by  tabular  amount;  $6.7275. 


nnuitie^s, 


An  Annuity  is  properly  a  sum  of  money  payable  annually;  but  tbe 
term  is  also  applied  to  payments  made  at  tlie  end  of  regular  intervals  of  time,  as 
monthly,  quarterly,  semi-annually,  etc.,  and  to  continue  for  a  given  period,  for  life, 
or  forever. 

Annuities  are  of  three  kinds,  viz. :   Certain,  Contingent,  and  Perpetual. 

A  Certain  Annuity  is  one  that  commences  at  a  specified  time  and 
continues  for  a  definite  period. 

A  Contingent  Annuity  is  one  whose  commencement  or  termination 
depends  on  some  contingent  event,  as  the  life  of  a  person  or  persons,  heiu^o  its 
continuance  is  also  uncertain;  when  depending  on  the  life  of  an  individual,  it  is 
called  a  Life  Annuity. 

A  Perpetual  Annuity,  or  Perpetuity,  is  one  which  is  to  continue 
forever. 

Each  of  these  three  kinds  of  Annuities  is  subject  to  the  three  following 
conditions,  viz. :  It  may  be, 

1st.  An  Annuity  in  Possession,  or  an  Immediate  Annuity,  when  it  begins  at 
once,  or  has  already  commenced. 

2d.     A  Deferred  Annuity,  when  it  is  to  commence  at  some  certain  future  time. 

3d.  An  Annuity  in  Reversion,  when  it  is  to  commence  at  the  death  of  one 
person  and  revert  to  another,  or  when  its  commencement  depends  on  the  occurrence 
of  some  other  certain  event. 

An  Annuity  Foreborne,  or  in  Arrears,  is  one  whose  periodical  pay- 
ments have  not  been  made  when  due,  but  allowed  to  accumulate  with  interest  on 
each  payment  in  arrears,  until  final  maturity. 

The  Amount  or  Final  Value  of  an  annuity,  or  the  value  of  one  in 
arrears,  is  the  sum  of  all  the  payments  due,  -with  interest  on  each,  from  the  time 
they  became  due  to  the  end  of  the  given  period,  or  until  the  termination  of  the 
annuity. 

The  Present  Worth  or  Yalue  of  an  annuity  is  eiiual  to  the  sum  of 
the  present  values  of  all  its  payments,  on  such  a  sum  as  would  amount  to  its  Jinal 
value,  at  the  given  rate  and  time.  The  present  worth  of  a  Perpetual  Annuity  is  a 
sum,  whose  periodical  interest  would  be  equal  to  the  annuity. 

When  simi)le  interest  is  allowed  on  annuties,  the  computations  may  be  direct 
on  the  several  payments,  or  by  the  principles  of  Arithmetical  Progression. 

When  compound  interest  is  allowed,  we  may  employ  the  principles  of  Geo- 
metrical Progression,  or  use  the  Tables  calculated  for  that  purpose. 

(920) 


*  ANNUITIES.  .  921 

Remarks  on  Annuities.  It  is  not  always  easy  to  determine  when  an 
annuity  begins.  It  should  be  remembered  that  it  begins,  not  at  tlie  time  the  first 
payment  is  made,  but  one  interval  of  time  before,  flence  in  problems  where  the 
annuity  is  paid  in  advance,  the  tabular  number  to  be  used  should  be  for  one  year 
more  than  the  number  of  annuities. 

Note. — The  interval  of  time  is  the  period  between  tlie  payments,  and  may  be  a  year,  half- 
year,  (juarter  or  month. 

Problem  I. 
Annuities  at  Simple  Interest. 

To  find  the  amount  of  an  annuity  foreborne  or  in  arrears,  at  simple  interest. 

1.     What  would  be  the  amount  of  an  annuity  of  $1000,  in  arrears  for  4  years, . 
payable  annually,  at  8  per  cent,  simple  interest?  Ans.  $4480. 

SOLUTION. 

Sum  of  four  annual  payments,  .$1000  X  4  =  |4000 

Interest  on  1st  payment  of  |l000   for  3  years  ." 
"   2d        ■"  "     1000  - 


3d  "  "     1000 


■  3  years  ) 
1       "      S 


Equals  interest  on  $1000   for  6  years  at  8  %  480 


Amount,  |4480 

Explanalion. — Since  the  first  payment  does  not  become  due  until  the  end  of  the  first  year,  it 
■nill  draw  interest  for  only  3  years  ;  the  second  payment  for  2  years;  the  third  payment  for  1  year; 
and  the  last  payment  not  bearing  interest;  the  sum  of  the  interests  on  the  several  payments,  for 
their  respective  times,  is  equal  to  the  interest  on  one  payment  for  the  sum  of  the  several  periods 
produciuj;  interest,  or  for  3  +  2  +  1  =  6  years,  which  at  8  per  cent  gives  S480  interest  to  be  added 
to  the  sum  of  the  4  annual  payments  of  $1000,  giving  for  the  amount  $4480. 

SOLUTION   BY  ARITHMETICAL  PROGRESSION. 

Problem  No.  IV,  Arithmetical  Progression,  for  finding  the  sum  of  a,  series,  may  be  used:  The 
first  payment,  $1000,  being  the  first  term;  the  amount  of  the  first  payment  for  the  number  of  years 
■or  periods  less  one,  the  last  term  ;  and  4  years  the  number  of  terms  ;  hence  we  have 

1000  +  1240  X  4  ^  1120  X  4  =  $4480,  the  .amount  or  sum. 

2.  A  person  deposited  in  a  savings  bank,  for  the  benefit  of  his  son,  $100  on 
each  birthday;  Avhat  is  due  him  on  his  21st  birthday,  and  how  much  when  he  is  21 
years  of  age,  interest  at  G  per  cent  ? 

SOLUTION. 

(100  +  220)  X  V-  =  $3360,  amount  due  on  21st  birthday. 
And        (100  +  226)  X  ^^  =  $3586  —  $100  =  $3486,  amount  due  when  21  years  of  age. 
Or,     $3360  +  interest  on  $2100  for  1  year  $126  =  $3486,  as  above. 


92  2  SOULE  S    raiLOSOPHlC    PRACTICAL    MATHEMATICS.  * 

ritOBLEM    II. 

Annuities  at  Compound  Interest. 

The  amounts  of  the  several  payments  of  an  annuity  at  compound  interest, 
form  a  desceiidiii<j  geometrical  series,  and  by  reversing,  or  considering  the  series  as 
ascending,  we  have  the  annuity  or  periodical  p.iyment,  for  the  first  term;  the 
amount  of  $1  for  one  year,  or  $1  plus  the  rate  of  interest  expressed  decimally,  for 
the  ratio;  the  time  in  years  or  other  intervals,  for  the  number  of  terms;  and  the 
required  amount  for  the  sum  of  the  series.  Hence,  to  find  the  amount  or  final  value 
of  an  annuity  in  arrears,  or  for  any  given  time,  at  compound  interest,  we  may  use 
the  formula  of  Problem  IV,  Geometrical  Progression,  as  illustrated  in  the  following: 

1.  What  will  a  certain  annuity  of  $800,  in  arrears  for  4  years,  amount  to  at 
5  per  cent  compound  interest  ?  Ans.  $3448.10. 

SOLUTION. 
1.05*  — 1 


X  $800  =  4.310125  x  $800  =  $3448.10        General  formula. 
1.05  —  1 

t;  _  A  X  (R»  -  1) 

K—  1 

Or,  by  a  modification  of  the  above  formula,  thus; 

(800  X  1.05M  —  800       4,Moin  1 

; . :=  $3448.10,  as  above. 

.05 

When  the  number  of  years  is  large,  instead  of  involving  the  ratio  to  that  power,  take  the 
amount  of  $1  for  the  given  number  of  years  and  rate,  from  the  ComiKiund  Interest  Tables,  pages- 
611-12,  which  corresponds  to  tlie  required  power  of  the  ratio.  When  Annuity  Tables  are  accessible, 
we  simply  multiply  the  amount  of  $1,  for  the  given  time  and  rate,  by  the  given  annuity;  thus, 

From  Table  No.  1,  page  925,  $4.310125  x  800  =  $3448.10 

2.  What  is  the  final  value  of  a  life  annuity  certain,  of  $500,  in  20  years, 
payable  annually  at  6  per  cent  compound  interest?  Ans.  .$18392.7055. 

3.  A  person,  at  the  age  of  23,  deposits  in  a  savings  bank  $250  each  year, 
until  he  is  60  years  old;  what  will  it  amount  to  at  7  per  cent  comiwund  interest? 

Ans.  $40084.35. 

Problem  III. 

To  find  the  Present  ^Vortll,  or  Value  of  a  Certain  Annuity. 

The  present  worth  or  value  of  a  certain  annuity,  is  that  principal  which,  at 
the  same  rate  of  interest  till  the  termination  of  the  annuity,  would  be  increased  to 
the  amount  of  the  annuity  foreborne  for  the  same  time. 

Therefore,  if  we  find  the  final  value,  or  amount  of  an  annuity  by  the  last 
problem,  and  divide  it  by  the  amount  of  $1  for  tlie  same  time  and  rate,  (as  found 
from  the  Compound  Interest  Tables,)  the  quotient  will  be  the  present  worth  of  the 
annuity.  Or,  having  Annuity  Tables,  we  simply  multiply  the  present  worth  of  $1 
for  the  given  time  and  rate,  as  found  in  Table  No.  2,  by  the  given  annuity. 

1.  What  is  the  present  value  of  an  annuity  of  $400,  payable  annually  for  10 
years,  at  6  per  cent  compound  interest  T  -  Ans.  $2944.0348. 


ANNUITIES.  923 


We  first  find  tlie  amount  of  tlie  annuity  by  Problem  I,  to  be  $5272.318;  tben  this  amount 
divided  by  tbo  amount  of  Jil  for  the  same  rate  and  time,  as  per  Compound  Interest  Table,  or  by 
1.06'°,  gives  the  present  Tvorth  ;  thus, 

1.790848  )  5272.318  (  §2944.0348,  present  worth. 
Or  by  Annuity  Table  No.  2,     $7.3G0087  X  400  =  $2944.0348 

2.  Wliat  is  tlie  pre.seiit  value  of  a  peusioii  of  $100,  payable  annually  for  25 
years,  at  4  iier  cent  coniiiound  interest!  Ans.  $1562.208. 

3.  W^hat  is  the  present  value  of  an  annuity  of  $500,  to  commence  10  j'ears 
heuce,  and  continue  20  years  thereafter,  at  C  per  cent  compound  Interest? 

Ans.  $3202.372, 

SOLUTION. 

Amount  of  $500  .annuity  for  20  years  =  $18392.7955. 
Amount  of  |1  for  30  years  =  $5.743491. 

Then,  $18392.7955  —  5.743491  =  $3202.372 

Or  by  Problem  V  for  annuities  in  reversion. 

Problem  IV. 

To  find  the  Present  Value  of  a  Perpetual  Annuity. 

The  present  worth  of  a  perpetuity,  or  perpetual   annuity,   is   that   sum   of 
money,  whose  interest  is  ecxual  to  the  given  annuity. 

1.     What  is  the  present  worth  of  a  perpetuity  of  $400  per  annum,  allowing 
compound  interest  at  8  per  cent?  Ans.  $5000. 

OPERATION. 


8     400 


Explanation. — At  8  per  cent,   it  Tvill  require 

100  §100  capital  to  produce  $8  interest,  and   hence 

$1  interest  will  require  ^  as  much  capital,  and 

40000  $400  interest  ■will  require  400  times  as  much,  or 

$5000   capita],    which   is   equal   to   the    present 

w^OOO  value  of  a  perpetual  annuity  of  $400. 

2.     What  is  the  present  worth  of  a  perpetual  annuity  of  $1500  a  year,  at  6 
per  cent  compound  interest  ?  Ans.  $25000. 

Problem  V. 

To  find  the  Present  Vi'orth  of  an  Annuity  Deferred,  or  in  Reversion. 

1.     What  is  the  j^resent  worth  of  an  annuity  of  $500,   to   commence  in   4 
years,  and  to  continue  6  years,  allowing  6  per  cent  compound  interest  ? 

Ans.  $1947.4906. 


924  souLE  s  ruiLOSornic  practical  mathematics. 

SOLUTION. 
$7.360087       fM)iii  Table  2  presiMit  worth  of  $1  for  10  years.  ■ 
3.4t)5106  "  "      "         "  "      "    "    "     4       " 


$3.894981        valiK^  of  $1  for  tlio  6  interveuiug  years. 
500   giveu  auuuity. 

$1947.490500   present  ■worth  of  the  annuity. 

Explntiation. — We  first  find,  from  the  Tahles,  the  present  worth  of  an  annuity  of  $1  for  the 
xrhole  i)eri<)(l.  or  until  the  annuity  teiniinates,  viz.,  for  4  -|-  6  ^  10  years,  we  tlien  find  from  t\\<i 
Table  the  jiresent  value  of  $1  to  the  eoninieneenient  of  the  reversion,  viz.,  lor  4  years;  wo  thea 
find  the  difference  between  these  present  values  of  an  annuity  of  |il,  and  multiply  it  by  the  num- 
ber of  dollars  in  the  given  annuity,  for  the  present  worth  required. 

2.    What  is  the  present  worth  of  a  reversionary  annuity  of  $1000,  deferred 
for  10  years,  and  to  continue  15  years  thereafter,  at  5  jier  cent  componntl  interest? 

Ans.  $0372.21. 

Problem  VI. 

To  find  the  Annuity,  when  the  Frcsent  Worth,  Time,  and  Rate  per  cent 

are  given. 

1.     What  annuity  continued  for  11  years,  at  6  per  cent  compound  interest, 
would  give  a  present  worth  of  $0309.50  ?  Ans.  $800. 

SOLUTION. 

$6309.50  —  7.886875  =  $800 

Since  the  present  worth  of  $1  multiplied  by  the  annuity  would  give  the  full  present  worth, 
it  follows,  that  if  we  divide  the  given  present  worth  by  the  present  worth  of  $1  (as  found  in  Table 
No.  2),  the  (juotient  will  be  the  annuity. 

Problem  VII. 

To  find  the  Annuity,  when  the  Amonnt,  Time,  and  Rate  per  cent  are  given. 

1.  What  annuity  will  amount  to  $5172.15,  in  4  years,  at  5  per  cent  compound 
interest?  Ans.  $1200. 

SOLUTION. 
$5172.15  —  4.310125  =  $1200 

AVe  here  divide  the  given  amonnt  by  the  amount  of  $1  (from  Table  No.  1),  for  the  same  time 
and  rate ;  the  reason  is  the  same  as  in  the  preceding  problem. 

The  subject  of  annuities  is  of  a  very  comprehensive  nature,  and  of  great 
practical  importance,  but  a  full  discussion  of  all  its  principles  and  applications 
would  be  too  extended  for  this  treatise.  Tlie  number  of  problems  here  given  is 
greater  than  is  usually  presented  in  arithmetics,  and  with  the  aid  of  the  accom- 
jianying  Tables,  which  give  the  present  and  final  values  of  $1  at  the  ordinary  rates 
of  interest,  from  1  to  50  years,  its  applications  to  reversions,  leases,  rents,  etc.,  may 
be  readily  comprehended  and  applied.  Its  employment  in  problems  of  life  insur- 
ance, life  estates,  dowers,  pensions,  etc.,  would  require  in  addition  the  use  of 
Mortality  Tables,  and  the  special  regulations  determined  by  actuaries,  to  whose 
works  the  student- is  referred  for  more  elaborate  dissertations  on  the  subject. 


ANNUITY   TABLE   NO.    I 


925 


TABLE    No.    1. 

SHOWING  THE  AMOUNT  OF  AN  ANNUITY  OF  ONE  DOLLAR  PER  ANNUM   MADE  AT  THE 

END  OF  THE  YEAR  AND  IMPROVED  AT  COMPOUND  INTEREST  FOR 

ANY  NUMBER  OP  YEARS  NOT  EXCEEDING  FIFTY. 

Note. — In  this  table,  the  rates  of  interest  may  be  regarded  as  either  annual,  semi' 
anjiual,  quarterly,  monthly,  or  ewenweekly;  provided  the  numbers  in  column  headed  "YEARS" 
betaken  accordingly;  that  is,  &s  so  many  years,  half-years,  quarters  or  months  or  weeks, 
respectively. 


Yrs. 

3  per  cent. 

3^  per  cent. 

4  per  cent. 

5  per  cent. 

6  per  cent. 

7  pcT  cent. 

1 

1.000  000 

1.000  000 

1.000  000 

1.000  000 

1.000  000 

1.000  000 

2 

2.030  000 

2.035  000 

2.040  000 

2.050  000 

2.060  000 

2.070  000 

3 

3.090  900 

3.106  225 

3.121  6(10 

3.152  500 

3.183  600 

3.214  900 

4 

4.183  627 

4.214  943 

4.246  464 

4.310  125 

4.374  616 

4.439  943 

.5 

5.309  136 

5.362  466 

5.41b  323 

5.525  631 

5.637  093 

5.750  739 

6 

6.468  410 

6.550  152 

6.632  975 

6.801  913 

6.975  319 

7.1.53  291   • 

7 

7.662  462 

7.779  408 

7.898  294 

8.142  008 

8.393  838 

8.654  021 

8 

8.892  336 

9.051  687 

9.214  226 

9. .549  109 

9.897  468 

10.259  803 

9 

10.159  106 

10.368  496 

10.582  795 

11.026  564 

11.491  316 

11.977  989 

10 

11.463  879 

11.731  393 

12.006  107 

12.577  893 

13.180  795 

13.816  448 

11 

12.807  796 

13.141  992 

13.486  351 

14.206  787 

14.971  643 

15.783  599 

12 

14.192  030 

14.601  962 

15.025  805 

15.917  127 

16.869  941 

17.888  451 

13 

15.617  790 

16.113  030 

16.626  838 

17.712  983 

18.882  138 

20.140  643 

14 

17.086  324 

17.676  986 

18.291  911 

19.598  632 

21.015  066 

22.5.50  488 

15 

18.598  914 

19.295  681 

20.023  588 

21.578  564 

23.275  970 

25.129  022 

16 

20.156  881 

20.971  030 

21.824  531 

23.657  492 

25,670  528 

27.888  054 

17 

21.761  588 

22  705  016 

23.697  512 

25.840  366 

28.212  880 

30.840  217 

18 

23.414  435 

24.499  691 

25.645  413 

28.132  385 

30.905  653 

33.999  033 

19 

25.116  868 

26.357  180 

27.671  229 

30.539  004 

33,759  992 

37.378  965 

20 

26.870  374 

28.279  682 

29.778  079 

33.065  954 

36.785  591 

40.995  493 

21 

28.676  486 

30.269  471 

31.969  202 

3.5.719  252 

39.992  727 

44.865  177 

22 

30.536  780 

32.328  902 

34.247  970 

38.505  214 

43.392  290 

49.005  739 

2! 

32.452  884 

34.460  414 

36  617  889 

41.430  475 

46.995  828 

53.436  141 

24 

34.426  470 

36  666  528 

39.082  604 

44  501  999 

50.815  577 

.58.176  671 

25 

36.459  264 

38.949  857 

41.645  9U8 

47.727  099 

54.864  512 

63.249  030 

26 

38.553  042 

41.313  102 

44.311  745 

51.113  454 

59.156  383 

68.676  470 

27 

40.709  634 

42.759  060 

47.084  214 

54.669  126 

63.705  766 

74.483  823 

28 

42.930  923 

46.290  627 

49.967  583 

58.402  583 

68.528  112 

80  697  691 

29 

45.218  850 

48.910  799 

52.966  286 

62.322  712 

73.639  798 

87.346  529 

30 

47  575  416 

51.622  677 

56.084  938 

66.438  848 

79.058  186 

94.460  786 

31 

50.002  678 

54.429  471 

59.328  335 

70.760  790 

84.801  677 

102,073  041 

32 

52.502  759 

57.334  502 

62.701  469 

75.298  829 

90.889  778 

110.218  154 

3;i 

55.077  841 

60.341  210 

66.209  527 

80.063  771 

97.343  165 

118.933  425 

34 

57.730  177 

63.453  152 

69.8."i7  909 

85.066  9.59 

104.183  7.55 

128.2.58  765- 

35 

60.462  082 

66.674  013 

73.652  225 

90.320  307 

111.434  780 

138.236  878 

36 

63.271  944 

70.007  603 

77.598  314 

95.836  323 

119.120  867 

148.913  460 

37 

66.174  223 

73.457  S69 

81.702  246 

101. (i28  139 

127.268  119 

160.337  400 

38 

69.1.59  449 

77.028  895 

85.970  336 

107.709  546 

135.904  206 

172.561  020 

39 

72.234  233 

80.724  906 

90.409  1.5(1 

114.0115  023 

145.0.58  458 

185  640  292 

40 

75.401  200 

84.550  278 

95.025  516 

120.799  774 

154.761  966 

199.635  112 

41 

78.663  298 

88.509  537 

99.826  536 

127  839  763 

165.047  684 

214.609  570 

42 

82.023  196 

92.607  371 

104.819  598 

135.231  751 

175.950  645 

230.632  240 

43 

85.483  892 

96.848  629 

110.012  :i82 

142.993  339 

187.507  577 

247.776  496 

44 

89.048  409 

101.238  331 

115.412  877 

151.143  006 

199,758  032 

266.120  851 

45 

92.719  861 

105.781  673 

121.029  392 

159.700  156 

212.743  514 

285.749  311 

46 

96..501  457 

110.484  031 

126.870  568 

168.685  164 

226.508  125 

306.751  763 

47 

100.396  501 

115.350  973 

132.945  390 

178.119  422 

241.098  612 

329.224  386 

48 

104.408  396 

120.388  297 

139.263  206 

188.025  393 

2.56.564  529 

353.270  093 

49 

108  540  648 

125.601  846 

145.833  734 

198.426  663 

272.958  401 

378.999  OnO 

50 

112.796  867 

130  999  910 

152.667  084 

1  209.347  976 

290.335  905 

406.528  929 

Note. — By  taking'  one  year  more  and  subtracting  $1.00,  we  will  have  in  the  remainder 
the  amount  of  $1.00  when  tlie  payments  are  made  at  the  beginning  of  the  year.  Thus  the 
Aviount  of  $1.00  payable  at  the  end  of  each  year,  for  12  years  5%,  is  $1.5.917127:  When  pay- 
able at  the  beginning  of  each  year,  the  Amount  is  $17.712983— $1.00=$1(). 712983.  See  above 
Table. 


926 


SOULE  S   PHILOSOPHIC   PRACTICAL  MATHEMATICS. 


TABLE    No.    2. 

SHOWING     THE     PRESENT     WORTH     OF    AN    ANNUITY    OP    ONE    DOLLAR    PER    ANNUM, 
TO   CONTINUE  FOR   ANY  NUMBER   OF   YEARS   NOT  EXCEEBING  FIFTY. 


Vis. 

3  per  cent. 

3i  per  cent. 

4  per  cent. 

4i  percent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

— ■■ =s 

8  per  cent. 

1 

0.970  874 

0  966  184 

0.961  538 

0.95694 

0.952  381 

0.943  396 

0.934  579 

0.9259 

2 

1.913  470 

1.899  694 

1.886  095 

1.87267 

1.859  410 

1.833  393 

1.808  017 

1.7833 

3 

2.828  611 

2.801  637 

2.775  091 

2.74896 

2  723  248 

2.673  012 

2.624  314 

2.5771 

4 

3.717  098 

3.673  079 

3.629  895 

3.587.53 

3.545  951 

3.465  106 

3.387  209 

3.3121 

5 

4.579  707 

4.515  052 

4.451  822 

4.38998 

4.329.477 

4.212  364 

4.100  195 

3.9927 

6 

5.417  191 

5.328  553 

5.242  137 

5.15787 

5.075  692 

4.917  324 

4.766  537 

4.6229 

7 

6.230  283 

6.114  544 

6.602  055 

5.89270 

5.786  373 

5.582  381 

5.389  286 

5.2061 

8 

7.019  692 

6.873  956 

6.732  745 

6.59.589 

6.463  213 

6.209  744 

5.971  295 

5.7466 

9 

7.786  109 

7.607  687 

7.435  332 

7.26879 

7.107  822 

6.801  602 

6.515  228 

6.2469 

10 

8.530  203 

8.316  605 

8.110  896 

7.91272 

7.721  735 

7.360  087 

7.023  577 

6.7101 

11 

9.252  624 

9.001  551 

8.760  477 

8.52S92 

8.306  414 

7.886  875 

7.498  669 

7.1390 

12 

9.954  004 

9.G63  343 

9.385  074 

9.11858 

8.863  252 

8.383  844 

7.943  671 

7..5361 

13 

10.634  955 

10.302  738 

9.985  648 

9.68285 

9.393  573 

8.852  6S3 

8.357  635 

8.9038 

14 

11.296  073 

10.920  520 

10.563  123 

10.22283 

9.898  641 

9.294  984 

8.745  4,53 

8.2442 

15 

11.937  935 

11.517  411 

11.118  387 

10.739,55 

10.379  6.58 

9.712  249 

9.107  898 

8.5595 

16 

12.561  102 

12.094  117 

11.652  296 

11.23402 

10.837  770 

10.105  895 

9.446  633 

8.8514 

17 

13.166  118 

12.651  321 

12.165  669 

11.70719 

11.274  066 

10.477  260 

9.7C3  206 

9.1216 

18 

13.753  513 

13.189  683 

12.659  297 

12.15999 

11.689  587 

10.827  603 

10.059  070 

9.3719 

19 

14.323  799 

13.709  837 

13.133  939 

12.59329 

13.085  321 

11.1.58  116 

10.335  578 

9.6036 

20 

14.877  475 

14.212  403 

13.590  326 

13.00794 

12.462  210 

11.469  921 

10.593  997 

9.8181 

21 

15.415  024 

14.697  974 

14.029  160 

13.40472 

12.821  153 

11.764  077 

10.8.35  527 

10.0168 

22 

15.936  917 

15.167  125 

14.451  115 

13.78442 

13.163  003 

12.041  582 

11.061  241 

10.2007 

23 

16.443  608 

15.620  410 

14.856  842 

14.14777 

13.488  574 

12.303  379 

11.273  187 

10.3711 

21 

16.935  542 

16.058  368 

15.246  963 

14.49548 

13.798  642 

12.550  358 

11.469  3.34 

10.5388 

25 

17.413  148 

16.481  515 

15.622  080 

14.82821 

14.093  945 

12.783  356 

11.653  583 

10.6748 

26 

17.876  842 

16  890  352 

15.982  769 

15. 14661 

14.275  185 

13.003  166 

11.825  779 

10.8100 

27 

18.327  031 

17.285  365 

16.329  586 

15.45130 

14.643  034 

13.210  .534 

11.986  709 

10.9352 

28 

18.764  108 

17.667  019 

16.663  063 

15.74287 

14.898  127 

13.406  164 

12.137  111 

11.0511 

29 

19.188  455 

18.035  767 

16.983  715 

16.02189 

15.141  074 

13..590  721 

12.277  674 

11.1584 

SO 

19.600  441 

18.392  045 

17.292  033 

16.28889 

15.372  451 

13.764  831 

12.409  041 

11.3578 

31 

20.000  428 

18.736  276 

17.588  494 

16. 54439 

15..592  811 

13  929  086 

12.531  814 

11.3498 

32 

20.338  766 

19.068  865 

17.873  552 

16.78889 

15.802  667 

14.084  043 

12.646  555 

11.4350 

33 

20.765  792 

19.390  208 

18  147  646 

17.02286 

16.002  549 

14.230  230 

12.753  790 

11.5139 

34 

21.131  837 

19.700  684 

18.411  198 

17.24676 

16.192  204 

14.368  141 

12.854  009 

11.5869 

35 

21.487  220 

20.000  661 

18.664  613 

17.46101 

16.374  194 

14.498  246 

12.947  672 

11.6546 

56 

21  832  252 

20.290  494 

18.908  282 

17.66604 

16.546  8.52 

14.620  987 

13.035  208 

11.7172 

37 

22.167  235 

20.570  525 

19.142  579 

17.86224 

16.711  287 

14.736  780 

13.117  017 

11.7752 

38 

23.492  462 

20.841  087 

19.367  864 

18.04999 

16.867  893 

14.846  019 

13.193  473 

11.8289 

39 

22.808  215 

21.102  500 

19.584  485 

18.22966 

17.017  041 

14.949  075 

13.264  928 

11.8786 

40 

23.114  772 

21.355  072 

19.792  774 

18.40158 

17.159  086 

15.046  297 

13.331  709 

11.9246 

41 

23.412  400 

21.599  104 

19.993  052 

18.56611 

17.294  368 

15.138  016 

13.394  120 

11.9672 

42 

23.701  3.-.9 

21.834  883 

20.185  627 

18.72355 

17.423  208 

15.224  543 

13.453  449 

12.0067 

43 

23.981  902 

22.062  689 

20.370  795 

18.87421 

17.545  912 

15.306  173 

13.506  962 

12.0432 

44 

24.2.-i4  274 

22.282  791 

20.548  841 

19.01838 

17.662  773 

15.383  182 

13.557  908 

12.0771 

45 

24.518  713 

22.495  450 

20.720  040 

19.15635 

17.774  070 

15.455  832 

13.605  523 

13.1084 

46 

24.775  449 

22.700  918 

20.884  6.^4 

19.28837 

17.880  067 

15.524  370 

13.650  020 

12.1374 

47 

25.024  708 

23.899  438 

21.042  936 

19.41471 

17.981  016 

15.589  028 

13  691  608 

12.1643 

48 

25.266  707 

23.091  244 

21.195  131 

19.53561 

18.077  158 

15.650  027 

13.730  474 

12.1891 

49 

25.501  657 

23.276  564 

21.341  473 

19.65130 

18.168  722 

15.707  572 

13.766  799 

12.2122 

60 

25.729  764 

23.4.55  618 

21.482  185 

19.76201 

18.255  925 

, 15.761  861 

13.800  746 

12.2335 

Note. — In  this  table,  the  rates  of  interest,  as  they  appear  at  the  heads  of  the  columns, 
xna.y  he  rega,vded  SiS  either  annua/,  seini-annua/,  guar/gr/y,  monthly,  or  even  weekly;  provided  the 
numbers  in  column  headed  "YEARS"  be  taken  aocording-ly;  that  is,  so  many  years,  half-years, 
quarters,  months  or  weeks,  respectively.  Thus:  What  is  the  Present  Worth  of  $200  due  at  the  end 
of  each  half  year  for  14  half  years'  (7  years)  interest  compounded  half  yearly  at  a  semi-amiual 
rate  of  3%  ? 

Operation:  11.296073  X  200  =  $2259.214600. 


ANNUITY   TABLE    NO.    3. 


927 


TABLE    No. 

TABLE    OF    PRESENT    WORTH,     OB     VALUE     OF 


3. 


DUE    AT     THE    END     OF    ANY 


NUMBER  OF  YEARS  NOT  EXCEEDING 

50,  ALLOWING 

COMPOUND  INTEREST 

iaiof 

Iper 

2  per 

3  per 

4  per 

5  per 

6  per 

?  per 

8  per 

9  per 

10  per 

Enil  ot 

"fear. 

cent. 

CL'llt. 

cent. 

ceut. 

cent. 

cent 

cent. 

cent. 

ceut. 

ceut. 

Ve;ir. 

1 

.9901' 

.9804 

.9709 

.961538 

.952381 

.943396 

.934579 

.925926 

.917431 

.909091 

1 

2 

.9803 

.9612 

.9426 

.924556 

.907029 

.889996 

.873439 

.857339 

.841680 

.826446 

2 

3 

.9706 

.9423 

.9151 

.888996 

.863838 

.839619 

.816298 

.793832 

.772183 

.751315 

3 

4 

.9610 

.9238 

.8885 

.854804 

.822702 

.792004 

.762895 

.735030 

.708425 

.683013 

4 

5 

.9515 

.9057 

.8626 

.821927 

.783526 

.747258 

.712986 

.680583 

.649931 

.620921 

5 

6 

.9420 

.8880 

.8375 

.790315 

.746215 

.704961 

.666342 

.630170 

.596267 

.564474 

6 

7 

.9327 

.8706 

.8131 

.759918 

.710681 

.665057 

.622750 

.583490 

.547034 

.513158 

7 

8 

.9235 

.8535 

.7894 

.730690 

.676839 

.627413 

.582009 

.540269 

.501866 

.466507 

8 

9 

.9143 

.8368 

.7664 

.702587 

.644609 

.591898 

.543934 

.500249 

.460428 

.424098 

9 

10 

.9053 

.8203 

.7441 

.675564 

.613913 

.588395 

.508349 

.463193 

.422411 

.385543 

10 

11 

.8963 

.8043 

.7224 

.649.581 

.584679 

.526788 

.475093 

.428883 

.387533 

.350494 

11 

12 

.8874 

.7885 

.7014 

.624597 

.556837 

.496969 

.444012 

.397114 

.355535 

.318631 

12 

13 

.8787 

.7730 

.6810 

.600574 

.530321 

.468839 

.414964 

.367698 

.326179 

.289664 

13 

14 

.8700 

.7579 

.6611 

.577475 

.505068 

.442301 

.387817 

.340461 

.299246 

.263331 

14 

15 

.8613 

.7430 

.6419 

.555265 

.481017 

.417265 

.362446 

.315242 

.274538 

.239392 

15 

16 

.8528 

.7284 

.6232 

..533908 

.458112 

.393646 

.338735 

.291890 

.251870 

.217629 

16 

17 

.8444 

.7142 

.6050 

.513373 

.436297 

.371364 

.316574 

.270269 

.231073 

.197845 

17 

18 

.8360 

.7002 

.5874 

.493628 

.415521 

.350344 

.295864 

.250249 

.211994 

.179859 

18 

19 

.8277 

.6864 

.5703 

.474642 

.395734 

.330513 

.276508 

.231712 

.194490 

.163508 

19 

20 

.8195 

.6730 

.5537 

.456387 

.376889 

.311805 

.258419 

.214548 

.178431 

.148644 

20 

21 

.8114 

.6598 

.5375 

.438834 

.3.58942 

.294155 

.241513 

.19,S656 

.163698 

.13.5131 

21 

22 

.8034 

.6468 

.5219 

.421955 

.341850 

.277.505 

.225713 

.183941 

.1.50182 

.122,S46 

23 

23 

.7954 

.6342 

.5067 

.40.5726 

.325571 

.261797 

.210947 

.176315 

.137781  1  .111678 

23 

24 

.7876 

.6217 

.4919 

.390121 

.310068 

.246979 

.197147 

.157699 

.126405 

.1(11526 

24 

25 

.7798 

.6095 

.4776 

.375117 

.295303 

.232999 

.184249 

.146018 

.115968 

.092296 

25 

26 

.7720 

.5976 

.4637 

.360689 

.281241 

.219810 

.172195 

.13,5202 

.106393 

.083905 

26 

27 

.7644 

.5859 

.4502 

.346817 

.267848 

.207368 

.160930 

.125187 

.0!I7608 

.076278 

27 

28 

.7568 

.5744 

.4371 

.333477 

.25.5094 

.195630 

. 150402 

.115914 

.080548 

.069343 

28 

29 

.7493 

.5631 

.4243 

.320651 

.242946 

.184557 

.140563 

.107328 

.082155 

.063039 

29 

30 

.7419 

.5521 

.4120 

.308318 

.231377 

.174110 

.131367 

.099377 

.075371 

.057309 

30 

31 

.7346 

.5412 

.4000 

.296460 

.220359 

.1642,55 

.122773 

.092016 

.069148 

.052099 

31 

32 

.7273 

.5306 

.3883 

.285058 

.209S66 

.154957 

.114741 

.0,S52()0 

.063438 

.(U7362 

33 

S3 

.7201 

.5202 

.3770 

.274094 

.199873 

.146186 

.107235 

.078889 

.058200 

.043057 

33 

34 

.7130 

.5100 

.3660 

.263552 

.1903.55 

.137912 

.100219 

.073045 

.053395 

.039143 

31 

35 

,7059 

.5000 

.3554 

.253415 

.181290 

.130105 

.093663 

.067635 

.048986 

.035584 

35 

36 

.6989 

.4902 

.34.50 

.243669 

.172657 

.122741 

.087535 

.062625 

.044941 

.032349 

36 

37 

.6920 

.4806 

.3350 

.234297 

.164436 

.115793 

.081809 

.057986 

.041231 

.029408 

37 

38 

.6852 

.4713 

.3253 

.225285 

.156605 

.109239 

.076457 

.053690 

.037826 

.026735 

38 

39 

.6784 

.4619 

.31.58 

.216621 

.149148 

.103056 

.0714,55 

.049713 

.034703 

.024304 

39 

40 

.6717 

.4529 

.3066 

.208289 

.142046 

.097222 

.066780 

.046031 

.031838 

.022095 

40 

41 

.66.50 

.4440 

.2976 

.200278 

.135283 

.091719 

.062412 

.042621 

.029209 

.020086 

41 

42 

.6584 

.4353 

.2890 

.192575 

.128840 

.086527 

.058329 

.039464 

.026797 

.oi.s2i;o 

43 

43 

.6519 

.4268 

.2805 

.185168 

.122704 

.081630 

.0.54513 

.036541 

.024584 

.0166H0 

43 

44 

.64r>4 

.4184 

.2724 

.178046 

.116861 

.077009 

.050946 

.033834 

.022555 

.01.5091 

44 

45 

.6391 

.4102 

.2644 

.171198 

.111297 

.072650 

.047613 

.031328 

.020692 

.013719 

45 

46 

.6327 

.4022 

.2.567 

.164614 

.105997 

.068538 

.044499 

.029007 

.018984 

.012472 

46 

47 

.6265 

.3943 

.2493 

.158283 

.100949 

.064658 

.041587 

.026859 

.017416 

.011338 

47 

48 

.6203 

.3865 

.2420 

.152195 

.096142 

.060998 

.038867 

.024869 

.015978 

.010307 

48 

49 

.6141 

.3790 

.23.50 

.146341 

.091564 

.057546 

.036324 

.023027 

.014659 

.009370 

49 

SO 

.6080 

.3715 

.2281 

.140713 

.087204 

.054288 

.033948 

.021321 

.013449 

.008519 

50 

Note  1. — ^The  t.ible  shows  wh.it  per  cent,  of  any  principal  the  pre.sent  worth  is:  Hence  to 
get  P.  W.,  multiply  by  the  number  given.  Thus  the  present  worth  of  $10000  due  iu  15  years  at 
i<}^  =  $10000  X  .555265  =  $5552.65. 

Note  2.— The  per  cent,  of  present  worth  for  any  number  of  years  may  be  found  by  dividing 
fl.OO  by  the  amount  of  Jl.OO  at  compound  interest  as  given  in  the  table,  on  page  611.  Thus  P.  \V. 
of  $1.00  for  55  years  at  3%  =  $1.00  ^  5.08214859. 

Note.— In  this  table,  the  rates  of  interest  may  be  regarded  as  either  annual,  /^"'J.', 
annual  quarterly,  monthly,  or  even  weekly;  provided  the  numbers  in  column  headed    'YEARS 
be  taken  accordingly;  that  is,    as  so  many  years,    half-years,   quarters,    months  or   weeks, 
respectively. 


ktric  System  of  Weights  and  Measures. 


The  Metric  System  of  Weights  and  Measures  derives  its  name  from 
the  metre,  -wliieli  is  tlie  i)riiiiary  base  or  unit  froiii  wbicli  all  of  the  otlier  units  of  the, 
system  are  derived.  This  metre,  -when  adopted  by  the  French  nation,  was  thought 
to  be,  and  is  very  nearly,  the  (toobVooo)  one  ten-millionth  jjart  of  the  distance  from 
the  equator  to  either  pole,  measured  on  the  earth's  surface.  Its  length,  in  American 
measure,  is  39.3708  inches,  very  nearly. 

Note. — The  term  meter  is  from  the  Greek  metron,  a.  me.asnre. 

The  Standard  Meter  is  a  bar  of  platinum  and  is  kept  among  the 
national  archives  in  Paris;  but  duplicates  of  it  have  been  furnished  to  the  United 
States  and  other  nations. 

HISTORICAL. 

The  Metric  System  is  one  of  the  products  of  the  French  revolution 
of  1789,  which  resulted  in  tlie  establishment  of  the  French  Republic,  on  the  22d 
day  of  September,  1792.  The  progressive  spirits  of  the  revolutionists  claimed  that 
every  thing  needed  reforming.  They  demanded  a  change  in  the  calendar,  a  new 
classification  of  the  seasons,  a  new  order  of  mouths,  with  new  names  and  a  change 
of  days,  a  new  arrangement  for  Sunday  and  festive  days,  and  above  all  they 
demanded  a  decimal  system  of  Aveights,  measures,  and  values. 

March  17,  1791,  a  report  was  presented  to  the  French  National  Assembly 
jiroposing  the  adoption  of  the  Metric  System.  By  the  laws  of  1793  and  1795,  a 
temporary  meter  and  kilogramme  were  adopted,  and  in  1795  a  commission  was 
appointed,  under  tlie  direction  of  the  Academy  of  Sciences,  for  the  purpose  of 
perfecting  the  system. 

The  first  and  most  important  duty  was  to  determine  an  invariable  standard 
unit  for  all  measures  of  length,  area,  solidity,  capacity,  and  weiglit.  For  the  con- 
summation of  this  objet't,  a  trigonometrical  survey  was  made,  by  two  eminent 
mathematicians,  Delambre  and  Mechain,  of  the  arc  of  the  meridian  through  Paris, 
from  Dunkirk,  France,  to  Mont  Jouy,  near  Barcelona,  Spain,  a  distance  of  9f 
degrees,  more  than  one-tenth  of  the  quadrant  of  the  meridian.  From  the  survey  of 
this  meridian,  the  length  of  tlie  quadrant  from  the  equator  to  the  pole,  measured  on 
the  earth's  surface,  was  computed. 

The  length  of  the  Paris  meridian  quadrant,  thus  obtained,  was  divided  into 
10,000,000  equal  parts,  and  one  of  these  i)arts  was  called  a  meter  and  taken  as  the 
primary  unit  of  tlie  French  system  of  measures.     This  meter  was  adopted  as  the 

.      (9-'8) 


METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES.  929 

unit  of  length,  niul  from  it  all  tlie  other  units  of  measure  are  derived  by  the  appli- 
cation of  the  decimal  scale.  It  is  equal  to  39.37079,  practically  39.37  inches  iii 
length,  and  at  the  time  of  its  adoption,  it  was  thought  to  be  exactly  the  one  ten- 
millionth  part  of  the  distance  from  the  equator  to  the  ])o\e,  on  the  meridian  of 
Paris.  But  by  more  accurate  measurement,  based  upon  the  fact  that  the  earth's 
equator  is  not  a  perfect  circle,  bnt  slightly  elliptical,  a  fact  not  considered  by  the 
French  mathematicians  and  astronomers,  it  is  found  to  be  ^^3  part  of  an  inch  too 
short.  This  very  small  error,  almost  imperceptible  in  a  single  meter,  amounts  to 
5124  feet  in  the  length  of  the  quadrant  measured.  But  this  slight  discrepancy  does 
no  injury  to  the  system.  The  meter  has  a  fixed  length  and  is  decimally  divided,  and 
that  is  all  that  is  required. 

In  1799,  the  French  nation  formally  adopted  the  revised  metric  system ;  but 
by  decree  of  February  12,  1S12,  the  old  system  was  re-established  and  continued  in 
use  until  July  4, 1837,  when  the  decree  of  February  12,  1812,  was  repealed  and  the 
metric  system  was  again  brought  in  use.  A  royal  decree,  April  1 7, 1839,  determined 
the  denominations,  form,  and  dimensions  of  all  instruments  and  measures,  for  trade 
and  use.  The  metric  system  is  now  compulsory  throughout  France.  It  has  also 
been  adopted  and  is  in  general  use  by  several  other  nations. 

Previous  to  the  adoption  of  the  metric  system  by  the  French  nation,  their 
units  of  weights  and  measures  were,  in  many  respects,  quite  similar  to  those  of  the 
English,  many  of  which  came  down  the  centuries  through  Egypt,  Greece,  and 
Eome,  and  were  originally  largely  derived  from  different  parts  of  the  human  body ; 
such  as  the  foot, — the  length  of  the  foot  of  Hercules,  or  of  a  king;  the  ulna  or 
yard, — the  distance  from  the  middle  of  the  chest  or  lips,  to  the  tip  of  the  middle 
finger;  the  palm  or  hand, — the  width  of  the  hand;  the  span, — the  distance  from 
the  tip  of  the  thumb  to  tip  of  the  little  finger  when  extended ;  the  digit  or  the 
finger, — the  length  of  the  first  finger;  the  cubit, — the  distance  from  the  elbow  to 
the  tip  of  the  middle  finger ;  the  fathom, — the  distance  from  the  tips  of  the  middle 
fingers  of  the  two  hands  when  extended  in  opposite  directions,  etc.  Other  units 
were  taken  from  familiar  objects  in  nature,  such  as  the  barley  corn,  the  grain  of 
wheat,  shells,  horns,  etc. 

These  once  variable  ancient  and  English  units  of  length  have  now  certain 
fixed  values  based  upon  the  Imperial  Standard  Yard,  the  length  of  which  is  36 
inches.  This  standard  yard  is  such  that  a  pendulum  equal  in  length  to  39.13929  of 
its  inches,  will  vibrate  seconds,  in  a  vacuum,  at  the  level  of  the  sea,  in  the  latitude 
of  London,  the  thermometer  being  62°  Fahrenheit.  The  Standard  Yard  of  the 
United  States  is  a  copy  of  the  English  Standard  Yard,  marked  upon  a  brass  scale, 
and  deposited  in  the  Treasury  Department  at  Washington.  Copies  of  this  yard 
have  been  supplied  to  all  the  States. 

By  reason  of  the  decimal  divisions  of  the  metric  system,  it  possesses  many 
advantages  over  all  other  systems  and  is  now  used  wholly,  or  in  part,  by  nearly  all 
civilized  countries. 


930  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

ADOPTED  BY  THE  UNITED  STATES. 

The  Metric  System  was  legalized  and  adopted,  but  not  made  compul- 
sory in  the  United  States,  by  an  act  of  Congress  passed  in  1866.  And  it  is  tlie 
only  system  ever  authorized  by  the  Government  of  the  United  States. 

It  is  adopted  by  the  United  States  Coast  Survey,  is  used  in  the  Mints  and 
the  Post  OfiQces,  and  largely  by  all  scientists,  colleges,  and  universities. 

•  The  Meter  is  the  unit  of  Length,  and  from  it  are  derived  the  Are,  the 
Stere,  or  Cubic  Meter,  tlie  Liter,  and  the  Gram.  From  these  five  units,  all  others 
are  formed. 

The  Are  (air),  is  the  unit  of  surface,  or  square  measure,  and  consists 
of  a  square  whose  side  is  10  meters;  hence,  it  contains  100  square  meters. 

Note. — The  Are  is  from  the  Latin,  area,  a  surface. 

The  Cubic  Meter,  or  Stere  (stair),  is  the  unit  of  solidity,  and  consists 
of  a  cube  whose  edge  is  one  meter. 

The  Liter  (Lecter),  is  the  unit  of  the  capacity  of  vessels,  etc.,  and  is 
a  vessel  whose  volume  is  equal  to  a  cube  whose  edge  is  one-tenth  of  a  meter. 

Tlie  Gram  is  the  unit  of  weight,  and  is  the  weiglit  of  a  cube  of 
distilled  water  (weighed  in  a  vacuum,  39.2°  F.,  or  4°  C.)  whose  edge  is  one-hundredth 
of  a  meter. 

Each  of  these  five  units  has  its  multiples  and  sub-multiples,  or  its  higher  and 
lower  metric  denominations. 

The  Multiple  Units,  or  Higher  Denominations,  are  formed  by  prefixing 

to  the  name  of  tlie  b<(se  units,  the  Greek  numerals,  Deka,  (10),  Hecto,  (100),  Kilo, 
(1000),  and  Myria,  (10000). 

They  are  used  as  follows : 

Dekameter,        10  meters.  Kilometer,        1000  meters. 

Hectometer,      100  meters.  Mjaiameter,    10000  meters. 

The  Sub-Multiple  Units,  or  lower  denominations,  are  formed  by 
prefixing  to  the  name  of  the  i«se  M?((7,s',  the  Latin  numerals,  Deci  (n,))  Genti  (xoo)) 
and  Milli  (tctoo)- 

They  are  used  as  follows  : 
Decimeter,  i^o,  meter.         Centimeter,  -jio  meter.        Millimeter,  -i-^oo  meter. 

Note. — The  student  should  memorize  these  Greek  and  Latin  prefixes,  and  also  the  names  of 
the  primary  or  base  units,  before  proceeding  further. 


*  METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES.  931 

METRIC   TABLES. 

■   TABLE  OF  LINEAR  MEASURE,  OF  WHICH  THE  METER  IS  THE  BASE  UNIT. 

Erjuivalenta  in 
Metric  TJnita.  Eoglish  Measures. 

1  Millimeter  Mtia  of  a  M.)  = .03937+  incL. 

10  mm.  =  1  Centimeter  (y^^  of  a  M.)  = .3937  -f  iuch. 

10  em.     =  1  Decimeter  (i^ii  of  a  M.)  = 3.93707-(- iucbea- 

10  dm.    =  1  Meter  (1  meter)  =  39.37079  in.,  ])racticallv  39.37  in.,  or    3.28089-j- feet. 

10  M.       =  1  Dekameter  (10  meters)  =  ...-■...  32.80899+ feet. 

10  Dm.    =  1  Hectometer  (100  meters)  = 19.88423+ rods. 

10  Hm.    =  1  Kilometer  (1000  meters)  = .62138+ mile. 

10  Km.    =  1  Myriameter  [Mm.]  (10000 

meters)  =  - 6.21382+  miles. 

The  Meter,  like  our  yard,  is  usetl  in  measuring  short  distances,  cloths, 
etc. 

The  Kilometer  is  used  iu  measuring  long  distances  and  is  about  |  of 
a  common  mile. 

The  Centimeter  and  Millimeter  are  used  by  artisans  and  scientists 
in  measuring  very  small  lengths. 

Methods  of   Reading.     The  number  25365.897  metres,    is    read,    in 
English, 

Twenty-five  thousand  three  hundred  and  sixty-five  metres,  and  807  thou- 
sandths of  a  metre.     But  iu  the  language  of  the  metric  system,  it  may  be  read, 

Two  myriametres,  5  kilometres,  3  hectometres,  6  decametres,  5  metres,  8 
decimetres,  9  centimetres,  and  7  millimetres.  It  may  also  be  read,  beginning  with 
the  lowest  denomination,  7  millimetres,  9  centimetres,  etc.,  etc. 

In  reading,  remember  that  the  unit  of  any  place  is  one-tenth  of  the  unit  in  the 
place  next  at  the  left,  and  ten  times  as  great  as  the  unit  of  the  place  next  at  the 
right.  Hence,  the  change  from  one  unit  to  another,  and  the  methods  of  reduction 
and  reading,  are  identical  with  those  in  the  system  of  decimal  currency. 

.  TABLE  OF  SURFACE  OR  SQUARE   MEASURE,    OF  WHICH   THE   ARE   IS   THE 

BASE  UNIT. 

100  sq.  Millimeters,  (sq.  mm.)  =  1  sq.  Centimeter,  (Bq.  cm.)  =  .155+  sq.  in. 

100  sq.  Centimeters  =  1  sq.  Decimeter,  (sq.  dm.)  =  15.5+  sq.  in, 

or   .1076+  sq.  ft. 
100  sq.  Decimeters  =  1  sq.  Meier,  =  1  Centare  (Ca.)  or  (sq.  M.)  =      1.19603+  sq.  yds. 

100  sq.  Meters  =  1  sq.  Dekameter,  (sq.  Dm.)  or  Are,  (A.)  =    119.6034+  sij.  yds. 

or  3.95383+  sq.  rods. 
100  sq.  Dekameters,  or  Ares,  =  1  sq.  Hectometer,  (sq.  Hm.  or  Hectare,)  (Ha.)  =      2.47114+  acres. 
100  sq.  Hectometers  or  Hectares,  ;=  1  sq.  Kilometer  =  .3861+  sq.  mile. 

Note. — This  measure  is  used  for  measuring  land,  flooring,  ceilings,  etc.     The  lower  denom- 
inations are  seldom  used. 


932  soule's  ruiLosoPHic  practical  mathematics.  * 

,     TABLE  OF  CUBIC,  OR  SOLID  MEASURE,  OF  WHICH   THE   CUBIC   METER   OR 

STERE  IS  THE  BASE  UNIT. 

1000  en.  Millimeters,  (cu.  mm.)  =  1  cu.  Centinicter,  (cu.  cm.)      =      .001027+  en.  ia. 
1000  cu.  Centimeters  =:  1  en.  Deeinieter,  (cu.  <lni.)  =  61.0270.")-|-  cu.  in. 

1000  cu.  Decimeters  =  1  Cubic  Meter,  (cu.  M.)  or  Strre,  (St.)  =  35.316."i84-  cu.  ft., 

or  1.30802+  cu.  yds.,  or    .2759+  cord. 

Note. — The  Cubic  Meter  is  the  unit  used  for  measuring  ordinary  solids,  as  boxes,  excava- 
tions, etc.     The  Cubic  Centimeter  and  Cubic  Millimeter  are  used  for  mtasuriug  very  minute  bodies. 

When  the  Cubic  Meter  is  applied  to  the  measurement  of  wood,  it  i.s  called  the 
Stere,  and  has  the  following  units  : 

table. 

1  Becistere,  (dst.)  =      3.5316.5+  cu.  ft. 

10  Decisteres,  (dst.)  =  1  Stere,  =    35.31058+ cu.  ft. 

10  Steres,  (St.)  =  1  Delsastere,  (Dst.)  =  353.16.58+  cu.  ft. 

or    13.0802+  cu.  yds. 

'.      TABLE   FOR   MEASURING   THE   CAPACITY   OF  VESSELS,    ETC.,    OF    WHICH 
THE  LITER  IS   THE  BASE   UNIT. 

1  Milliliter,  (yo^in-  of  a  liter)  =  .06102+  cu.  in. 

10  ml.  =  1  Centiliter,     (n^„  of  a  liter)  =  .61027+  " 

10  cl.  =  1  Deciliter,  ( 1^0  of  alitor)  =  6.1027+  " 

10  dl.  =  1  Liter,  (1  liter)  =  61.02705+  " 

10  L.  =  1  Dekaliter,  (10  liters)  =  610.2705+  " 

10  Dl.  =  1  Hectoliter.  100  liters)  =  6102.70.502+  " 

10  HI.  =  1  Kiloliter,  (1000  liters)  =  61027.05024+  " 

10  Kl.  =  IMyrialiter  Ml.  (lOOOOlit.)  =610270.5024+  " 

Note. — The  Liter,  or  the  cube  of  ii  decimeter,  is  the  vnit  of  capacity  for  both  Liquid  and 
Dry  Measures,     The  Hectoliter  is  the  unit  for  measuring  liquids,  grain,  or  fruits. 

The  following  table  shows  the  equivalents  of  the   Liter  units,   in  United 
States  measures : 

TABLE  OP  EQUIVALENTS, 

Metric  XtenominatioDS.  Dry  Measure.  Liquid  Measure. 

1  Milliliter  =        .001816+  pts.  =         .0338  fl.  oz. 

or   .00845+  gi. 
1  Centiliter  =        .01816+    pts.   =  .338  11.  oz., 

or   .084539+  gi. 
1  Deciliter  =        .18162,5+  pts.  =  .84539+  gi. 

1  Liter  =        .908128+  qts.   =        1.0.56745+  qts. 

1  Dekaliter  =      9.08128+    qts.   =        2.64186+ gals. 

1  Hectoliter  =      2.8379+      bus.   =      26.4186+      " 

1  Kiloliter,  or  Stere    =    28.379+        bus.  =    264.186+        " 
1  Myrialiter  =  283.79+  bus.  =  2641.86+  " 

TABLE   FOR  MEASURING  WEIGHT,   OF   WHICH   THE   GRAM  IS    THE    BASE 

UNIT. 

1  Milligram  (Trfojr  of  a  gram)  =  .015432+  gr.  Troy. 

10  mg.  =  1  Centigram  (t^u  of  a  gram)  =  .15432+         " 

10  eg.  =  1  Decigram  (,J(f  of  a  gram)  =        1.54324+         " 

10  dg.  =  1  Gram  (1  gram)  =      15.43248+         " 

10  G.  =  1  Dekagram  (10  grams)  =         ,3.5273+    oz.  Avoir. 

10  Dg.  =  1  Hectogram  (100  grams)  =        3,52739+ 

10  Hg.  =  1  Kilogram  or  Kilo  (1000  grams)  =        2.20462+  lbs.  Avoir. 

10  Kg.  =  1  Mvriagram  (10000  grams)  =      22.04621+         " 

10  Mg.  or  100  Kg.  =  1  Quintal  (100000  grams)  =    220,46212+         " 

10  Q.  or  1000  Kg.  =  1  Tonueau  or  Ton  (T.)  (1,000,000  grams)  =  2204  62124+         " 

or  1.10231+  Tons. 


METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES. 


IJJ 


,    The  Gram  is  tbe  U7iit  used  in  weighing  gold,  silver,  jewels,  andlettersj 
and  in  coniponiiding  medicines.     It  is  a  little  less  than  15^  grains  Troy. 

The  Kilogram,  or  Kilo,  is  the  unit  used  in  weighing  common  articles 
in  trade;  as  grain,  sugar,  butter,  etc.  It  is  a  very  little  more  than  2i  lbs.  Avoir- 
dupois. 

The  Tonneau,  or  Ton,  is  used  for  weighing  very  heavy  articles,  and 
is  a  little  in  excess  of  -i\,  more  than  the  United  States  ton. 

NoTB. — The  pound  of  Germany,  Austria,  and  Denmark  is  J  of  a  Kilogram. 

TABLE  OP  EQUIVALENTS. 

By  which  Units  of  the  American  measure  may  be  easily  changed  to 
metric  units,  and  vice  versa. 


1  inch 

=    2.54  cm. 

1  dm. 

.328  ft. 

1  foot 

=    3.0487  dm. 

IM. 

= 

1.0936   yards    =   3.2808    feet   = 

1  yard 

=      .9144  M. 

39.3707  in. 

Irod 

=    5.029  Dm. 

IDm. 

= 

1.9884  rods. 

1  mile 

=    1.6095  Km. 

1  Km. 

= 

.6213  mi. 

1  sq. in. 

=    6.4516  sq.  cm. 

1  sq.  cm. 

= 

.155  sq.  in. 

1  sq.  ft. 

=    9.2936  sq.  dm. 

1  sq.  dm. 

:= 

.1076  sq.  ft. 

Isq.  yd. 

=      .8361  sq.  M.  or  centare. 

1  sq.  M. 

= 

1.196  sq.  yds. 

1  sq.  rd. 

=      .2531  sq.  Dm.  or  Ave. 

lAre 

= 

3.9538  sq.  rds. 

1  acre 

=      .4048  Hectare,  or  sq.  Hm. 

=  119.6034  sq.  yds. 

1  sq.  mi. 

=    2. .59  sq.  Km. 

1  Hectare 

= 

2.4711  acres. 

1  cu.  in. 

=  16.3934  cu.  cm. 

1  S(|.  Km. 

= 

.3861  sq.  mi. 

1  cu.  ft. 

=  28.3286  cu.  dm. 

1  cu.  cm. 

= 

.061  cu.  iu. 

1  cu.  yd. 

=      .7645  cu.  M. 

1  cu.  dm. 

= 

.0353  cu.  ft. 

1  cord. 

=    3.6281  Steres. 

1  cu.  M. 

= 

1.308  cu.  yds. 

1  11.  oz. 

=    2.9585  cl. 

1  Stere 

= 

.2759  cord. 

1  1.  qt. 

=      .9463  L. 

Icl. 

= 

.338  11.  oz. 

1  gal. 

=      .3785  DI. 

1  L. 

^ 

1.0567  1.  qts. 

1  dry  qt. 

=    1.1012  L. 

1  1)1. 

^ 

2.6418  gals. 

1  peck 

=      .8809  Dl. 

1  L. 

^ 

.9081  dry  qts. 

1  ))ushel 

=      .3523  HI. 

1  Dl. 

= 

1.1351  pecks. 

1  Troy  gr. 

=  64.935  mg. 

1  HI. 

= 

2.8379  bushels. 

1  oz.  Troy 

=  31.1.526  G. 

1  mg. 

:= 

.0154  gr.  Troy. 

1  Troy  lb. 

=      .3732  Kilo. 

1  Gram 

= 

15.4324  grs.  Troy. 

1  oz.  Av. 

=  28.409  G. 

= 

.0321  oz.  Troy. 

1 11).  Av. 

=      .4536  Kilo. 

1  Kilo 

= 

2.6791  lbs.  Troy. 

ITon  (200011)8.) 

=      .9072  Tonneau. 

1  Gram 

= 

.0352  oz.  Av. 

ITon  (2240  lbs.) 

=    1.01605  Tonneau. 

1  Kilo 

= 

2.2046  lbs.  Av. 

1  lb.  Av. 

=      .9070  German  lb. 

1  Tonneau 

= 

1.1023  ton  =  2204.6212  lbs  Av. 

1  cm. 

=        .3937  iu. 

1  Ger.  lb. 

= 

1.1023  lbs.  Av. 

GENERAL  PRINCIPLES  AND  DIRECTIONS. 

Since  the  metric  system  is  based  upon  the  decimal  scale,  wherein  10 
units  of  a  lower  order  or  denomination  make  1  of  the  next  higher,  therefore  all 
operations  in  addition,  subtraction,  multiplication,  and  division  are  performed  in 
the  same  manner  and  are  governed  by  the  same  numerical  law  as  the  operations  iu 
decimals  or  with  dollars,  cents,  and  mills. 


934  SOULE  S    PHILOSOPHIC    rRACTICAL    MATHEMATICS.  *" 

Abbreviations  of  the  base  unit  and  of  its  higher  denominations  bepjin  with 
a  capitid  ;  tlie  lower  denominations  begin  with  a  small  letter.  All  abbreviations 
should  be  read  in  full.  Thus  0  M.,  7  dm.,  6  cm.,  3  mm.,  should  be  read  G  meters.  7 
decimeters,  5  centimeters,  3  millimeters. 

The  units  of  length,  capacity,  and  weight  increase  and  decrease  accord><ig 
to  the  decimal  scale;  hence,  each  order  of  units  must  occupy  one  place,  wheu 
written  in  full. 

Thus,  127.4215  meters  would  be  written,  1  Hm.,  2  Dm.,  7  M.,  4  dm.,  2  cm.,  1.5 
mm. 

The  scale  of  square  or  surface  measure  is  (10  x  10)  100;  hence,  each  order 
of  units  must  have  two  places  of  tigures. 

Thus,  43  Ila.,  8  A.,  0  ca.  may  be  written  43.0806  Ha. ;  and  read,  43  hecta*-es 
and  806  centares.  Or,  they  may  be  written  4308.06  A.,  and  read,  4308  ares  and  0 
ceutares.     Or  thus,  430806  ca.,  and  read,  430806  centares. 

The  units  of  cubic  or  solid  measure  increase  by  the  scale  of  (10  x  10  x  10) 
1000  ;  hence,  each  order  of  luiits  must  have  three  places  of  figures. 

Thus,  36  cu.  M,,  8  cu.  dm.,  25  cu.  cm.,  may  be  written,  36.008025  cu.  M. ;  or 
thus,  36008.025  cu.  dm. ;  or  thus,  36008025  cu.  cm. 

TO  WEITE  AND  EEAD  METEIG  KUMBEES. 

1.     Write  and  read  in  the  unit  of  meters  and  the  lower  denominations^ 
127.4215  meters.  Ans.  127  M.  4  dm.  2  cm.  1.5  mm. 

2.  Write  and  read  in  higher  units  210524683  millimeters. 

Ans.  21  Mm.  0  Km.  5  Hm.  2  Dm.  4  M,  6  dm.  8  cm.  3  mm. 

3.  Write  and  read  46286  mm.  as  decimeters.  Ans.  462.86  dm, 

4.  Write  and  read  46286  mm.  as  dekameters.  Ans.  4.6286  Dm, 

5.  Write  in  lower  units  5.46732  Kg. 

Ans.  5  Kg.  4  Hg.  6  Dg.  7  G.  3  dg.  2  eg. 

6.  Write  5.46732  Kg.  as  grams.  Ans.  6467.32  G, 

7.  Write  604  L.  in  higher  units.  Ans.  6  HI.  0  Dl.  4  L. 

8.  Write  604  L.  as  centiliters.  Ans.  60400  cl. 

9.  Write  604  L.  as  myrialiters.  Ans.  .0604  Ml. 

10.  Write  6  Dm.  3  dm.  4  mm.  as  meters  and  decimals.  Ans.  00.304  M. 

Note. — When  there  are  omissions  in  any  of  the  denominations  of  the  number  given,  their 
places  must  be  tilled  with  naughts. 

11.  Write  as  Kilograms  and  decimals,  434278  dg.  Ans.  43.4278  Kg. 

TO  EEDUCE  METEIG  XUMBEES  FEOM  HIGHEE  mSTITS  OE  DENOMIN- 
ATIONS TO  LOWEE. 

■  1,    Eeduce  8  meters  to  millimeters. 

OPERATION. 

8  M.  Explanation. — Remembering  the  table  for  the 

10  Metric  Linear  Measuns,  we  reason   as   follows: 

10  Since  1  meter  =  10  dm.,  there  are  10  times   a» 

10  many  dra.  as  M. ;   then  since   1   dm.  =  10  cm., 

there  are  10  times  as  many  cm.    as   dm. ;   then 

8000  mm.     Ans.  since  1  cm.  =  10  mm.,  there  are  10  times  as  many 

mm.  as  cm.,  which  is,  as  shown  by  the  opera- 
tion, 8000  mm.  Or  we  might  reason  thus:  Since  1  M.  =  1000  mm.,  there  are  1000  times  as  many 
mm.  as  met*srs. 


InETRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES, 


935 


GENERAL   DIRECTIONS. 

Prom  the  foregoiiijr  elucidation,  we  derive  the  following  general  direc- 
tions, to  reduce  Metric  Numbers  from  Higher  to  Lower  denominations: 

For  the  Linear,  Capacity,   and    Weight  Measures,   multiply  by  10  (annex  1 
naught)  for  each  loirer  denomination  to  which  the  given  number  is  to  be  reduced. 

For  Surface  Measure,  multiply  by  100  instead  of  10;   and  for  Cubic  Measure, 
multiply  by  1000  instead  of  10. 

Note. — In  reducing  the  denominations  of  tbe  Stere,  in  wood  measure,  10  is  the  multiplier. 

PROBLEMS. 


2.  Reduce  42  Ilni.  to  meters. 

3.  Reduce  33  Dm.  to  decimeters. 

4.  Reduce  8  L.  to  milliliters, 
f).  Reduce  1  Kl.  to  liters. 

C.  Reduce  4  G.  to  milligrams. 

7.  Reduce  21  Kg.  to  centigrams. 

5.  Reduce  9.82  M.  to  millimeters. 
9.  Reduce  16  A.  to  centares. 

10.  Reduce  25  sq.  M.  to  sq.  millimeters. 

11.  Reduce  14  cu.  M.  to  cu.  millimeters. 

12.  Reduce  3  cu.  dm.  to  cu.  centimeters. 

13.  Reduce  22  Ha.  to  centares. 

14.  Reduce  54  Ds.  to  decisters. 

15.  Reduce  1  millier  or  tonneau  to  milligrams. 


Ans.  4200  M. 

Ans.  3300  dm. 

Ans.  8000  ml. 

Ans.  1000  L. 

Ans.  4000  mg. 

Ans,  2100000  eg. 

Ans.  9820  mm. 

Ans.  ICOO  ca. 

Ans.  25000000  sq.  mm. 

Ans.  14000000000  cu.  mm. 

Ans.  3000  cu.  cm. 

Ans.  220000  ca. 

Ans.  5400  ds. 

Ans.  1000000000  mg. 


TO  REDUCE  METRIC  NUMBERS  FROM  LOWER  UNITS  OR  DENOMINA- 
TIONS  TO   HIGHER. 


1.    Reduce  44505  mm.  to  meters. 


OPERATION. 

I  44505 
10 
10 
10 


I  44.505  M.     Ans. 
reason  thus :  Since  1000  mm.  =  1  M.  there  are  yuu 


ExpJnnalion. — Here  again  remembering  the 
Metric  Linear  Table,  we  reason  as  follows: 
Since  10  mm.  =  1  cm.  there  are  -r^j  as  many  cm. 
as  mm. ;  then  since  10  cm.  =  1  dm.  there  are  V,f 
as  many  dm.  as  cm. ;  then  since  10  dm.  =  1  M. 
there  are  -]V  as  many  M.  as  dm  .  which  is,  as 
shown  by  the  operation,  44.505  M.  Or  we  may 
i  as  many  M.  as  mm. 


GENERAL  DIRECTIONS. 

From  the  foregoing  elucidation,  we  derive  the  following  general  direc- 
tions to  Reduce  Metric  Numbers  from  Lower  to  Higher  denominations  : 

For  the  Linear,  Capacity,  and  Weight  Measures,   divide   by  10   (point  off  1 
placej  for  each  higher  denomination  to  which  the  given  number  is  to  be  reduced. 


93^  soule's  riiiLosoPHic  practical  mathematics.  * 

For  Surface  Measure  point  off  2  places  instead  of  1. 

And  for  CtihiG  Measure lioint  off  3 places  instead  of  1,  for  each  higher  denom- 
ination to  ichich  the  given  number  is  to  he  reduced. 

Note. — In  reducing  the  ileuominatinns  of  the  Stere,  in  wood  measure,  10  is  the  divisor. 

PROBLEMS. 

2.  Reduce  48753  cm.  to  meters.  Ans.  487.52  M. 

3.  Reduce  12307  dm.  to  Km.  Ans.  1.2307  Km. 

4.  Reduce  120  M.  to  Ilm.  Ans.  1.2  Hm. 
6.  Reduce  333444  cl.  to  liters.  Ans.  3334.44  L. 

6.  Reduce  10234  dl.  to  Kl.  Aus.  1.0234  Kl. 

7.  Reduce  24  ml.  to  ]\I1.  Ans.  .0000024  Ml. 

8.  Reduce  484500  mg.  to  G.  Ans.  4S4.5  G. 

9.  Reduce  2389  eg.  to  kilograms.  ■  Ans.  ,02389  Kg. 

10.  Reduce  10  G.  to  Mg.  Ans.  .001  Mg. 

11.  Reduce  1  mg.  to  tonne.aus.  Ans.  ,000000001  T. 

12.  Reduce  34148C4  sq.  mm.  to  sq.  M,  Ans.  3.414864  sq.  M. 

13.  Reduce  8.27  sq.  cm.  to  sq.  dm.  Ans.  ,0827  sq,  dm. 

14.  Reduce  655444333G6  cu.  ram.  to  cu.  M.  Ans.  55.544433366  cu.  M. 

15.  Reduce  897508  cu.  dm.  to  steres.  Ans.  897.508  S. 

16.  Reduce  2245  cu.  M.,  or  Steres,  to  dekasteres.  Ans.  224.5  Ds. 

17.  Reduce  447  ds.  to  dekasteres.  Ans.  4.47  Ds. 

TO  ADD  METRIC  NUMBERS. 

1.     What  is  the  sum  of  462  mm.  28  cm.  406  dm.  and  16  Dm. 

Ans.  201.342  M.,  or  201342  mm. 

Explanation. — In  all  problems  of  this  kind, 
the  numbers  should  be  written  in  the  base  unit 
of  the  table  and  added  as  in  decimals.  Accor- 
dingly, as  shown  in  the  first  operation,  we  wrote 
the  numbers  in  meters  and  decimals  of  meters 
201.342  M.     Ans.  201342  mm.  and  then  added. 

In  the  second  operation,  we  wrote  and  added  the  numbers  as  mm. 

GENERAL  DIRECTIONS. 

From  the  foregoing  elucidation,  we  derive  the  following  general  direc- 
tions for  Adding  Metric  Numbers  : 

Write  the  mmihers  to  he  added  in  the  BASE  UNIT,  and  decimals  of  the  same,  of 
the  table  to  ichich  they  belong  and  then  add  as  in  decimals.     Art.  445,  page  223. 

PROBLEMS. 

2.  Add  08  M.  42  dm.  3204  mm.  and  03  Hm.  Ans.  6375.404  M. 

3.  Add  7  Mm.  2  Km.  5  Hm.  7  Dm.  8  M.  3  dm.  4  cm.  and  8.07  mm. 

Ans.  72578.34807  M. 


FIRST   OPERATION. 

SECOND   OPERATION 

.462 

462 

.28 

280 

40.6 

40600 

160. 

160000 

*  METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES.  937 

4.  Find  the  sum  of  21  L.  16  dl.  4  cl.  and  27  HI.  Aiis.  2722.04  L. 

5.  A  grocer  has  four  boxes  containing  as  follows  :  1st,  8.5  L. ;  2d,  7  Dl. ;  3d, 
21  cl. ;  4th,  1  M.  and  8  ml.    How  many  L.  in  all  f  Ans.  79.718  L. 

6.  What  is  the  sum  of  12  G.  3  dg.  9  eg.  and  6  mg  ?  Ans.  12.396  G. 

7.  What  is  the  weight  in  Kg.  of  3  bales  of  cotton  which  weigh  respectively: 
204.6  Kg. ;  205  Kg.  4  Hg. ;  and  208  Kg.  8  G.  ?  Ans.  618.008  Kg, 

8.  Add  in  Kg.,  14  T.  6  Q.  8  Mg.  7  Kg.  4  Hg.  5  Dg.  9  G. 

Ans.  14687.459  Kg. 

9.  Add  the  above  problem  in  the  unit  of  Tonneans.  Ans.  14.687459  T. 

10.  What  is  the  sum  of  124  sq.  M.,  6  sq.  dm.,  and  37  sq.  cm.  ? 

Ans,  124.0637  sq,  M. 

OPERATION   INDICATED. 
124. 
.06 

.0037 

11.  Add  42,8  sq,  M.,  21.65  sq.  M.,  28  sq.  dm.,  and  4  sq.  cm. 

Ans.  64.7304  sq.  M. 

12.  Add  in  the  unit  of  Ares,  39.5  A.,  25  Ha.,  and  84  ca.        Ans.  2540.34  A. 

13.  What  is  the  sum  of  14.5  cu.  M.,  23  cu.  dm.,  123  cu.  cm.,  and  24  cu.  mm.  f 

Ans.  14.523123024  cu,  M, 

OPEKATION   INDICATED. 

14.5 
.023 

.000123 
.000000024 

14.  Add  in  the  unit  of  Steves,  22  S.,  12  Ds.,  and  15  ds,  Ans.  143.5  S. 

TO  SUBTRACT  METRIC  NUMBERS. 

1.    What  is  the  difference  between  75.6  M.  and  85  mm. ! 

Ans.  75.515  M. 

OPERATION. 

75.6  =  M.  Explanation.- — In  snhtractioii    .ns   in   addition, 

.085  =  M.  we  write  the  numbers  to  be  subtracted  in  the 

base  unit   of   the   measnre   table   to   which    the 

75.515  M.    Ans.  imnibers  belong,  and  subtract  as  in  decimals. 

GENERAL  DIRECTIONS. 

'     From  the  foregoing  elucidation,  we  derive  the  following  general  direc. 
tions  for  Subtracting  Metric  Numbers : 

Write  the  numbers  to  he  suhtraeted  m  the  base  unit  and  decimals  thereof,  of  the 
table  to  which  the  numbers  belong,  and  then  subtract  as  in  decimals.  Art,  447,  page 
224. 

PROBLEMS. 

2.     From  4324.08  Km.  take  123.5  M.  in  the  unit  of  M.  and  also  of  Km. 

Ans.  4323956.5  M.     4323.9565  Km. 


938  soule's  philosophic  practical  mathematics.  * 

3.  Find  the  difference  between  274.25  L.  and  44.5  el.  Ans.  273.805  L. 

4.  A  barrel  contained  151.44  L.,  and  6  L.  and  7  dl.  leaked  out.     How  many 
liters  still  remain  1  Ans.  144.74  L. 

5.  What  is  the  difference  in  L.  between  1  Ml.  and  1  ml.  1 

Ans.  9999.999  L. 

6.  From  2G4.5  G.  take  2S.4  eg.  Ans.  264.216  G. 

7.  From  1428  Kg.  take  16.5  Tig.  Ans.  1426.35  Kg. 
S.     What  is  the  difference  in  T.  and  Kg.  between  2  tonneaus  and  2  Kg.? 

Ans.  1.998  T.;  1998  Kg. 
9.     What  is  the  difference  between  88.21  sq.  M.  and  38  sq.  dm.  ? 

Ans.  87.83  sq.  M. 

10.  A  plantation  contains  2471.14  Ha.     If  2471.14   A.  are  sold,  how  many 
hectares  will  remain  ?  Ans.  2446.4286  Ha. 

11.  From  1528  cu.  M.  take  1  cu.  M.  1  cu.  dm.  1  en.  cm.  and  1  cu.  mm. 

Ans.  1526.998998999  en.  JI. 

12.  A  wood  dealer  bought  150  steres  of  wood;   he  sold  70.25  cu.  M.     How 
much  has  he  remaining?  Ans.  79.75  S.,  or  cu.  M. 

TO  MULTIPLY  METRIC  NUIVIBERS. 

1.     How  many  meters  are  there  in  11  pieces  of  cloth,  each  containing 
36.25  meters?  Ans.  398.75  M. 

OPERATION. 

36.25  Explanation. — In   niultiplioation    of   Metric 

11  Numbers,  we  proceed  in  the  laiiie  manner  as  in 

simple  and  decimal  numbers.     Hence  no  extend- 

398.75  M.     Ans.  ed  elucidation  and  general  directions  are  deemed 

necessary. 

2.  What  will  204.5  meters  cost  at  $2  per  meter  ?  Ans.  $409. 

3.  If  a  man  walks  20000  meters  per  day,  how  many  kilometers  will  he  walk 
in  60  days  ?  Ans.  1200  Km. 

4.  What  will  484  dm.  cost  at  $1.50  per  meter?  Ans.  $72.60. 

5.  What  will  484  dm.  cost  at  $1.50  per  dm?  Ans.  $726. 

6.  AVhat  cost  75.2  liters  of  milk  at  5,(r  per  L.  ?  An.s.  $3.76. 

7.  What  cost  200.52  HI.  of  corn  at  $1.60  per  HI.  ?  Ans.  $320.83. 

8.  A  fruit  dealer  bought  6  HI.  of  pecans,  at  $9  per  HI.,  and  sold  them  at  15/ 
per  liter.  How  much  did  he  gain  ?  Ans.  $36. 

9.  Sold  26.4  Kg.  of  grapes  at  45/  per  kilo.    How  much  was  received  for 
them?  Ans.  $11.88, 

10.  What  cost  27.25  tonneaus  of  hay  at  $20.50  per  T.  ?  Ans.  $558,625. 

11.  What  cost  416.3  Kg.  of  sugar  at  20/?  Ans.  $83.26. 

12.  Bought  362.3122  G.  of  silver  at  3 J/  per  gram.    What  did  it  costt 

Ans,  $12.08+, 

13.  How  many  kilograms  in  49000  pills,  the  weight  of  each  being  .5  dg.  ? 

Ans.  2.45  Kg. 


*  METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES.  939 

14.  How  many  sq.  meters  in  a  yard  40.5  M.  long  and  15.24  M.  widet 

Ans.  617.22  sq.  M. 

15.  A  plantation  is  2.42  Km.  long  and  1500  M.  wide.  How  many  hectares 
does  it  contain  ?  Ans.  3G3  Ha. 

10.  How  many  cubic  meters  in  a  box  2.8  M.  long,  2.1  M.  wide,  and  8.5  dm. 
deep  ?  Ans.  4.998  cu.  M. 

17.  What  will  be  the  freight  on  4  boxes,  each  measuring  3  M.  by  2.6  M.  by 
.9  M.  at  $5.25  per  cu.  M.  ?  Ans.  $147.42, 

IS.  How  many  cubic  meters  of  earth  in  a  levee  140  M.  long,  2.3  M.  deep,  and 
20  M.  wide  at  the  base  and  15.2  M.  wide  at  the  top  ?  Ans.  5C67.2  cu.  M. 

19.  How  many  steres  in  a  pile  of  wood  28.5  M.  long,  3.2  M.  high,  and  4.3 
M.  wide  1  Ans.  392.16  S.,  or  cu.  M. 

20.  How  many  cubic  meters  of  earth  will  it  take  to  fill  a  lot  .5  meter  deep, 
the  lot  being  60.2  meters  long,  and  25  meters  wide  ?  Ans.  752.5  cu.  M. 

TO  DIVIDE  METRIC  NUMBERS. 

1.    A  man  Malted  1600  meters  in  20  minutes.    How  many  meters  did 
he  walk  in  1  minute  ?  Ans.  80  M. 

OPERATIOX. 

1600  —  20  =  80  M.     Ans. 

Explanation. — In  division  of  Metric  Numbers,  we  proceed  witli  the  operations  in  the  same 
manner  as  in  simple  and  decimal  numbers.  Hence  no  extended  elucidation  and  general  directions 
are  deemed  necessary. 

2.  The  steamer  Katie  ran  500  Km.  in  22  hours.  How  many  kilometers  did 
she  run  per  hour?  Ans.  22.727+  Km. 

3.  If  6.5  meters  make  a  suit,  how  many  suits  can  be  made  from  195  meters? 

Ans.  30  suits. 

4.  In  425  liters,  how  many  hectoliters?  Ans.  4.25  HI. 

5.  Bought  45  liters  of  strawberries,  for  $3,375.     What  was  the  price  per  L.  ? 

Ans.  7  J/. 

6.  If  you  divide  ISO  hectoliters  of  potatoes  equally  among  24  persons,  what 
will  each  receive?  Ans.  7.5  HI. 

7.  21  kilograms  of  sugar  cost  $4.62,     What  was  the  cost  of  1  Kg.  ? 

Ans.  22/.     ■ 

8.  How  many  days  will  42.5  T.  of  coal  last  a  family,  if  they  burn  150  Kg- 
per  day?  Ans.  283.33+  days. 

9.  A  druggist  has  2.45  Kg.  of  medicine  which  he  wishes  to  make  into  pills, 
each  to  contain  .5  dg.    How  many  pills  will  there  be  ?  Ans.  49000  pills. 

10.  A  garden  contain,s  900  sq.  M.,  and  is  22.5  M.  wide.     How  long  is  it  ? 

Ans.  40  M. 

11.  A  side- walk  is  80.4  M.  long  by  4.2  M.  wide.  How  many  tiles,  each  2.4 
dm.  long  and  1.2  dm.  wide,  will  be  required  to  pave  the  side- walk? 

Ans,  11725  tiles. 


940 


soule's  philosophic  practical  mathematics. 


12.  IIow  many  ares  in  a  piece  of  land  60  meters  long  and  42.2  meters  wideT 

Ans.  25.32  A. 

13.  A  box  is  2.5  dm.  lonjr,  2  dm.  wide,  and  1.5  dm.  deep.     How  many  of  such 
boxes  may  be  jmt  in  a  larger  box  -wliich  is  2.5  M.  long,  2  M.  wide,  and  1.5  M.  deep  f 

Ans.  1000  boxes. 

14.  If  yon  buy  5.0  dekasteres  of  ■wood  and  use  1.1  cu.  meters  per  day,  how 
long  will  it  last  ?  Ans.  50j^  days. 

TO  KEDUCE  METRIC  TO  AMERICAN  WEIGHTS  AND  IMEASURES. 


1.    Reduce  2.6  meters  to  feet. 

FIRST   OPERATION. 

1  M.  =  39.37+  inches,  practically* 
2.6 


12  )  102.362  inches. 

8.530+  feet.     AnSc, 


or  thus: 


I  39.37 
12    2.6 


Ans.  8.53+  feet. 

Explanation. — Referring  to  the  table  of  equiv- 
alents, -we  find  that  1  M.  =  (practically)  39.37 
in.  We  then  reason  as  follows:  Since  1  M.  = 
39.37  in.,  2.6  M.  are  equal  to  2.6  times  as  many, 
-which  is  102.362  in. ;  then  since  12  in.  =  1  ft., 
there  are  -jV  as  many  feet  as  inches,  ■which  is 
8.53+  feet. 


SECOND   OPERATION. 

3.2808+  =  feet,  the  equivalent  of  a  meter. 
2.6 


8.53008+  =  feet,     Ans. 


GENERAL  DIRECTIONS. 


From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 
tions for  Reducing  Metric  to  American  Weights  and  Measures : 

Multiply  the  equivalent  vahie  of  the  metric  unit  given,  hy  the  metric  number  to  he 
reduced,  and  then,  if  necessary,  reduce  the  product  to  the  denomination  required. 

Note. — In  -working  the  follo-ning  prnlilpms,  four  flecimal  places  have  been  taken  as  a  stand- 
ard for  all  unit  equivalents  of  more  than  that  many  decimals.  The  calculations  are  made  directly 
upon  the  equivalency  of  the  denominations  to  be  reduced. 


2.  Reduce  408.2  kilometers  to  yards. 

3.  In  24.5  cm.  how  many  inches  ? 

4.  Reduce  70  M.  to  yards. 

5.  Reduce  25.5  liters  to  dry  quarts. 

0.  In  40  kiloliters  how  many  gallons  ? 

7.  In  200  hectoliters  how  many  bushels? 

8.  Reduce  25  grams  to  Troy  grains. 

9.  Reduce  75.2  kilograms  to  pounds  Avoirdupois. 

10.  In  444  tonneaus  how  numy  tons? 

11.  Reduce  5.5  mg.  to  Troy  grains. 


Ans.  446407.52  yds. 

Ans.  9.6456+  in. 

Ans.  76.552+  yds. 

Ans.  23.1565+d.  qts. 

Ans.  10507.44+  gals. 

Ans.  567.58+  bus. 

Ans.  385.81+  grs. 

Ans.  165.7859+  lbs. 

Ans.  489.4256+  tons. 

Aus.  .0847+  grs. 


METRIC    SYSTEM    OF    WEIGHTS    AND    MEASURES. 


941 


12.  Reduce  24  arcs  to  square  rods. 

13.  Eeduce  85.5  hectares  to  acres. 

14.  lu  500  hectares,  lio-sv  many  sq.  miles  ? 

15.  Eeduce  150  cu.  meters  to  cu.  feet. 

16.  Eeduce  7.9  cu.  mm.  to  cu.  inches. 


Aus.  94.8912+  sq.  rds. 

Ans.  211.279+  acres. 

Ans,  1.9305+  sq.  mi. 
Ans.  5297.475+  cu.  ft- 
Ans.  .0004819+  cu.  in. 


TO  EEDUCE  AMEEIGAN  TO  METEIG  WEIGHTS  AND  MEASURES. 


1.    Eeduce  40  feet  to  meters. 

OPERATION. 

40  ft. 
12 
39.37)  480  in.   (12.192+ M. 


or  thus : 


39.37 


40 
12 


Aus.  12.192+  M. 


Explanation. — Eemembering  tliat  a  meter  = 
39.37  inches,  practically,  we  first  reduce  the 
feet  to  inches,  and  then  reason  as  follows :  Since 
in  39.37  in.  there  is  1  meter,  in  480  in.  there  are 
as  many  meters  as  480  in,  are  times  =  to  39.37 
in.,  which  is  12.192+. 


SECOND    OPERATION. 

3.048     decimeters  =  the  equivalent  of  1  foot. 
40 


121.920   decimeters,  which  divided  by  10  =  12.192  meters,     Ads. 

2.     Eeduce  1  mile,  8  rods,  10  feet,  and  6  inches  to  kilometers. 

Ans.  1.652781+  Km. 

OPERATION. 


Imi.,  8  rds.,  10  ft.,  6  in. 
320 


328   rods. 
16i 


39.37)65070  inches. 

1652.781+  M. 
1.652781+  Km. 


5422    feet. 
12 


65070  in.  x  2.54  cm.  =  165278.1+  cm.  =  1.652781+  Km. 


65070    inches. 


Explanation. — In  this  prohlem,  we  first  reduce  the  given  number  to  inches,  then  to  meters,  as 
above  explained ;  and  then  to  kilometers  by  dividing  by  1000,  the  number  of  meters  in  a  kilometer. 

GE]SrEEAL   DIRECTIONS. 


^  From  the  foregoing  elucidations,  we  derive  the  following  general  direc- 

tions for  reducing  American  to  Metric  Weights  and  Measures : 

1°.  Eeduce  the  given  nnmher  to  the  lowest  nnlt  named,  or  to  a  convenient 
denomination,  of  which  the  equivalent  value  of  the  metric  unit  is  known  ;  then  divide 
the  same  by  the  equivalent  value  of  the  metric  unit,  and  when  necessary  reduce  the 
quotient  to  the  required  denomination. 

Or  2°.  Multiply  the  given  number,  reduced  to  its  lowest  denomination,  by  the 
equivalent  value  of  such  denomination  in  the  metric  unit. 


942  SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

3.     Ill  70  yards,  how  jnaiiy  meters?  Aus.  C4.00S+  M. 

i.     lieduee  35  {gallons,  3  quarts,  and  1  pint  of  molasses  to  liters. 

Alls.  135.8001+  liters. 

5.  Keduce  5  bushels,  2  pecks,  5  quarts,  and  1  pint  to  liters, 

Aus.  109.8678+  liters. 

6.  Ill  20  gallons  of  milk,  how  many  liters?  Aus.    7.'5.70-l-|-  L. 

7.  Ill  400  bushels,  how  many  hectoliters?  Ans.  140.9493— III. 
S.  In  3  gills,  how  many  centiliters?  Ans.  35.486+  cl. 
9.     Eeduce  200  pounds  Avoirdupois  to  kilograms.            Ans.  90.7194+  Kg. 

10.  Eeduce  15  ounces  Avoirdupois  to  hectograms.  Ans.  4.2525+  Hg. 

11.  Eeduce  10  ounces  Troy  to  grams.  Ans.  311.0339+  G. 

12.  Eeduce  50  tons  to  touneaus.  Ans.  45.3597+  T. 

13.  Eeduce  20  grains  Troy  to  centigrams.  Aus.  129.6176+  eg. 

14.  Eeduce  240  sq.  yards  to  sq.  meters.  Ans.  200.G689—  sq.  M. 

15.  In  500  sq.  feet,  how  many  sq.  decimeters?  Ans.  4646.8401+  sq.  dm. 

16.  In  200  acres,  bow  many  hectares  ?  Ans.  80.9356  Hectares. 

17.  Eeduce  1610  cu.  feet  to  cu.  meters.               •  Ans.  45.5877+  cu.  M. 

18.  Eeduce  2500  cu.  yards  to  cu.  meters.  Ans.  1911.3149+  cu.  M. 

19.  In  190  cords,  how  many  steres  ?  Ans.  688.6553+  Steres. 

MISCELLANEOUS  PEOBLEMS. 

1.     A  yard  is  21.4  M.  long  and  15.12  M.  wide.     How  many  bricks  will 
it  require  to  pave  the  yard,  each  brick  being  10.32  cm.  long,  and  9.16  cm.  wide? 

2.  How  many  cubic  meters  in  a  piece  of  marble  3.5  M.  long,  3.21  M.  wide, 
and  1.S5  M.  thick  ? 

3.  How  many  liters  in  a  cylindrical  vessel  2.3  M.  deep  and  8.2  dc.  in  diame- 

4.  What  is  the  weight  of  22  bags  of  wheat  averaging  68.42  Kg.  ? 

5.  If  4  men  eat  9.5  Kg.  a  week,  how  much  would  25  men  eat  at  the  same 
rate  ? 

6.  If  4  men  can  dig  245  M.  of  a  ditch  in  2^  days,  how  long  will  it  take  10 
men  to  dig  3200  M.  at  the  same  rate  ? 

7.  A  room  is  6.5  M.  long  and  4.25  M.  wide.  How  many  square  meters  of 
carpet  will  be  required  to  carpet  the  floor,  the  carpet  being  75  cm.  wide,  no  allow- 
ance being  made  lor  matching  patterns  or  in  turning  under? 

S.  A  room  is  8.4  M.  long,  6.25  M.  wide,  and  4.5  M.  high.  How  many  liters  of 
air  does  it  contain  ? 

9.  If  the  breathing  of  one  person  pollutes  the  air  of  a  room  at  the  rate  of 
.1855  cbm.  per  minute,  how  long  will  it  take,  at  the  same  rate,  for  50  jiersons  to 
pollute  the  air  of  a  room  30  M.  long,  20  ]M.  wide,  and  8.25  M.  high  ? 

10.  If  42  Kg.  of  cotton  are  required  to  make  150  M.  of  cloth  .75  M.  wide, 
how  many  kilometers  of  cotton  will  it  take  to  make  5000  M.  of  cloth  1.2  M.  wide? 

11.  How  many  cubic  meters  in  a  cylindrical  piece  of  marble  2.1  M.  in 
diameter  and  6.42  M.  long? 

12.  How  many  square  meters,  board  measure,  in  24  pieces  of  timber  9.32 
M.  long,  48  cm.  wide,  and  32  cm.  thick? 


ter? 


combinations  or  Choice. 


^ 


Combinations  are  the  different  groups  which  can  be  made  of  n  things 
taken  m  in  a  group,  m  being  equal  to  or  less  than  n. 

Or  Combinations  are  the  various  ways  that  different  things,  the  number  of 
■which  is  known,  can  be  chosen  or  selected  taking  them  one  by  one,  two  by  two, 
three  by  three,  etc.,  without  regard  to  their  order. 

PKOPOSITION. 

Any  number  of  things  being  given,  to  determine  in  how  many  ways 
they  may  be  combined  two  and  two,  three  and  three,  etc.,  without  regard  to  the 
order. 

PROBLEMS. 

1.  If  there  are  four  routes  1,  2,  3,  i  leading  to  New  York,  in  how  many 
diff'erent  ways  can  a  person  go  and  return  ?  Ans.  16  ways. 

Explanation. — Since  a  person  has  the  choice  of  going  by  any  route  and  returning  by  the  same 
or  another  route,  four  choices  are  oflered  to  him,  and  since  he  has  the  same  number  of  choices  with 
each  route  (there  being  four  of  them)  he  must  therefore  have  the  choice  of  16  ways,  viz.: 

Go.      Eeturn.      Go.       Return.      Go.       Return.      Go.       Return. 
11213141 
12223242 
13233343 
14243444 


4  +4  +4  +4=  16 

2.  Suppose  in  the  above  problem,  the  tourist  does  not  desire  to  return  by  the 
same  route,  how  many  combinations  or  choices  of  route  has  he  ?  Ans.  12. 

Explanation. — If  he  goes  by  any  route,  then  he  has  the  choice  of  returning  by  three  others 
or  three  ways  of  making  the  round  trip  for  each  ditt'erent  route  leading  to  New  York.  Since  there 
.are  four  routes  leading  there,  he  has  3  x  4  =  12  choices  iu  the  matter,  viz. : 

Return.  Go.      Return.  Go.      Return. 

1                   (1  (1 

3  3(2  4^         2 

__t                 (    _i  (  3 

3+  3+  3+  3  =  12 

From  the  above  the  following  principle  is  evolved : 

If  one  thing  can' be  done  in  a  different  ways,  and  (when  it  has  been  done  in 
any  one  of  these  ways)  another  thing  can  be  done  in  b  different  ways,  then  both  can 
be  done  in  a  x  b  different  ways. 

(943) 


Go.      Eetura. 

Go. 

\      2 

1.          3 

944  souLE  s  rniLosoPHic  practical  mathematics.  * 

3.  In  how  many  ways  can  tlie  letters  from  the  word  marble  be  taken  Avith 
the  letters  from  the  ■word  home  f  Ans.  24  ways. 

Explanation. — Each  Jotter  of  the  word  marlile  (there  are  6)  can  be  comhineil  with  each  letter 
of  the  wold  home  (there  are  4).  Therefore  we  have  6  x  4  =  24  ways  they  may  be  taken.  Exteiul- 
iuR  the  j)rece(liii};  principle,  we  tlerive  the  follo\vin<;:  If  one  thing  can  be  done  iu  a  ways,  and  theu 
a.KecoiKl  thiuf;  can  be  <l()ne  in  h  ways,  and  a  third  in  c  ways,  and  a  fourth  in  d  ways,  etc.,  the 
number  of  ways  of  doing  all  the  things  will  he  a  X  b  X  c  X  d,  etc. 

4.  lu  how  many  ways  can  5  books  be  given  to  five  persons  ?  Ans.  120. 

Explanation. — Since  the  first  book  may  be  given  to  any  one  of  the  five  persons,  there  are  five 
ways  of  bestowing  it. 

The  second  book  may  he  given  to  .any  one  of  four  persons;  the  third  book  to  any  one  of 
three  persons ;  the  fourth  book  to  any  one  of  two  ])er8ons,  and  the  last  book  to  the  fifth  person. 
Hence  the  books  may  be  bestowed  in  one  of  5x4x3x2x1  =  120  ways. 

5.  In  how  many  ways  can  a  vowel  and  a  consonant  be  chosen  out  of  the 
word  logarithms  ?     Work  according  to  first  principle.  Ans.  21  ways. 

6.  In  how  many  ways  can  4  dolls  be  given  to  3  girls  ?  Ans.  81  ways. 

Explanation. — Each  doll  can  be  given  to  any  one  of  the  three  girls.  Therefore  the  number  of 
■ways  of  disposing  of  them  is  3x3x3x3  =  81  ways. 


PEHMUTATIONS. 


In  the  definition   of  combinations,   Ave   found  that   the  order   of  the 
elements  in  combination  was  disregarded. 

A  rermutation  of  any  number  of  elements  or  things  means  a  group  of  thafc 
■number  of  elements  or  things  put  together  icith  reference  to  their  order  or  sequence. 

PROBLEMS. 

1.  In  how  many  ways  can  the  letters  in  the  word  hound  be  arranged,  taken 
all  at  a  time  Ans.  120  ways. 

Explanation. — The  reason  given  in  problem   4   in   combinations   .applies   here,  and  hence  is^ 
omitted. 

Formula  1.     The  jiumber  of  permutations  of  n  diifferent  elements  taken  all 
at  a  time  is 

n  (n  —  ])  (n  — 2) x  2  x  1. 

The  continued  product  of  this  formula  is  the  whole  number  of  permutation » 
of  n  elements  taken  all  at  a  time. 

For  convenience  this  formula  may  be  written    |  n  and  is  read  factorial  n. 

2.  In  how  many  ways  can  10  different  colored  lights  be  arranged,  taken  3  at 
a  time?  Ans.  720  ways. 

Formula  2.     n  (n  —  1)  (n  —  2) to  r  factors  or  n  (u  —  1)  (n  —  2) 

(n  —  r  +  1).     Substituting  the  figures  for  the  letters  we  have  10  (10  —  1)  (10  —  S 
+  I)  =  10x9x8  =  720,   Ans. 


*  PROBABILITIES    OR    CHANCE.  945 

In  the  formula,  ?i  =  number  of  things  given  to  be  arranged  and  r  =  number 
of  tilings  taken  at  a  time. 

Formulas.     The  number  of  arrangements  of  7i   elements,   of  which  j>   are 
alike,  q  others  are  alike,  and  r  others  are  alike  etc.,  is 

I  n 


P  X    \  q  y. 


3.     How  many  permutations  can  be  made  out  of  the  letters  of  the  word 
tintinnabulation  ?  Ans.  12,108,096,000. 

OPERATION. 

I  16 


|4    |3    13    |3 


4.     How  many  arrangements  or  permutations  can  be  made  out  of  8  books,  3 
being  Arithmetics,  2  being  Histories,  2  being  Grammars,  and  1  being  a  Dictionary  ? 

Ans.  1680. 

OPERATION. 

LFTfli 

Eor  other  and  higher  cases  of  permutations,   the  learner  is  referred  to 
Wentworth's  Complete  Algebra. 


PROBABILITIES  OR  CHANCE. 


This  subject  has  engaged  the  attention  of  writers  on  mathematics  since 
Galileo  first  wrote  on  it  in  1642.  Many  mathematicians  lioweTer,  were  prejudiced 
against  it  on  account  of  its  ready  applicability  to  games  of  chance  with  cards, 
dice,  etc. 

When  an  event  can  happen  in  a  certain  number  of  ways,  and  it  fails  in  a 
certain  other  number  of  ways,  the  chance  of  its  happening  can  be  expressed  by  a 
fraction  whose  numerator  expresses  the  number  of  chances  favorable  to  the  event 
happening  and  whose  denominator  expresses  the  whole  number  of  cases. 

Simpson  defines  the  probability  of  an  event  happening  to  be  the  ratio  of  the 
chances  by  which  the  event  in  question  can  happen  to  all  the  chances  by  which  it 
cannot  happen  or  fail.  If  in  +  n  be  the  whole  number  of  cases  that  can  happen,  and 
m  represents  the  number  of  cases  favorable  to  the  happening  of  the  event,  the 
probability  of  the  event  happening  would  be  expressed  by  the  fraction  — -— ;  and 


94^  soule's  philosophic  practical  mathematics.  * 

the  chances  against  its  happeniner  would  be  — - — .    Therefore  we  see  that  the  sum 

lie  »tt  -)-  »t 

of  the  chances  of  an  event  happening  or  not  happening  is  unity  or  certainty. 

In  tossing  a  coin  once,  what  chance  is  there  that  it  will  turn  up  a  head  or  a 
tail !  Since  there  are  only  two  things  possible  to  turn  up,  either  a  head  or  a  tail,  it 
js  evident  that  the  probability  of  its  being  a  head  is  ^         =  J  and  the  probability 

that  it  will  be  a  tail  is  =  J.     The  sum  of  the  chances  for  and  against  is  i  +  .J 

=  1  =  certainty.    That  is,  it  is  certain  that  one  or  the  other  will  turn  up. 

What  chance  is  thei-e  in  two  throws  of  turning  a  head  wyi  twice? 

According  to  the  subject  of  combinations  or  choice,  we  find  that  with  2 
throws  there  will  be  4  combinations  possible. 

1st  explanation  for  Problem  1,  in  Combinations,  is  ai>plicable,  viz: 

1  head  and  head. 

2  head  and  tail. 

3  tail  and  head. 

4  tail  and  tail. 

Of  the  four  ways  in  which  it  is  possible  for  the  coins  to  fall,  there  is  only  1 
Way  in  which  the  event  will  happen.  The  probability  of  the  event  happening  is 
hence  ^.  It  might  be  shown  that  the  chance  of  throwing  3  heads  or  3  tails  in 
succession,  will  be  J,  or  J  x  J  X  J.  The  chance  of  two  or  more  single  events  being 
known,  the  chance  of  their  all  taking  place  together,  may  be  found  by  multiplying 
the  probabilities  of  the  events,  considered  singly. 

To  find  the  chance  of  throwing  in  one  throw  all  heads  or  all  tails  with  2,  3,  or 
4  pieces,  may  be  found  as  above.  When  two  pieces  are  tossed  up  there  are  4  com- 
binations possible,  hence  the  chance  is  J.  When  three  pieces  are  thrown  up  there 
are  8  combinations,  for  each  coin  may  fall  in  one  of  two  ways,  and  there  being  3 
coins  there  are  2x2x2  =  8  combinations  possible.  The  chance  is  therefore  ^. 
Tossing  up  three  pieces  at  once,  is  just  the  same  as  tossing  up  the  same  piece  3 
times. 

1.  What  odds  should  be  offered  to  make  an  even  bet  that  in  2  throws  a 
bead  will  fall  at  least  once?  Ans.  3  to  1. 

As  we  have  seen  before,  in  two  throws  there  is  only  1  throw  in  which  we  find 
BO  head,  while  we  have  a  head  in  three  of  them. 

2.  What  chance  is  there  in  one  throw  of  throwing  an  ace  with  one  die  f 

Ans.  I 

3.  What  chance  is  there  in  one  throw  of  throwing  double  aces  with  two 
dice!  Ans.  aJg.   . 


PROBABILITIES    OR    CHANCE. 


947 


Table  to   snow  how  many  ways  a  given  sum  may   be  thrown 

WITH    ONE,    TWO,    THREE,    OR   MORE   DICE. 


Explanation  of  TalU.—li  it  be  required  to  find  how 
many  ways  8  could  be  thrown  with  4  dice,  we  first  look 
in  the  1st  column  on  the  left  for  8,  and  at  the  top  of 
the  table  for  4,  and  in  the  square  below  opposite  to  the 
8  we  find  35,  the  number  of  ways  in  which  8  may  be 
thrown  with  4  dice.  The  probability  of  throwing  8 
points  with  4  dice  is  y^f^.  The  probability  of  throwing 
any  number  of  points  with  a  given  number  of  dice  is 
found  by  dividing  the  number  of  ways  the  points  cau 
be  thrown  with  the  given  number  of  dice  by  that 
power  of  6  indicated  by  the  number  of  dice. 


Number  oC  dice. 
12       3       4         5       fi 

03 

o 
u 
.a 

a 
;5 

1 

2 

3 

4 

5 

6 

7 

8 

9 

10 

11 

12 

1 

2 
3 
4 
5 
6 
5 
4 
3 
2 
1 

1 
3 
6 
10 
15 
21 
25 
27 
27 
25 

1 

4 

10 

20 

35 

56 

80 

104 

125 

1 

5 

15 

35 

70 
126 
205 
305 

1 

6 

21 

56 

126 

252 

456 

PROBLEMS 

1.  Two  persons  sit  down  to  play  for  a  certain  snra  of  money;  and  agree 
that  lie  who  first  gets  three  games  shall  be  the  winner.  One  of  them  won  2  games 
and  the  other,  won  1 ;  but  being  unwilling  to  continue  their  play,  they  resolve  to 
divide  the  stake ;  how  much  ought  each  person  to  receive  ? 

If  the  play  were  to  be  continued,  and  the  second  should  win,  then  each  one 
having  only  1  game  to  win  out,  would  be  entitled  to  half  of  the  money. 

But  if  the  first  one  should  win  the  game,  the  whole  money  (|)  deposited 
would  be  his  ;  and  if  he  lost  it,  he  would  have  a  right  to  the  half.  But  since  one 
case  is  as  probable  as  the  other,  the  first  player  has  a  right  to  one-half  of  the  sum 
of  the  two  parts  (|  +  i  =  |)  =  |,  and  the  second  jdayer  only  ^. 

2.  Suppose  the  first  player  needs  one  game  to  win,  and  the  second  three,  how 
should  the  money  be  divided  ? 

If  the  first  i)layer  should  win  one  game,  he  would  be  entitled  to  all  (|)  the 
money;  and  if  he  lost  the  game  he  would  be  entitled  to  |,  according  to  the  case 
above.  But  as  both  are  equally  probable,  the  first  player  should  have  J  of  the  two 
sums  (I  +  I  =  \)  =  I,  and  the  second  player  J.     ■ 

3.  Suppose  the  first  player  needs  two  games  to  be  out,  and  the  second  three, 
how  should  the  money  be  divided  ? 

If  the  first  player  won  the  game,  he  would  be  entitled  to  I  as  shown  in  the 


948  soule's  philosophic  practical  mathematics.  * 

problem  above.  If  the  second  playe;-  won,  each  would  be  entitled  to  J.  But  as 
both  are  equally  probable,  the  first  player  is  entitled  to  i  of  the  sum  of  the  two 
parts  (i  +  i  =  tl)  =  "1 6>  ^n*!  t^®  second  to  -i^g. 

4.  If  4  cards  are  drawn  from  a  pack  of  52  cards,  wliat  is  the  chance  that 
four  kings  will  be  drawn  1 

FIRST     OPERATION. 

OJ  X  o     X X ^  __  270725  =  number  of  ways  four  cards  can  be  drawn  from  tlie  pack. 

1X2X3X4 

BECOND  OPERATION. 

Four  cards  can  be  selected  so  as  to  be  kings  in  4x3x2x1=  24  ways.     Therefore  the 
«hance  of  drawing  four  kings  from  a  pack  is  jy^fji^  or  1  chance  in  11280. 

The  first  operation  depends  upon  the  principle,  that  "Out  of  n  different 
elements,  the  number  of  selections  of  r  elements  is  equal  to  the  number  of  arrange- 
ments of  n  elements  divided  by   |  r."    The  formula  for  which  is 

g  ^  n(»-l)  (n- 2)  ...■■■  (n  —  r  +  1) 
ilL 

The  second  operation  depends  upon  the  principle,  that  "The  number  of 
arrangements  or  permutations  of  n  different  elements  taken  all  at  a  time  is  n  {n  — 
1)  (n  —  2) X  2  X  1."     This  formula  is  represented  by  |  n. 

5.  If  2  letters  are  selected  at  random,  out  of  the  alphabet,  what  is  the 
chance  that  both  will  be  vowels  f  Ans.  1  in  32. 

OPERATION. 

Two  letters  can  be  selected  from  26  in  ^  '     =  325  wavs :   and  2  vowels  can  be  selected 

1X2  •     ' 

5x4 
from  5  vowels  in _  =  10  ways.     Therefore  the  chance  of  drawing  2  vowels  is  ji^  =  -^^  or  1 

chance  in  32. 

6.  In  a  box  of  chess  are  16  pawns,  4  knights,  4  bishops  and  2  queens.  What 
is  the  chance  of  drawing  one  of  each  in  4  draws  ?  Ans.  1  in  700. 

OPERATION. 

4  pieces  of  chess  can  be  selected  from  2fi  pieces  in  "     "^        "^  ~    '^  "     =  358800  ways.     Also 

1X2X3X4  ^ 

1  pawn  can  be  selected  from  16  pawns  in  16  ways ;   1  knight  from  4  knights  in  4  ways ;   1  bishop 
from  4  bishops  in  4  ways,  and  1  queen  from  2  qiicens  in  2  wavs.     Hence  1  pawn,  1  knight,  1  bishop 
ind  1  queen  can  be  drawn  in  16  X  4  X  4  x  2  =  512  ways.     Hence  the  chance  is  ^s%j  or  1  in  700.  ' 


ftontinued  Fractions. 


A  continued  fraction  is  a  fraction  with  one  for  a  nnmerator,  and  whose 
denominator  is  an  integer  plus  a  fraction  with  one  for  a  numerator,  and  whose 
denominator  is  an  integer  plus  a  fraction,  etc. 

Note. — For  a  full  elucidation  of  fractions,  see  pages  164  to  167,  and  216  to  219. 

Continued  fractions  were  first  used  by  Cataldi,  in  1613,  and  by  Lord  Brounker, 
in  1670,  Continued  fractious  are  used  to  express  approximately,  ratios  that  by 
reason  of  the  large  numbers  used,  do  not  carry  with  them  ajipreciated  values. 
They  are  also  used  to  determine  approximations  to  roots  of  surds,  and  are  also 
applied  to  the  solution  of  indeterminate  problems. 

Reduce  If  to  a  continued  fraction. 


3  +  1 

6  +  1 


1  +  i 

Directions. — Divide  both  numerator  and  denominator  by  27  and  we  have 
;  divide  the  numerator  and  the  denominator  of  the  second  fraction  by  4,  and 


3  +  jV' 

1 

we  have  3  +  l ;  divide  the  numerator  and  the  denominator  of  the  third  fraction 

6  +  f 

by  3,  and  we  have  i_ 


3  +  1 


6  +  1 

r+1    Ans.      This  completes  the  work,  as  the  numerator 


of  the  last  fraction  is  unity. 

Ecsduce  the  continued  fraction 
1 


3+1 


6  +  1 

1  -j.  ^  to  a  common  fraction. 

First  Method. 

Reduce  the  last  two  terms  of  the  continued  fraction  ^-qj^  to  a  simple  fraction 
by  multiplying  the  numerator  and  the  denominator  by  3,  adding  to  the  denominator 

(949) 


95o  soule's  philosophic  practical  mathematics.  * 

the  numerator  of  the  last  fraction,  and  we  obtain  J.    Take  this  result  and  the  preceding 
partial  fraction,  and  we  have  t-—  ;   multiply  as  before,  and  we  have  the  simple 

fraction  /^.    Take  this  result  and  the  preceding  partial  fraction,  and  we  have  y 

multiply  as  before,  and  we  have  fj,  the  simple  fraction. 

Second  Method. 

1st    step  =  i  =  1st  approximation. 

2d     step  =  5— -7  =  T9  =  2d  approximation. 

>>  +  7 

3d     step  =      ■        =  -^2  =  3d  approximation. 

ly  -f"  T 

4th   step  =  „,,  .   ■  =  ii  =  original  fraction. 

Explanation. — Write  the  first  term  (i)  of  tlie  continued  fraction  as  the  first  approximate 

fraction.     Write  this  approximate  fraction   ■with   the   next  term   of  the   continued   fraction   and 

reduce  to  a  simple  fraction,  and  we  have  •  =  -1%   second   approximate    fraction.     Write    this 

3  +  ^ 

fraction  with  the  next  term  of  the  continued  fraction  and  yve  have  ;   multiply  this  fraction 

19  +  f 
■by  the  denominator  of  the  second  term  as  above,  and  add  to  hoth  terms  the  corresponding  terms  of 

the  first  approximate  fraction  (J),  and  we  have 1_^  =  -,Jj  the  third  approximate  fraction. 

Write  the  third  approximate  fraction  with  the  last  term  of  the  continued  fraction,  and  we  have 

7 .  multiply  both  terms  liy  3,  and  add  to  the  product  the  corresponding  terms  of  the  second 

23  +  i  ' 

approximate  fraction  (t^),  and  we  have  _: "^        =  H,  the  answer. 

To  illustrate  the  application  of  continued  fractions,  let  us  find  the  approxi- 
mate ratios  of  a  circle  to  a  circumscribed  square.  The  ratio  is  -iWoV  Keducing 
this  to  a  continued  fraction,  we  have 


1  +  1 

3  +  1 


1  +  1 


1  ■ 


15  +  1 


1  +  1 


1  +  1 


2  +  i 

Kow  obtain  the  approximate  values  according  to  the  1st  or  2d  methods,  and 
«ro  Loire  J.  a   <    •'XI  lis.  XS.S.  a.s.5.  .s.s.s..  sjls.z    tsb 4_ 

we  nave    ij^jTJSjTO    ZTs;    233;    462>    llS??    6000>    lOOOO- 

Note. — Fractions  should  be  reduced  to  their  simplest  form  before  beginning  the  operation. 


iscellaneous  Problems. 


^^N 


1.    The  product  of  three  numbers  is  71757 ;  one  of  the  numbers  is  51, 
and  another  21.    What  is  the  third  number?  Ans.  67. 

2.  What  is  that  number  ■which  being  divided  by  25,  the  quotient  increased 
by  6^  +  2,  the  sum  diminished  by  the  difference  between  41  and  25,  the  remainder 
multiplied  by  9,  and  the  product  divided  by  27,  the  quotient  will  be  10? 

Ans.  200. 

3.  Henry  owes  James  $28.50  which  he  wishes  to  pay  with  an  equal  number 
of  picayunes,  dimes,  quarters,  halves  and  dollars.  How  many  of  each  will  it 
require?  Ans.  15. 

4.  What  is  the  fraction,  which  being  divided  by  ^-V,  will  produce  |  ? 

Ans.  -i,-. 

5.  What  number,  multiplied  by  §,  will  give  a  product  of  22  ?  Ans.  33. 

6.  What  fraction,  to  which  if  you  add  ^,  the  sum  will  be  \^  ?  Ans.  •^, 

7.  What  quantity,  to  which  if  you  add  §  of  itself,  the  sum  will  be  75  ? 

Ans.  45. 

OPERATION. 

f  +  1  =  If  =  §,  then  if  5  =  'i^i  i  '^vill  =  15  and  J  =  45.     Ans. 

8.  What  quantity,  from  which  if  you  subtract  |  of  itself,  the  remainder  will 
be  15?  Ans.  24. 

OPERATION. 

1  —  I  =  f,  then  15  —  i  =  24.     Ans. 

9.  Divide  18  oranges  between  A.  and  B.  so  that  A.  will  have  ^  more  than 
B.    What  number  will  each  have?  Ans.  A.  10.    B.  8. 

OPKRATION. 

B.  A. 

18  I  18 

9    4  9  |4 

1    assumed  for  B.  —    —  4  i  o 

1}        "  "   A.  8  —  i 

I    10 
2i  =  3  the  sum  of  the  proportions  according  to  the  conditions  ol  the  question 

10.  Divide  18  oranges  between  A.  and  B.  so  that  A.  will  have  ^  less  than 
B.    What  number  will  each  have?  Ans.  A.  7»    B.  lOf. 


OPERATION. 

B.  A. 


18 

4  7 

4 

10^  — 


7 
1    assumed  for  B. 

i        "  "  A. 

7'> 
IJ  =  J  the  sum  of  the  proportions  according  to  the  conditions  of  the  jiroblem. 

(951) 


18 

4 

3 


952  soule's  philosophic  practical  mathematics.  * 

11.  Divide  $2000  among  A.,  B.  and  C.  so  that  A's  part  will  be  to  B's  as 
2  to  3,  and  that  C.  will  have  as  much  as  A.  and  B.  together,  less  $200.  What 
amount  will  each  receive  t 

SOLUTION. 

As  C.  is  to  h.avo  as  much  as  A.  and  B.  together,  less  1200,  A.  and  B.  will  therefore  have  +  of 
S2000  -f  +  of  $200,  which  is  $U00.  Then,  as  A.  and  B.  are  to  receive  this  in  the  j)roportion  to  2  and 
3,  A.  will  therefore  receive  5  and  B.  ?  of  $1100.  s  of  «1100  =  $440  A's  share;  I  of  $1100  =  $660 
B's  share;  and  la.stlv.  as  C.  is  to  receive  as  much  as  A.  and  B.  together,  less  $200,  he  will  therefore 
receive  $1100  —  200  =  $900. 

12.  Divide  125J  acres  among  A.,  B.  and  C.  giving  C.  TJ  acres  more  than 
B.  and  B.  12 J  acres  more  than  A. 

OPERATION   INDICATED. 

7i  4-  12|  ^  20    =  number  acres  C.  had  more  than  A. 

Y2%  =       *'  '^     B.     '*      **        "      *' 

32i  =  number  acres  over  an  equal  division  of  the  balance. 
125^  —  Slf  =  92i  balance. 

92.7.5  -f-  3  =    30.911  acres,  A's  share. 
12.75  +  30.911  =    43.661      "      B's      " 
20  +  30.91J  =    50.91J      "      C's      " 


•     125. .50  acres,  total. 

13.  A.  and  B.  have  $9000.  5  times  A's  money  is  to  6  times  B's  money  as  5 
to  9.    How  much  has  each. 

OPERATION   INDICATED. 

5  X  A's  share  =  §^  x  6  times  B's  share.     1  x  A's  share  =  J  of  B's  share. 

1  =  B's  share  of  the  whole.     J  =  A's  share  of  the  whole. 

1  -|-  }  ^  IJ  =  sum  of  A.  and  B's  shares.     Then, 

f  :  1  :  :  $9000  :  $5400  B's  share.     ^- :  f  :  :  $9000  :  $3600  A's  share. 

14.  A  merchant  sold  384  bbls.  flour,  part  at  $7.25  and  part  at  $5.50  per  bbl. 
He  received  for  the  whole  $42  more  than  if  he  had  sold  all  at  $C|  a  bbl.  How 
many  bbls.  were  sold  at  each  price  1 

OPERATION  INDICATED. 

$7.25  =  value  1st  quality  =        -  $7.25 

5.50=      "      2d         "         =        -  5.50 

$1.75  =  difference  in  price  of  1st         2  |  12.75 

quality  over  2d.  $6.37}  =  average  price  if  h.tlf  of  each 

quality  were  sold. 

2  I  384 

192  =  half  of  amount  sold. 

1.75)42.00  I  24  bbls.  mon<  of  1st  quality  sold  than  192. 

192  +  24  =  216  Vibls.  1st  quality. 
192  —  24  ^  168     "      2d         " 


MISCELLANEOUS  PROBLEMS. 


953 


15.  A  man  engaged  to  work  a  year  for  $540,  and  a  suit  of  clothes.  He 
•worked  9  mouths  and  received  $39G  anil  the  suit  of  clothes.  What  was  the  suit 
■worth  ? 

OPERATION   INDICATED. 

12  months  =  $540  +  1    suit. 

1  month    =      45  -|-  ,V     " 

9  mouths  =  §396  -f  1       " 

1  month    =      44  +  ^       " 

$45  -f-  i^,  monthly  average  for  twelve  months. 

44  +  1  "  "  "    nine  " 

tV  =  3*5-     Hence  $1  =  ^^  value  of  the  suit  and  J|  = 


I  value  of  suit. 


845  — 44  =  $1.     ^■ 

16.  A  man  rides  a  certain  distance  at  the  rate  of  6  miles  an  hour,  and 
■walks  back  the  same  distance  at  the  rate  of  3J  miles  an  hour.  If  it  takes  him  4f 
hours  to  go  both  ways,  what  is  the  distance  between  the  two  places? 

OPERATION  INDICATED. 

1  hour's  ride  =  6  miles. 


1  hour. 

2 

6 


or  3i  :  6  :  :  1  :  1?  hours. 


14  =  hours  to  walk  back. 
1    =  hour  to  ride  there. 


6    miles. 

7 

19 


or  2^  :  6  :  :  4i  :  lOJ 


lOi  =  number  miles  places  are  apart. 


24  =  hours  to  ride  and  walk. 

17.  Two  trains,  one  210  feet  long  and  the  other  230  feet  long,  were  going  in 
the  same  direction  on  parallel  tracks.  Having  started  together  with  engines 
opposite,  one  of  them  passed  the  otlier  in  15  seconds.  When  going  in  opposite 
directions  at  the  same  rate  of  speed  they  passed  each  other  in  3|  seconds.  How 
fast  were  they  going  t 

OPERATION    INDICATED. 

When  the  longer  train  is  the  faster. 


When  the  shorter  train  is  the  faster. 

210  feet  =  length  of  faster  train. 
230  feet  =  length  of  other  train. 

440  =  number   of    feet   heads   of   engines   were 

apart  at  end  of  34  seconds. 
Seconds.     Feet.     Feet. 
3i        440  I  440 

15  X  15    4 

15 

1760  =  feet  travelled  by 

trains  in  15  sec. 


1760 

210: 

2  I  1550 


;  feet  faster  train  went  over  the  other. 


775  =  feet  travelled  by  slower   train   in   15 
210  seconds. 

985  =  feet  travelled  by  faster  train  in   15 

seconds. 
Mi.  Mi. 


775 
15    60 
5280    60 

35^  mi.  per  hour 
slower  train  runs. 


985 
15    60 

5280    60 

44ij  mi.  per  hour 
faster  train  runs. 


1760 
230  =  feet  faster  train  went  over  the  other. 

2  I  1530 

765  =  feet  travelled  by  slower   train    in   15 
230  seconds 

995  =  feet  travelled  by   faster  train  in   15 
seconds. 


Mi. 
765 
15     60 
5280     60 

34i5  miles  per  hour  slower  train  runs. 


Mi. 
995 
15    60 
5280    60 

45^;  miles  per  hour  faster  train  runs. 


954  soule's  philosophic  practical  mathematics.  * 

18.  A  column  of  troops,  25  miles  long  is  ordered  to  a  point  25  miles  distant. 
A  courier  starts  simultaneously  with  the  rear  of  the  column,  and  reaches  the  head 
thereof.  Eeturning,  he  meets  the  rear  of  the  column  at  the  point  wliere  the  head 
originally  was.  Eoth  the  troops  and  the  courier  are  to  travel  at  a  uniform  rate  of 
speed.    How  many  miles  does  the  courier  travel? 

It  is  evident  from  the  conditions  of  the  problem  that  the  courier  had  to  travel  the  -whola 
length  of  the  column  (25  miles)  plus  the  distance  which  the  head  of  the  column  moved  while  the 
courier  was  overtaking  it  (  x  miles).  It  is  also  evident  that  w  hen  the  courier  reached  the  head  of 
the  column,  the  distance  that  the  rear  of  the  column  travelled  must  have  been  the  extra  distance 
the  courier  had  to  travel  to  overtake  the  head  of  the  column,  viz. :  x  miles.  Then,  since  the 
courier,  in  returning  arrives  at  the  end  of  the  column  when  it  had  reached  the  point  where  the 
head  first  lay  he  must  have  travelled  x  miles  iu  returuing  from  the  head  of  the  column. 


25   MII-E:S  >         X  MIIL£S 


<- 


X  MILEIS 


^^         X  MII-E.&  -^ 25vJmII-ES )| 

*<  25  MII-E.&  X  25is/iii_e:s  > 


B 


A  =  rear  of  column. 

O   =  head  of  column. 

B    =  point  head  of  column  must  reach. 

C    ^  point  courier  reached  when  he  overtook  the  head  of  the  column. 

C  =  point  where  head  of  column  is  when  overtaken  by  the  courier. 

D    =  point  rear  of  column  reached  when  courier  had  reached  the  head  of  the  column. 

From  the  above  explanations  and  diagram,  we  obtain  the  following  proportion 
which,  when  simplified,  gives  the  desired  result : 

25  +  a; :  I :  :  X  :  25  —  x 
x^  =  625  —  x" 
2  x^  =  625  ' 

x«  =312.5 
X    =  17.6776+  miles. 

Since  x  =  the  extr.a  distance  the  courier  travelled  to  overtake  the  head,  and  as  he  had  to 
return  the  same  distance  to  meet  the  rear,  he  must  have  travelled  (2  x  17.6776-1-)  -j-  25  ;=  60.3552 
miles. 

19.  If  a  man  counts  $1  every  second  and  counts  12  hours  every  day,  without 
stopping,  how  many  days  will  it  take  him  to  count  $1000000,  and  how  many  to 
count  a  billion  dollars  ? 

Ans.  23-2*7  days  to  count  a  million. 

231'lS.,-7-  days  =  63  years,  137i-o%  days  to  count  a  billion. 

Note. — 365i  days  were  considered  a  year,  in  reducing  the  days  to  years. 

20.  How  many  years,  of  365J  days  each,  will  it  require  to  count  a  trillion,  by 
counting  12  hours  every  day,  and  counting  $2  every  second? 

Ans.  31C88  years,  32/j-  days. 


MISCELLANEOUS    PROBLEMS. 


955 


21,     I  of  A's  money  and  J  of  B's  money  is  $14,  and  A's  money  is  $4  more 
than  B's  money.     Hovr  much  Las  each  ?  Aus.  A.  $40 ;  B.  $36. 


$13.20, 


OPEUATIOX. 

i  of  |4  =  80c.  part  of  A's  $4  excess  iu  the  $11;   $14.00  —  80c.  =  $13.20;   then  |  +  ^  =  J  J  : 


$ 


11 

5 


13.20 
30 

1    =  A's  part. 


11 
6 


$7.20  +  .80  A's  excess  =  $8  =  i  of  A's  money, 
which  is,  therefore,  $40. 


13.20 

30 

1    =  B's  part. 

$6  =  J  of  B's  money,  -which  is, 
therefore,  $36. 


22.  What  is  the  interest  on  £50  12s.  6d.  for  93  days  at  6  per  cent,  allowing 
365  days  to  the  year?  Ans.  15s.  Cd. 

Note, — See  interest  on  English  Money. 

23.  A  benevolent  man  gave  $3  to  each  of  several  beggars  and  had  $15 
remaining  on  hand.  Had  he  given  each  $10,  it  would  have  taken  all  the  money  he 
Lad.     How  many  beggars  were  there  ?  Ans.  3. 


OPERATION. 


Ans. 


24. 
receive  ? 


A.,  B.  and  C.  are  to  receive  $26  iu  proportion  to  J,  J,  J.     What  will  each 


Ans.  A.  $12 ;  B. 


C.  $6, 


OPERATION. 
A. 


B. 


13 

2 

$26 
12 

13 
3 

$26 
12 

13 
4 

$26 
12 

$12 

$8 

$6 

i  +  i  +  i  =  li 


23.  A.  and  B.  bought  at  auction  an  invoice  of  merchandise  for  $800,  of 
•which  sum  A.  paid  $500  and  B.  $300.  They  then  sold  to  C.  J  interest  in  the 
merchandise  for  $400.  How  much  of  the  $400  must  A.  and  B.  receive  respectively 
iu  order  to  constitute  each,  A.,  B.,  C.  J  owner  of  the  goods? 

Ans.  A.  $350;  B.  $50. 
Note. — For  tlie  solution  of  this  and  similar  questions,  see  pages  872  and  873. 

26.  A  California  miner  has  a  spherical  ball  of  gold  2  inches  in  diameter 
which  lie  -wishes  to  exchange  for  spherical  balls  1  inch  iu  diameter.  How  many  of 
the  smaller  spheres  should  he  receive  1  Ans.  8. 

Note. — For  the  solution  of  all  similar  questions,  see  pages  377  and  394. 

27.  A  boy  bought  a  certain  number  of  peaches  at  the  rate  of  4  for  5  cents 
and  paid  for  them  with  apples  at  the  rate  of  2  for  3  cents.     How  many  peaches  did 


he  buy,  providing  it  required  300  apples  to  x^ay  for  the  peaches? 


Ans.  360. 


300 
3 


OPERATION. 
P. 

5  I    4 
I  $4.50 


Or, 


r.50  value  of  apples. 


360  Ans, 


300 

3 

4 

360 


956  soule's  philosophic  practical  mathematics.  * 

28.  Two  little  girls,  Susie  and  Katie,  were  employed  by  a  fruit  dealer  to 
peddle  oranges  ;  the  first  day  tbey  received  30  oranges  each,  Susie's  being  of  superior 
quality,  whicb  tbey  sold  as  follows  : 

Susie  sold  2  for  5  cents,  and  received         75   cents. 

Katie    "    3   "   5      "  "  50       " 


Total  receipts,  $1.25 

The  jiext  d;iy  Susie  was  sick,  and  Katie  took  tbe  60  oranges  and  sold  tbem  at 
the  rate  of  5  for  10  cents.  At  night,  M'hen  she  made  her  returns  to  tbe  fruit  dealer, 
she  bad  but  $1.20;  and  as  60  oranges  brought  $1.25  tbe  day  before,  when  sold  2  for 
5  cents  and  3  for  5  cents,  or  seemingly  5  for  10  cents,  the  fruit  dealer  charged  little 
Katie  with  stealing  tbe  5  cents.  Did  little  Katie  steal  the  5  cents,  and  if  not,  bow 
do  you  account  for  its  loss  ? 

Ans.  Little  Katie  did  not  steal  it.     It  was  lost  because  tbe  ratio  of  5  for 
10  cents  ceased  when  10  sales  bad  been  eflected. 

SOLUTION. 

As  Snsio  sold  2  oranges  at  a  sale,  it  is  evident  that  to  sell  30  she  made  15  sales,  which,  at  5 
cents  a  sale,  amnunted  to  75  cents;  and  as  Katie,  tbe  first  day,  sold  3  oranges  at  a  sale,  it  is  evident 
that  to  sell  30  she  made  10  sales,  which,  at  !>  cents  a  sale,  amounted  to  50  cents,  making,  as  above, 
SI. 2.5  for  60.  Tbe  second  day  Katie  mixed  tbe  (iO  oranges,  and  by  selling  5  at  a  sale  made  12  sales, 
which,  at  10  cents  a  sale,  amounted  to  $1.20.  By  this  we  see  that  the  5  cents  was  lost  by  the 
manner  of  effecting  sales;  and  hence,  little  Katie  is  innocent  of  the  crime  with  which  she  is 
charged.  Tbe  error  in  making  the  sales  was  in  mixing  and  selling  all  of  the  oranges  at  the  rate 
of  5  for  10 cents;  for,  as  shown  above,  the  inferior  oranges,  3  for  5  cents,  were  all  sold  in  10  sales, 
and  hence  tbe  r.itio  of  3  inferior  and  2  superior  making  .5  for  10  cents,  then  ceased,  and  tbe  remain- 
ing 10  oranges  should  have  been  sold  2  for  5  cents.  By  selling  tbe  60  oranges  at  the  rate  of  5  for 
10  cents,  there  must  have  been  2  sales  which  consisted  wholly  of  superior  oranges. 

29.  If  5i  times  5i  yards  of  cloth,  which  is  IJ  times  IJ  yards  wide,  cost  2| 
times  25  dollars,  what  will  10|  times  10§  yards  cost  that  are  1;^  times  IJ  yards 
wide?  *  Ans.  $25. 

30.  A  planter  hired  a  laborer  for  a  term  of  120  days,  on  condition  that  for 
every  day  be  worked  be  should  receive  $1.00  and  his  board,  and  that  for  every  day 
be  did  not  work  be  should  pay  50  cents  for  his  board.  At  the  expiration  of  tbe  120 
days  a  settlement  was  made  and  he  received  $76.50.     How  many  days  did  be  work  ? 

Ans.  91. 

OPERATION. 

Had  he  worked  tbe  120  days  he  would  have  received  (at  $1  per  day)        $120. 
But  as  be  only  received  .........  76.50 

He  lost  by  reason  of  not  working  .......  $43.50 

And  as  he  lost  each  day  that  he  did  not  work  ($1.00,  amount  of  services,  +  50  cents,  amount 
paid  for  board)  =  $1.50,  it  is  clear  that  he  lost  as  many  days  as  $43.50  are  times  equal  to  $1.50, 
which  is  29 ;  and  if  he  lost  29  days,  he  therefore  worked  120  —  29  =  91  days. 

31.  A.  can  do  a  piece  of  work  in  6  days,  and  B.  can  do  tbe  same  work  in  8 
days.     How  many  days  will  it  take  for  both  to  do  tbe  work  1  Ans.  3f . 

OPERATION. 

If  a.  can  do  the  work  in  6  days,  in  1  day  be  can  do  i  of  it;  and  if  B.  can  do  the  work  in  8 
d.iys,  in  1  day  he  can  do  |  of  it;  therefore  ^  +  i  =  s'l  of  tbe  work  is  done  in  1  day  ;  and  if  y^  of 
the  work  is  done  in  1  day,  by  transposition  ^4  of  the  work  require  one  day,  and  if  jV  of  the  work 
require  1  day,  5^,  will  require  |  of  1  day,  and  H,  or  the  whole  work,  will  require  24  times  as  many 
days,  which  is  3^. 


*  MISCELLANEOUS    PROBLEMS.  95/ 

32.  A.  and  B.  can  do  a  piece  of  work  in  10  days ;  A.  alone  can  do  it  in  15 
days.    How  many  days  will  it  take  B.  to  do  it?  Ans.  30  days. 

OPERATION. 

iV  —  i^f  =  aV  of  the  work  clone  in  1  day  by  B. 
1    -;-  jV  =  30  days.     Ans. 

33.  A.  and  B.  can  do  a  piece  of  work  in  14  days.  A.  can  do  f  as  much  aa 
B.     How  many  days  will  it  take  each  to  do  it,  working  alone? 

Ans.  24J  days  for  B.     32§  days  for  A. 

OPERATION. 

(B.)  J  4-  (A.)  J  =  I  work  done  by  A.  and  B.  in  one  day. 
Hence  in  1  day  B.  does  }  of  the  work. 
And      "   1     "    A.     "     f      "  " 

14  -H  ^  =  24i  days  for  B. 
14  ^  S  =  32J  "       A. 

34.  A  cistern  has  3  pipes  hy  which  it  may  be  filled,  and  2  by  which  it  mfiy 
be  emptied.  The  first  of  the  three  filling  pipes  can  fill  it  in  12  hours,  the  second  in 
20  hours  and  the  third  in  2  days.  The  first  of  the  emptying  pipes  can  empty  it  in 
24  hours  and  the  second  in  30  hours.  If  all  of  the  jjipes  are  left  open,  how  many 
hours  will  it  require  to  fill  the  cistern  ?  Ans.  12{f  hours. 


OPERATION. 

iV  +A-  +  .-V  =  /4V 

h. 

1 

19    240 

K-<     +  A-                  =  T^iT 

AV  —  tIb          =  A^o 

12}  5    honrs. 

Ans. 

35.    If  it  takes  5  boys  20  days  to  do  a  piece  of  work,  and  6  girls  30  days  to 
do  the  same  work,  how  many  days  will  it  take  a  boy  and  a  girl  to  do  it? 

Ans.  64f  days. 


OPERATION. 

d. 

5  X  20  =  100. 

TTO  +  jhs  =  TSTJ 

1 

7     450 

6  X  30  =  180. 

— 

I    644   days. 

36.  A  barrel  of  flour  will  last  a  man,  wife  and  servant  50  days;  but  when 
the  man  is  absent,  it  will  last  his  wife  and  servant  90  days.  How  long  would  it  last 
the  man  ?  Ans.  112J  days. 

OPERATION. 

d. 

11 

sV  —  A  =  tIt  2  I  225 

I  112i 

37.  A  merchant  bought  5  bbls.  flour  for  $40  and  sold  the  same  at  a  gain  of 
25%.  He  then  bought  5  bbls.  more  for  $40,  and  sold  the  same  at  a  loss  of  25%. 
Did  he  gain  or  lose,  and  if  so,  how  nuieli  ?  Ans.  He  neither  gained  nor  lost. 

38.  A  merchant  sold  5  bbls.  of  inferior  flour  for  $40  and  gained  25%.  He 
then  sold  5  bbls.  of  superior  flour  for  $40  and  lost  25%.  Did  he  gain  or  lose  in  the 
two  transactions,  and  if  so  how  much  ?  Ans.  He  lost  $5^. 


958  soule's  thilosophic  practical  mathematics.  * 

a  notk  with  collateral  security  and  a  margin. 

39.  A  merchant  discounted  his  note  for  $1200  with  a  banker  and  gave  him 
insurance  stock  as  collateral  security. 

The  market  value  of  the  stock  -was  $112.  A  margin  of  20%  was  to  be  kept 
good  on  the  stock.  How  many  shares  of  stock  did  the  merchant  deliver  and  what 
instrument  of  writing  did  he  execute  and  deliver  to  the  banker  besides  the  note? 

Ans.  1.  14  shares  of  stock. 

2.  A  special  jiower  of  attorney  to  sell  the  stock. 


FIRST    OrERATION. 


I  $100 
80    1200 


I  $1500  =  ainonnt  of  collaterals. 
luOO  -:-  112  =  131^  shares,  practically  14  shares. 


SECOND   OPERATION. 

$112       marVet  value  of  shares. 
22.40  =  20%  margin. 


$  89.00  net  security  per  share. 

$1200  —  89.60  =  13iJ  shares,  practically  14  shares. 

NoTK. — It  is  the  cnstom,  -when  giving  stock  collaterals  to  give  a  special  power  of  attorney 
to  sell  the  stock,  so  that  in  case  the  stock  should  decline  and  the  margin  should  not  he  made  good, 
a  sale  of  the  stock  can  he  made  to  i)rotect  the  i)ayment  of  the  note. 

40.  A.  can  do  |  of  a  piece  of  work  in  10  days ;  B.  can  do  ^  of  it  in  6  days, 
and  C.  can  do  f  of  it  iu  3  days.  How  many  days  Mill  it  take  them  all  to  do  the 
work ?  Ans.  3^^^^  days. 

OPERATION. 

^  X  A  =    A  the  part  of  the  work  done  hy  A.  in  1  d.ay.  d. 

f  X    i   =     i         "  "  "  B.   "  1     "  11 

I  X   i  =    ,V       "  "  "  C.  "  1     "  113  I  360 

m  the  part  of  the  work  done  hy  A.,  B.  and  C.  iu  1  day.  |  3,V3-    Ans. 

41.  Three  persons,  A.,  B.  and  C,  do  a  piece  of  work ;  A.  and  B.  together  do 
2  of  it,  and  B.  and  C.  do  -^  of  it.    What  part  of  the  work  is  done  by  B.  ? 


-^ns.   ^g« 


SOLUTION. 


As  A.  and  B.  do  ?,  of  it,  it  is  clear  that  C.  does  the  remaining  § ;  and  as  B.  and  C.  do  i^r  of  it, 
it  is  clear  that  A.  does  the  rem.aining  -|\.  Then,  as  A.  does  ,V  and  (J.  §,  they,  together,  do  -jV  +  § 
=  -Q§ ;  and  if  A.  and  C.  do  §5  of  ^  piece  of  work  doue  by  A.,  B.  aud  C,  it  is  clear  that  13.  does  the 
difiereuce  between  ^g  aud  §5,  which  is  J^. 

42.  A.,  B.,  C.  and  D.  agree  to  do  a  piece  of  work  for  $5500.  A.,  B.  and 
•C.  can  do  it  iu  20  days;  B.,  O.  and  D.  in  24  days;  C,  D.  and  A.  in  30  days;  and 
D.,  A.  and  B.  in  36  days.  In  how  many  days  can  all  do  it,  working  together;  iu 
how  many  days  can  each  do  it,  working  alone ;  and  what  part  of  the  pay  ought  each 
to  receive  ?  Ans.  For  the  solution  see  page  879. 

43.  A  planter  being  asked  how  many  cattle  he  had,  replied  that  he  had  ^  of 
them  in  one  pasture,  J  in  another,  and  25  in  a  third  jiasture.    How  many  had  he  ? 

Ans.  60. 

SOLUTION. 

i  +  i  =  I'j  the  number  in  the  first  and  second  pastures.  Hence  /./  from  all  of  his  cattle,  j| 
z^  i-i,  the  number  in  the  third  pasture ;  25  is  therefore  f'i  of  his  cattle ;  and  if  ,*;;  are  25,  A  '»  the  i 
j)»»t,  which  is  5,  and  Yi,  "''  t^o  whole  number,  are  12  times  5,  which  is  60. 


*  MISCELLANEOUS    PROBLEMS.  gSq 

44.     If  i,  i  and  |-  of  a  mau's  money  is  $780,  what  amount  has  he  T 

Ans.  $800. 

SOLUTION. 

i  +  J  +  5  =  3?.     Tlien  it  U  equal  $780,  -/j  is  the  39th  part,  which  is  $20,  and  J2,   or  the 
■whole  amount,  is  equal  to  40  times  i\,,  or  $20,  which  is  f  "" 


45.  A  grocer  bought  a  barrel  of  molasses,  and  after  ^  of  it  had  leaked  away 
he  drew  out  6  gallons,  when  it  was  found  to  be  |  full.  How  many  gallons  did  the 
barrel  hold  ?  Ans.  40  gals. 

SOLUTION'. 

J  —  J  =  J ;  then  f  —  f  =  j'.t,  then  i^j  =  6  gallons,  and  ^V  =  4  of  6  gallons,  which  is  2,  and  ?3 
=  20  times  as  much,  which  is  40. 

*         46.    A  speculator  after  losing  J  of  his  money,  and  J  of  the  remainder,  ha4 
$1750.     How  many  dollars  had  he  at  first.  Ans.  $7000. 

SOLUTION. 

i  =:     his  first  loss. 
^  X  i  =  i  =       "   second  loss. 

J  -|-  i  =  }    "   total  loss,  which,  deducted  from  his  whole  amount, 
leaves  J,  hence  $1750  is  }  of  his  money,  and  the  whole  amount,  4  times  as  much,  which  is  $7000. 

47.  A  patron  of  the  "wheel  of  fortune"  (misfortune)  in  the  lottery,  lost  by 
his  first  investment  ^  of  his  money;  by  his  second  investment,  J  of  the  remainder, 
and  by  his  third  investment,  J  of  the  second  remainder ;  he  then  donated  $10  to 
aid  in  the  destruction  of  such  disreputable  institutions,  and  had  $75  left.  What 
sum  had  he  at  first  1  Ans.  $680. 

SOLUTION. 

Having  lost  hy  the  1st  investment  J  of  his  money,  he  had  |  remaining;  then  h.aving  lost  by 
the  2d  investment,  J  of  the  remainder,  he  therefore  lo.st  J  of  |  =  J;  and  }  deducted  from  J  =  J  as 
the  second  remainder;  then  having  lost  by  his  3d  investment  }  of  tlio  2d  remainder,  he  therefore 
lost  f  of  i  =  f ;  and  |  deducted  from  i  =  i  remaining  on  hand  after  his  3d  loss.  Hence,  the  $10 
donated  +  the  $75  on  hand  =  $85,  is  i  of  his  first  sum,  and  consequently  the  whole  amount  was  8 
times  $85  =  $680. 

48.  A  merchant  deposited  his  money  in  4  banks ;  in  the  first,  he  deposited 
f ;  in  the  second,  ^;  in  the  third,  |;  and  in  the  fourth,  $2400  more  than  ,-V  of  the 
whole.     How  much  money  had  he?  Ans.  $72000. 

OPERATION. 

I  +  i  +  i  =  Tif  =  proportion  deposited  in  three  banks,  hence  the  difference  between  .'j  and 
-}5  =  A  is  the  proportion  deposited  in  the  fourth  banli.  And  as  the  amount  deposited  in  the  fourth 
■bank  was  $2100  more  than  ,^i  of  the  whole,  the  diB'erence  between  ,\^  and  ^K,  is  therefore  tho  pro- 
portion that  $2400  is  to  the  whole  sum.  /„  —  ,J^,  =  ^o,  hence  $2400  is  ^g  of  the  whole  sum,  and  jj 
or  the  whole  sum  is  30  times  $2400,  which  is  $72000. 

49.  If  3  men  and  5  women  can  do  a  piece  of  work  in  14  days,  6  women  and 
8  boys  can  do  the  same  work  in  10 J  days,  and  4  men  and  7  boys  can  do  it  in  12| 
days.    How  long  will  it  take  2  men,  10  boys  and  3  women  to  do  it? 

Ans.  9j^  days. 

Note. — For  the  solution  of  this  problem,  see  page  331. 


96o 


SOULE  S    PHILOSOPHIC    TRACTICAL    MATHEMATICS. 


50.  A  farmer  sold  CO  fowls,  a  part  were  turkeys  and  a  part  chickens;  for  the 
turkeys  be  received  $1.10  a  jnece,  and  lor  the  chickens,  60  cents  a  piece,  and  for 
the  whole,  he  received  $51.00;  how  many  were»they  of  each?     Aiis.  24  chickens. 

3(J  turkeys. 

SOLUTION. 

Had  all  tlio  fowls  Tiecn  turkeys  ami  sold  at  $1.10  each,  ho  would  liave  received  ($1.10  X  60), 
$fi6.00;  but  because  some  were  chickeus,  and  sold  at  50  cents  each,  he  only  received  $.")!. 60,  bene© 
the  difference,  ($G6  —  $51.60)  =  $14.40  is  the  delicit  by  reason  of  ha vinj;  sold  some  chickeus,  and  as  the 
turkeys  were  sold  at  $1.10,  and  the  chickens  at  50  cents,  there  is  a  defieit  on  each  chicken  of  ($1.10 
—  50c.  =  60) ;  therefore  there  were  as  many  chickens  as  $14.40,  are  equal  to  60c.  which  is  ($14.40 
-i-  60o.)  =  24 ;  and  as  there  were  60  in  all,  it  is  clear  that  OO  —  24  =  36,  is  the  uumber  of  turkeys. 

51.  A  planter  being  asked  how  many  head  of  mules  he  had,  replied,  that  if 
he  had  as  many  more,  one-half  as  many  more,  and  seven  mules  and  a  half,  he  would 
then  have  100.     How  many  had  he?  Ans.  37. 

SOLUTION. 

According  to  the  conditions  of  the  question  the  100  head  is  the  result  of  as  many  more,  one- 
half  as  man;/  more,  and  seven  and  one-half  added  to  the  true  number;  and  hence  it  is  evident,  that 
by  deducting  7i  from  100  we  shall  have  in  the  remainder  the  true  number,  jilus  as  many  more,  and 
one-half  as  many  more.  100  —  7i  =  ir2i  ;  uow  as  i)2i  is  the  true  uumber  idus  the  true  uumber,  and 
one-half  of  the  true  uumber,  it  is  evident  that  it  is  2J  or  -5  times  the  true  number,  and  hence  92^ 
-:-  2^  =  37,  the  true  number. 

52.  A  father  bought  a  certain  number  of  oranges,  which  he  divided  among 
the  members  of  his  family,  as  follows  :  To  his  wife  he  gave  J  of  all  that  he  had,  and 
J  an  orange  more;  to  his  daughter  he  gave  J  of  the  remainder,  and  ^  an  orange 
more;  to  his  son  he  gave  J  of  the  second  remainder,  and  J  an  orange  more;  and  to 
his  servant,  he  gave  1,  and  had  1  left  for  himself;  how  many  oranges  did  he  buy 
and  how  many  did  each  receive? 

SOLUTION. 

(Problems  of  this  character  are  most  easily  solved  by  working  in  the  reverse  direction  or 
order  from  which  the  occurrences  of  the  jiroblem  took  place). 

It  is  evident,  on  inspection,  that  he  had  2  oranges  after  b.aving  made  the  third  division  (with 
his  son),  and  as  he  gave  his  son  i  au  orange  more  than  i  of  what  be  bad  before,  making  the  third 
division,  it  is  evident  that  be  must  have  liad  2  times  2  (the  remainder)  jilus  1,  which  is  5  ;  then  a.* 
■  5  was  what  he  had  before  making  the  third  division,  it  is  what  he  had  after  making  the  second 
division;  and  as  he  gave  in  the  second  division  A  an  orange  more  than  i  of  what  he  had  before 
making  the  second  division,  it  is  also  evident  that  lie  had  2  times  5,  plus  1,  which  is  11.  Then  as 
H  was  what  he  had  left  before  m.aking  tlie  Kccon<l  division,  it  is  what  he  had  after  making  the  first 
division;  and  as  he  gave  in  the  lirst  division  |  an  orange  more  than  i  of  what  he  had  before 
making  the  lirst  division,  it  is  likewise  evident  that  he  had  2  times  11,  plus  1,  which  is  23. 

Then  23  —  2  =  Hi  -f  i  =  12,  the  wife's  part. 

23  —  12  =  11 ;  11  —  2  =  5i  -(-  i  =  6,  the  daughter's  part. 

11  —    6  =    5 ;     5  -^  2  =  2i  -f  i  =  3,  the  sou's  part. 

5  —    3=    2;    2—1  =  1,  the  servant's  part. 

2  —    1  =    1  remaining,  the  father's  part. 

53.  A.  and  B.  had  the  same  income  ;  A.  saved  i  of  his,  but  B.  by  spending 
$50  more  per  annum  than  A.  found  himself  at  the  end  of  ten  years  $300  in  debt; 
how  much  was  the  income  of  each.  Ans.  $100. 

SOLUTION. 

B.  having  spent  $300  in  10  years,  at  $50  per  year,  it  is  plain  that  he  spent  ,^n  of  it  in  one 
year;  ,\  of  $300  is  $30.  Now  as  B.  spent  $50  more  per  year  than  A.,  the  difference  between  what 
B.  spent  per  year,  and  $50  is  what  A.  saved.     This  difference  is  50  —  30  =  $20. 

Therefore  $20  is  i  of  their  yearly  income,  and  g  or  the  whole  income  is  8  times  $20,  which  is 
$160. 


*  MISCELLANEOUS    PROBLEMS.  96 1 

54.  There  is  a  fish  -whose  head  is  9  inches  long,  his  tail  is  as  long  as  his  head 
and  half  of  his  body,  whilst  the  body  is  as  long  as  his  head  and  tail  both;  what  is 
the  length  of  the  fish?  Ans.  72  inches. 

SOLUTIOf. 

Since  the  tail  is  as  lonR  as  tlie  head  and  i  tlie  hndy,  it  is  eqnal  to  9  inches  -f  i  of  the  hody, 
hut  as  the  body  =  tlie  head  and  tail,  it  (the  body)  =  9  'inches  -f  9  inches  +  +  of  the  body.  That 
is  the  body  =  18  inches  +  i  the  body.  Hence  i  the  body  is  18  inches,  and  I  or  the  whole' body  36 
inches;  and  as  the  hody  is  i  the  IcDgth  of  the  fish,  |  or  the  -whole  length  is  2  times  36  inches, 
r.hich  is  72  inches. 


A  hare  is  40  leaps  before  a  hound  and  takes  4  leaps  -while  the  honnd 
it  2  of  tlie  hound's  leaps  are  equal,  to  3  of  the  hare's;   ho-s\'  many  leaps 


5".. 

takes  3,  but 

must  the  hound  take  to  catch  the  hare  ?  Ans.  240. 


SOLUTION. 

Since  2  leaps  of  the  hound  =  3  leajis  of  the  hare,  1  leap  of  the  honnd  =  li  leaps  of  the 
hare;  then,  3  leaps  of  the  houud  =  4i  of  the  hare;  hence,  in  making  3  leaps  tlie  hound  gains  ^ 
leap  of  the  hare,  and  in  making  1  leap,  (J  —  3)  j, ;  then  6  leaps  of  the  hound  gain  1  leap  of  the 
hare,  and  6  X  •10  =  240,  number  of  leaps. 

50.  A  general  formed  his  men  into  a  square,  that  is,  an  equal  number  in 
rank  and  file,  and  foutid  that  he  had  59  men  over;  and  increasing  the  number  in 
both  rank  and  file  by  1  man,  lie  -wanted  84  men  to  conijilete  the  square;  bow  many 
men  had  he  ?  Ans.  5100. 

SOLUTION. 
To  increase  the  number  in  both  rank  and  file  by  1  man,   it  is  evident,  that  it  would  require 
twice  the  number  of  men  in  rank  or  file  at  first,  plus  1  (for  the  man  occupying  the  corner).     And 
as   it  require  to  efl'ect  this,  59  -|-  84  =;  143  men  ;    then  (143  —  1)  -H  2  =  71,  the  number  of  men 
in  rank  and  file  at  first.     Therefore,  71''  -f-  59  =  5100  men  in  the  army. 

57.  Three  footmen.  A.,  B.  and  C.  start  together  and  travel  in  the  same  direc- 
tion arouud  an  island  72  miles  in  circumference ;  A.  travels  6  miles  a  day,  B.  8,  and 
C.  12.    In  -what  time  will  they  all  be  together  again?  Ans.  36  days. 

SOLUTION. 

72  —    6  =  12  days  required  for  A.  to  travel  around  the  island. 
72—    8=    9       '   '•  "        B.     "  "  " 

72  — 12=    6  "  "       C.     "  "  "    ■ 

Then  the  least  common  multiple  of  12,  9  and  6  is  the  number  of  days  required. 

3  )  12    9    6 

2)432 

2    3    1 

3x2x2x3  =  36  days,  Ans. 

or,  thus : 

B.  gains  on  A.  2  miles  a  day,  and  hence  will  pass  A.  every  ''^  days,  =  36  d.ays.  C.  gains  on 
A.  6  miles  a  day,  and  hence  -svill  pass  A.  every  ''^  days  =  12  days.  Therefore,  the  least  common 
multiple  of  36  and  12  will  give  the  number  of  days  that  B.  and  C.  -will  first  pass  A.  together. 

12  )  36    12 


1 
12  X  3  =  36  days,  Ans. 


962 


SOULE  S    rHILOSOPHIC    PRACTICAL    MATHEMATICS. 


58.  A  lake  is  56  miles  in  circumference.  A.  traveling  30  miles  per  day,  B. 
24.  C.  16,  D.  12  and  E.  10,  all  commence  to  travel  at  tbe  same  time  and  in  the  same 
direction  around  tlie  lake.    In  how  many  days  will  tbey  all  be  together  again  ? 

Ans.  28  days. 

SOLUTION. 


A.  gains  on  E.  20  miles  per  day. 

B.  "      E.  U     "  " 

C.  "      E.    6    "  " 

D.  "      E.    2     "  " 


56 
56 
56 
56 


20  =    2J  =    2\i  =  f  i 

14  =    4    =    4      =  f ^ 

6=    9i=    9,5,' =  W 

2  =  28    =28      =^5*^ 


2  )  fl     ??    W     W 


3  )  21 

30 

70 

210 

7  )  7 

10 

70 

70 

10  )  1 

10 

10 

10 

111      1 

2X3X7X10=  V.!*^  =  28  days,  Ans 


59.  How  many  gallons  in  a  tank  of  tbe  following  description  and  dimen- 
sions :  Tbe  ujjper  portion  is  in  tbe  form  of  a  vertical  rectangular  prism,  13  ft.  5  inches 
deep,  11  feet  9  inches  long,  and  9  feet  3  iiiches  wide.  The  bottom  is  a  segment  of  a 
horizontal  cylinder,  the  chord  being  9  feet  3  inches,  and  tbe  height  or  versed  sine 
4  feet?  Ans.  13372.404  gallons. 

Note. — For  the  solution  of  tliis  problem,  see  page  399. 

60.  A  live  stock  dealer  sold  two  horses,  each  at  the  same  price.  On  one 
horse  he  gained  25  per  cent,  and  on  the  other  he  lost  25  per  cent.  His  net  loss  on 
tbe  two  sales  Avas  $12.     What  was  the  cost  of  each  horse  ? 

Ans.  $72  cost  of  1st  horse.     $120  cost  of  2d  horse. 

Note. — For  the  solution  of  this  problem,  see  page  477. 

61.  A  man,  at  his  death,  left  an  estate  amounting  to  $35000,  to  his  wife  and 
two  children,  a  son  and  a  daughter.  His  children  being  absent  in  Europe,  he  directed 
by  will,  that  if  his  son  returned  his  wife  should  have  J  of  the  estate,  and  the  sou 
the  remainder ;  but  if  the  daughter  returned,  his  wife  should  have  §  and  the 
daughter  the  remainder.  Now  it  so  happened  that  they  botli  returned.  What  sum 
should  each  receive?  Ans.  $20000  son.     $10000  wife.    $5000  daughter. 


Note. — For  the  solution,  see  jiage  876. 


62,  A  man  had  2  sons  and  3  daughters ;  to  the  youngest  son  be  gave  J  of  1 J 
times  $3600,  which  was  f  of  the  elder  sou's  share,  and  what  the  elder  son  received 
was  f  of  I  of  tbe  whole  estate ;  tbe  balance  was  divided  among  bis  3  daughters  in 
reciprocal  proportion  to  their  ages,  8,  10,  12  years ;   what  was  each  daughter's 


share  ? 


Ans.  $7114.87  for  the  one    8  years  old. 
$5691.89       "      "       10  " 

$4743.24      "      "       12  « 


Note. — For  tlie  sulutiou,  see  page  874. 


MISCELLANEOUS    PROBLEMS. 


963 


63.  A.,  B.  and  C.  were  to  receive  $6000,  in  proportion  to  i,  ^  and  ^ ;  but  B. 
having  died,  it  is  required  to  divide  the  luouey  between  A.  and  C.  What  sum 
should  each  receive  ?  Ans.  A.  $4000.    C.  $2000. 

Note. — For  tho  solution,  see  page  875. 

64.  A.  and  B.  paid  $600  ($300  each)  for  300  acres  of  land.  In  dividing  the 
land  according  to  quality,  it  was  agreed  that  A.  should  have  the  best  quality,  and 
pay  therefor  75  cents  per  acre  more  thau  B.  How  many  acres  did  each  receive,  and 
what  price  per  acre  did  each  pay  1 

Ans.  A.  received  123.8  acres  and  iiaid  $2,443  per  acre. 
B.         "         177.2      "  "  1.G93       " 

Note. — For  the  solution,  see  page  878. 

65.  A  merchant  receives  from  France,  an  invoice  valued  at  port  of  shipment 
at  4550.25  francs. 

In  the  invoice  are  100  dozen  pairs  of  kid  gloves  that  cost  18.50  francs  per 
dozen.  Freight  and  insurance  on  the  whole  invoice  to  New  Orleans,  amounted  to 
$87.82.  Duty  was  paid  on  the  gloves  at  33J  per  cent.  For  what  price  per  pair 
must  the  merchant  mark  these  gloves  so  that  he  may  discount  10  jier  cent  from  his 
asking  ijrice  and  yet  gain  16§  per  cent  on  full  importing  cost  ?  Ans.  55+  /. 

OPERATION   INDICATED. 


%■  of  charges  on  invoice. 

$  import  cost  1  dozen  gloves. 

87.82 

.193 

4550.25 

18.50 

,193 

100 

$3.57 

10%  practically. 

1.19  =  duty  .It  33J. 

.36  =  proportion  of  charges  at  10%. 

$  selling  price  per  pair. 

— 

5.12                                                  $5.12  full  N.  0.  cost  per  dozen. 

6 

7 

9 

10 

12  I 

I  55+  c. 

66.  August  30, 1S95,  No.  7,  coffee  was  quoted  in  Eio  at  13$850,  exchange 
lOgd.  What  is  the  equivalent  value  per  pound  in  New  Orleans,  the  rate  of  English 
Exchange  being  $4.88,  no  allowance  being  made  for  time,  freight  or  insurance? 

Ans.  9.24+  /. 

FIRST    OPERATION.  SECOND   OPERATION. 


32.38 

240 


13$850 

10| 

4.88 

9.24+  c.     Ans. 


13$850  X  lOid  =  147.1o625d. 

147.15625  H-  32.38  =  4.545  —  d  per  pound. 

(4.545  —  d  X  $4.88)  ^  240  =  9.24+  c.     Ans. 


Explanation. — The  rate,  13$850,  is  13  milries  and  850  reis  per  arroba,  which  is  equal  to  32.38 
pounds  avoirdupois.  Tho  10|d.  ex-;hango  is  the  English  gold  value  of  a  inilreis  of  silver  of  Brazil 
including  excUuuge.     Tho  $4.88  is  N.  O.  gold  rate  of  exchange  on  England. 

67.  A.  and  B.'  together  agree  to  dig  100  rods  of  ditch  for  $100.  The  part  of 
the  ditch  on  which  A.  was  employed  was  more  difBcult  of  excavation  than  the  yiart 
on  whicli  B.  was  employed;  and  it  was  therefore  agreed  that  A.  suould  receive  for 
each  rod  25  cents  more  than  B.  received  for  each  rod  that  he  dug.  How  many  rods 
must  each  dig  and  at  what  price  so  that  each  may  receive  just  $50? 

Ans.  B.  receives  $  ,8903  +  jier  rod,  and  digs  50.16  rods. 
A.  receives  $1.1403  per  rod,  and  digs  43.84  rods. 
Note. — For  the  solution,  see  piige  878. 


964  'soule's  philosophic  practical  mathematics.  * 

68.  A  person  being  asked  the  hour  of  the  day,  said,  the  time  past  uoon  is  f 
of  the  time  from  now  to  midniglit.     "What  is  the  hour  ? 

Alls.  3  hrs.     25  m.     42f  s. 

OPERATION. 

12  lirs.  =  1  +  f  =  J  of  time  to  midnight. 

If  12  =  1,  ^  =  }  X  12  and  g  =  ?  X  12  =  2,1  hrs. 
S^^  =  3}  hrs.  =  3  hrs.  25  m.  42f  s.     Ans. 

Explanation. — Since  the  time  p.ast  noon  is  |  of  the  time  to  midnight,  we  may  consider  the 
time  to  midnight  as  a  whole,  or  3,  and  the  two  periods  of  time  taken  together  =  f  +  j-  =  ?,.  Tlieii 
as  the  time  from  noon  to  nii<lnight  =  12  hrs.,  we  have  12  hrs.  =:  j  of  tlie  time  to  midnight;  aud'if 
I  =  12  hrs.,  i  =  I  X  12,  and  f  =  f  X  12  =  s,i  =  3  hrs.  25  m.  42?  s.     Ans. 

69.  What  time  after  5  o'clock  will  the  hour  aud  minute  hands  of  a  watch  be 
together  ?  Ans.  27  m.     IGAj-  s. 

OPERATION. 

,V  X  60  =  5n-    m.  =  numhcr  minute  sp.aces  passed  by  lioiir  liand  at  first  crossing. 

5  X  5i\  =  27  fc  m.  =  number  minute  sjjaces  after  5  crossings,  i.  e.,  Ijetwecn  5  and  6  o'clock. 

27A  m.  =  27  m.  16  A-  s. 

Explanatio-n. — Since  the  hands  are  together  at  12  jr.,  the  l)eginning  of  the  hour  circle,  .and 
cross  each  other  11  times  during  the  passage  of  the  hour  iiand  (with  a  uniform  njotion)  on(-e  around 
the  circle  of  60,minute  spaces,  we  see  that  the  first  crossing  of  the  hour  liaiid  will  require  ,',  of  the 
60  spaces,  or  5n  spaces,  and  at  the  fifth  crossing,  occurring  between  5  aud  G  o'clock,  the  hour  hand 
will  hare  passed  over  5  times  5i^i  m.  =  21^^  m.  =  27  m.  16,^,  s.     Ans. 

70.  On  the  dial  of  a  town  clock  the  end  of  the  minute  hand  requires  4 
minutes  and  42  seconds  to  jiass  over  a  sj^ace  of  6  inches.     What  is  its  length  ? 

Ans.  12.lt)  +  inches. 

OPERATION. 

42  8.  =  II  =  i^i,  of  a  minute.  Explanation. — We  first  find  what  part  of  the 

dial  circle  is  4  m.  and  42  s.     Tlio  circle   being 

<  1   ,,,    ■i^  47    „f  „  nWn^a  divided  into  fiO  spaces  (i.  e.,  minutes)  wo  find  it 

so        """  to  be  bVi)  ei  tlie  wliole.     This  traction  represents 

6  inches  of  length  of  perimeter.     If  f^J„  =  6  in., 
600  ^^  =  4I,  of  6  and   ^Sg,    or   whole   circle,  =  600 

^  times  /,  of  6  or  ^""7  ^.     Having  thus  found  the 

circumference,    we    find   the   radius,    i.    e.,    the 

12  19056  in      Ans  minute  hand,  by  dividing  by  3.1416  aud  taking 

i  the  quotient. 

71.  A  lady  being  asked  how  old  she  was  at  the  time  of  lier  marriage, 
replied,  that  the  age  of  her  oldest  sou  was  13  years;  that  he  was  born  2  years  after 
her  marriage ;  that  when  she  married,  the  age  of  her  husband  was  three  times  her 
own,  and  that  now  her  husband  was  twice  as  old  as  herself.  How  old  were  she  and 
lier  husband  when  married  ?  Ans.  15  and  45. 

^  SOLUTION. 

Since  the  son,  aged  13,  was  born  2  years  after  tlie  marriage,  she  is  15  years  older  tli.au  when 
married.  Her  husband's  age  then  was  to  hcr's  as  3  :  1 ;  with  15  years  added  to  both  it  was  as  2  :  1, 
or  4  :  2,  t.  e.,  she  is  now  twice  as  old  as  when  married.  But  as  the  addition  of  15  produced  this 
duplication,  15  years  must  have  been  her  ago  and  3  X  15  =  45,  the  age  of  her  husband  at  marriage. 


47 
3.1416 


*  MISCELLANEOUS  PROBLEMS.  965 

72.  A  father  was  34  years  old  when  his  son  was  boru;  his  son  is  now  one- 
half  as  old  as  his  father,  less  4  years.     How  old  is  he  ?  Ans.  26  years. 

SOLUTION. 

AVe  see  that  if  the  son  were  4  years  older,  lie  ■would  be  |  the  father's  age,  therefore  twice  his 
age  plus  twice  4  years  would  e(iual  the  father's  age;  hut  as  the  father's  age  equals  34  -)-  the  sou's 
age,  2x4  =  8,  must  he  the  difference  between  34  and  the  son's  age,  or  34  —  8  =  26.     Ans. 

73.  A  person  being  asked  his  age,  rejilied,  that  if  his  age  were  increased  by 
§  of  itself  and  J  of  itself,  and  16  more,  the  sum  would  equal  3  times  his  age.  What 
was  his  age  ?  27  yrs.,  5  mos.,  4|  days,  Ans. 

SOLUTIOX. 

Assuming  unity  or  1  as  his  age,  and  increasing  this  by  f  +  f  we  have  1  +  f  +  |-  ^  2,\.  Since 
the  addition  of  16  to  this  sum  equals  thrice  his  age,  or  3,  we  see  that  16  must  be  the  difference 
between  2fi  and  3,  i.  e.,  3  —  2,A,  or  j\  =  16.  If  16  is  ,'2  of  his  age,  j^;^  =  f  x  16,  and  the  whole,  or 
■}j  =  12  X  7-  X  16  =  -4^  =  27|  years,  which  reduced  gives  27  years,  5  months,  4f  days.     Ans. 

74.  A  cistern  drains  a  roof  04  feet  by  42,  how  many  gallons  will  be  caught, 
supposing  a  rain-fall  of  J  an  inch  ?  Ans.  837.81  gallons. 

OPERATION. 

2  I  64  Explanation. — We  first  find  the  area  of  roof  in 

I  12  sq.  inches,  then  as  the  fall  is  I  inch,  there  will 

231    42  be  J  as  many  cuh.  inches  of  water  as  there  are 

I  12  sq.  inches  of  surface.     Dividing   this   quotient 

by  231  cub.  inches  =  1  gallon,    we  obtain  the 

I  837.81  +  required  gallous. 

75.  I  can  insure  a  iiiece  of  property  worth  $25000,  by  paying  1 1%  per  cent 
premium  annually,  or  effect  a  permanent  insurance  by  jiaying  13  annual  premiums 
down ;  money  being  worth  7  per  cent,  which  ought  I  to  prefer,  and  how  nuich  will 
I  gain?  Ans.  $85,714  +. 

OPERATION. 

25000  X  lu,  %•  =  300  =  annual  premium. 
300  —  7  5o  =  4285.714  +  =  principal. 
(300  X  13)  +  300  =  4200. 

4285.714  —  4200  =  85.714  +  =  difierence  in  favor  of  permanent  insurance. 

Explanation. — In  order  to  pay  every  year  a  premium  of  $300,  a  sum  must  be  placed  at  interest 
at  7  per  cent,  sufficient  to  produce  it,  i.  e.,  4285.714;  the  difference  between  this  and  13  annual 
premiums  (  =  3900  )  plus  the  premium  for  the  first  year  in  advance,  (i.  e.,  3900  -)-  300  =  4200)  will 
be  the  gain  in  favor  of  paying  in  one  payment. 

76.  A.,  B.,  and  C.  bought  a  grindstone,  each  paying  J  of  the  cost.  The 
diameter  of  the  stone  is  32  inches,  allowing  6  inches  waste  for  aperture,  how  many 
inches  must  each  grind  off,  to  get  an  equal  share. 

Ans.  1st,  2.821  +  in.     2d,  3.621  +  in.     3d,  6.556  +  in. 


966 


soule's  philosophic  practical  mathematics.  * 

SOLUTION. 

Diameter  of  grindstone,  32  in.,  and  32'^  =  1024  proportional  area. 
"  "    aperture,        6  "        "      6^  =      36  "  " 

3  )  988  proportional  area  for  nw. 

Eacli  person's  proportion,     -        -        329^   square  inches. 

And        1024    square  of  A's  diameter,  32  A's  diameter. 

Less  329J  A's  portion, 

Equali)    694J  square  of  I5's  diameter,  and  VG94j  =  26.3565  B's  diameter. 
Diflereuce  bet.  A's  and  B's  diameter  divided  by      2)  5.6434 

gives  thickness  A.  grinds  off,       ...        2.8217  +  inches. 

And         694|{  square  of  U's  diameter. 

Less       329j  B's  portion,  26.3565  B's  diameter, 

Equals  365J  square  of  C's  di.ameter,  and    VSBoJ  ^  19.1136  C's        " 
Difference  bet.  B's  and  C's  diameter,  divided  by     2)  7.2428 

gives  thickness  B.  grinds  off,       ...        3.6214  inches. 

Lees        19.1136  C's  diameter. 

And;  6.  diameter  of  aperture. 

2  )  13.1136  difference  between  C's  diameter,  and  aperture.^  2, 

Gives       6.5568  inches,  thickness  C.  grinds  off. 

Explanation. — Since  the  areas  of  circles  are  proportional  to  the  squares  described  on  their 
diameters,  we  can  most  easily  operate  by  using  the  squares  of  the  diameters,  instead  of  the  exact 
areas  of  the  circles  themselves.  The  remainiug  operations  will  be  apparent  from  an  inspection  of 
the  solution  as  given  above. 

77.  A  cylindrical  column  is  40  feet  high  and  6J  feet  in  circumference;  what 
length  of  rope  will  it  require  to  wind  round  this  column  from  the  bottom  to  the  top, 
by  leaving  a  space  of  2  feet  between  each  coil,  making  no  allowance  for  the  thick- 
ness of  the  rope  ?  Aus.  136.014  +  feet. 

SOLUTION. 
2  )  40      feet  of  cylinder. 

20     number  of  coils  of  rope. 
6J    feet  circumference  of  cylinder. 

130    sum  of  bases  of  the  20  rectangles  made  by  the  rope. 
Then     ViO^  =      16900   square  of  base. 
Aud        40^  =        IGOO  square  of  altitude. 


Vl8500  =  136.014  +  feet,  hypotenuse  or  length  of  rope. 

Explanation. — A  rope  wound  round  a  cylinder  in  a  spiral  line,  will  trace  at  each  circnit,  the 
typotenuse  of  a  right  angled  triangle,  or  diagonal  of  a  rectangle,  of  which  the  base  would  be  the 
circumferenco  fif  tlie  cylinder,  .and  the  altitude,  the  distance  between  the  colls  of  the  rope.  This 
is  easily  demonstrated  Viy  taking  a  rectangular  piece  of  paper  and  drawiug  a  diagonal  with  a 
pencil.  Then  by  joining  the  two  opposite  sides  of  the  paper,  we  have  a  sj)iral  line  around  a 
cylinder,  aud  as  there  will  be  as  many  such  dicTgonals  as  there  are  coils  round  the  cylinder,  or  as 
many  as  the  number  of  times  the  height  of  the  cylinder  is  greater  than  the  height  of  one  coil; 
therefore,  if  we  divide  the  altitude  of  the  cylinder  by  the  distance  between  the  coils,  we  will  have 
the  number  of  eircuits  made  by  the  rope,  then  the  circumference  of  the  cylinder,  multiplied  by  the 
number  of  circuits,  will  give  the  sum  of  the  bases  of  all  the  rectangles.  Then  if  to  the  square  of 
this  base,  we  add  tlie  square  of  the  height  of  the  cylinder,  and  extract  the  square  root  of  the  sum, 
the  result  will  be  tlio  sum  of  all  the  diagonals  or  length  of  the  rope. 


*  MISCELLANEOUS    PROBLEMS.  967 

78.  A  stone  ■\^•eighs  121  ])ouiids;  what  is  the  least  number  of  pieces  into 
which  it  may  be  broken,  and  what  is  the  weight  of  each,  in  order  that  we  can  weigh 
with  the  aid  of  a  pair  of  scales  and  number  of  pounds,  from  1  to  121  T 

Aus.  1,  3,  9,  27,  81. 

SOLUTION. 

In  any  geometrioal  scries  proceeding  in  a  triple  ratio  each  term  is  1  more  tlian  twice  the  sum 
of  all  the  preceding,  and  hence,  tlie  above  series  might  proceed  to  any  extent. 

In  using  the  weights,  they  ninst  be  put  in  one  or  both  sides  of  the  scales,  asmay  he  required ; 
thus  to  weigh  2  ])Ounds,  we  would  put  1  pound  in  one  side  and  3  pouuds  in  the  opposite  side. 

79.  A  nurseryman  had  COOO  trees  whicli  he  wished  to  set  out,  in  a  rectangular 
orchard,  whose  sides  were  as  3  to  5 ;  if  he  placed  them  5  yards  apart,  each  way,  how 
many  trees  wimhl  there  be  in  each  row,  any  how  many  square  yards  of  ground 
would  they  stand  on  ? 

OPERATION. 

3  X  5  =  15  trees  in  proportional  rectangle. 
Then     15)  6000 

400  =  multiple   of  proportional  numher  of  trees  or  area; 
And  y    400  =  20  mnltiple  of  ratio  of  sides. 

Then     20  X     3  =  60  number  of  trees  on  shorter  side; 
and        20  X     5  =  100  nuraljer  of  trees  on  longer  side. 
Proof  100  X   60  =  6000,  whole  number  of  trees. 
Also    (60  —  1)  X  5  =  295  yards  wide  ; 
.and   (100  —  1)  X  5  =  495  yards  long. 
Then  495  X  295  =  146025  square  yards. 

Explanation. — Since  the  numher  of  trees  on  each  side  are  in  the  ratio  of  3  to  5,  if  we  take  3 
trees  one  way,  and  5  the  other,  we  will  have  3  X  5  =  15  trees  in  the  proportional  rectangle;  but 
there  are  6000  trees  to  be  set  out,  hence  the  rec|uired  rectangle  will  bo  as  many  times  greater  than 
the  proportional  rectangle,  as  6000  is  greater  than  15,  or  400  times  greater;  and  sinoe  the  siiles  of 
similar  rectangles  are  as  the  square  roots  of  their  areas,  by  extracting  the  square  root  of  400,  we 
have  20  for  the  multiple  of  the  given  ratios  3  and  5,  "giving  60  trees  on  one  side  and  100  on  the 
other.  Then  to  find  the  length  of  each  side,  since  the  number  of  spaces  is  one  less  than  the  number 
of  trees  iu  a  row,  we  multiply  the  distance  5  yards  by  the  numher  of  trees  less  one. 

SO.  A  drover  bought  100  head  of  sheep,  for  which  he  paid  $100,  paying  $3 
a  piece  for  the  wethers;  $1.50  a  piece  for  the  ewes;  and  60  cents  a  piece  for  the 
lambs  ;  how  many  of  each'  did  he  buy? 

Note. — For  the  solution,  see  page  801. 

81.  If  12  oxen  consume  3J  acres  of  grass  in  4  weeks,  and  21  oxen  consume 
10  acres  in  9  weeks,  how  many  oxen  will  consume  21  acres  in  18  weeks,  the  grass 
being  at  first  equal  on  every  acre,  and  growing  uniformly  and  constantly  ? 

Ans.  3G, 

The  analysis  of  this  problem  naturally  divides  itself  into  two  distinct  parts, 
Tiz  :  1st,  the  number  of  oxen  the  grass  already  groirn  will  support  for  a  given  time- 
and  2d,  the  number  of  oxen  the  increase  only  will  support.  In  the  first  case  the 
number  of  oxen  supported  varies  directly  as  the  number  of  acres  grazed  and  inver- 


968  SOULe's    philosophic    rUACTICAL    MATHEMATICS.  * 

sely  as  the  time.     In  the  second  case  the  number  of  oxen  supported  varies  directly 
only  as  the  number  of  acres  grazed,  time  being  irrelevant. 

SOLUTION. 

If  3J  acre.s  ■with  4  Tvceljs  incre<a8e  Biipport  12  oxen  for  4  weeks,  10  acres  -would  support  for  the 
s.ame  time  10  X  12  ^  36  oxen,  ami  if  10  acres  (with  4  weeks  increase)  support   36  oxen  for  4  weeks 

W~ 
they  woulil  support  for  9  weeks  J  X  36  =  16  oxen.  But  by  the  problem,  10  acres  with  9  weeks 
increase  would  sii]>]Kirt  21  oxen,  hence  the  5  weeks  tlifference  of  increase  must  support  21  —  16  :=  .5 
oxen  clurinu'  the  t)  weeks,  and  the  whole  9  weeks  increase  of  10  acres  would  support  H  x  5  =  9  oxen 
for  same  time.  Then  since  the  grass  already  grown  jilus  the  9  weeks  increase  supports  21  oxeu  and 
the  increase  alone  supports  9  oxen,  the  uumber  sui)ported  by  that  already  grown  will  bo  21  —  9  -■= 
12  oxen. 

Now  since  the  grass  already  grown  of  10  acres  will  support  12  oxen  for  9  weeks,  the  grass  of 
24  acres  will  support  ;  J  X  12  =  28*  oxeu  for  9  weeks,  and  for  18  weeks  i  x  28}  =  14 J  oxen. 

And  again,  since  the  increase  of  10  acres  supports  9  oxen,  the  increase  of  24  acres  will,  by 
the  principal  in  the  second  case,  support  f  J  x  9  =  21§  oxeu  auy  length  of  time. 

Summing  )ip,  we  have  the  already  grown  grass  of  24  acres  supporting  14|  oxen  for  18  weeks, 
and  thp  increase  of  24  acres  suijportiug  213  oxen  for  same  time;  by  addition  14|  +  21'  =  36  oxen, 
Ans. 

82.  If  two  wagon  wheels,  one  5i  and  the  other  C  feet  in  diameter  be  fixed 
upon  an  axle,  seveti  feet  apart,  and  moved  round  on  their  resulting  line  of  direction ; 
how  large  circles  will  they  describe?  Ans.  168  feet  diameter  of  large  circle. 

15-4        "  "     small    " 

.■SOLUTION. 

6  —  oj  =:  i  foot,  difference  of  diameter. 
Then     i   :   6    :  :  7  :  84  radius  X  2  =  168  feet  diameter  of  Large  circle, 
and  less       7 

Or,        I  :   5J  :  :  7  :  77  radius,  X  2  =  154  feet  diameter  of  small  circle. 

Explanation. — Since  the  wheels  are  fixed  on  the  axle,  tbey  will  make  the  same  number  of 
revolutions,  and  since  one  is  smaller  than  the  other,  they  will  describe  concentric  circles,  and  the 
circumferences  of  these  circles  will  be  proportional  to  the  circumferences  of  the  wheels,  but  the 
diameters  of  the  wheels  are  proportion.al  to  their  circumferences,  hence  we  m.ay  operate  directly 
from  the  diameters.  Therefore,  if  we  <lraw  ,a  line  from  the  top  of  the  larger  wheel  to  the  top  of 
the  smaller,  and  extend  it,  it  will  touch  the  ground  at  the  centre  of  the  circles,  and  form  the 
hypotenuse  of  two  right  angled  triangles,  whose  liases  are  the  radii  of  the  circles,  and  altitudes 
tile  diameter  or  height  of  the  wheels;  then  if  we  draw  a  straight  line  from  the  top  of  the  lower 
wheel  to  the  higher  wheel,  parallel  to  the  ground  or  base,  we  will  have  .a  similar  triangle,  whose 
base  is  the  distance  between  the  wheels,  and  altitude  the  difference  between  the  heights  of  the 
wheels.  Then,  since  the  sides  of  similar  triangles  are  pronjirtioual,  wo  have  the  proportion  as 
given  above,  viz:  The  difference  in  the  height  of  the  wheels,  is  to  the  height  of  cither  wheel,  as 
the  distance  between  the  wheels,  is  to  the  distance  from  that  wheel  to  the  centre  of  the  circle. 
From  the  above  demonstration  we  see,  that  in  questions  of  this  kind,  the  diameter  of  the  circles  is 
equal  to  twice  the  jiroduct  of  the  height  of  the  corresponding  wheel.'}  by  the  distance  between 
them,  divided  by  the  diii'ereuco  in  their  diameters. 

83.  Three  persons  bought  a  sugar  loaf  in  the  form  of  a  perfect  cone  25  inches 
high,  and  agreed  to  divide  it  equally  by  sections  parallel  to  the  base;  what  was  the 
slant  height  of  each  one's  share,  the  base  being  13  inches  in  diameter  f 

Ans.  Upper  section,  17.826  +  inch. 
Middle        "         4.633  +     " 
Lower         "         3.250  +     " 


*  MISCELLANEOUS    PROBLEMS.  969 

The  analysis  of  this  problem  is  based  upon  the  Geometrical  principles  that 
the  volumes  of  similar  cones  vary  as  the  cubes  of  the  radii  of  their  bases  and  also 
as  the  cubes  of  their  altitudes. 

SOLUTION'. 
Since  tlie  altitude  of  t)ie  cone  constituting  the  upper  J  of  the  loaf  deducted  from  the  whole 
altitude  gives  the  height  of  the  lower  section,  we  have  for  the  altitude  of  the  upper  cone  V  3  :  V  2 
=  25  :  required  altitude,  or  required  altitude  =  V  ^  x  25,  from  which  we  have, 

1.25992  X  25  -i-  1.44224  =  21.839  +  in.  altitude  of  f. 
25  —  21.839  =  3.161  +  in.  altitude  of  lower  section. 
In  same  way  for  the  cone  forming  the  upper  section  we  have, 
3 .  _ .    3  ,_ 
V  J   •   V  I  ^  25  :  required  altitude,  or, 

3,_ 

Required  altitude  =  v  ,  X  25,   which  gives  25  x  1.44224  =  17.334  +  inches,   altitude    of 

upper  section. 

And  21.839  —  17.334  =  4.505  +  altitude  of  middle  section. 

Then  by  means  of  the  ratios  of  the  cubes  of  the  radii  we  find  in  the  same  manner: 
Slant  height  upper  section  =  17.826   inches. 
"  "       middle     "        =    4.633 

"  "       lower       "        =    3.250        "  Ans. 

84.  A  father  leaves  a  number  of  children  and  a  certain  sum  of  money  to  be 
divided  among  them  as  follows : 

The  first  is  to  receive  $100  and  i^o  of  the  remainder;  the  second  is  to  receive 
$200  and  yV  part  of  what  theii  remains ;  again,  the  third  is  to  receive  $300  and  -^g 
of  the  residue,  and  so  on ;  each  succeeding  child  is  to  receive  $100  more  than  the 
one  immediately  preceding,  and  then  -^^  part  of  what  still  remains.  At  last  it  is 
found  that  all  the  children  have  received  equal  sums.  What  was  the  amount  left, 
and  how  many  children  were  there  ? 
Note. — For  the  solution,  see  page  876. 

85.  The  diameter  of  a  cylindrical  cistern  is  6  feet,  8  inches,  what  must  be 
its  height  to  hold  3000  gallons?  Ans.  137.SG7G  in.  height. 

OPERATION. 

3000     gallons. 
231    cu.  in.  1  gallon. 


693000     en.  in.  in  a  cistern  to  hold  3000  gallons. 

80    in.  diameter  of  cistern. 

80      " 

6400     square  of  diameter. 

.7854    ratio  of  the  area  of  a  square  to  the  area  of  a,  circle. 


5026.5600     area  of  the  cistern. 

693000  -^  5026.56  =  137.8676  inches  height 

80 

80 

or  thus:           .7854 

3000   ■ 
231 

137.8676  in.  =  11  feet,  5.867( 

970  soule's  philosophic  practical  mathematics.  * 

86.    The  height  of  a  cylindrical  cistern  is  7  feet,  6  inches,  what  must  be  the 
diameter  so  that  it  will  hold  2000  gallons  ?  Ans.  80.84  inches  diameter. 


OPERATION. 


fi535.9477 
V6535.9477  =  80.84  +  inches.     Ans. 


90  I  2000 
.7854     231 


Note. — For  more  extended  work  of  this  character,  see  the  Measurement  of  Solids,  pages  370 
to  418. 

87.  A  merchant  sold  goods  to  the  amonnt  of  $10,  and  in  payment  therefor 
the  purchaser  handed  him  a  $.^0  bill,  but  not  having  change,  he  stepped  into  thfe 
adjoining  store  and  had  the  bill  changed.  lie  then  returned  $40  to  the  purchaser 
and  placed  $10  in  his  cash  drawer.  A  few  hours  afterwards  the  party  that  had 
changed  the  $50  bill  rettirned  it  as  a  counterfeit,  and  to  redeem  it  the  merchant 
took  the  $10  from  the  drawer  and  borrowed  $10  from  a  third  party.  How  much  did 
he  lose  by  reason  of  the  counterfeit  bill  ? 

Ans.  $10  worth  of  goods  and  $40  in  money. 

SOLUTION. 

By  the  question,  vre  see  that  he  first  gave  $40  in  money  and  $10  in  goods  to  the  man  who  gave 
hira  the  counterfeit  hill,  then  borrowed  $40  from  a  third  party  and  paid  $50  to  the  man  that 
changed  the  hill. 

When  the  merchant  sold  the  goods  and  made  change  through  a  third  party  there  was  no  loss 
sustained,  for  he  gave  hut  an  equivalent  for  a  hill  supposed  to  he  good. 

When  he  redeemed  the  counterfeit  bill  he  lost  $40,  the  amount  borrowed,  and  the  $10  received 
for  the  goods,  but  as  the  $10  was  received  for  the  goods,  it  was  not,  therefor,  a  loss  of  $10  in  money 
but  of  $10  worth  of  goods.     Hence  his  loss  was  $40  in  money  and  $10  in  goods,  or  thus : 

After  redeeming  the  counterfeit  bill  the  merchant  owed  $40  borrowed  money  and  had 
disposed  of  $10  worth  of  goods.  And  as  he  hiid  no  money  on  hand  before  selling  the  $10  worth  of 
goods,  or  after  redeeming  the  counterfeit  bill,  it  is  hence  clear,  that  ho  lost  $40  in  money  and  $10 
•worth  of  goods. 

88.  The  price  of  corn  is  SO  cents  per  bushel.     What  is  it  worth,  jier  cental? 

Ans.  $1.42f. 

Note. — For  the  solution  of  this  and  the  following  problem,  see  page  279. 

89.  The  i^rice  of  wheat  is  $2.50  per  cental.     What  is  it  worth  per  bushel  ? 

Ans.  $1.50. 

90.  A  merchant  in  Mobile  has  a  balance  of  $5000  due  him  in  Louisville,  and 
he  instructs  the  Louisville  merchant  to  remit  by  a  draft,  payable  CO  days  after 
sight,  exchange  being  |  per  cent  premium.  The  Louisville  merchant,  through  eiTor, 
remitted  a  sight  draft,  which,  when  received  in  Mobile,  is  accepted  and  paid  after 
the  exiiiration  of  three  days  of  grace.  2^ow  if  the  Mobile  merchant  loans  this 
money  at  8  per  cent,  and  allowing  that  money  is  worth  9  per  cent  in  Louisville,  will 
he  gain  or  lose  by  the  error  of  the  Louisville  merchant?  Ans.  Loses  $12.87. 


MISCELLANEOUS    PROBLEMS. 


971 


OPERATION 

To  fiud  the  face  of  the  siglit  draft  sent. 

$ 

I  100 
100.50  I  5000.00 

I  $4975.124 

OPERATION 

To  find  the  value  of  the  sight  draft  at  the 
expiration  of  60  ds.  at  8  per  cent. 


$ 

I  4975.124 
360  I  8 
60 


$66.3349  interest. 
4975.124 

$5041.4589 


OPKRATIOX 


To  find  the  face  of  a  60  day  draft,  sight  exchange 

heing  i  per  cent  premium  and  money 

■worth  9  per  cent. 


360 


100 

9 

63 

$1.57i 


$100 
1.57i 

$98.42^ 
.50 


=  i   ° 


$98.92i 


197.85 


;„  i)reniiuni. 


100 

2 

5000.00 


$5054.33 

OPEKATIOX 

To  find  the  loss. 


V.alue  of  60  day  draft, 

"       "   sight"draft, 

Loss,     - 


$5054.33 
5041.46 


$    12.87 


Note. — Sight  Drafts  in  Mobile  aie  entitled  to  3  days  grace. 

For  more  extended  explanations  on  this  character  of  problems,  see  pages  730  to  736. 


91.     Sold  a  watcli  for 
What  was  the  cost  ? 


L7.1G  aud  lost  as  much  per  cent  as  the  vatch  cost. 
1st  Aus.  $78.       2d  Aus.  $22. 


Note  1. — It  is  generally  believed  that  an  arithmetical  problem  can  h.ave  bnt  one  correct 
answer;  but  this  belief  is  clearly  erroneous.  For  if  we  apply  the  conditions  of  the  sale  to  both  of 
the  answers  to  this  problem,  we  prove  their  correctness. 

Note  2. — All  problems  of  this  kind,  provided  the  selling  price  does  not  exceed  $25  or  25  of 
any  other  monetary  unit,  may  be  solved  and  the  two  answers  obtained  by  the  following  method, 
which  is  derived  from  an  Algebraic  solution  : 

Subtract  the  selling  price  from  25,  and  to  the  square  root  of  the  remainder  add  5;  then 
multiply  this  sum  by  10  and  the  product  will  be  the  greater  .answer.  Or,  from  5  subtract  the  square 
root  above  obtained  aud  multiply  the  difi'erence  by  10;  the  result  will  be  the  lesser  answer. 

operation, 

25  —  17.16  =  7.84;  then  the  square  root  of  7.84  is  2.8;  then  2.8  +  5  =  7.8;  7.8  x  10  =  $78 
the  greater  cost.     And  5  —  2.8  =  2.2;  then  2.2  X  10  =  $22  the  lesser  cost. 


$78  cost  —  (78%"  =  $60.84)  =  $17.16  selling  price. 
$22  cost  —  (22»^  =  $4.84)  =  $17.16  selling  price. 

92.     Sold  goods  for  $12.75  aud  lost  as  much  i>er  ceut  as  they  cost.    What  did 
they  cost?  Aus.  $85;  or  $15. 


972  SOULE  s  rmi.osoPHic  practical  mathematics.  ♦ 

93.  Sold  goods  for  $39  and  gained  as  much  per  cent  as  they  cost.    How 
much  did  they  cost  ?  Ans.  $30. 

Note. — All  jnolilcms  of  this  kind  may  bo  solved  by  the  following  method: 
To  the  sellinjt  jirice  add  ifL'5;  from  the  S(|naro  root  of  this  sum  subtract  5;  then  multiply  the 
remainder  by  10,  aud  the  product  'will  be  the  cost. 

OPEIiATIOX. 

$39  -f  $25  =  $64  ;  the  square  root  of  $64  is  $8;  $8  —  $5  =  $3;  $3  X  10  =  $30  cost,  Ans. 

94.  Sold  goods  for  $50,904,  and  gained  as  much  per  cent  as  they  cost.    AYhat 
did  they  cost  1  Ans.  $40J. 

95.  If  an  article  had  cost  20  per  cent  less,  the  gain  would  have  been  30  per 
cent  more.     What  was  the  per  cent  gain  ?  Ans.  20%. 

Note. — This  and  the  foUowini;  probli>m  are  purely  Algebraic,  and  can  only  be  -worked  arith- 
metically, by  processes  deduced  or  abbreviated  from  Algebraic  equations.  Heuce,  iu  the  operatious 
giveu,  no  arithmetical  reasoning  is  presented. 

OPEKATION. 
$100  cost  assumed. 
20%  less. 

$20.00  =  20",;  less. 

100.00  =  cost  as  above.  | 

$80.00  =  decreased  cost.  20  |  100 

30%      more.  — 


$24.00  =  30%  more. 
20.00  =  20%  less. 


I  gain  =  20%  gain.     Ans. 


$  4        —  gain  on  $20  less. 


96.    If  the  cost  had  been  12  per  cent  more,  the  gain  would  have  been  15  per 
cent  less.     What  was  the  per  cent  gain  ?  Ans.  40%. 


OPERATION. 
100  cost  assumed. 
12%  more. 

$12.00  =  12%  more. 

100.00  =  cost  as  above.  $ 

'  I  4.80 

$112.00  =  increased  cost.  12.00  |  100 

■  15%       less. 


$10.80  =  15";;  less. 
12.00  =  12"'  more. 


40%  gain,  Ans. 


$  4.80  =  gain  on  $12  more. 

97.     What  is  the  value  of  30  —  (2.4  —  S.37  +  21.025)  -f  3.40  x  .12? 

Ans.  14.7002. 

OPERATION. 

2.4  —  8.37  =  —  5.97;  —  5.97  +  21.625  =  15.6.55,  the  result  of  the  parenthesis  ;   3.46  X  .12  =: 
.4152.     Then  30  —  15.655  =  14.345;  14.345  -f  .4152  =  14.7602,  Ans. 

Or, 

2.4  +  21.625  =  24.025;  24.025  —  8.37  =  15.655,  tbe  result  of  the  parenthesis.     30  —  15.655  = 
14.345;  3.46  X  .12  =  .4152;  14.345  +  .4152  =  14.7602,  Ans. 

Note. — The  above  and  the  following  problem  are  Algebraic  equations,   aud  the  Algebraic 
principles  involved  in  their  solution  are  as  follows: 


MISCELLANEOUS    PROBLEMS. 


973 


In  ;ill  Algebraic  ami  aritbinetical  equations  or  operations  indicated  liy  tlie +,  — ,  X,  or  ^- 
signs,  and  in  simplifying  fractions,  the  application  of  the  sifliis  of  operation  should  he  well  under- 
stood, as  w;is  ex])lained  on  page  134.  But  to  iiuderstaud  thoroughly,  the  elucidation  of  the  pre- 
ceding problem,  us  well  as  that  of  the  following  and  all  others  of  a  similar  character,  we  deem  a 
repetition  of  the  principles  necessary  and  state  them  as  follows: 

The  signs  -j-  -ind  —  aftect  only  the  number  or  expression  which  immediately  follows  either 
of  them. 

The  signs  X  and  -:-  indicate  an  operation  to  bo  performed  with  the  numbers  or  expressions 
between  which  either  may  be,  and  this  operation  must  be  iierformed  before  that  of  any  -f-  or  — 
which  may  immediately  precede  or  follow. 

The  parenthesis,  (   ),  or  the  vinculum, ,  indicate  that  the  quantities  which  they  include 

must  be  simplilied  into  one  expression  before  they  can  be  connected  to  any  other  quantity. 

The  operation  of  combining  and  simplifying  always  begins  with  the  (   )  or if  any,  and 

then  with  the  operations  indicated  by  the  X  and-^  sigus.  The  operations  of  -(-  and  —  are  then 
performed  and  are  generally  considered  from  left  to  right. 

It  will  bo  observed  that  the  result  of  the  parenthesis  in  the  above  equation  was  subtracted 
from  the  quantity  before  it,  as  indicated  by  the  minus  sign  between  them ;  this  was  done  because 
the  result  of  the  "parenthesis  was  a  plus  quantity.  Had  it  been  a  minus  quantity,  i.  €.,  had  the 
result  been  a  quantity  jireceded  by  a  minus  sign,  it  would  then  have  Iveen  added  to  the  quantity 
before  the  parenthesis,  for  the  reason  that  the  minus  sign  before  the  parenthesis  would  cancel  the 
minus  sign  before  the  result  of  the  parenthesis  and  thereby  produce  a  plus  quantity.  Should  a 
plus  sign  precede  the  parenthesis,  the  operation  to  be  performed  with  the  result  of  the  parenthesis 
would  be  in  accordance  with  the  sign  before  the  resulting  quantity  and  independent  of  the  -|-  sign 
before  the  (  ). 


9S.     20  —  30  -f  2  X  (5  =  wbat  ? 
99.     Show  that  H  i^  greater  than 


Il- 


ia—3 


-  and  less  than  ^„  ,  „ 


100.     Express  deciniallv  the  value  of  ,    1',, — ^-^ 

i0l(5—  lb) 


101.     Simplify  (1  -f 


l+^\ 


-^(^  +  rTl) 


Ans.  2. 

Aus.  2.429';+. 
Ans.  -.;%-. 


102.  A  cyliiulrieal  oil  tank,  2,3  feet  11  inches  long,  and  having  an  inside 
diameter  of  G  feet  4f  inches,  was  full  of  oil.  When  the  tank  was  nieasured  it 
was  found  that  oil  to  the  depth  of  6  inches  of  the  diameter  had  leakeil  out.  How 
many  gallons  of  oil  were  lost? 


OPERATION. 


76.75  inches  diameter 
70.75  inches  full. 


76.75)  6.0000000  (.07817+ 


6       inches  empty. 


.078  =  .02835 
.079  =  .02889 


.00054 
".17 


.02835 
.001109 

.02844 


4 

4 

231 


.0000918 


311 
307 
307 
.7854 


231 


.02844 
76.75 
76.75 
311 

225.544  gallons  leaked  out. 


6228.680-f  gallons  if  tank  were  full. 
KoTE. —  See  Problem  1,  page  398,  for  explanation  of  similar  problems. 


974  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  * 

103.  The  diameter  of  a  safety  valve  is  4J,  inches,  and  it  is  placed  4^  inches 
from  the  fnlcnun.  Tlio  weight  of  the  lever  and  safety  valve  is  140  pounds.  What 
weight  must  be  attached  to  the  end  of  a  48  inch  lever,  operating  the  safety  valve,  ta 
withstand  a  jjiessure  of  100  jjounds  to  the  square  inch?        Ans.  98.505+  pounds. 


OPKRATION    INDICATED. 


9 
9 

.7854 

15.90435  squ-ire  inclics  in  surface  of  valve. 


15.90435 

IGU 

9 

11451.132  lbs.  pressure  ou  valve. 


11451.132  -;-  48  =  238. .565+  poniuls  Tveight  to  witliRtand  pressure  of  160  pounds  to  the  square 
incli.     238.565  —  140  =  98.565+  ]iouiid.s  to  be  added  to  the  eud  of  the  lever. 

Note. — See  Mechanical  Powers,  page  419,  and  problem  2,  page  422. 

104.  Allowing  the  journals  of  a  car  to  bo  8  inches  long,  and  the  projected 
surface  on  the  brass  to  be  7^  inches  long  by  3i  inches  wide,  what  is  the  actual  weight 
per  squaie  inch  ou  each  of  the  8  journals  of  a  car  which  weighs  20000  pounds  and 
is  laden  with  32  tons  of  freiglit?  Ans.  414.7+  pounds  per  sq.  in. 

FIRST  OPERATION.  SECOND   OPERATION. 

7J  X  3i  =  274  sq.  in.  surface  bearing  on  each  journal.  I  90000 

31  I  4 
64000  +  26000  =  90000  pounds,  car  and  freight.  7     2 

8  I 
90000  —  (274  X  8)  =  414.7+  pounds  per  sq.  in.     Ans. 


414.7+  lbs.     Ans. 

CURIOUS  AND  AMUSING  QUESTIONS. 

1524.     1.     A  sea  captain  on  a  voyage,  had  a  crew  of  30  men,  half  of  whom 

were  blacks.     Being  becalmed  on  the  passage  for  a  long  time,  their  provisions  began 

to  fail,  and  the  captain  became  satisfied,  that  unless  the  number  of  men  was  greatly 

diminished,  all  would  iierish  of  hunger  before  they  could  reach  any  friendly  i^ort. 

lie  therefore  jiroposed  to  the  sailors  that  they  should  stand  in  a  row  on  deck,  and 

that  every  ninth  mail  should  1)6  thrown  overboard  until  one-half  of  the  crew  were 

thus  destroyed.     To  this  they  all  agreed.     Eow  should  they  stand  so  as  to  save  the 

whites  ?  Ans.  wwwwbbbbbwwbwwwbwbbwwbbbwbbwwb. 

Note. — This  result  might  have  been  easily  ascertained  before  liand  by  the  captain  by  the 
following  method.  Place  30  letters,  as  the  letter  w,  it  being  the  initial  of  white  in  a  row;  then 
commence  counting  off  every  ninth  one  which  may  be  cancelled,  when  15  have  in  this  way  beea 
cancelled,  their  jjlaces  may  be  supplied  with  the  letter  b,  the  initial  of  black. 

2.  A.  and  B.  have  an  eight  gallon  cask  full  of  wine,  and  wish  to  make  an 
equal  division  of  it,  but  have  only  two  empty  vessels  with  which  to  do  it,  one  of 
which  contains  5  gallons,  and  the  other  3.  By  what  means  shall  they  accomplish 
this  division  ? 

SOLUTION. 

Fill  the  3  and  pour  it  into  the  5;  then  fill  it  again  and  from  it  fill  up  the  5,  which  will  leave 
one  gallon  in  the  3  gallon  keg,  empty  the  5  into  tbe  eight  and  jiour  the  one  from  the  3  into  the  5; 
fill  the  3  again  and  emjity  it  into  the  5.  Theu  there  will  bo  4  gallons  iu  the  5  gallon  keg,  and  the 
same  left  iu  the  8. 


"*  CURIOUS    AND    AMUSING    QUESTIONS.  9/5 

3.  A  countryman  having  a  fox,  a  goose  and  a  peck  of  corn,  came  to  a  river, 
where  it  so  happened  that  he  could  carry  but  one  over  at  a  time.  ISfow  as  no  two 
were  to  be  left  together  that  might  destroy  each  other,  he  was  at  his  wit's  end,  for 
says  he :  "■  Though  the  corn  can't  eat  the  goose,  nor  the  goose  eat  the  fox ;  yet  the 
fox  can  eat  the  goose,  and  the  goose  eat  the  corn."  How  shall  he  carry  them  over, 
that  they  shall  not  destroy  each  other? 

SOLUTION. 

Let  liim  first  take  over  the  goose,  leaving  the  fox  and  corn,  then  let  him  take  over  the  fox 
and  hring  the  goose  back,  then  take  over  the  corn,  and  lastly  take  over  the  goose  again. 

4.  A  man,  his  wife,  and  two  sons  desire  to  cross  a  river.  They  have  a  boat 
that  will  carry  but  100  pounds.  The  man  weighs  100  pounds,  the  woman  weighs 
100  pounds  and  each  boy  weighs  50  pounds.  How  can  they  all  cross  the  river  iu 
the  boat? 

OPERATION. 

1.  Let  the  2  hoys  go  over,  and  1  hoy  return  with  the  hoat. 

2.  Let  the  man  go  over,  and  the  hoy  return  ■with  the  boat. 

3.  Let  the  2  boys  go  over,  and  1  boy  return  with  the  boat. 

4.  Let  the  woman  go  over,  and  the  boy  return  with  the  boat, 

5.  Let  the  2  boys  go  over. 

5.  Three  jealous  husbands  with  their  wives  are  to  pass  over  a  river  in  a  boat 
which  can  carry  but  two  at  a  time.  How  can  these  six  jiersons  row  themselves 
over  two  at  a  time,  so  that  none  of  the  wives  may  be  found  iu  the  company  of  one 
or  two  men  unless  her  husband  be  present? 

SOLUTION. 

This  may  be  effected  in  two  or  three  ways ;  the  following  may  be  as  good  as  any  :  Let  A.  and 
wife  go  over;  let  A.  return ;  let  B's  and  C's  wives  go  over;  A's  wife  returns;  B.  and  C.  go  over;  B. 
and  wife  return  ;  A.  and  B.  go  over;  C's  wife  returns,  and  A's  and  B's  wives  go  over;  then  C.  comes 
back  for  his  wife.  i 

Simple  as  this  question  may  appear,  it  is  found  iu  the  works  of  Alcinm,  who  flourished  a 
thousand  years  ago,  hundreds  of  years  before  the  art  of  printing  was  invented. 

C.  How  can  you  arrange  the  9  significant  figures  so  that  when  added  the 
sum  will  be  100  ?  Ans.  75f . 

4 

8|. 
9 

100. 

7.  Prove  that  a  pound  and  a  half  a  cheese  weighs  more  than  2  pounds  of 
butter. 

8.  How  much  heavier  is  a  pound  of  iron  than  a  pound  of  gold  ? 

Ans.  1240  grains  Avoirdupois.     1020Jf  grains  Troy, 

9.  How  can  5  be  taken  from  5  and  leave  a  remainder  of  5  ? 


976  soule's  philosophic  practical  mathematics.  * 

10.  How  can  45  be  taken  from  45  and  leave  a  remainder  of  45  ? 

OPKT5ATION. 

9  +  8  +  7  +  6  + 5 +  4  +  3  +  2  +  1=  45 
1  +  2  +  3  +  4  +  5  +  6  +  7  +  8  +  9  =  45 

8  +  6  +  4  +  1  +  9  +  7  +  5  +  3  +  2  =  45     Remainder. 

11.  What  can  yon  prefix  to  IX  to  make  it  C.  Ans.  S. 

12.  A  student  of  mathematical  logic  proves  that  a  cat  has  3  tails  by  the 
following  process  of  reasoning: 

1.  No  cat  has  two  tails ; 

2.  One  cat  has  1  tail  more  than  no  cat ; 

3.  Now  since  no  cat  has  2  tails  and  since  1  cat  has  1  tail  more 

than  no  cat,  therefore  1  cat  has  3  tails. 
AVhere  is  the  error  in  the  logic  ? 

13.  A  young  hoodlum,  a  modern  evolvement  of  the  human  race,  stole  a  basket 
of  peaches  and  divided  tliem  among  3  bi'other  hoodlums  and  himself  as  follows:  To 
the  first,  he  gave  ^  of  the  whole  number  and  ^  of  a  peach  more ;  to  the  second,  be 
gave  J  of  what  remained  and  J  of  a  peach  more ;  to  the  third,  he  gave  J  of  what 
remained  and  J  of  a  peach  more.  The  stealer  retained  what  was  then  left  for 
himself,  M'hicli  was  i  the  number  he  gave  to  the  first  hoodlum.  What  was  the 
number  of  peaches  stolen,  and  how  many  did  each  hoodlum  receive? 

Ans.  7  peaches  were  stolen. 
1st  hoodlum  received  2. 
2d         "  "        2. 

3d         "  "         2. 

The  stealer  had  1. 

SOLUTION. 

Ill  all  problems  of  this  kind  the  following  principle  prevails:  That  when  the  fractional  parts 
of  the  successive  portions  are  consecutively  increasing,  each  of  the  parts  or  shares  of  the  several 
persons  are  equal,  exccjit  the  last  one,  which  from  the  nature  of  things  must  he  one  less  than  the 
average.  Hence  in  the  given  problem,  the  fact  stated  that  the  last  one's  share  is  half  of  the  first 
(and  each  of  the  others  also)  and  from  the  above  principle  that  it  nnist  be  one  less,  it  must 
necessarily  be  one,  and  since  each  of  the  others  is  one  greater,  they  must  be  2  each,  and  since  there 
are  three  persons  having  ecjual  jiortious,  there  will  be  6  +  the  1  for  the  last,  making  7  in  all. 
Hence,  to  solve  problems  of  this  kind  : 

Add  1  to  the  last  number,  (one),  and  multiply  by  the  number  of  i)er»ons,  and  then  subtract 
1  from  the  product. 

14.  If  3  men  eat  3  dozen  oysters  in  3  minutes,  how  many  dozen  oysters  can 
100  men  eat  in  100  minutes?  Ans.  3333^  doz.  oysters. 

15.  If  1-h  cats  catch  14  rats  in  IJ  miiuites,  how  many  cats  will  it  take  to  catch 
200  rats  in  50  minutes  ?  Ans.  6  cats. 

10.  If  IJ  hens  lay  li  eggs  in  li  days,  how  many  eggs  will  6  hens  lay  in  7 
tliiys  ?  Proportional  Answer,  28  eggs.     Logical  Answer,  24  eggs. 

Note. — For  the  solution  of  the  three  preceding  problems,  see  page  320. 


A.ppErNniji:. 


luilding  and  Loan  Associations. 


Building  aud  Loan  Associations  are  co-operative  corporations,  institu- 
ted for  the  purpose  of  raising  a  fund  by  periodical  subscriptions  or  deposits, 
interest  on  loans,  premiums  or  discounts,  etc.,  for  the  following  purposes:  1.  To 
assist  members  in  building  or  purchasing  homes,  or  to  loan  to  members  on  mortgages 
for  any  purpose.  2.  To  serve  as  a  depository  for  such  members  as  desire  simply 
to  invest  their  money  in  its  shares  and  to  participate  in  the  profits. 

Note. — In  some  of  the  States,  Building  Associations  are  permitted,  under  certain  restrictions, 
to  loan  money  to  persons  who  are  not  members.  It  is  claimed  by  some  of  the  -wisest  linancial  men 
in  these  Associations  that  this  privilege  should  be  general,  under  certain  equitable  regulations  in 
behalf  of  members. 

Associations  of  this  character  are  known  under  various  names,  such  as 
Building  Associations,  Loan  aud  Investment  Companies,  Mutual  Homestead  Asso- 
ciations, Saving  Fund  Associations,  Mutual  Benefit  Societies,  etc.,  etc. 

These  institutions  are  practically  Co-oijerative  Banking  Institutions,  having 
within  their  charter  rights,  special  privileges  and  advantages  as  to  the  numner  of 
loaning  their  funds  at  high  rates  of  discount  or  premium,  and  of  taking  mortgage 
securities,  etc. 

The  usury  laws  do  not  apply  to  Building,  and  Loan  Associations  in  many  of 
the  States  of  the  union. 

These  institutions  had  their  origin  in  England  in  1795,  were  established  in 
Philadelphia  in  1831,  and  have  become  eminently  popular  throughout  the  United 
States  and  Europe.  They  have  done  and  are  doing  a  great  financial  and  economic 
work  for  private  citizens,  and  through  them,  as  self-supporting,  tax-paying  and 
happy  people,  the  resultant  good  to  States  and  Nations  is  clearly  apparent. 

Through  these  Associations,  nnuiy  persons  of  very  moderate  earnings  are 
encouraged  and  enabled  to  profitably  invest  their  savings,  and  by  the  aid  of  the 
Associations  to  buy  or  to  build  homes. 

The  Capital,  or  the  authorized  capital  of  an  Association,  is  the  sum  of 
the  shares  as  set  forth  in  the  Charter,  the  capital  clause  of  which  generally  reads 
as  follows :  The  capital  of  this  corporation  shall  be  Five  Hundred  Thousand  Dollars, 
and  shall  be  issued  in  shares  of  $200  each,  making  2500  shares. 

Note. — The  amount  of  capital  or  authorized  capital  varies  in  dirt'erent  Associations  from 
550000  to  §50000000. 

The  Members  of  au  Association  are  the  parties  who  subscribe  for 
sliares,  and  are  of  two  classes,  BORROWING  and  depositing  members. 

(977) 


978 


SOULE  S    I'lIILOSOPHIC     I'KACTICAL    MATHEMATICS. 


Borrowing  Members  are  those  subscribers  who  borrow  money  from 
tlie  Association. 

Depositing  Members  are  tliose  -vvlio  subscribe  for  sliares  as  an  invest- 
ment 

The  Shares  are  generally  issued  in  series  of  a  specified  sum  for  eacli 
series,  and  at  periodic  times  according  to  the  demands  of  tlie  business  and  tlie 
judgment  of  the  Board  of  Directors.  Each  series  issued  is  generally  closed  when- 
ever the  installments  and  i)rofit8  Lave  brought  the  shares  of  the  series  to  i>ar.  In 
some  cases  the  series  is  closed  before  the  shares  reach  par.  "When  a  new  series  is 
issued  no  more  shares  of  a  ^irior  series  are  issued. 

In  some  associations  the  full  paid  shares  are  licld  as  deposits,  and  draw  a 
sjiecified  rate  of  interest  in  favor  of  the  owner. 

Members  are  Ihnitcd  to  a  si)eciiied  number  of  shares,  which  each  may  own ; 
in  some  associations  they  are  subject  to  Jine  for  certain  delays  in  making  payment 
of  subscriptionf,  and  for  failure  to  make  payments  of  installments  and  interest  for 
a  sj)ecified  time,  their  shares  may  become  forfeited  to  the  Association. 

NoTK. — Re<;ardiiig  fines  anil   the   forfeitnre   of  shares,    tlie   Charters   and   By-Laws   of  the 
various  associations  jiruscribe  diti'ei'ent  regulations. 

The  Dues  are  payments  for  subscrijitions  for  shares,  and  are  generally 
made  payable  Aveekly  or  monthly.  Monthly  payments  are  $1  per  share,  and  weekly 
payments  are  25  cents  or  50  cents,  on  each  share  of  $200  of  stock.  In  some  of  the 
plans,  members  may,  however,  pay  any  multiple  of  the  monthly  or  weekly  iiayment, 
or  any  part  of  their  subscriptions  in  advance.  Such  i)ayments  will  be  credited 
on  the  shares  and  will  participate  in  the  profits. 

Premium  or  Discount  is  the  percentage  that  the  borrowing  members 
bid,  and  under  regulations,  varying  in  different  associations,  jiay  as  a  bonus  for  tlie 
privilege  and  preference  of  borrowing  and  using  the  funds  of  the  Association. 

Some  associations  fix  a  minimum  premium  below  which  rate  the  funds  will 
:iot  be  loaned.  But  loans  are  generally  made  by  auction,  and  all  members  have  an 
equal  chance  to  bid  for  the  loan. 

The  following  figures  taken  from  a  table  of  the  literature  of  a  prosperous  ami 
well  managed  association,  will  illustrate  how  premiums  or  discounts  are  made  on 
loans: 

Table  shon-ing  discount  and  net  loan  on  one  atid  on  five  shares  at  15, 
18,  20,  22\  and  35  per  cent,  also  the  amount  of  weekly  dues  and  u-celdy  interest. 


ON   ONE   SHARE. 


ON  yiVE   SHARES. 


Rate 

Discount. 

Xet  Loan. 

Weekly 
Dues.               Int. 

Discount. 

Net  Loan. 

Weekly 
Dues.             Int. 

per  cent. 

15 

$30 

$170 

-5  ceuts. 

23  ceuts. 

$150 

$850 

,$1.25 

$1.15 

18 

36 

164 

180 

820 

20 

40 

160 

200 

800 

22i 

45 

155 

225 

775 

25 

50 

loO 

250 

750 

Note  1. — The  full  table  exteuds  from  15  to  25  per  cent  by  J  per  cent  increase  of  rate. 
Note  2. — To  find  the  net  loan  on  any  number  of  shares,  multiply  the  net  on  one  shark  at 
the  given  rate  by  the  mimber  of  shares. 


BUILDING    AND    LOAN    ASSOCIATIONS.  979 

The  Charters  of  some  Building  Associations  fix  the  minimum  premium  at  15 
per  cent.  In  some  States,  there  is  no  minimum  limit  to  the  premiums ;  this  seems 
more  business  like  and  more  ethical,  for  thereby  the  rate  of  premium  M-ill  adjust 
itself  according  to  the  law  of  supply  and  demand. 

Divideuds  are  the  profits  of  the  Association,  declared  monthly,  quar- 
terly, semi-annually  or  annually,  and  distributed  among  all  shares  in  force.  Various 
regulations  are  made  by  diflerent  associations  regarding  tlie  manner  of  declaring 
dividends,  according  to  the  kind  of  shares,  "Current"  or  "Full  Paid,"  the  dues 
l>aid  thereon,  the  cash  ^alue  of  the  capital,  profits,  etc. 

Withdrawals.  Members  may  -withdraw,  under  stated  regulations  of 
the  Association,  partially  paid  or  full  paid  shares  on  the  surrender  of  their  certifi- 
cates and  receive  in  return  the  an:ount  paid  in  as  dues,  and  a  fair  share  of  the 
jjrofits,  less  any  delinquencies  or  pro  rata  of  expenses  that  they  may  owe. 

NoTK. — There  is  a  great  diversity  of  method  or  plan  hy  Associations  in  diflerent  States,  and 
also  in  the  same  state,  governing  ivithdrawals. 


TEEMmATING,  SERIAL,  PERMANENT  AND  DAYTON  PLANS. 

There  are  four  plans  for  Building  and  Loan  Associations  which  are  known  as 
the  Terminating,  the  Serial,  the  Permanent,  and  the  Dayton  Plans. 

The  Teriiiinating  Plan,  as  originally  adopted  and  carried  out,  required 
all  the  stock  to  be  issued  at  the  beginning  of  the  Association  ;  or  in  case  shares 
were  subscribed  at  any  time  during  the  progress  of  the  Association,  the  shares 
should  date  back  to  the  time  of  first  payment  of  dues,  and  the  subscriber  should 
liay  in  a  sum  equal  to  all  the  dites  on  the  shares  subscribed,  for  the  full  time  elapsed 
since  the  date  of  organization,  together  with  an  additional  sum  equal  to  a  pro  rata 
share  of  the  accumulated  earnings  of  the  Association,  up  to  the  date  of  his  subscrip- 
tion. By  this  plan,  all  profits  and  earnings  were  retained  until  such  time  as  they 
together  with  the  payments  made  by  the  subscribers  for  dues,  amounted  to  the 
$200  liar  value  of  the  stock.  Wlien  this  result  was  reached,  the  Association  and 
membership  terminated,  with  a  division  of  the  capital. 

The  Serial  Plan  is  a  modification  or  extension,  or  evolution  of  the 
terminating  jjlan.  While  it  is  conducted  in  the  main  on  the  same  general  principles^ 
it  diflers  in  the  manner  of  issuing  shares  subsequent  to  the  organization.  This 
plan  reqiures  the  same  monthly  payments  of  dues  and  the  same  observance  of 
charter  regulations,  but  the  distinguisluTig  feature  is  the  issuing  of  stock  at  different 
periods  during  the  life  of  the  Association.  Each  series  issued  bears  the  date  of  the 
issue,  and  while  the  shares  of  each  separate  series  are  equal  in  value,  they  differ  in 
A'alue  from  the  shares  of  every  other  series.  New  members  pay  only  from  the  date 
of  the  series  they  enter.  It  will  be  remembered,  that  in  the  terminating  plan,  all  the 
shares  have  at  all  times  an  equal  value. 

By  the  Serial  Plan,  each  issue  or  series  runs  its  course  independently,  and 
without  reference  to  others  issued  before  or  after  it.     Thus,  through  the  serial  issues, 


980  soule's  philosophic  practicai.  mathematics.  * 

tlie  association  may  continue  iu definitely,  or  as  long  as  tliere  are  persons  who  desire 
to  subscribe  for  the  shares. 

The  periods  for  the  issue  of  the  different  series  may  be  monthly,  quarterly, 
semi-annually,  annually  or  biennially,  according  to  the  amount  of  business  and  the 
judgment  of  the  directors. 

The  Pcrnianent  Plan  is  closely  related  to  the  serial  plan.  It  is  claimed 
to  be  the  highest  development  of  the  Co-operative  Building  Association  idea  yet 
attained.  Ey  this  plan,  there  are  no  stated  periods  for  the  issue  of  its  stock ;  stock 
is  subscribed  and  i)ayments  made  at  any  time.  Each  member's  shares  ai'e  treated 
as  a  separate  series.  Dividends  are  declared  on  the  actual  amount  of  dues  paid  in 
by  each  member,  and  as  each  member's  stock  matures  or  reaches  par,  it  is  paid  ofif^ 
One  of  the  features  of  this  jdan  is  the  net  premium  system,  by  "which  only  earned 
l>remiums  become  the  visible  profits. 

The  Dayton  or  Ohio  Plan  is  the  same  in  all  essential  particulars  as 
the  Permanent  Plan.  The  stock  face  value  (usually  $100)  is  issued  at  any  time, 
and  generally  no  fines  or  forfeitures  are  prescribed  for  non-payment,  the  members 
making  payments  at  such  times  and  in  such  amounts  as  may  suit  their  own  con- 
venience. One  of  the  special  features  of  this  plan  is,  that  the  borrowing  and  the 
non-borrowing  members  pay  the  same,  the  weekly  payments  of  the  former  being 
first  applied  to  the  payment  of  the  interest  upon  the  loan,  the  balance  being  placed 
as  a  credit  ou  the  stock;  credits  commence  to  draw  dividends  at  once  and  are 
compounded  semi-annually.  The  stock  is  matured  by  the  payment  of  the  weekly 
dues  and  the  dividends.  There  is  no  jiremium  or  bonus  in  this  plan,  and  it  becomes 
more  nearly  a  Banking  Institution  than  any  of  the  others,  and  has  less  restrictions 
upon  its  members.  The  popularity  of  this  system  seems  to  be  growing  in  all  parts 
of  the  country,  and  in  modified  forms,  has  been  adopted  by  many  of  the  Building 
Associations. 

Note. — Thore  is  much  discussion  among  building  association  men  regarding  tlie  respective 
merits  of  tlieso  different  systems ;  aud  changes  and  modifications  tliereot'  arc  made  Ijy  dift'erent 
associations. 


DIFFERENT  PLANS  OF  CHARGING  PREMIUMS  AND  INTEREST. 

There  are  four  general  plans  of  charging  i)remiums  and  interest  on 
loans,  which  constitute  the  chief  items  of  profit. 

In  the  equitable  adjustment  of  the  profits  among  the  different  serial  accounts 
constituting  the  capital,  it  is  necessary  to  understand  the  nature  and  effects  of 
these  dififerent  plans.  They  are  as  follows:  The  Gross  Plan,  the  Net  Plan,  the 
Installment  Plan  and  the  Interest  in  Advance  Plan. 

The  Gross  Plan  is  that  in  which  the  interest  is  charged  on  the  gross 
amount  of  the  loan  and  the  payment  of  the  interest  is  required  to  be  paid  in 
advance.  Tlins,  a  member  borrows  on  1  share  $200  at  20  i)er  cent  i)remium,  lie 
receives  the  net  sum  of  $100,  and  pays  interest  on  the  gross  sum  of  $200.  Hence 
the  name.  Gross  Plan. 


*  BUILDING    AND    LOAN    ASSOCIATIONS.  98 1 

The  Net  Plan  is  that  in  -wliicli  the  interest  is  charged  only  on  the  net 
sum  received  by  the  borroiccr,  bnt  as  in  the  Gross  Plan,  it  is  i)aid  or  deducted  in 
advance.     Ileuce  the  name,  Net  Plan. 

The  Installment  Plan  is  one  of  recent  adoption,  and  is  a  modification 
of  the  Gross  and  Xct  Plans. 

By  this  plan,  the  borrower  receives  the  full  ultimate  value  of  $200  jier  share, 
and  pays  the  premium  in  monthly  or  wecJcly  installments,  in  the  same  manner  that 
dues  and  interest  are  paid.  From  this  fact,  it  is  named  the  Installment  Plan.  It  is 
more  favorable  to  the  borrower  than  the  Gross  or  Net  Plan,  and  it  is  claimed  that 
its  merits  entitle  it  to  universal  adoption.  By  it,  many  of  the  complications 
resulting  from  the  deduction  of  premiums  paid  in  advance  are  avoided,  and  as  the 
premium  is  paid  only  as  it  falls  due,  there  is  no  premium  to  return  to  a  repaying 
borrower  for  unexpired  time,  and  no  discount  on  premiums  bid  on  loans  advanced 
upon  stock  of  the  older  series.  Thus  the  premium  account  by  this  plan  shows  the 
actual  profit  from  this  source. 

The  Interest  in  Advance  Plan.  This  plan  has  been  recently  introduced 
into  many  associations  throughout  the  country.  It  does  not,  like  the  Gross  Plan, 
require  the  premium  in  advance,  but  it  substitutes  interest  in  advance.  Eegarding 
this  plan  Mr.  Wrigley  remarks : 

"The  iuterest  is  made  payable  as  many  months  in  advance  of  the  date  of  the  loan,  as  the 
borrower  chooses  to  bid,  and  the  one  bidding  the  greatest  number  of  months  in  advance  will  be 
the  successful  bidder.  This  interest  ia  the  same  on  each  share  for  each  month  ($1.00)  as  under  the 
Gross  Plan,  the  difference  consisting  only  in  paying  it  in  advance  for  a  given  time.  The  mode  of 
obtaining  a  loan  on  this  plan  is  as  follows  : 

"A  borrower  wishing  to  realize  about  $800  net,  and  bidding  100  months  interest  in 
advance,  would  take  eight  shares,  which  at  $200  per  share  would  be     -----         $1600 
Deduct  100  months'  interest  in  advance  ---------  800 

Net  sum  realized        ------------  $800 

"Having  paid  the  interest  in  advance  for  100  months,  or  eight  years  and  four  months,  of 
course  he  has  no  interest  to  pay  in  monthly  installments,  as  by  the  Gross  Plan ;  and  there  being  no 
premium  to  pay  or  deduct,  there  remains  but  the  $8  per  month  dues  to  pay  on  the  eight  shares  he 
has  borrowed  upon.  He  therefore  pays  no  more  per  month  as  a  borrower  than  he  would  pay  as  a 
non-borrower.  If  the  shares  should  mature  in  eight  years,  the  borrower  having  paid  four  months' 
interest  in  advance  of  the  time  of  winding  up,  the  four  months'  excess  of  interest  would  be 
returned  to  him  with  his  siitisfied  mortgage.  On  the  other  hand,  if  the  shares  should  exceed  eight 
years  and  four  months  in  running  out,  and  reach  as  far  as,  say  nine  years,  the  borrower  would,  on 
and  after  the  101st  month,  have  to  pay  monthly  interest  in  addition  for  eight  months  longer. 

"If  one  borrows  on  an  old  series,  he  is  allowed  a  deduction  or  rebate  from  his  interest  for 
each  month  of  the  past  age  of  the  series. 

"If  he  desires  to  repay  a  loan,  he  is  allowed  a  return  of  interest,  paid  in  advance,  for  each 
unexpired  mouth. 

"The  foregoing  is  an  outline  of  the  plan  of  working. 

"The  apparent  gain  of  the  association  by  this  method  results  from  the  greater  amount  of 
profit  deducted  at  once  on  making  a  loan  as  compared  with  the  Gross  Plan. 

"For  instance:  On  a  loan,  under  the  new  plan,  on  eight  shares  at  100  mouths'  interest  in 
adv.ance,  $800  would  be  retained  as  a  profit,  the  borrower  receiving  the  net  sum  of  $800;  while 
nnder  the  Gross  Plan  of  deducting  premiums  on  a  loan  of  five  shares  at  20  per  cent  premium,  but 


982  soule's  philosophic  practical  mathematics.  * 

$200  will  bo  retained  as  jiroiit,  wbilo  the  borrower  receives  the  same  net  amount  of  money  (|800) 
as  the  borrower  under  the  new  plan  receives. 

"  It  will  thus  be  seen  that  the  association  apparently  gains  the  advantage  of  $800  immediate 
profit  under  the  jiew  plan,  against  $200  under  the  (iross  Plan,  while  the  borrower  receives  the  same 
amount  of  money  in  each  case;  and,  under  the  Interest  in  Advance  Plan,  pays  but  $8  a  month, 
against  $10  a  month  under  the  Gross  Plan.  " 

Other  advantages  are  also  claimed  for  this  ))lan. 

STATE  AND  UNITED  STATES  LEAGUES. 

A  Building  and  Loan  Association  League  is  a  cliartered  organization 
or  confederation  of  difl'ereiit  associations,  for  the  ijroinotiou  of  joint  interests,  com- 
mon purposes,  and  mutual  support. 

State  Leagues  are  those  formed  by  associations  domiciled  within  the 
respective  States. 

A  United  States  League  is  one  formed  by  associations  domiciled  in 
the  various  States.  These  leagues  are  of  great  benefit  to  the  cause  in  which  they 
serve.  Through  them,  all  matters  relating  to  the  cause  are  discussed  by  the  ablest 
minds  engaged  in  the  service,  and  the  interest  of  each  is  made  the  good  of  all. 

NATIONAL  BUILDING  ASSOCIATIONS. 

~  The  name  "Nationals"  is  applied  to  a  class  of  Building  and  Loan 
Associations  that  is  patterned,  organized,  and  conducted  somewhat  according  to 
the  general  methods  of  the  original  or  parent  Building  and  Loan  Associations  as 
above  described.  They  are  a  growth  (branches  or  children,  as  it  were)  of  the  parent 
institutions,  and  have  individualized  themselves  by  incorijorating  some  new 
features,  the  practical  advantages  of  which  are  seriously  questioned  by  the  original 
Associations.  The  new  leading  features  of  the  Nationals  are  the  establishment  of 
agencies  in  different  cities  and  States  and  the  loaning  of  money  to  whomsoever 
may  wish  to  borrow  on  their  terms.  By  reason  of  these  features,  they  are  denomin- 
ated "Nationals"  in  contradistinction  to  the  parent  associations  which  loan  money 
only  in  their  respective  localities,  and  as  a  rule  only  to  their  respective  members, 
and  are  hence  denominated  "  Locals. "  The  "  Nationals  "  are  of  quite  recent  birth, 
and  in  pressing  their  claims  for  patronage  have,  in  some  cases,  introduced  non- 
ethical  methods  in  their  management,  and  used  decoying  statements  through  their 
agents  and  in  their  circular  and  advertising  literature. 

In  many  cases,  they  masquerade  and  do  business  on  $20,000,000  or  $150,000,000 
authorized  capital,  when  in  truth  they  have  not  as  many  thousand  in  net  assets.  To 
open  offices  throughout  the  country,  they  incur  expenses  for  salary  of  agents,  rent 
of  offices,  commission  of  solicitors,  and  advertising  that  consume  much  of,  and 
sometimes,  all  of  the  profits  and  all  the  dues,  and  then  dishonorable  bankruptcy 
follows.  The  leading  special  features  of  the  Nationals  make  some  of  them  of 
greater  service  to  their  army  of  officials  than  to  their  subscribers.    These  officials 


*  BUILDING    AND    lOAN    ASSOCIATIONS.  983 

of  the  "Nationals"  are  often  protected  by  a  special  expense  fund,  wliicli  is  raised 
by  a  direct  tax  on  tbe  members.  It  is  believed  by  those  who  have  had  the  largest 
experience  with  Co-operative  Building  Associations,  that  the  business  of  loaning 
money  is  conducted  ou  too  small  a  nuirgiu  of  profit  to  admit  of  the  various  expenses 
necessary  to  carry  ou  Building  Associations  ou  the  expeusive  agency  plau  of  the 
Nationals. 

Great  prudence  should  be  exercised  by  parties  who  wish  to  receive  the 
benefits  of  a  well  and  ethically  conducted  Building  Association.  "All  is  not  gold 
that  glitters."  And  experience  proves  that  some  locals  as  well  as  numy  Nationals 
have  proved  unfaithful. 

PLANS  OF  LOANING  MONEY  OE  PREMIUM  PLANS. 

The  six  thousand  Building  and  Loan  Associations  in  the  United  States 
adopt  diflFerent  plans  with  various  combinations,  of  loaning  their  money.  The 
Commissioner  of  Labor,  Washington,  D.C.,  in  his  Ninth  Annual  Eeportou  Building 
and  Loan  Associations,  1S94,  gives  C8  different  plans  covering  38  pages.  It  is  not 
the  function  of  this  work  to  present  these  plans.  Those  who  are  interested  in  the 
subject,  we  refer  to  the  report  of  the  Commissioner  of  Labor,  to  be  found  at  almost 
every  Building  and  Loan  Association. 

PLANS  OF  WITHDEAWAL. 

Different  Associations  adopt  different  plans  for  the  withdrawal  of 
members.  This  subject  is  also  fully  treated  by  the  Commissioner  of  Labor,  in  his 
Ninth  Annual  Eeport,  1894,  wherein  he  presents  12  plans,  with  over  100  variations. 
We  refer  those  interested  in  the  subject  to  said  report. 

PLANS  OF  DISTEIBUTION  OF  PEOFIT. 

To  adjust  the  profits  and  to  find  the  value  of  each  series  aiul  of  each 
share  at  the  close  of  each  gain  declaring  period,  is  one  of  the  most  iniijortant  and 
complex  opei'ations  which  arises  in  the  accounting  work  of  Building  and  Loan 
Associations.  As  in  the  case  of  premium  plans  and  plans  of  withdrawing,  there 
are  various  plans  for  the  distribution  of  profits.  Of  these  numerous  plans,  the  Com- 
missioner of  Labor,  in  his  Ninth  Annual  Eeport  on  Building  and  Loan  Associations, 
1894,  presents  25  plans  with  nuniei'ous  variations,  covering  35  pages. 

We  give  space  and  credit  for  the  1st,  Sth  and  9th  plans  presented  in  said 
report : 

Plan  1. — {Partnership  F(an). 

"  This  plan  aiiportioiis  tbe  profits  among  series  just  as  profits  among  partners  are  apjiortioned 
in  a  firm  where  the  partners  enter  at  difi'erent  dates,  each  series  representing  a  partner. 


984 


SOULE  S    rillLOSOPHIC    PRACTICAL    MATHEMATICS. 


1.  Multiply  tho  duos  paid  in  on  the  shares  in  force  in  each  series  hy  the  equated  time  of 
'uvestmeiit. 

2.  Take  tho  sum  of  these  products  and  then  find  what  fractional  part  each  product  is  of  the 
sum. 

3.  These  fractions  are  the  parts  of  the  total  net  profits  belonging  to  each  series. 

To  illustrate  the  rule,  let  us  suppose  that  an  association  whoso  monthly  dues  are  $1  per  share 
had  three  scries  in  force  at  tho  end  of  the  third  year,  and  that  the  number  of  shares  in  each  series 
and  their  value  per  share  were  as  follows  : 

First  series,  500  shares,  value  per  share  $38.87;  second  series,  600  shares,  value  per  share 
$25.27;  third  series,  400  shares,  value  per  share  f  12.32;  that  a  fourth  series  of  500  shares  is  then 
issued ;  the  net  profits  for  the  fourth  year  are  $3,000,  and  tho  total  net  profits  for  the  four  years  are 
$5,325.     Kequircd  :  The  value  of  a  share  of  each  series  at  the  end  of  the  fourth  year. 

The  first  series  above  alluded  to  has  run  four  years,  or  forty-eight  months.  Forty-eight  $1 
payments  have  therefore  been  made  on  each  share  of  stock.  The  first  dollar  paid  has  been  invested 
forty-eight  months ;  the  second  dollar  paid,  forty-seven  months;  the  third  dollar  paid,  forty-sis 
juonths,  etc.,  the  last  dollar  of  the  forty-eight  having  been  invested  one  month.  The  times  of 
iivestonent  thus  form  .a  decreasing  arithmetical  series,  with  forty-eight  for  the  first  term,  one  for 
"^.he  last  term,  and  forty-eight  for  tho  number  of  terms.  The  total  investment  is  thus  equal  to  $1 
invested  for  1,176  months  (the  sum  of  the  series),  equivalent  to  $48  invested  for  24i  months. 

Treating  the  other  series  in  the  same  way,  we  find  that  $36  paid  per  share  in  the  second 
series  has  been  invested  for  ISJ  months;  $24  paid  per  share  in  the  third  series,  for  12i  months;  and 
$12  paid  per  share  in  the  fourth  series,  for  6i  months ;  then — 

$48  X  500  X  24^  =  $588,000,  first  series'  investment  for  one  month. 

$36  X  600  X  18i  =  $399,600,  second  series'  investment  for  one  month. 

$24  X  400  X  12i  =  $120,000,  third  series'  investment  for  one  month. 
$12  X  500  X    6|  =       $39,000,  fourth  series'  investment  for  one  month. 

$1,146,600,  total  investment  for  one  month. 

Hence  the  total  net  profits  are  divided  as  follows : 

'iWeWo  or  i^/i°r  of  the  total  profits  belong  to  the  first  series. 
AVe^soir  or  i^^iV  of  the  total  profits  belong  to  the  second  series. 
AWn'oHf  or  T^'i'V  of  the  total  profits  belong  to  tho  third  series. 
rflssoiJ  or  t|?t  of  the  total  profits  belong  to  the  fourth  series. 

Total  profits  to  bo  divided  are  $5,325. 

An'r  of  $5,325  =  $2,730.77,  first  series'  share  of  the  profits. 

"l^a^A"  of  $5,325  =  $1,855.81,  second  series'  share  of  the  profits. 

^W",-  of  $5,325  =     $557.30,  third  series'  share  of  the  profits. 

-,  Iff  of  $5,325  =     $181.12,  fourth  series'  share  of  tho  profits. 

$2,730.77  —  500  =  $5.46,  profit  of  a  share  of  the  first  series. 

$1,855.81  -r  600  =  $3.09,  profit  of  a  share  of  the  second  series. 
$557.30  H-  400  =  $1.39,  )>rofit  of  a  share  of  the  third  series. 
$181.12  —  500  =  $0.36,  ])rofit  of  a  share  of  the  fourth  series. 
$48.00,  dues  paid,  -f  $5.46,  profit,  =  $53.46,  value  of  a  share  of  the  first  series. 
$36.00,  dues  ])aid,  -j-  .$3.09,  profit,  =  $39.09,  value  of  a  share  of  the  second  series. 
$24.00,  dues  ])aid,  -|-  $1.39,  profit,  =  $25.39,  value  of  a  share  of  the  third  series. 
$12.00,  dues  paid,  -j-  $0.36,  profit,  =  $12.36,  value  of  a  share  of  the  fourth  series. 

In  the  above  example,  all  the  net  profits  made  during  the  four  years  have  been  apportioned 
to  the  several  scries,  but  some  associations  apportion  only  each  year's  profits  in  this  way.  Other 
associations  using  this  plan  simplify  the  process,  but  obtain  the  same  results,  by  dividing  the  total 
Investment  for  one  month  (or  for  one  week,  as  the  case  may  be),  into  the  profits  to  be  apportioned, 
for  the  profit  on  $1  invested  for  one  month,  which  is  then  multiplied  successively  by  the  sum  of  the 
number  of  weeks,  months,  or  other  periods  of  time  for  which  each  dollar  of  dues  in  each  series 
has  been  invested.     The  products  will  be  the  amount  of  the  profits  belonging  to  a  share  in  each 


*  BUILDING    AND    LOAN    ASSOCIATIONS.  985 

A  few  associations  have  Ijcen  found  that  arrive  at  the  same  results  hy  using  the  following 
method,  ■which  is  known  as  Clark's  plan : 

1.  Multiply  the  number  of  shares  in  force  in  each  series  by  the  quotient  obtained  by 
dividing  the  sum  of  tlie  number  of  weeks,  mouths,  or  other  periods  of  time  for  which  each  dollar 
of  dues  in  each  series  has  been  invested  by  the  product  obtained  by  multiplying  the  dues  paid  in 
on  one  share  during  the  first  year  by  the  average  time  of  investment,  for  the  equalized  results  for 
each  series. 

2.  Take  the  sum  of  these  results  and  divide  it  into  the  total  jirofits  since  the  beginning  of 
the  associations,  for  the  rate  per  cent  of  profit. 

3.  Multiply  the  quotients  already  found  by  the  rate  per  cent  of  jnofit,  for  the  profit  of  a 
share  in  each  series. 

We  have  before  seen  that  $iS  dues  per  share  in  the  first  series  have  been  invested  for  24^ 
months,  which  is  equal  to  $1  invested  for  1,176  months.  In  like  manner  $36  dues  per  share  in  the 
second  series  have  been  invested  for  18^  months,  which  is  equal  to  $1  invested  for  666  months;  $24 
dues  per  share  in  the  third  series,  for  12J  mouths,  which  is  equal  to  $1  invested  for  300  months;  and 
$12  dues  per  share  in  the  fourth  series  for  6J  months,  which  is  equal  to  $1  invested  for  78  months. 
The  average  time  of  first  year's  payments  is  simply  half  the  time  of  investment,  which  is  6 
months.  Twelve  dollars  invested  for  an  average  period  of  6  months  is  equal  to  $1  invested  for  72 
months. 

Then : 

1,176  -f-  72  =  16.333. 
666  —  72  =    9.250. 
300^72=    4.166. 
78-^72=    1.083. 
500  X  16.333  =  8,166.50,  equalized  result  for  the  first  series. 
600  X    9.250  =  5,550.00,  equalized  result  for  the  second  series. 
400  X     4.166  =  1,666.40,  equalized  result  for  the  third  series. 
500  X     1.083  =     541.50,  equalized  result  for  the  fourth  series. 

15,924.40,  equalized  result  for  all  series. 
$5,325,  the  total  profits,  —  15,924.40  =  33.4392,  the  rate  per  cent  of  profit. 
$0.334392  X  16.333  =  $5.46,  profit  of  a  share  of  the  first  series. 
$0.334392  X    9.250  =  $3.09,  profit  of  a  share  of  the  second  series. 
$0.334392  X    4.166  =  $1.39,  profit  of  a  share  of  the  third  series. 
$0.334392  X    1.083  =  $0.36,  profit  of  a  share  of  the  fourth  series. 

For  the  value  of  each  share,  add  the  dues  as  above. 

There  is  a  modification  of  plan  1,  which  follows  the  same  general  method  as  that  shown  in 
the  first  illustration,  but  difl'ers  in  certain  particulars  and  gives  a  diflerent  result.  The  modifica- 
tion is  as  follows:  Instead  of  finding  the  exact  equated  time  of  investment,  many  associations 
arrive  at  an  approximate  equated  time  by  taking  one-half  the  number  of  months  a  series  has  run. 
Using  the  same  data  as  in  the  above  illustrations,  we  get  24,  18,  12,  and  6  as  the  average  number 
of  months  the  series  have  run.  It  is  this  modification  that  is  commonly,  but  erroneously,  called 
the  partnership  plan. 

ILLUSTRATION. 

$48  X  500  X  24  =  $576,000,  first  series'  investment  for  one  month. 
$36  X  600  X  18  =  $388,800,  second  series'  investment  for  one  month. 
$24  X  400  X  12  =  $115,200,  third  series'  investment  for  one  month. 
$12  X  500  X    6  =    $36,000,  fourth  series'  investment  for  one  month. 


$1,116,000,  total  investment  for  one  month. 


986  SOULe's    riULOSOPHlC    PRACTICAL    MATHEMATICS.  * 

The  totul  net  jirolits  are  then  divided  in  proportion  to  each  series'  investment  for  one  month, 
thus  : 

■iViH>°o"m(  or  I'i'i  of  $5,325  =  $2,748.39,  first  series'  share  of  the  profits. 
iVi'iriHMJ  or  1%  of  |i5,325  =  |H,855.16,  second  series'  share  of  the  profits. 
iWs^'l;  or  iV-V  of  $5,325  =     $549.08.  tlurd  series'  share  of  the  profits. 
tHSSoij  or  TS5-  of  $5,325  =     $171.77,  fourth  series'  share  of  the  profits, 
$2,748.39  —  500  =  $5.50,  profit  of  a  share  of  the  first  series. 
$1,855.16  —  600  =  $3.09,  profit  of  a  share  of  the  second  series. 
$549.68  —  400  =  $1.37,  profit  of  a  share  of  the  third  series. 
$171.77  -H  500  =  $0.34,  profit  of  a  share  of  the  fourth  series. 

Tliis  modification  of  plan  1  has  been  simplified,  the  principle  consisting  in  casting  out  com'^ 
mon  factors  in  the  process  of  multiplication. 

The  first  series  has  run  48  months;  the  second,  36  months;  the  third,  24  months;  and  the 
fourth,  12  months.  The  average  time  of  investment,  as  we  h.ave  before  seen,  is  24,  18,  12,  and  6r 
respectively.     Then  we  proceed  thus  : 

48,  age  in  months,  x  24,  average  time,  X  500  shares. 
36,  age  in  months,  X  18,  average  time,  X  600  shares. 
24,  ago  in  months,  X  12,  average  time,  X  400  shares. 
12,  age  in  months,  X    6,  average  time,  X  500  shares. 

It  will  he  readily  seen  that  12  is  a  factor  common  to  all  the  numbers  of  the  first  column,  and 
that  6  is  a  factor  common  to  all  the  numbers  in  the  second  column.  Casting  out  these  factors,  we 
have — 

4  X  4  X  500  =  8,000.  Hence  -f^^  of  the  total  profits  belong  to  the  first  series. 
3  X  3  X  600  =  5,400.  -^sV  of  the  total  profits  belong  to  the  second  series, 

2  X  2  X  400  =  1,600.  -iVs-  of  the  total  profits  lieloug  to  the  third  series. 

1  X  1  X  500  =     500.  xfy  of  the  total  profits  belong  to  the  fourth  series. 


Tot.al,   15,500. 

Briefly  put,  then,  the  simplification  is  as  follows:  Multiply  the  number  of  shares  in  force  in 
each  series  by  the  square  of  the  time  of  investment  expressed  in  terms  or  periods  corresponding  to 
the  intervals  between  the  series,  and  then  divide  the  profits  in  proportion  to  these  products. 

The  foregoing  simplification  has  been  still  further  simplified  by  finding  the  profit  of  a  share 
in  each  series  directly,  instead  of  finding  eiich  series'  share  of  the  profit,  as  follows: 

1.  Multiply  the  number  of  shares  in  force  in  each  series  by  the  square  of  the  time  of  invest- 
ment expressed  in  terms  or  periods  corresponding  to  the  intervals  between  the  series. 

2.  Divide  the  sum  of  these  products  into  the  results  obtained  by  multiplying  the  total  net 
profit  by  the  square  of  the  time  of  investment  expressed  as  above. 

The  total  of  the  products,  as  in  the  last  illustration,  is  15,500;  then— 

$5,325,  total  profits,  X  4  X  4  -H  15,500  =  $5.50,  profit  of  a  share  of  the  first  series. 
$5,325,  total  profits,  X  3  X  3  -;-  15,500  =  $3.09,  profit  of  a  share  of  the  second  series. 
$5,325,  total  profits,  X  2  X  2  H-  15,500  =  $1.37,  profit  of  a  share  of  the  third  series. 
$5,325,  total  profits,  X  1  X  1  H-  15,500  =  $0.34,  profit  of  a  share  of  the  fourth  series. 

A  share  of  the  first  series  receives  16  times  as  much  profit  as  a  share  of  the  fourth  series;  a 
share  of  the  second  series,  9  times  as  nnich  ;  and  a  share  of  the  third  series,  4  times  as  much.  This 
method,  therefore,  reveals  the  fact  that,  by  multiplying  the  number  of  sh.arcs  in  force  in  each 
series  by  the  square  of  the  time  each  series  has  been  invested,  expressed  in  years,  half  years,  quarter 
years,  etc.,  corresponding  to  the  intervals  between  the  series,  a  correct  basis  of  calculation  is 
reached.  These  simplifications,  however,  are  practicable  only  where  series  are  issued  at  regular 
intervals,  as  fractions  complicate  the  operation.     This  simplification  is  known  as  Rice's  rule. 


BUILDING    AND    LOAN    ASSOCIATIONS. 


987 


A  few  associations  arrive  at  tlie  same  results  by  dividing  the  total  investment  f(ir  one  month 
into  the  profits,  for  a  rate  per  cent  of  profit,  and  then  applying  the  rate  to  each  series'  investment 
for  one  month  for  each  series'  share  of  the  jirolits.  The  jiroccss  is  also  varied  in  the  following 
manner:  Find  what  annual  rate  of  interest  the  profits  are  equivalent  to  on  the  amount  of  dues 
paid  for  oue-half  the  time  that  all  the  dues  have  been  invested,  and  apply  this  rate  on  the  dues 
paid  per  share  for  one-half  the  time  of  investment,  for  the  profit  of  a  share  in  any  series. 

Other  variations  are  the  following  : 

1.  The  profits  are  distributed  on  the  amount  of  dues  actually  ]Kiid  in  on  the  shares  in  force 
in  each  series  (not  what  the  regular  payments  should  have  amounted  to),  multiplied  by  one-half 
the  time  of  investment. 

2.  The  profits  are  distributed  on  the  total  amount  of  dues  standing  to  the  credit  of  the 
shareholders  in  the  loan  fund  multiplied  by  one-half  the  time  of  investment. 

3.  The  series  are  not  allowed  to  i)articipate  in  the  profits  for  the  term  in  which  they  were 
issued. 

4.  The  profits  are  distributed  on  the  amount  of  dues  actually  paid  in  on  all  shares  in  force 
that  are  three  months  old  or  over,  multiplied  by  one-half  the  time  of  investment,  shares  less  than 
three  months  old  not  jiarticipating. 

5.  The  profits  are  distributed  to  the  free  shares  only,  dues  on  shares  borrowed  on  being 
credited  on  loans. 

6.  Profits  arising  from  withdrawals  are  divided  equally  among  the  shares  of  the  respective 
series  from  which  the  shares  were  withdrawn. 

7.  Profits  arising  from  entrance  fees  are  divided  equally  among  the  shares  of  the  respective 
series  in  which  the  shares  are  taken. 

8.  A  profit  of  $1  is  given  to  all  shares  sis  months  old  or  over.  The  remainder  of  the  profits 
is  distributed  on  the  dues  jiaid  in  on  the  shares  in  force  six  months  old  or  over  multiplied  by  one, 
half  the  time  of  investment. 

9.  Profits  arising  from  premiums  are  divided  equally  among  all  the  shares  in  force  at  the 
end  of  the  period  during  which  the  loans  were  made.  Profits  from  all  other  sources  are  distributed 
in  accordance  with  the  modified  rule. 

10.  A  fixed  rate  of  interest  is  given  on  the  total  amount  of  dues  paid  on  the  shares  in  force 
at  each  apportionment.  This  interest  is  deducted  from  the  profits  for  the  term  and  the  remainder 
distributed  according  to  the  modified  rule. 

11.  A  fixed  rate  of  interest  is  given  on  the  value  of  the  shares  in  force  as  declared  by  the 
last  report.  This  interest  is  deducted  from  the  profits  for  the  term  and  the  remainder  is  distributed 
according  to  the  modified  rule. 

12.  A  portion  of  the  total  amount  of  premiums  received  by  and  due  the  association  is 
arbitrarilj'  determined  upon,  and  held  in  reserve  to  bo  applied  in  future  dividends;  the  amount 
thus  determined  upon  is  deducted  from  the  total  profits,  and  the  remainder  of  the  profits  is  dis- 
tributed as  follows:  The  interest  and  dividends  allowed  on  free  shares  withdrawn  are  added  to  the 
dues  paid  in  on  such  shares,  and  the  sum  total  of  said  interest  and  dividends  is  deducted  from  the 
amount  of  distributable  profits ;  the  balance  is  distributed  among  all  the  shares  in  accordance  with 
the  foregoing  modified  rule. 

There  is  still  another  modification  of  plan  1,  as  follows:  Multiply  each  series'  investment 
(that  is,  the  dues  paid  in  on  the  shares  in  force)  by  one-half  the  number  of  months  invested  plus 
one  and  apportion  the  profits  in  proportion  to  these  products. 

Using  the  same  data  as  before  we  proceed  as  follows  : 

ILLUSTRATION. 

$48  X  500  X  25  =  $600,000,  first  series'  investment  for  one  month. 
$36  X  600  X  19  =  $410,400,  second  series'  investuieut  for  one  month. 
$24  X  400  X  13  =  $124,800,  third  series'  investment  for  one  month. 
$12  X  500  X     7  =    $12,000,  fourth  series'  investment  for  one  month. 

$1,177,200,  total  investment  for  one  month. 


988  soule's  philosophic  practical,  mathematics.  * 

Then,  proceeding  as  before,  we  find  that — 

•iW»%%  or  OS?  of  $5,325  =  $2,714.07,  the  first  series'  share  of  the  profits. 
■iW^''o'^  or  fit  of  $5,325  =  $1,856.42,  the  second  series'  share  of  the  profits. 
T^iWa^oi  or  lu  of  $5,325  =     $564.53,  the  third  series'  share  of  the  profits. 
tH9S3tt  or  /,\  of  $5,325  =     $189.98,  the  fourth  series'  share  of  the  profits. 
$48,  dues,  +  ($2,714.07  —  500)  =  $53.43,  value  of  a  share  of  the  first  series. 
$36,  dues,  +  ($1,856.42  —  600)  =  $39.09,  value  of  a  share  of  the  second  series. 
$24,  dues,  +  (    $564.53  —  400)  =  $25.41,  value  of  a  share  of  the  third  series. 
$12,  dues,  +  (    $189.98  —  500)  =  $12.38,  value  of  a  share  of  the  fourth  series." 

Note  1. — 1280  local  associations  and  81  Nationals  throughout  45  States  adopt  this  plan  with 
some  of  its  variations. 

Note  2. — 1  association  in  Louisiana  adopts  this  method  with  variations. 

Plan  S.— {Brooks^  Plan). 

1.  Find  the  legal  rate  of  interest  on  the  values  of  the  shares  as  declared  hy  the  last  report 
and  deduct  it  from  the  profits  of  the  term  for  the  net  profit. 

2.  Divide  the  net  profit  hy  the  sum  of  the  dues  paid  in  during  the  term  and  the  interest  on 
the  previous  series,  for  the  rate  per  cent  of  profit. 

3.  Multiply  the  sum  of  the  interest  and  dues  for  the  term  on  one  share  of  each  series  hy  the 
rate  per  cent  of  profit,  for  the  profit  on  one  share. 

4.  Add  the  previous  value  of  a  share  of  each  series,  the  legal  interest  on  this  value,  the 
dues  paid  iu  during  the  term,  and  the  profit  on  the  share  to  find  the  present  value  of  a  share  of 
each  series. 

Let  the  legal  rate  of  interest  he  6  jier  cent,  and  using  the  same  data  as  iu  the  preceding 
illustrations,  we  proceed  thus : 

ILLU.STRATION. 

$38.87  X  .06  =  $2.33,  interest  on  a  share  of  the  first  series. 
$25.27  X  .06  =  $1.52,  interest  on  a  share  of  the  second  series, 
$12.32  X  .06  =  $0.74,  interest  on  a  share  of  the  third  series.     ' 

$2.33  X  500  =  $1,165,  interest  on  first  series. 

$1.52  X  600  =     $912,  interest  on  second  series. 

$0.74  X  400  =     $296,  interest  on  third  series. 

.$2,373,  total  interest  on  old  series. 

$3,000,  the  profits  for  the  term,  —  .$2,373,  =  $627,  the  net  profits. 

$12  X  2,000,  the  number  of  shares  iu  force,  =  $24,000,  the  dues  paid  during  the  term, 
$24,000  +  $  2,373  =  $26,373,  the  active  capital. 

$627  ^  $26,373  =  2.3774,  the  rate  pet  ceut  of  profit. 
($12  +  $2.33)  X  .023774  =  $0.34,  the  profit  of  a  share  of  the  first  series 
($12  +  $1.52)  X  .023774  =  $0.32,  the  profit  of  a  share  of  the  second  series. 
($12  +  $0.74)  X  .023774  =  $0.30,  the  profit  of  a  share  of  the  third  series. 
$12  X  .023774  =  $0.29,  the  profit  of  a  share  of  the  fourth  series. 
$38.87  +  $2.33  +  $12  +  $0.34  =  $53.54,  value  of  a  share  of  the  first  series. 
$25.27  +  $1.52  +  $12  +  $0.32  =  $39.11,  value  of  a  share  of  the  second  series. 
$12.32  -)-  $0.74  4-  $12  4-  $0.30  =  $25.36,  value  of  a  share  of  the  third  series. 
$12  +  $0.29  =  $12.29,  value  of  a  share  of  the  fourth  series. 

Some  associations  using  this  rule  vary  it  hy  allowing  less  than  the  legal  rate  of  interest  on 
the  old  values,  but  no  separate  classification  of  such  associations  has  been  made. 

Another  variation  from  the  rule  consists  iu  allowing  the  legal  rate  of  interest,  not  only  on 
the  values  of  the  old  series  at  the  beginning  of  the  term,  but  also  on  the  equated  amount  of  dues 
paid  in  on  all  the  shares  during  the  term. 


*  BUILDING    AND    LOAN    ASSOCIATIONS.  989 

Another  variation  from  the  rule  is  as  follows:  After  the  legal  rate  of  interest  for  the  term  ou 
the  values  of  the  okl  series  has  been  deducted  from  the  prolits  for  the  term,  the  remainder  of  the 
profit  is  divided  hy  the  sum  of  the  vahio  of  all  shares  at  the  heginning  of  the  term,  the  dues  paid 
iu  during  the  term,  and  the  interest  for  the  rate  per  cent  of  ]>rofit.  This  rate  is  then  applied  to  the 
sum  of  the  previous  value  of  each  share,  the  dues  paid  in  during  the  term,  and  the  interest  ou  the 
share  for  the  profit  on  one  share. 

Note. — 25  associations  adopt  this  plan. 

Plan  9. 

The  profits  are  divided  equally  among  all  the  shares  iu  force.  Tins  plan  is  used  principally 
T)y  terminating  associations. 

The  follo-n-ing  variations  of  this  rule  have  been  found  : 

1.  The  profits  for  each  term  are  divided  equally  among  all  the  shares  iu  force. 

2.  The  profits  for  each  term  are  divided  equally  among  all  shares  three  mouths  old  or  over, 
six  months  old  or  over,  or  nine  months  old  or  over,  as  the  rules  of  the  association  may  provide.  In 
such  cases  shares  less  than  three  months,  sis  months,  or  nine  months  old  do  not  participate  in  the 
profits. 

3.  Shares  less  than  a  year  old  are  given  such  portion  of  the  profits  as  the  hoard  of  directors 
may  allow.  The  remainder  of  the  profits  is  divided  equally  among  all  the  shares  one  year  old  or 
over. 

4.  The  profits  for  each  term  are  divided  equally  among  the  free  shares. 

5.  Shares  one  year  old  or  over  receive  equal  amounts  of  the  profits  for  the  term.  Shares 
nine  months  old  receive  three-fourths  as  much  as  those  a  year  old ;  shares  six  months  old  one-half 
as  much,  etc. 

6.  The  profits  made  liy  each  series  are  liept  separate,  and  are  divided  e(iuari}'  among  all  the 
shares  of  the  series. 

7.  Profits  arising  from  preminms  and  fines  are  divided  equally  among  the  shares  of  the 
series  in  which  such  2)rofits  were  made.     All  other  profits  are  divided  equally  among  all  the  shares. 

Note  1. — 1062  locals  and  2  Nationals  adopt  this  plan  with  variations. 
Note  2. — 2  associations  iu  Louisiana  adopt  this  method  with  variations. 


PROBLEJr  BY  THE  GROSS  PLAN. 

To  elucidate,  in  detail,  tbe  manner  of  adjusting  jirofits  and  finding  the 
value  of  the  different  series  and  the  separate  shares  by  tlie  Gross  phni,  we  present 
tlie  following  problem  and  work  the  same  in  full,  according  to  the  method  of  Edmund 
Wrigley. 


An  association  ■which  issues  yearly  series  and  loans  on  the  Gross  plan,  conv 
mencing  business  with  an  issue  of  500  shares  of  $200  each,  and  collecting  dues  25 
cents  per  week,  would,  during  the  first  fiscal  year,  receive  dues  amounting  to  $6500, 

Presuming  that  the  dues  had  been  loaned  at  25  per  cent  premium,  that  the 
gains  by  interest  etc.,  were,  at  the  close  of  the  fiscal  year,  $400,  and  the  exiieuses 


990  soule's  philosophic  practical  mathematics.  * 

$300,  wliat  would  be  tlie  7tet  gain  for  the  term,  tlie  value  of  the  series,  and  the 
value  per  share  1  Aus.  See  operation. 

OPERATION. 

500  shares  at  25  cents  for  52  weeks  =  $13  per  share,  =  $6500  dues. 

$6500  loaned  at  25  i>er  cent  premium  = -         -         $1625  prem. 

Gains  by  interest ,  etc.  .--.......  400 

Total  gain,  $2025 

Less  expenses,  say, 300 

Net  gain  for  the  term, -        -  1725 

Total  \aliie  of  first  series,  $8225 

$8225  -^  500  shares  =  $16.45  value  per  share. 

SECOND    TERM. 

The  association  issued  at  the  beginning-  of  the  second  year  or  term,  a  second 
series  of,  say,  800  shares. 

The  following  statement  shows  the  business  of  the  second  term  or  year : 

Passive  capital  $8225.  Dues  on  first  series  for  second  year  $6500.  Dues  on 
second  series  for  first  year,  52  weeks  at  25  cents  on  800  shares,  =  $10400.  Total 
dues  $10900.  Interest  on  $8225  passive  capital  or  value  of  first  scries  at  6  per  cent 
for  second  term;  $493.50.  Premium  on  $0500  dues  of  first  series  at  25  per  cent 
$1625.  Premium  on  $10400  dues  on  second  series  at,  say,  30  per  cent,  $3120.  Total 
premium  $4745. 

Average  interest  on  active  capital,  $22138.50,  at,  say  6  per  cent  for  J  term  or 
average  time,  $064.15.  Total  gain,  $5902.05,  expenses,  say  $402.05.  Net  gain  for 
second  term,  $5500.  What  is  the  value  per  share  of  each  series  and  what  is  the 
value  of  each  series  1 

OPERATION. 

Dues  for  Second  Term. 

Dues  for  second  term  on  first  series,  500  shares,  52  weeks  at  25  cents  =        -        -        -        -  $6500 

Dues  for  first  term  on  second  series,  800  shares,  52  weeks  at  25  cents  =        -         -        -         -  10400 

Total  dues  for  both  series,  second  term,  ....----        $16900 

Gains  for  Second  Term. 

Interest  on  $8225,  passive  capital,  at  6  per  cent  for  second  term,      -         -  $493.50 

Premium  on  $6500,  at  25  jier  cent  fur  second  term, 1625.00 

Premium  on  $10400,  at  30  per  per  cent  for  first  term,         .        .        -        .  3120.00 
Average  interest  on  active  capital,  $22138.50  at  6  }>er  cent  for  i  term, 

or  average  time,           - 644.15 

Total  gain,  $5882.65 

Less  expenses, 402.65 

Net  gain  for  second  term,       --- $5480.00 


*  BUILDING    AND    LOAN    ASSOCIATIONS.  991 

PRESENT   VALUE   OF    ONE   SHARE   OF   FIRST   SERIES   AND   TOTAL  VALUE   OF   FIRST  SERIES. 

A'alue  of  one  share,  first  series,  at  beginning  of  term,  §16.45.     Interest  on  same  at  6  per  cent 
for  the  term  is  98.7  cents. 

.987  X  500  shares  first  series  ^  $493.50,  as  shown  above. 
$5480  net  gain  —  $493.50  =  $4986.50  gain  to   he   divided   among   the   1300   shares   of  both 
series.     $4986.50  —  1300  =  $3.83576+  jicr  share. 

Dues  on  one  share  of  first  series  for  two  years,    $26 
Gain  on  one  share,  first  term,  ($1725  —  500)     -         3.45 
Gain  by  interest,  second  term,     -  .987         ^     4.82276+ 

Gain  by  interest,  first  term,  -        3.83576+  ) 

Total  gain  in  two  terms  or  years,        -        $34.27276+  value  of  each  share. 
$34.27276  X  500  shares  =  $17136.38  value  of  first  series. 

PRESENT  VALUE    OF   ONE   SHARE   OF   SECOND   SERIES   AND   TOTAL  VALUE   OF   SECOND   SERIES. 

Dues  on  one  share,  second  series,  for  one  term  or  year,  -         -        $13. 

Gain  per  share  for  second  term,  as  above,       -----  3.83576+ 


Value  of  each  share  of  second  series,      ------        $16.83576+ 

$16.83576  X  800  shares  =  $13468.61  value  of  second  series. 

$17136.38  +  $13468.61  =  $30604.99  total  value  of  both  series,  1300  sh.ares,  at  the  close  of  second 
term  or  year. 

Note  1. — 257  associations  adopt  this  plan  with  v.ariations.  The  variations  are  shown  on 
page  437  of  the  report  of  the  Commissioner  of  Labor,  1894,  and  pages  83  to  100  of  Wrigley's  How  to 
Manage  Building  Associations. 

Note  2. — 8  associations  in  Louisiana  adopt  this  method  with  variiitions. 


METHODS   ADOPTED   BY   SOME   OF   THE   NEW    ORLEANS    BUILDING 

AND  LOAN  ASSOCIATIONS. 

In  response  to  inquiries  made  of  several  Building  and  Loan  Associations 
in  New  Orleans,  regarding  their  method  of  determining  and  adjusting  iirofits,  we 
received  replies  from  four  of  the  most  pojiular,  prosperous  and  ably  managed 
associations  in  the  city.     They  are  as  follows: 


OFFICE  OF 

Mutual  Building  and  Homestead  Association, 

No.   337  (Old   No.   87)   ST.  CHARLES  STREET,   MASONIC  TEMPLE. 

New  Orleans,  Aug.  2,  1895. 
Geo.  Soul£,  Esq., 

City. 
Dear  Sir  : 

Replying  to  your  favor  of  the  29th  ult.,  I  am  pleased  to  furnish  you  with  the  information 
requested  by  you,  and  hope  it  may  prove  serviceable. 

Our  profits  are  apportioned  upon  the  partnership  plan,  and  the  dividend  is  a  jiercentago  based 


992  SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS.  ♦ 

iij)ou  tlm  amount  each  sbareliolikT  has  iiivcsteil  for  the  time  it  is  invested.    While  ■we  issue  shares 
iu  what  Ave  nominally  term  series,  said  shares  do  not  mature  at  the  one  time,  hecause  the  amounts 
de])ositod  thereon  may  exceed  the  minimum  installment,   and  hence  the  accumulated  capital 
invested,  and  consequent  profits  vary  and  cause  shares  to  mature  at  uneven  dates. 
The  rule  we  follow  may  he  formulated  thus: 

1.  Find  the  net  profits  to  lie  apportioned. 

2.  Find  the  cash  value  at  the  commencement  of  the  period  just  ended,  of  all  shares  then  in 
force,  which  are  still  iu  force. 

3.  Find  the  total  of  installments  paid  in  on  all  shares  within  the  period  just  ended. 

4.  Add  to  item  (2),  the  amount  of  item  (3),  reduced  to  its  equivalent  capital  for  an  invest- 
ment covering  the  whole  period. 

5.  Divide  item  (1),  hy  item  (4),  and  the  result  will  he  the  per  cent  of  profit. 

6.  At  the  rate  found  apportion  each  share  (or  hatch  of  shares  under  one  account)  with  its 
individual  profit,  following  in  each  case  the  method  described  ahove, 

EXAMPLE. 

On  10  shares,  the  dues  for  52  weeks  at  $2.50  equals, $130.00 

These  being  paid  monthly,  it  is  the  equivalent  of  Sixty-five  (65)  Dollars  invested  for  one 

year,  and  if  profit  is  six  (6)  per  cent,  the  dividend  equals,  -        -        .        .        .  3.9ff 

Cash  value  of  shares  at  second  year, $133.90 

Add  dues  for  year, :  130.0» 

The  capital  then  invested  is, $133.90 

Plus  average  of  $130,  65.00 

Equivalent  capital  for  year, $198.90 

On  which  a  six  (6)  per  cent  dividend  would  be,      -.---....  11.93 

Cash  value  of  shares  at  third  year, $275.83 

Add  dues  for  year, - 130.00 

Following  same  method,  $275.83,  plus  $65,  equals  $340.83  at  six  (6)  per  cent,      ...  20.44 

Cash  value  at  fourth  year,        - $426.27 

And  so  on. 

Very  respectfully, 

E.SPY  AV.  H.  "WILLIAMS,  Secretartj. 
p.  S, — This  company  also  issues  shares  on  the  D.ayton  plan.  These  shares  are  $100  face 
value,  payable  iu  installments  of  not  less  than  25  cents  per  week,  lint  as  much  more  as  desired  by 
holder.  The  shares  are  issued  at  any  time,  and  begin  to  participate  in  profits  at  once.  Full  paid 
shares  are  issued  on  payments  of  $100  each,  and  dividends  are  paid  thereon  in  cash  every  six 
mouths. 

Loans  are  also  made  without  any  premium  or  bonus,   and  without  bidding,  simply  upoa 
application  filed. 

E.  W.  H.  W.,  Sec'y. 

+ 

OFFICE  OF 

UNION  HOMESTEAD  ASSOCIATION, 

NEW  336  CARONDELET  STREET. 

New  Orleans,  August  7,  1895. 
Cou  Geo.  SouLfi, 

City. 
Dear  Sir  : — In  response  to  your  circular  letter  of  July  20th,  I  note  with  interest,  that  you 
Intend  giving  represent.ition  in  your  forthcoming  work  on  "Philosophic  Practical  Mathematics," 
to  tho  problems  of  Building  and  Loan,  or  Homestead  Associations.        »»»••• 
As  to  our  own  method  of  distributing  profits  to  members,  I  reply  as  follows: 
The  member  stands  to  us  iu  the  relation  of  a  series,  of  which  you  speak.     Our  system  con- 


*  BUILDING   AND   LOAN    ASSOCIATIONS.  993 

templates  the  calculation  of  dividend  on  every  pass-book,  independently  of  every  otLer.  No  two 
weed  Ijo  exactly  the  same.  We  do  not  issue  stock  in  series,  except  nominally  ;  v.e  call  all  the  issue 
of  any  one  month  the  series  of  that  month,  and  give  it  a  number;  hut  the  books  or  shares  in  that 
series  are  each  independent  of  every  other. 

For  example:  In  series  38,  issued  last  January,  books  521,  529,  and  536,  each  call  for  10 
shares,  on  -nhich  the  regular  deposit  -nonld  be  $5.00  per  month.  But  on  June  30th,  the  standing  of 
the  three  books  -n-as,  respectively,  $75,  $30  and  §70. 

We  do  not  declare  a  dividend  of  so  much  jier  share,  but  a  percentage  dividend,  this  time  5 
per  cent  for  the  six  months.  This  enables  us  to  give  to  each  of  these  books  its  fair  (right)  portion 
of  earnings.  Where  the  deposits  are  made  regularly,  month  by  month,  we  take  half  the  amount 
as  the  average  investment ;  so  on  book  529,  on  which  6  payments  of  $5  were  made,  total  §30,  we 
have  an  average  investment  of  $15  at  5  per  cent  dividend,  75  cents.  On  book  521,  there  was  a 
deposit  of  $50  in  February,  on  which  we  allow  interest  from  the  time  it  was  put  in,  and  give  that 
book  §2.G8  dividend. 

This  method  of  averaging  is  not  mathematically  exact,  but  is  close  enough  for  all  practical 
purposes,  especially  in  view  of  the  fact  that  the  surplus  or  undivided  profit,  is  held  and  belongs  to 
the  stockholders,  and  may  be  subsequently  divided  among  them,  when  ordered  by  the  directors. 

In  case  of  a  book  in  force  prior  to  January,  for  example.  No.  412,  issued  July  1894,  it  showed  a 
value,  Jan.  1,  of  $76.85,  (6  X  12.50  plus  dividend)  and  during  the  six  months  six  payments  of  $12.50 
were  made.  We  add  to  the  capital,  at  the  beginning  of  the  term,  half  the  deposits  of  the  term, 
$.'i7.50,  and  (dropping  fractional  parts  of  $1)  have  an  invested  capital  for  the  term  of  $114,  at  5  per 
cent,  gives  dividend  $5.70,  and  cash  value  of  the  book,  July  1,  1895,  $157.55. 

In  case  but  5  monthly  payments  had  been  made  during  the  term,  I  would  allow  but  5/6th8 
of  the  regular  dividend. 

This  method  is  simple,  and  can  be  made  as  accurate  as  desired.  Its  beauty  lies  in  its 
flexibility,  adapting  itself  to  the  needs  of  every  book,  or  every  share,  or  every  series,  if  applied  to 
the  series  capital.  It  can  be  made  as  exact  as  any  method  in  use,  if  one  wishes  to  determine  to  a 
minute  fraction  the  rate  of  earnings  or  dividend.  It  is  applicable  to  any  so-called  "permanent'' 
associations,  the  specific  name  adopted  by  most  associations  issuing  stock  continuously,  and 
treating  each  pass-book  independently. 

Trusting,  that  from  the  foregoing,  you  will  understand  clearly  our  method  of  division  of 
profits,  and  holding  myself  in  readiness  to  answer  promjitly  any  questions  you  wish  to  ask  in 
further  connection  with  this  subject,  I  am, 

Very  truly., 

JNO.  E.  HUFFMAN, 

Secretary/. 


THE 

SECURITY  BUILDING  AND  LOAN  ASSOCIATION. 

Mo,  5  Commercial  Place,  Room  1. 

New  ORLEAN.S,  August  12,  1895. 
GEor.GE  Soui.ii,  Esq. 

City. 
Deah  Siu: 

Your  favor,  29th,  was  duly  received,  and  allow  me  to  apologize  for  not  answering  sooner. 

Our  association,  the  Security  Building  and  Loan  Association  does  not  issue  stock  in  series. 
The  shares  are  $100  each,  and  are  issued  at  any  time  to  suit  the  subscriber.  Our  method  of  divid- 
ing the  profits  is  much  more  simple  than  that  of  serial  associations,  as  we  conduct  our  business  on 
the  cash  basis,  or  Dayton  plan. 

The  following  extract  from  Article  4,  of  our  Charter  will  help  to  explain: 

"  On  January  1,  and  July  1,  of  each  year,  the  Board  of  Directors  shall  ascertain  the  earnings 


994 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


of  the  association  for  the  preceding  six  months,  and  out  of  such  earnings  they  shall  (1)  deduct  all  the 
expenses  of  the  association  for  such  time ;  (2)  set  asido  to  the  fund  for  contingent  losses  such  an 
amount  as  they  may  deem  best,  until  such  contingent  fund  equals  5  per  cent  of  the  entire  out- 
standing mortgage  loans,  which  fund  sli:in  lie  invested  as  directed  by  the  board;  (3)  the  balance 
of  said  earnings  shall  be  declared  as  a  dividend  to  tlie  shareholders,  in  j)roportion  to  the  value  of 
stock  of  each  member,  at  the  time  the  preceding  dividend  was  declared,  and  all  payments  made 
between  such  dividend  periods  shall  begin  to  earn  dividends  from  the  beginning  of  the  next  semi- 
annual term." 

"The  dividends  declared  on  'running  stock'  shall  bo  credited  on  the  pass  books  of  the 
members  semi-annually,  and  dividends  on  'paid  up  stock'  shall  be  paid  in  cash  on  the  fourth 
Monday  of  January  and  July  of  each  year." 

The  Security  has  declared  its  third  semi-annual  dividend  of  4  per  cent. 

I  beg  to  enclose  copy  of  our  charter,  also  our  last  semi-annual  statement. 

The  serial  system  is  not  being  adopted  by  the  new  associations;  it  is  complicated  and  not 
easy  for  the  average  stockholder  to  understand. 

Any  further  information  that  I  can  furnish  will  bo  cheerfully  given.  ' 

Very  truly, 

W.  E.  DODSWOETH, 

Secretary, 


OFFICB  OF 


Third  District  Building  Association, 

Corner  Royal  street  and  Lafayette  avenue. 


Prof.  Geo.  Soul6, 


New  Okleans,  August  13,  1895. 


City. 


E.XAMPLE. 

ries 

Ko.  Shares. 

Paid  in  on 

Entitled  to  full 
divideud  for 

A 

708 

$9204.00  12 

months  or  52  weeks,  one  year, 

B 

109 

981.00    9 

"       "  39      "        f  of  year, 

C 

327 

2125.50    6 

"       "  26      "        i     " 

D 

841 

2943.50   3 

"       "  13      "        i     " 

Dear  Sir: 

Enclosed  please  to  find  example  of  the  method  of  distribution  of  profits  as  adopted  by  this 
association.     Trusting  same  will  bo  of  use  to  you,  I  remain, 

Yours  truly, 

JOS.  L.  SPORL,  Secretary. 

Pro  rata  on 

$9204.00 

735.50 

1062.75 

735.87 

Total  paid  in  $15254.00  '  '         $11738.12 

The  not  profits  for  distribution  are  $2969, 
Divide  net  profits  $2969  x  100  by  amounts  prorated,  |;11738,  gives  25.29-{-  %,  multiply  $9204 
amount  paid  in  by  series  A  by  25.29  %,  gives  $2327.69  gain  for  the  whole  number  of  shares  of  series 
A.    $2327.69  -i-  708  shares  of  series  A  =  $3.28  dividend  per  share  of  series  A.  , 

Then  to  find  the  dividend  per  share  for  each  of  the  other  series,  we  proceed  thus  : 
Series  B  |  735.50  prorated  value  X  25.29  %  =  $186.01  -;-  109  No.  shares  =  $1.71  dividend  per  share, 
"     C     1062.75        "  "       X  25.29"^=    268.77-^327    "        "      =      .82-1-       "  " 

'•     D      735.87        "  "       X  25.29  %  =    186.10  ~  841    "        "      =      .22-1-       "  " 

Note. — The  payments  on  stock  in  this  company  are  25  cents  per  week,  or  $13  jier  year,  of  52 
weeks.     Hence  the  difl'erent  series  are  credited  as  follows  : 


• 


BUILDING    AND    LOAN    ASSOCIATIONS.  ggS 


Series  A  is  credited  with  the  full  $13.00  on  each  share  as  paid  in. 

"     B     "        "         "      only  9.00      "       "  "  "        "      instead  of  $9.75  hccause  some  are 

"     C     "        "  "     the  full      6.30      "       "  "  "        "  '  [delinquent. 

"     D    "        "  "  3.50      "       "  ""        "      instead  of  5i3.23  because  some  are 

[paid  in  advance. 

According  to  the  foregoing  work,  the  value  of  the  stock  or  series  at  the  end  of  the  year  would 
be  as  follows : 

Series  A,  amount  paid  in  §9204.00+  dividend  .$2327.69  = $11531.69 

"       I?,         "          "      "     -981.00+        •'             186.01=  -■-....  1167.01 

"       C,         "          "      "    2125.50+        "             268.77  =  ......  2394.27 

"       D,         "          "      "    2943.50+        "             186.10  =  - 3129.60 

Total  value,  -- $18222.57 

Balance  undivided  by  reason  of  the  non-exactness  of  the  gain  per  cent,         -  .43 

$18223.00 
MISCELLAlirBOUS  PRACTICAL  PROBLEMS. 

To  find  the  rate  per  "ent  of  interest  receired  by  a  non-horroicing  member  on  maturing 

shares. 

1554.  What  rate  of  interest  is  grained  on  shares  of  buildinn:  associations,  the 
maturity  value  of  which  is  $200,  the  dues  being  $1  per  month,  payable  in  advance, 
supposing  the  series  to  mature  in  S}^  years  ?  Ans.  22.39  —  %. 

OPERATION. 

Si  years  =  102  months  at  $1  =  $102  dues  paid. 

$200  maturity  value  of  share  —  $102  =  $98  gain  on  the  monthly  investment  of  $1. 

102,  first  term  +  1,  last  term  =  103. 

103  X  51,  half  the  number  of  terms  =  52.">3  months  or  (5253  —  12)  437f  years,  equated  time 
of  interest  on  §1 ;  or  conversely,  interest  ou  $437|  for  1  year. 

Then  since  $4371  gain  $98  in  1  year,  the  rate  per  cent  is  4371  :  98  :  :  100  :  22.39  —  %,  Ans. 
Or  thus,  to  find  the  rate  per  cent : 

$ 

I  98 
437i  I  100 

I  22.39  —  ,"0',    Ans. 

SECOXD   OPER.4TION   TO   FIND    R.\TE    PER   CENT. 

Interest  on  $4371  for  1  year  at  1  per  cent  is  $4.37}. 
Then  as  $4.37i  :  $98  :  :  1  per  cent  :  22.39  —  5^  Ans. 

Explanation. — Since  $1  dues  were  paid  each  month  for  102  months,  the  first  dollar  paid  bears 
interest  in  favor  of  the  investor,  for  102  months ;  the  second  payment  is  at  interest  101  months, 
etc.,  to  the  last  payment,  which  is  at  interest  for  1  mouth.  Hence,  the  interest  on  the  diflerent 
payments  is  equivalent  to  the  interest  on  $1  for  a  number  of  months  equal  to  the  sum  of  an  arith- 
metical series,  whose  finst  term  is  102,  last  term  1,  and  number  of  terms  102.  This  number  is,  as 
shown  in  the  operation,  5253  months,  or  437i  years.  Then  since  the  interest  on  $1  for  437J  years  is 
the  same  as  the  interest  on  $4371  for  1  year,  we  may  use  either  statement  in  computing  the  interest 
or  in  finding  the  rate  per  cent. 


99^  soule's  philosophic  practical  mathematics.  * 

To  find  the  rate  of  interest  paid  hy  a  borrower,  on  the  Net  Plan. 

1555.  A  member  bought  a  loan  in  a  building  association  conducted  on  the 
net  plan,  and  paid  $C0  pieniium  on  exchange  of  8  shares,  uew  issue.  Wliat  will  be 
liis  rate  per  cent  of  interest  if  the  series  expires  in  SJ  years,  and  G  per  cent  interest 
is  charged  on  the  loan?  Aus.  6.25+  %. 

OPERATION. 

8  shares  .at  $200,  =  -         -         -        $1600  par  of  shares. 

8  shares  at    60  premium  ^     -         -  480  premium. 

Amount  of  loan,       -        -        -        $1120 

Interest  on  $1120  for  1  month  is  (^^^O  X  <'%')  _  jsgo 

Dues,  1  month  on  8  shares  at  $1,  =  -        -  8.00 

Monthly  p.ayment  is  .        -        .        .        .      $13.60 

$13.60  X  100  months  period  of  loan  =  $1360  total  payment. 

Then  the  interest  on  the  $13.60  monthly  payments  for  the  tlifferent  periods,  is  equivalent  to 
the  interest  on  $1  for  a  number  of  months  equal  to  the  sura  of  an  arithmetical  series,  whose  first 
term  is  100,  last  term  1,  and  number  of  terms  100,  which  is,  100  +  1  =  101  x  50,  half  the  number 
of  terms,  5050  months.  Interest  on  $13.60  for  5050  months  is  $343.40  +  $1360  =  $1703.40,  aggregate 
payments. 

$1703.40  —  $1120  =  $583.40  interest  on  the  loan  for  SJ  years. 
i.40  -i-  8i  =  $70,008  interest  for  1  year. 


Then  what  per  cent  of  $1120  is  $70,008?  It  is  ($70,008  X  100)  ~  $1120  =  .0625+  or  6.25+% 
Ans. 

Note. — By  the  net  plan  the  premium  is  deducted  from  the  par  valne,  and  interest  is  charged 
on  the  net  amount  of  the  loan. 

To  find  the  gain  per  cent  realized  by  a  B^lilding  and  Loan  Association,  on  its  invest- 
ments of  dues,  at  simple  interest. 

1556.  If  the  dues  of  a  Building  and  Loan  Association  are  $1  per  month,  and 
the  shares  mature  in  SJ  years,  what  would  be  the  annual  average  gain  per  cent  of 
the  association,  provided  it  invested  monthly  the  dues  as  received,  at  6  per  cent 
simple  interest?  Ans.  6.06  —  %. 

OPERATION. 

Si  years  =  102  months  at  $1  =  $102  dues  received  in  8}  years. 
The  interest  on  the  $1  monthly  dues,  or  receipts  for  the  102  months,  is  equiv.alent  to  the 
interest  on  $1  for  the  equated  time,  which  is  (102  +  1  =  103  X  51)  5253  months  or  437i  years. 
Interest  on  $1  for  437f  years,  at  6  per  cent,  is  $26.26i. 
This  $26.26i  interest  is  the  gain  on  $1  monthly  dues  for  8}  years. 
Then  $26.26i  ^  8i  years  =  $3.09  gain  per  year  on  $102  dues,  or  average  investment  of  ($102 
-i-  2)  $51. 

$3.09  —  $51  =  $.0606  —  or  6.06  —  %  average  gain  per  annum. 

Or  thus,  to  find  the  per  cent  as  per  page  599 : 

Interest  on  $51  for  1  year,  at  1  per  cent,  is  51  cents. 


^  BUILDING    AND    LOAN    ASSOCIATIONS.  997 

Then  since  1  per  cent  for  the  time  gives  51  cents  interest,  conversely  51  cents  reqnireil  1  per 
cent;  and  since  51  cents  interest  required  1  per  cent,  1  cent  will  require  the  51st  part  ami  30y  cents 
■will  recjuire  309  times  as  much,  which  is  6.06  —  %. 

Or  thus,  51  :  100  :  :  3.09  :  6,06  —  °^. 


Interest  on  $51  for  8i  years  at  6.06  —  %  =  $26.27  — . 

To  find  the  per  cent  realized  by  a  Building  and  Loan  Association  on  its  investments  oj 
dues,  at  simple  interest  invested  weekly, 

1557.  If  the  dues  of  an  association  are  50  cents  a  week,  and  the  shares  mature 
in  4i  years,  wliat  would  be  tlie  annual  average  gain  per  cent  to  the  association,  if 
it  invested  weekly  the  dues  received  at  6  i)er  cent,  simple  interest  1 

Ans.  6.03—  %. 

OPERATION. 

ii  years  x  52  weeks  =  234  -weeks  at  50  cents  =  $117  dues  received  in  4^  years. 
The  interest  on  the  50  cents  weekly  dues  for  the  234  weeks  is  equivalent  to  the  interest  on 
the  50  cents  for  the  equated  time,  which  is  (234  +  1  =  235  X  117)  27495  weeks,  or  528|i  years  of  52 
weeks  each. 

The  interest  on  50  cents  for  27495  weeks,  at  6  per  cent,  is  $15.86^. 
This  $15. 86^  interest  is  the  gain  on  50  cents  monthly  dues  for  4^  ye.ars.     Then  15.8625  —  4i  = 
$3,525  gain  per  year  on  $117  dues  or  average  investment  of  (117  -H  2)  $58.50. 

Then  $3,525  -H  $58}  :=  .0603  —  or  6.03  —  ^q  average  gain  per  annum  of  52  weeks. 
Or  thus,  58i  :  100  :  :  3.525  :  6.03  —  %. 
Note. — In  the  above  operation  52  weeks  are  used  as  a  year. 

To  find  the  average  annual  gain  per  cent  realized  on  ireeMy  payments^  simple  interest 

being  aUoutd  on  the  u-eeMy  payments  for  each  year,  and  the  annual  amvnnts 

to  be  put  at  compound  interest  to  the  close  of  the  period. 

1558.  Using  the  figures  of  the  preceding  problem,  what  would  be  the  com. 
pound  interest,  and  the  average  annual  gain  per  cent,  allowing  6  per  cent  simple 
interest  on  the  weekly  payments  for  each  year  of  52  weeks,  and  the  annual  amounts 
to  be  put  at  compound  interest  at  6  per  cent,  to  the  close  of  the  period  or  the 
maturity  of  the  shares,  4J  years  of  52  weeks  each  ? 

OPERATION. 

The  dues  being  50  cents  per  week,  the  sum  received  and  loaned  is  $26  for  the  first  year,  at  6 
per  cent. 

The  interest  on  the  50  cents  weekly  payment  of  dues  for  52  weeks,  of  the  first  year,  is  equiva- 
lent to  the  interest  on  50  cents  for  the  equated  time,  which  is  (52  +  1  =  53  X  26)  1378  weeks. 

The  interest  on  50  cents  for  1378  weeks,  at  6  per  cent,  is  79.5  cents.  Then  26  +  79.5  cents  = 
$26,795  yearly  value  of  the  50  cents  weekly  payments. 

This  $26,795  is  a  yearly  recurring  amount  to  be  placed  at  compound  interest;  thus  forming  a 
aeries  of  annuities,  each  term  of  which  is  $26,795. 


998  soule's  philosophic  practical  mathematics.  * 

Then  to  find  the  sum  of  tlicse  iuinuities,  we  have  tlie  following  formula: 

u  _  A  X  R»  —  A  _  a  _  $26.795  x  I.Od*  —  $26.795    _  $26.795  X  1.2624770  —  $26.795  _ 

"■     ~         R  —  1  ~  .06  "~  .06  ~ 

$117.-1  =:  the  sum  of  an  annuity  of  $1  at  compound  interest  for  4  years  at  G  2)er 
cent. 

Note. — The  1.2624770  is  the  fourth  power  of  1.06,  juul  is  obtained  from  the  Compound  Interest 
Table,  page  612. 

To  this  $117.21  we  must  add:  (1).  The  interest  thereon  for  6  months,  at  6  per  cent,  which  is 
$3.51.  (2).  Tlie  $13  dues  for  26  weeks.  (3).  The  interest  on  the  $13  dues  for  tho  equated  time, 
■which  is  20  cents.     Thus  making  a  total  siini  nf  $],'!3.93. 

See  the  close  of  the  first  solution  for  the  full  details  of  finding  gain  and  gain  per  cent. 

Second  Solution  by  Compound  Annuity  Table. 

$4.374616 
26.795 


$117.217835720 
3.51 
13.00 
.20 


$133,927+ 

To  find  the  total  cost  of  a  Joan  to  a  horrotrer  at  C  per  cent  simple  interest.  Gross  plan, 

monthly  payments,  ^1  per  share. 

1559.  A  member  of  a  building  association  bought  a  loan  on  10  shares  in  a 
new  series,  at  $15  and  the  stated  premium,  with  G  per  cent  interest.  What  will  be 
the  full  cost  of  his  loan  if  the  series  matures  in  110  months?  Ans.  $2810.50. 

operation. 

Note. — By  the  Gross  and  Installment  Plans,  the  dues  and  the  interest  at  6  per  cent  on  eacli 
share  of  $200,  are  each  $1  per  month. 

$1  dues  +  $1  interest  =  $2  monthly  payment  per  share. 

$2  X  10  shares  =  $20  monthly  installments  to  he  made  and  which  earn  interest  as  follows: 

The  first  installment  will  earn  interest  for  110  months,  the  second  installment,  109  months, 
and  so  on,  until  the  payment  of  the  last  inst.allment  which  earns  no  interest.  Hence  the  interest 
on  the  110  installment  payments  for  the  dift'erent  periods  of  time  is  equivalent  to  the  interest  on 
$20,  one  installment  for  the  equated  time  which  is  (110  -f-  1  =  111  x  55)  6105  months. 

The  interest  on  $20  for  6105  months,  at  6  per  cent,  is  $610.50.  $20  each  payment  X  110  payments 
=  $2200,  sum  of  payments. 

$2200  +  $610.50  =  $2810.50  total  cost  of  loan  to  the  borrower. 

Note. — This  method  is  known  as  the  6  per  cent  method. 

To  find  the  actual  premium  charged  on  loans  and  refunded  on  their  payment, 

1560.  A  member  bought  a  loan  on  8  shares  of  stock  1  year  and  4  months  old^ 
and  paid  $80  premium.  What  is  the  actual  premium  if  the  value  of  a  share  was 
lis  at  the  last  annual  report  ?  Ans.  $71.20. 


'  BUILDING    AND    LOAN    ASSOCIATIONS.  999 

OPERATION'. 

$18,  T.ihie  of  1  share  at  Last  report-)- $-1  tines  for  four  mouths  =  .$22  present  .accumulated  value 
of  each  share. 

§200  unaccumulatotl  value  —  $22  accumul.ated  value  =  $178  uuacoumulated  value.  Hence 
iJ-g  of  the  jiar  is  uuaccuimilateil ;  therefore  ^J»  of  the  |80  premium  bid  is  the  actual  premium  paid 
by  the  borrower.     fJ-J  of  $80  =  $71.20,  actual  premium. 

Note  1. — To  find  the  net  Joan,  deduct  the  actiial  premium. 

Note  2. — To  lind  the  total  cost  of  a  loan,  multiply  monthly  payments  by  the  number  of 
months. 

To  find  the  average  annual  gain  per  cent  realized  on  monthly  payments,  simple  interest 

being  alloived  on  the  monthly  payments  for  each  year,  and  the  annual  amounts 

to  he  put  at  compound  interest  to  the  close  of  the  period. 

1561.  1.  If  tlie  dues  of  a  Building  and  Loan  A.ssociation  are  $1  per  month, 
and  the  shares  mature  in  8.J  years,  what  wouhl  be  the  compound  interest;  and  the 
average  annual  gain  per  cent  allowing  6  per  cent  simple  interest  on  the  monthly 
payments  for  each  year,  and  the  annual  amounts  to  be  put  at  compound  interest  at 
6  per  cent  to  the  close  of  the  period  or  the  maturity  of  the  shares  1 

Ans.  Compound  interest  $30.42 ;  7.0164-%. 

Remarks. — The  solution  of  this  problem  forms  the  basis  for  solving  many 

problems  of  interest  to  Building  and  Loan  Associations,  and  to  all  parties  concerned 

in  loaning  or  borrowing  money.    We  therefore  give  three  solutions  of  the  problem 

in  order  to  fully  elucidate  the  principles  involved,  and  to  make  clear  the  methods  of 

solution. 

first  solution. 

lat  year.  12  months  at  $1  =  $12  dues  received  1st  year.  One  dollar  the  first  receipt  was  at  interest 
for  12  months,  the  second  receipt  of  $1  was  at  interest  for  11  months,  etc.  Hence  the 
interest  on  the  receipts  for  dues  for  the  12  ditt'ereut  periods,  is  equivalent  to  the  interest  on  SI 
for  a  numlicr  of  months  represented  by  the  sum  of  an  arithmetical  series,  whose  tirst  term  is 
12,  last  term  1,  and  niHuber  of  terms  12.  Thus  12  -]-  1  =  13  X  6,  half  the  number  of  terms, 
=  78  months  equated  time  of  interest  on  $1  at  6  per  cent  gives,  -        -         -         -        $     .39 

Dues  received  and  loaned  the  1st  year,  ----.....  12.00 

Amount  to  loan  the  2d  year,  -  .-        $12.39 

Sdyear.     Interest  on  $12.39  for  12  months,  at  6  per  cent,         - .74 

Dues  2d  year  and  interest  for  78  months,  equated  time,  on  $1  as  in  the  first  ye«^  12.39 

Amount  to  loan  the  3d  year,     ..-.-....---        $25.52 

Sdyear.    Interest  on  $25.32  for  12  months,  at  G  per  cent  = -  1.53 

Dues  the  3d  year  and  interest,  78  months,  on  $1  as  in  the  first  year,    -        -        -        .  12.39 

Amount  to  loan  the  4th  year,  .-..-.--..-        $39.44 

4ih  year.    Interest  on  $39.44,  12  months,  at  6  per  cent  = 2.37 

Dues  and  interest,  4th  year,  as  in  the  first  year,     ---.---.  12.39 

Amount  to  loan,  5th  year,         ..-.-.-..--.        $54.20 

StJi  year.    Interest  on  $54.20,  12  months,  at  6  per  cent  =         ----...  3.25 

Dues  and  interest,  5th  year,  as  in  first  year,  ........  12.39 

Amount  to  loan,  6th  year,        -- --...       $69.84 


looo  soule's  •philosophic  practical  mathematics.  * 

6th  year.     Iiitcrostnn  $09.84,  12  months,  at  fi  per  ccut,  -        $4.19 

i)ue.s  and  interest,  6th  year,  lis  ill  lirst  year,  --------  12. 3& 

Amount  to  loan,  7th  year, -        -        .        .$86.42 

7lh  i/ear.     Interest  on  $86.42,  12  months,  at  6  per  cent,  5.19 

Dues  and  interest,  7th  year,  as  in  lirst  year,  -- 12.39 

Amount  to  loan,  8th  year,  - - $104.00 

Sth  year.     Interest  on  $104.00,  6  months,  at  6  per  cent, 6.24 

Dues  and  interest,  8th  year,  as  in  lirst  year,  --------  12.39 

Amount  to  loau,  one-half  of  9tli  year,  ---.-.--.      $122.63 

S^th  year.     Interest  on  $122.63,  6  months,  at  6  per  cent, 3.68 

Dues  and  interest,  for  6  months  are  equivalent  to  the  interest  on  $1  for  the  equated 
time,  which  is  (6  +  1  =  7  X  3)  21  months.  Interest  ou  $1  for  21  months,  at  6  per 
cent,  is  $.11  +  |6  dues  = 6.11 

Compound  amount  of  $102  for  8i  years,  at  6  per  cent, $132.42 

Dues  deducted, - 102.00 

Interest  received  in  SJ  years,  ----- $30.42 

$30.42  -r  8i  =  $3.5782-|-  gain  per  year  on  an  average  investment  of  (102  -^  2)  $51. 

51  :  100  :  :  3.5782+  :  7.016+  ^^  average  gain. 

SECOND   SOLUTION'. 

By  the  Compound  Interest  Table,  page  611. 

The  payments  being  $1  per  month,  the  sum  received  and  loaned  is  $12  for  the  first  year.  The 
interest  on  the  $1  monthly  dues  or  receipts  for  the  12  months  of  the  lirst  year  is  equivalent  to  the 
interest  ou  $1  for  the  equated  time,  which  is,  as  fully  shown  in  the  first  solution,  (12  +  1)  =  13  X 
6  =  78  months.  The  interest  on  $1  for  78  mouths  at  6  per  cent,  is  39  cents.  $12  +  39c.  =  $12.39 
the  yearly  value  of  the  monthly  payments. 

This  $12.39  is  a  yearly  recurring  amount  to  be  placed  at  compound  interest;  thus  forming  a 
series  of  annuities,  each  term  of  which  is  $12.39. 

Then  to  find  tlie  sum  of  these  annuities,  we  have  the  following  formula: 

AxR«  — A     t,_  $12.39  X  1.06'  —  $12.39  _  $12.39  x  1.5938481  —  $12.39 

'^    =   — R_i        or    *    -  ^6  -  M = 

$122.62962852  the  sum  of  an  annuity  of  $1  at  compound  interest  for  8  years,  at  6 

per  cent. 

Note. — The  1.5938481  is  the  8tU  power  of  1.06,  and  is  obtained  from  the  Compound  Interest 
Table,  page  612. 

To  this  $122.63  we  must  add  :  (1).  The  interest  thereon  for  6  months  at  6  per  cent,  which  is 
$3.68.  (2).  The  $6  dues  for  6  months.  (3).  The  interest  on  the  $6  dues  for  the  equated  time,  which 
is,  as  shown  in  the  first  solution,  11  cents.     Thus  making  a  total  sum  of  $132.42. 

See  the  close  of  the  first  solution  for  the  full  details  of  finding  gain  and  gain  per  cent. 

Note. — See  the  remarks  on  Annuities,  on  page  921. 

THIRD   SOLUTION. 

By  the  Annuity  Table,  page  925. 

We  first  produce  the  annuity  of  $12.39  as  shown  in  the  first  and  second  solutions,  then  to 
find  the  sura  of  the  annuities  at  the  end  of  the  given  time,  we  have  but  to  multiply  tlie  amount  of 
the  annuity  of  $1  per  annum,  at  comiiound  interest  for  8  years,  by  the  annual  annuity  $12.39. 


BUILDING    AND    LOAN    ASSOCIATIONS.  lOOI 

OPERATION. 

$9.897468  =  sum  of  an  annuity  of  $1  per  annum  at  componntl  interest  for  8  years  at  6  per 
12.39  [cent.     See  Table,  page  925. 


$122.62962852  =  the  sum  of  an  annual  annuity  of  $12.39  for  8  years,  at  6  per  cent  compound  interest. 
To  this  sum  $122.63,  we  add  interest  and  dues  for  6  months,  as  shown  at  the  close  of  the 
first  and  second  solutions. 

Note  1. — See  the  close  of  the  first  solution,  for  the  full  details  of  finding  gain  and  gain  per 
cent. 

Note  2. — See  the  remarks  on  Annuities,  page  921. 

rOUKTH    SOLUTION 

By  Logarithms. 

All  problems  of  this  kind  can  be  worked  by  the  Algebraic  formula  for 
finding  the  amount  of  Sinking  Funds,  or  the  amount  of  Annuities. 

The  formula  is  A  = — — .     Applying  this  formula  to  the  above  problem, 

Sl'^  39  CL  0&^ 1) 

and  using  figures  instead  of  letters,  we  have  amount  =  '^-^ — ^-^ 

To  raise  $1.06  to  the  8th  power  hy  the  use  of  Logarithm  Tables  proceed  as  follows :  (1).  Find 
the  log  of  the  number  1.06  =  .025306.  (2).  Multiply  this  log  by  8  =  .202448.  (3).  Look  for  this 
log  In  the  Table  of  Logarithms.  If  you  cannot  find  it  exactly,  find  the  log  justless  than  it  (.202216) 
and  the  one  just  greater  than  it  (.202488).  The  difference  between  the  two  extremes  is  272,  and 
the  difference  between  the  given  log  (.202448)  and  the  lesser  log  (.202216)  is  232.  Write  these 
differences  into  a  fraction  as,  |fj.  (4).  Find  the  number  corresponding  to  log  .202216  =  1.59300; 
find  the  number  corresponding  to  log  .202488  =  1.59400.  (5).  T.ake  the  iff  part  of  100  =  85-f  and 
add  it  to  1.59300,  writing  it  in  the  place  of  the  naughts  =  1.59385.  Multiply  difference  between 
1.59385  and  1  =  .59385  by  12.39  =  7.3578015.  (6).  Divide  this  product  by  .06  =  $122.63  =  amount 
for  8  years. 

Now  having  the  compound  amount  for  8  years,  we  proceed  for  the  remaining  6  months  as 
shown  in  the  first  and  second  solutions. 

If  the  monthly  dues  in  the  above  problem  were  to  be  compounded  monthly^ 
the  following  would  be  the  formula  for  finding  the  amount  at  the  end  of  the  102nd 
month : 

S(R»  +  1  — 1)        o        $l(1.005i»'-l)        J.,  .         S(Rn  — 1)        Sl(1.005'°'-1) 

^  =  -r ^^  =  M5 -•^^'  "''^^  = r =  ^505 

=  $132.63,  to  this  amount  add  interest  for  one  month  =  fiC/  =  $133.29. 

OPERATION   FOR   FIRST  FORMULA   USING   LOGARITHMS   TO   FACILITATE   MULTIPLICATIONS. 

Log  1.005  =  .002166  X  103  =  log  .22.3098.  The  number  corresponding  to  this  log  is  $1.67147. 
From  this  subtract  $1  =  $.67147.  Divide  this  amount  by  $.005  =  $134.29.  Less  $1  =  133.29,  final 
value. 

OPERATION   BY  SECOND   FORMULA   USING   LOGARITHMS   TO  FACILITATE   MULTIPLICATION. 

Log  1.005  =  .002166  X  102  =  log  .220932.  The  number  corresponding  to  this  log  is  $1.66315. 
Subtract  $1  =  $.66315.  Divide  this  by  $.005  =  $132.63.  Add  i  per  cent  on  $132.63  =  66c.  =  $133.29f 
final  value. 


I002 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


2.  A  Laud  and  Investment  Company  sells  to  B.  a  piece  of  property  for 
$•4000,  at  8  per  cent  interest,  on  condition  that  B.  shall  pay  for  the  property  at  the 
rate  of  $90  per  month,  vhicli  monthly  payments  are  to  include  the  interest  on  the 
successive  monthly  balances.  How  long  will  it  take  B.  to  pay  for  the  property, 
the  first  payment  being  made  one  month  after  the  sale?  How  many  payments  did 
B.  make ?  What  was  the  total  interest?  What  was  the  average  rate  jier  cent,  per 
annum  on  the  $4000  ?  What  did  the  company  gain  and,  B.  lose  in  excess  of  the  S 
per  cent  interest  charged  on  the  successive  monthly  balances  ? 

Answers.  (1).  It  requires  53  months  to  pay  for  the  property.  (2).  B.  made 
53  payments,  52  of  $90  each  and  1  of  $79.CC.  (3).  B.  paid  $4759.66.  (4).  Total 
interest  was  $759.06.  (5).  The  average  rate  per  annum  on  the  $4000  was  4.3  per 
cent.  (C).  The  company  gained  and  B.  lost  iii  excess  of  the  8  per  cent  on  the 
successive  monthly  balances  which  bore  interest,  the  interest  (simple,  annual  or 
compound,  according  to  manner  of  making  Investments),  on  each  $90  monthly 
liayment,  except  the  last  payment  of  $79.66,  from  the  time  each  payment  was 
made  until  the  expiration  of  53  months. 

PARTIAL    OPERATION 

By  the  United  States  P.artial  Payment  Method. 


OPERATION 

For  the  first  two  payments. 

Amt.  at  S  "(f  int.  for  the  1st  month, 
Int.  ou  same  for  the  1st  mouth,     - 

Amonnt  due  at  close  of  1st  month, 
1st  payment  for  the  1st  muutU, 

Bal.  at  8  %■  int.  for  the  2d  month. 
Int.  ou  same  for  the  2d  mouth, 

Amount  due,  at  close  of  2d  montli, 

2d  jiLiymcut,     ------ 


$4000.00 
2(5.67 

§4026.67 
90.00 

$3936.67 
26.24 

§3962.91 
90.00 


OPERATION 

For  the  last  two  payments. 

Bal.  at  8  »o  int.  for  the  52d  month,      -        $168.01 
Int.  ou  the  same  for  the  52d  mouth,        -  1.13 


Amt.  due  at  the  close  of  52d  mouth, 
52d  payment,  -        .        -        . 


Bal.  at  8  %  int.  for  the  53d  month. 
Int.  on  same,  for  the  53d  mouth,     - 

Amt.  due  at  the  close  of  53d  mouth, 
53d  jjaymeut,    -        -        •        -        . 


$169.13 
90.00 

$79.13 
.53 

179.66 
79.66 


Bal.  at  8  ,%'  int.  for  the  3d  month,  -    $3872.91 

We  have  omitted  in  this  partial  operation  the  several  items  of  interest, 
amount,  payments  and  balances,  from  the  third  item  of  interest  to  the  51st  payment, 
both  inclusive.  The  beginning  and  closing  work  of  the  problem  which  we  have 
presented  is  sufBcient  to  clearly  elucidate  the  same. 

A.  made  52  payments  of  $90  each  =  $4680,  and  1  iiayment,  the  last,  of  $79.66, 
making  a  total  of  $4759.66,  and  53  payments  in  53  months  or  4  years  and  5  months. 
The  interest  paid  was  $759.06.  $759.66  interest  ou  $4000,  for  53  months  =  4.3 
per  cent  per  annum,  as  per  Article  1166,  page  599. 

Though  the  average  annual  interest  is  but  4.3  per  cent  on  the  $4000,  the 
company  made  and  B.  lost  8  per  cent  on  the  successive  monthly  amounts  that  bore 
interest.  And  as  the  $90  monthly  payments  were  received  by  the  company  and 
paid  out  by  B.  the  use  or  interest  on  said  payments  to  the  company  would  be  equal 
to  about  4.3  per  cent,  thus  making  nearly  8.6  jier  cent. 


It 


BUILDING    AND    LOAN    ASSOCIATIONS.  IOO3 


It  will  be  reipemberecl  that  by  tbe  United  States  system  of  partial  payments, 
the  shorter  the  intervals  of  time  between  payments,  the  greater  the  gain"is  to  the 
party  receiving  the  payments. 

3.  A  Loan  and  Land  Company  effects  loans  or  sells  lands  on  the  following 
conditions :  1.  To  charge  8  per  cent  interest  on  the  monthly  principal  loaned,  or  on 
the  amount  due  for  land.  2.  To  divide  the  sum  loaned  or  the  amount  due  for  land 
by  the  number  of  months  the  loan  is  to  run,  and  to  take  notes,  secured  by  mortgage 
payable  in  successive  months,  each  for  an  amount  equal  to  the  quotient  obtained 
by  this  division,  plus  the  monthly  interest  on  the  successive  principals  loaned. 

According  to  this  method,  what  would  be  the  face  of  each  note  for  a  loan  of 
$3600  for  6  years  at  8  per  cent  ?  Ans.  See  partial  operation. 

PARTIAL    OPERATION. 

$3600  loaned  -;-  72  inontlis  =  $50  partial  face  of  each  of  tlie  72  notes. 

Amount  at  interest  the  first  month,        ..-.-------        $3600.00 

Interest  on  same,  8  per  cent,  1  month,  ..-.-.----  24.00 

Amount  (Ine  at  the  close  of  the  first  month, ...-        $3624.00 

Face  of /rsJ  note,  $50 +  $24  =  $74,        -        - 74.00 

Balance  at  interest  the  second  month,  ......----        $3550.00 

Interest  on  same,  8  per  cent,  1  month,  ......-.--  23.67 

Amount  dne  at  the  close  of  the  second  month,      .-..----         -        $3573.67 
Face  of  second  note,  $50  +  §23.67  =  $73.67, -  73.67 

Balance  at  interest  the  third  month,        -----------        $3500.00 

Interest  on  same,  8  per  cent,  1  month,  --.-...-.-  23.33 

Amount  at  the  close  of  the  third  month,         -----...--        $3523.33 
Face  of  third  note,  $50  +  $23.33  =  $73.33,      -        -        -        - 73.33 

Balance  at  interest  the  fourth  month,  ..--...-•.        $3450.00 

,     And  in  like  manner  we  continue  the  operation  to  find  the  face  of  the   72 
notes. 

It  will  be  observed  that  the  operation  involves  the  United  States  method  of 
partial  payments  as  elucidated  in  the  preceding  problem. 

To  find  the  interest  and  the  rate  per  cent  of  interest  realized  by  a  Loan  and  Investment 

Company  on  certain  conditions. 

1562.  1.  A  Loan  and  Investment  Company  loans  $3000  for  5  years,  at  8  per 
cent  per  annum.  The  interest  ($1200)  is  added  in  advance,  and  the  whole  amount, 
$4200,  is  made  payable  in  GO  notes  of  $70  each,  payable  monthly,  without  interest. 
Allowing  that  the  company  may  re-invest  the  monthly  collections  of  the  notes  at  8 
per  cent  simple  interest,  what  would  be  the  total  interest  received  by  the  company, 
and  what  would  be  the  rate  per  cent  per  annum  1 

Ans.  $2026  total  interest;  13ff  rate  per  cent  per  annum. 

Note.— For  the  solution  of  this  and  the  two  following  prohlems,  see  page  675. 


I004  soule's  philosophic  practical  mathematics.  * 

2.  Suppose  iu  problem  1,  above,  that  the  moutbly  note  aud  intei-est  collec- 
tions were  re-invested  at  simple  interest  as  therein  stated,  what  ■would  liave  been 
the  respective  interest  gain  for  each  year? 

3.  Suj^pose  in  problem  1,  above,  that  the  monthly  collection  of  interest  on 
the  notes  and  the  interest  ou  such  interest  were  used,  at  the  close  of  each  year,  as 
a  i)art  of  the  dividend  fund,  what  would  be  the  amount  of  interest  earned  yearly? 

4.  April  8,  1804,  a  party  purchased  from  a  real  estate  holder  a  house  and  lot 

for  $7500,  on  the  following  terms  and  conditions :  $300  was  paid  in  cash.    36  notes 

of  $200  each  bearing  7  per  cent  interest  were  given ;  three  of  the  notes  were  made 

payable  in  3  months,  July  8,  1894.     The  other  33  notes  were  made  payable  monthly 

after  July  8,  1894.    After  the  three  notes  payable  July  8,  and  19  notes  payable 

monthly  after  July  8  had  been  paid,  the  holder  sold  the  14  remaining  notes  at  7  per 

cent  per  annum  discount.     Counting  30  days  to  the  month  without  grace  or  discount 

day,  what  were  the  proceeds  of  the  14  notes,  Bankers'  Discount  Method  T 

Ans.  $3136.74. 
Note. — For  the  solution  of  this  problem,  see  page  590. 

5.  A  party  holds  $2700  of  stock,  which  pays  $80  per  month  revenue,  (35| 
per  cent  interest  per  annum).  The  stock  has  27  years  to  run.  What  is  the  jiresent 
value  of  the  stock,  the  current  rate  of  interest  being  6  per  cent,  allowing  6  percent 
interest  on  all  the  monthly  payments  of  the  i-eyenue  or  income  ?  Ans.  $12614.02. 

Note. — For  the  solutiou  of  this  and  the  following  problem,  see  page  591. 

6.  Suppose  in  the  above  problem,  interest  is  allowed  only  on  the  annual  pay- 
ments of  the  monthly  revenue  instead  of  the  monthly  payments,  what  would  be  the 
present  value  of  the  stock  ?  Ans.  $12575.34. 

7.  A  Homestead  Association  sold  to  Levin  Cooper  Soule,  a  house  and  lot 
for  $5000,  on  the  following  conditions  of  payment:  Cash  $1000;  the  remainder  in 
80  notes  of  $50  each,  bearing  7  per  cent  interest  and  payable  monthly,  as  the 
months  expire,  in  80  consecutive  months. 

In  this  transaction,  what  is  the  rate  per  cent  interest  that  the  homestead 
makes  and  what  per  cent  interest  does  Levin  Cooper  Soule  j^ay  on  the  $4000? 

Ans.  The  homestead  makes  and  Levin  Cooper  SouM  loses, 
3.54|%  +  3.45|%  =  7%  +  .5512+  %  use  of  inter- 
est  on  interest  payments,  =  7.5512+  %. 

Note  1. — For  the  solution  of  this  problem,  see  jiages  592  to  596. 

Note  2. — The  following  are  valuable  works  ou  Building  and  Loan  Associations:  How  to 
Manage  Building  Associations,  by  Edmund  Wrigley.  Treatise  on  Building  Associations,  by  N.  C. 
Thompson.  H.  S.  Rosenthal's  Manual  for  Building  and  Loan  Associations.  G.  A.  Eudlish  on  the 
Laws  of  Building  Associations.     Ninth  Annual  Report  of  the  Commissioner  of  Labor. 


M 


®^y|» 


CHECK   FIGURE   SYSTEM   OF   PROVING   POSTINGS. 


1005 


Multiples  of  IS. 

To  be  memorized 

or  used  ou 

a 

card. 

13 

351 

689 

26 

364 

702 

39 

377 

715 

52 

390 

728 

65 

403 

741 

78 

416 

754 

91 

429 

767 

104 

442 

780 

117 

455 

793 

130 

468 

806 

143 

481 

819 

156 

494 

832 

169 

507 

845 

183 

520 

858 

195 

533 

871 

208 

546 

884 

221 

559 

897 

234 

572 

910 

247 

585 

923 

260 

598 

936 

273 

6U 

949 

286 

624 

962 

299 

637 

975 

312 

650 

988 

325 

663 

338 

676 

THE  THIRTEEN  CHECK  FIGURE    SYSTEM   OF    PROV- 
ING POSTINGS  OR  TRANSFERS. 


GENERAL    DIRECTIONS   FOR   USING    THIS   SYSTEM. 

1.  Separate  the  number  when  consisting  of  more  than  three  figures, 
into  periods  of  threes. 

2.  When  there  is  more  than  one  period  of  odd  or  of  even  periods  in  the 
number,  add  the  like  periods  together. 

3.  AVhen  the  number  contains  less  than,  or  only  three  figures,  then 
subtract  therefrom  some  multiple  of  13  which  is  less  than  the  number. 

4.  When  the  odd  period  or  the  sum  of  the  odd  periods,  is  more  than  13  less 
than  the  eren  period  or  the  sum  of  the  even  periods,  then  increase  the  odd  period  or 
periods  by  adding  thereto  some  multiple  of  13,  so  as  to  produce  a  sum  not  exceeding 
13  greater  than  the  snin  of  the  even  period  or  jieriods.  Then  subtract  the  even 
period  from  the  odd  and  the  difference  trill  be  the  Check  Figure. 

5.  When  the  even  period  is  more  than  13  less  than  the  odd  period,  then  add 
some  mutiple  of  IS  thereto  so  as  to  produce  a  sum  less  than  the  odd  period  and 
which,  tvhen  subtracted  from  the  odd  period,  will  leave  a  remainder  less  than  IS, 
which  tvill  be  the  Check  Figure. 


A  SHORT  PRACTICAL  METHOD  OF  FINDING  THE  13  CHECK  FIGURE. 


Knltiples  of   13 

To  be  memo- 
rized. 

13 

26 

39 

52 

65 

78  . 

91 
104 
117 
130 


In  practice  it  is  necessary  to  memorize  only  the  first  ten  multiples  of  13, 
and  to  acquire  quickness  in  seeing  the  products  thereof  when  multiiilied  by 
the  numbers  from  2  to  10.  By  this  knowledge,  the  various  multiples  of  13 
may  be  seen  instantly  by  the  mind  and  applied  in  a  manner  as  shown  in  the 
following  illustrative  examples: 

1st.  When  there  is  but  one  period,  mentally  see  what  multijile  of  13 
must  be  subtracted  from  it  to  leave  a  remainder  less  than  13;  this  remain 
der  is  the  Check  Figure. 

Thus :  To  find  the  Check  Figure  of  57,  we  subtract  52  from  57  and  obtain 
a  remainder  of  5,  which  is  the  Check  Figure. 


The  Learner  should  also 
memorize  the  products  of  the 
most  of  these  multiples  by 
2,  3,  4,  5,  6,  7,  8,  9  and  10. 


2d.  When  there  are  two  periods,  odd  and  even,  mentally  see 
what  multiple  of  13,  if  any,  must  be  added  to  the  lesser  jjeriod  so 
that  the  dift'erence  between  them  will  be  less  than  13. 

Thus :  1st.     To  find  the  Check  Figure  of  27453,  we  mentally 
Thus :  65  X    3        195  '  "''^'^  ^^^  *°  ^'^'  *^®  even  period,  making  443,  which  we  subtract  from 

453,  the  odd  period,  and  produce  10,  as  the  Check  Figure. 

Or  2d.  To  find  the  Check  Figure  of  352865,  we  add  520  to  the 
even  period,  making  872  ;  from  which  we  subtract  the  odd  period 
and  produce  7  remainder;  then  as  872  was  larger  than  the  odd 
period,  we  subtract  7  from  13  and  produce  6  as  the  Check  Figure.  If  we  had  added  507  to  the 
even  period,  the  operation  would  have  been  shorter.  But  to  produce  507  from  the  table  of  the  first 
10  multiples  of  13,  more  mental  work  is  required  than  when  we  use  520. 

Find    the  Check  Figure  of  242,723,492,168.  (891)=   sum  of  odd  periods.      734  =  sum  of 
even  periods,  plus  156  (2  times  78)  =  890.     1  Check  Figure. 


65  X  3 

195; 

78  X  4 

312; 

91  X  5 

455; 

52  X  10 

520,  etc 

ioo6 


SOULE  S    PHILOSOPHIC    I'RACTICAL    MATHEMATICS. 


Find  tlie  Check  Figure  of  the  following  numbers  by  mental  operatioiis. 
figures  to  be  written  except  the  Check  Figure. 


No 


(1) 

(2) 

13 

15 

(^) 

129 

(4) 
(5) 
(6) 

84 

142 

508 

(7) 

794 
907 

(9) 


(10) 
(11) 

(12) 

(13) 
(14) 


(15) 

(IG) 
(1') 

(18) 


NOS. 


2,640 

0.028 
2,972 

53,415 

178,213 
249,007 


7,210,580 

324,808,410 
1,540,017 

90,244,503 


(19) 

25,000 

(20) 

1,302 

(21) 

100 

(22) 

$   3.00 

(23) 

$  300.00 

(24) 

$  42500 

(25) 

$  204.50 

(20) 

$0028.15 

(27) 

$   400.00 

(28) 

$    .43 

OPERATIONS. 

13  from    13  -  -  -  -  - 

13  from    15  -  -  -  -  - 

20  from    29  -  -  -  -  - 

78  from    84  -  -  -  -  - 

].)0  from  142  ..... 

."".20(]()  times  52)4-39=559  from  508         =     - 

780  (10  times-78)  +  13  =  793  from  794         =     - 

780  (10  times,  78)  +  117=897    "     907         =     - 

or  tlius  :  910  less  807=3,  and  3  from  13  =     - 
052  (10  times  05+2  in.  2(1  period),  less 

040  =  12,  iind  12  from  13,      ...     -       =     - 

or  thus:  520  +  2+117=039  from  640       =     - 
13  +  0=19  from  28        -  -         -  =     - 

<jl()+05+2  =  !t77  less  972=5  &  5  from  13  =      - 

or  thus:  912+52=904  fnmi  972  =     - 

351  (3  times  117)  +  53  in  2(1  period=404 

and  404  from  415  -  -         -         =     - 

178+20  =  204  from  213       .         .        .         =     . 
234(2timesll7)  +  7+13=2541ess245        =     - 

or  thus  :  234+7  =  241  from  249=8  and 

S  from  13 =     - 

or  thus:  200— 13+7=254  less  249         =     - 
390   (10  times   39)  +  210,    iu   2d  period, 

=000   less  593  (sum  of  odd  periods) 

=  7  from  13 =     - 

or  thus:   351    (3  times  117)  +  210+26 

=587  and  587  from  593  -         -         =     - 

734  (sum    of   odd    periods)  +  78  =  812 

less  even  i)eii(Kl,    •     •         -         -         -    =     - 
18  (sum  of  odd  periods)  +  520  (10  times 

52)  =  538  from  540=2;  and  2  from  13     =     - 

or  tlius:  18+533=551  less  540       -         =     - 
599  (sum  of  odd  periods)  less  (244+351) 

595  from  599  ....=- 

26  (2  times  13)  less  25  (even  period)  =     - 

312   (4  times   7S)+1=313  less   302=11 

and  11  from  13       -         -         -        -         =     - 

91  from  100 =     - 

351  -         -         -         -         .         =     9cf. 

39 =     3  cf. 

455  +  42=497  -         -         -         =     3  cf. 

410  +  13+20=449  from  450  -         =     1  cf. 
(208+(;02)  from  815       -         -         =5  cf. 

52  less  40    -         -         -         -         =  12  cf, 

39  from  43  -        •        -        =     4  cf. 

By  memorizuig  )ierfectly  the  first  10  multiples  of  13,  aiul  hecomins;  quick  in  producinfj mentally 
the  products  tlicreof,  by  the  numliers  13  to  10,  any  ])erson,  wlio  can  add  and  subtract  with  facility, 
may  see  mentally  and  instantly  the  successive  stei)s  which  produce  the  Check  Fi{;ure. 

An  hour's  practice  per  day,  for  a  few  days,  on  the  foregoing,  and  on  similar  problems  which 
may  be  assumed  or  taken  from  any  book  of  account,  will  enable  any  bookkeeper  or  clerk  to  write 
the  Check  Figure  almost  at  sight. 

This  knowledge  of  the  13  Check  Figure  system  is  of  great  value  to  all  bookkeepers  and  to 
office  men. 


0  Check 

Fig. 

.> 

u 

3 

u 

6 

(( 

12 

i. 

9 

a 

1 

i. 

10 

u 

10 

'' 

1 

u 

ii 

1 

.. 

9 

u 

8 

a 

8 

u 

11 

a 

9 

i. 

5 

4. 

5 

U 

5 

u 

G 

ii 

6 

(< 

4 

li 

11 

,u 

11 

ii 

4 

ii 

1 

u 

5 

ii 

' 

9 

ii 

1 

CHECK    FIGURE   SYSTEM    OF    PROVING    POSTINGS,    ETC. 


1007 


APPLICATlOiSr  OF  THE  13  CHECK   FIGURE    METHOD    OF    LOCATING 
EKllOKS  AND  OF  PliOVIXG  FOSTINGS  AND  TEANSFERS. 

Post  or  transfer  the  following  items  from  the  Sales  Book  to  the  Ledger : 

LEDGER  ACCOUNTS. 
J.  P.    Hayter. 


SALES  BOOK  ITEMS. 
J.  P.  Hayter,  -         -     $     120  00 

D.  M.  Smith,    -         -         -  2400  0(1 

F.  B.  Denson,  -         -       172S4  50 

J.  C.  Rimes,     -        -        -  79 


Total, 


619S05  29  IS 


$120  00 


rig. 

1 


D.  M.  Smith. 


Check  Fig.  of  the  total  is         -  5 

"         "       "  IS.  the  sum 
of  the  Cheek  Figures  is  -  5 

which  proves  the  cor- 
rectness of  the  addition. 
By  inspection,  we  observe  that  the 
respective  sums  of  the  Check  Figures  in 
the  Sales  Book  and  on  the  debit  of  the 
Ledger  are  the  same,  IS,  and  that  5  is  the 
Check  Figure  of  each. 

This  proves  that  the  items  were  ]iosted 
to  the  correct  side  of  the  Ledger  Ac- 
counts. 


$2400  00 


<;k 

Fig. 


F.  B 

.  Bensok. 

$17284  50 

Ck. 
Fig. 

9 

J.  C.  Rimes. 


Ck. 

rig. 

IS 


sum  of  c.  fs. 


In  like  manner  the  po.sting  of  the  Invoice  Book,  Cash  Book,  Journal  or  any 
other  book  that  is  used  as  a  posting  medium,  would  be  made  and  the  work  proved 
by  the  13  Check  Figure. 

Note. — When  jio.sting,  set  the  Ledger  check  figures  on  a  slip  containing  debit  and  credit 
check  figure  columns,  and  make  this  test  for  each  book  posted. 

TO  PROVE  ADDITION  BY  THE  13  CHECK  FIGURE  METHOD. 

First.  Find  the  check figuie  of  each  of  the  numbers  added.  2d.  Find  the 
check  tigure  of  tliese  cheuk  figures.  3d.  Find  the  check  flguie  of  the  sum  of  the 
numbers.  If  the  check  figure  of  the  sum  of  the  numbers  is  the  same  as  the  check 
hgure  of  the  check  hgures  of  the  difl'ereut  numbers,  the  work  is  correct  as  far  as 
this  method  of  proof  can  determine. 


PROBLEM. 

Prove  the  addition  of  the  follow- 
ing numbers  by  the  13  Check  Figure 
method. 


345  =  7  c.  f. 

S7  =  9  c.  f. 

4357  =  2  c.  f. 

796  =  3  c.  f. 


21  —  13  =  S  c.  f.  V 


5585  =  8  c.  f.  ■/ 
TO  PROVE  SUBTRACTION  BY  THE   13  CHECK  FIGURE  METHOD. 

First.  Find  the  check  figures  of  the  minuend,  subtrahend  and  remainder. 
2d.  Subtract  the  check  tigure  of  the  subtrahend  from  the  check  tigure  of  the 
minuend  and  if  the  difference  is  the  same  as  the  check  figure  of  the  remainder,  the 
work  is  correct  as  far  as  this  method  of  proof  can  determine. 

NoTK  1. — Increase  the  check  figure  of  the  minuend  by  13,  when  it  is  less  than  the  check 
figure  of  the  subtrahend. 

Note  2. — By  adding  the  check  figures  of  the  subtrahend  .and  the  remainder,  the  sum  should 
ef|nal  the  check  figure  of  the  minuend,  and  thus  prove  the  work.  When  this  sum  is  in  excess  of 
13,  then  subtract  13  therefrom  and  the  ditlereuce  will  equal  the  check  figure  of  the  minneud. 

PROBLEMS. 

Minuend,     -    857  =  12  c.  f.     $230  50  c.  f.  =  l  )  .^      ~ 
Subtrahend,     465  =  10  c.  f.         87  25  c.  f.=2  )    -'  ^-  ^' 


Prove  the  subtraction 
of  the  following  numbers 
by  the  13  Check  Figui'e 
method. 


Remainder,      392=  2  c.  f.      $143  25    = 


12c.£ 


looS  soule's  rHiLosoi'iiic  practical  mathematics.  * 

TO  PEOVE  MULTirLICATIOX  BY  THE  13  CHECK  FIGURE  :-ETUOD. 

First.  Find  the  clieck  figure  of  tlio  niiiltiplicaiid,  tlie  nmltiplier,  and  tlie 
product.  I'd.  Find  tlie,  check  figure  of  the  product  of  the  check  figures  of  the 
multiplicand  and  the  multiplier.  If  this  check  figure  is  the  same  as  the  check 
figure  of  the  pioduct,  the  operation  is  correct,  as  far  as  this  proof  can  determine. 


PEOBLEMS. 


Prove  the  iiiiiltipli-:Mulliplic;uid,      C4L'S  =  f)  )   _  ,  „        „ 


cation  of  the  follow- 
ing numbers  hy  the  13 
Check  Fiffiire  method 


.Multiplier,     -     43.")7  =  -!  ) 


Product,      L'SO()67U0  =  V2  c.  f. 


451  = 
(;4  = 


j^  \  =10S=4  c.  f. 


2.s«ti4  =  4  c.  f. 


TO  PEOVE  DIVISION  BY  THE  13  CHECK  PIGUEE  METHOD. 

First.  Find  the  check  figures  of  the  divisor,  dividend,  quotient,  and  remain- 
der, if  any.  2d.  Multiply  together  the  check  figures  of  the  divisor  and  quotient, 
and  find  the  check  flguie  of  the  product.  3d.  To  this  check  figure  add  the  check 
figure  of  the  remainder  and  find  the  check  figure  of  this  sum;  if  this  check  figure 
is  the  same  as  the  check  figure  of  the  dividend,  the  work  is  correct,  as  far  as  this 
method  of  proof  can  determine. 

PROBLEM. 

Prove  the   division  of  divisor.                          dividend.                 quotient. 

the  following  number  by  (9  c.  f.)  3207                )  4675S2  =  (11  c.  f.)     ( 145  =  2  c.  f. 

the    1 3     Check    Figure  

method.  Eemainder,  2567  =  6  c.  f. 

0x2  =  IS  =  5  c.  f.  -f  6  =  11,  same  as  c.  f.  of  the  dividend. 

Note. — There  lire  certain  errors  thiit  cannot  be  iletecteil  by  the  13,  the  11  anil  the  9  Check 
Figure  methods  of  proof. 

THE  CENT  ET  USE,  OR  101  METHOD  FOR  FINDING  THE  CHECK  FIGURE  IN  POSTING. 

Sejiarate  the  given  number  into  periods  of  two  figures  or  places  each,  from  the  right. 

From  the  sum  of  the  odd  place  periods,  subtract  the  suui  of  the  even  place  periods,  adding 
101  to  the  sum  of  the  odd  place  periods,  when  siiialler  than  the  sum  of  the  even  place  periods. 

If  tlie  remainder  contains  three  or  more  figures,  proceed  as  before,  until  there  remains  not 
more  than  two  figures.  Then  subtract  the  tens  figure  from  the  units  figure,  adding  10  to  the 
units  figure  when  the  smaller,  as  iu  oriliuary  subtraction.  The  remainder,  or  final  dilferential, 
will  be  the  cheek  figure  required.  The  two  places  or  figures  for  cents  must  always  be  considered 
the  first  odd  period. 

THE  MILLS  ET  UXE,  OR  1001  METHOD  FOR  FINDING  THE  CHECK  FIGURE  IN  POSTING. 

Separate  the  given  number  into  periods  of  three  figures  each,  frcmi  the  riglit.  Then  from  the 
sum  of  tlie  odd  jdace  jieriods,  subtract  the  sum  of  the  even  place  ])eriods,  the  (lifl'erence  wmU  be  the 
remaiuder.  if  the  whole  number  had  been  divided  by  1001.  Then  reduce  the  remainder  by  the  9, 
11,  13,  or  101  method  lor  the  final  differential  or  check  figure. 

THE  99  METHOD  OF  FINDING  THE  CHECK  FIGURE. 

Separate  the  number  into  periods  of  two  figures  from  the  right.  Then  add  all  the  periods 
and  reduce  the  sum  liy  the  same  method  to  two  figures;  then  reduce  to  one  figure  by  adding, 
as  for  y's.     The  tiual  result  will  be  the  check  figure. 

THE  999  METHOD  OF  FINDING  THE  CHECK  FIGURE. 

Separate  the  number  into  periods  of  three  figures  from  the  right.  Then  add  all  the  periods, 
and  reduce  the  sum  to  two  figures  by  the  99  method,  and  reduce  this  result  by  the  9  method,  for 
the  final  check  figure. 

See  pages  590  to  596  of  Sonl6's  New  Science  and  Practice  of  Accounts  for  full  explanation 
and  application  of  the  9  and  11  check  figure  systems. 


EQUATION   OF   PAYMENTS   WITH   SPECIAL   FEATURES. 


1009 


Problem  1. 

Jones  bought  of  Brown  the  following  bills  of  goods  on  CO  days  credit  and 
2%  discount  in  10  days,  -with  the  privilege  of  60  days  additional  credit,  in  case  the 
discount  privilege  is  not  availed  of:  Jau'y  4,  '03,  $300;  Jan'y  20,  $150;  Feb'y  10, 
$800  ;  Mar.  20,  $1200  ;  Apr.  1,  $90  ;  May  25,  $500. 

[f  payment  of  this  account  is  made  May  30,  '03.  what  is  the  amount  due  to 
Brown,  the  rate  of  interest  being  6%  ?  Ans.     $3019.73. 

Note. — See  pp.  627  to  653,  for  full  elucidation  of  Equation  of  Accounts. 


FIRST  OPEEATIOK— FiitsT  Step 

To  equate  the  sales  and  find  wlieu   tlie   time 

expires  for  the  2%  discount  privilege. 

1903. 

Jan'y    4.  $  300x  0  =  0 

"    20.  150  X  23  =  3300 

Feb'y  10.  800  x  37=  29600 

Mar.    20.  1200  x  75  =  90000 

Apr.      1.  90  X  87  =  7830 

May    25.  500x141  =  70.500 


$3040 


)  201230  (66  ds 


SECOND  STEP  OF  FIRST  OPERATION. 

201230  ds.  ^  $3040  =  06  days. 

66  ds.  equated  date  of  sales  after  Jan. 

4,  1903. 
CO  ds.  credit  ou  each  sale. 
10  ds.  discount  privilege. 


136  ds.  after  Jan'y  4,  '03,  is  May  20,  '03, 
the  Equatcil  Date  i'or  the  pay- 
ment of  all  the  purchases  at 
2%  discount.  But  as  pay- 
ment was  not  made  until  May 
30,  '03,  the  2%  tUscount  i  s  not 
allowed,  but  CO  ds.  additional 
credit  is  allowed  iu  place 
thereof. 


SECOND  OPERATION— First  Step 
To  find  the  Equ.ated  Date  of  the  account,  allow- 
ing 120  days  credit  on  all  sales. 

1903. 

$  300x120=  36000 

1.50x142=  21300 

800x157=  125000 

1200x195=  234000 

90x207=  180.30 

500x201=  130500 


Jau'y  4. 
"  20. 
Feb'y  10. 
Mar.  20. 
Apr.  1. 
May    25. 


$3040  )  5CC030  (186  ds. 

500030  ds.  -H  $3040  =  ISO  days  after 

Jau'y   4,   '03,  gives  July  9,  '03,  as  the 

Equated  Date  for  the  payment  of  the 

account  without  discount. 

Note. — By  adding  120  days  to  the  66  d.ays. 
equated  date  of  the  sales,  as  shown  in  the  sec- 
ond step  of  the  first  oiieration,  wo  could  have 
found  the  186  days  credit  on  the  whole  account, 
without  making  the  second  operation. 


Explanation. — From  the  above  we  see  that 
the  discount  privilege  ceased  May  20,  '03;  that 
the  equated  date  of  the  account,  allowing  120 
days  credit  on  each  sale,  is  July  9,  '03;  and 
that  the  account  was  settled  May  30,  '03. 
Hence  Jones  i.s  entitled  to  interest  ou  the 
amount  purchased  from  May  30  to  July  9,  03, 
which  is  40  days. 


SECOND  STEP  OF  SECOND  OPERATION 

To  find  the  interest  and  balance  due 

to  Brown. 


00 


.1040 
40 


.$3040       Amount. 
20.27  Int. 


$20.2CS=iut.     $3019.73  due  Browii. 


lOIO 


SOULE  S    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Peoblem  2. 

A.  sold  to  B.  the  Ibllowiug  invoices  of  goods  :  May  10,  $.J00  ;  May  20,  $700; 
June  1,  $400  ,  July  3,  $1000.  The  terms  are  GO  days  credit  and  2%  discount  in  10 
days,  or  60  days  additional  credit,  in  case  the  In/o  discount  is  not  availed  of. 

B.  desires  to  settle  the  account  on  July  20,  1904.  What  amount  does  he  owe 
A.  on  this  date  ?  Aus.     $2530.90. 

2d  Operation  to  fiiul   amount  due  .July  20,  '04i 
when  the  settlement  was  made. 

By  the  above  work  we  see  that  the 
discount  privilege  extended  to  Aug.  15, 
'04,  and  as  the  settlement  was  made 
July  20,  '04,  B.  is  entitled  to  the  2%  dis- 
count, and  to  the  interest  on  the  balance 
from  July  20,  to  Aug.  15,  '04.     Thus  : 

$2000    Amount  of  purchases. 
52=27^  discount. 

$2548      Balance. 

11.04  lut.  from  July  20,  to  Aug. 
15,  '04,  26  days  at  6%. 


1st  Operal 

ion  to  ef|uate  the 

sales  and  find  the 

date  that  the  2%  discount 

))rivilege  ceases. 

1904. 

May  10. 

$500  X     0  = 

0 

"     20. 

700  X  10  = 

7000 

June  1. 

400  X  22  = 

8800 

July    3. 

1000  X  54  = 

64000 

$2536.96   Net  balance  due  A.  July 
20,  '04. 


$2(J00  )  09800  (27  days 

To  this  27  days  credit  after  May  10,  '04, 
■neadd  60  days  credit  on  each  sale, 
and  also  10  days  credit  for  dis'nt  option 
and  thus  — 

produce  97  days  credit  after  May  10,  '04, 
which  gives  Aug.  15,  '04,  as  the  date  on 
which  the  2%  discount  jirivilege  expires. 

Note. — By  adding  120  days  credit  to  the 
above  27  days,  -vve  produce  147  days  credit  after 
May  10,  which  gives  Oct.  4,  '04,  as  the  equated 
date  of  the  whole  account,  which,  however,  is 
not  required  in  this  problem,  as  the  settlement 
was  made  within  the  2%  discount  limit. 

Problem  3. 
Suppose  in  the  preceding  problem  that  B.  had  settled  the  account  August 
28,  '04,  what  amount  would  have  been  due  to  A.  ?  Ans.     $2583.97. 

Explanalion. — From  the  work  of  the  preceding  problem,  we  see  that  the  privilege  of  2%  dis. 
count  expired  Aug.  15,  '04.     Then,  by  the  terms  of  the  contract,  B.  was  entitled  to  120  days  credit 
which  gave  Oct.  4,  '04,  as  the  equated  date  of  the  account,  without  considering  the  element  of 
2"o  discount.     Hence,  if  he  settles  the  account  Aug.  28,  '04,  he  is  entitled  to  the  interest  thereon 
from  Aug.  28,  to  Oct.  4,-37  ds.  ®  6%  =  $16.03  from  $2600  =  $2583.97  balance. 

Problem  4. 

The  firm  of  E.  P.   Singeltary,  sold  to  A.  Barbier  &  Co.,  the  following 
invoices  of  merchandise : 

Tlie  terms  for  each  sale  are  as  follows :  60  days 
credit  and  2%  discount  in  10  days,  or  60  days  additional 
credit  in  case  the  2%  discount  privilege  is  not  accepted 
by  the  purchaser. 

A  settlement  of  this  account  was  made  Sept.  1, 1904. 
(1).  What  is  the  date  on  which  the  2%  discount  privi- 
lege expires  ?  (2).  What  is  the  average  date  of  the 
account?  (3).  What  is  the  amount  due  to  E.  P.  Sin- 
geltary on  settlement,  allowing  (ifo  interest  ? 

Ans.  1,  May  22/04.    Ans.  2,  July  11/04. 
Aus.  3,  $1522.43. 


1904. 

Eeb.  in. 

$372.30 

"     23, 

53.41 

Mar.    3, 

181.39 

"     10, 

305.79 

"     25, 

240.25 

"    29, 

109.12 

Apl.    7, 

57.09 

>'    20, 

130.00 

$1509.35 


EQUATION   OF   PAYMENTS  WITH   SPECIAL   FEATURES. 


lOII 


Problem  5. 

Smith  sold  to  Jones  on  GO  days  credit,  with  the  privilejre  of  3%  discount  in 
10  days,  the  folh)winy  invoices  of  goods:  May  10,  $500 ;  May  20,  $700;  June  1, 
$400;  Julys,  *1000. 

Jones  has  paid  on  account,  as  follows  :  June  9,  $800  ;  Aug.  25,  $1)00.  A  final 
settlement  was  made  Sejjt.  15,  '01.  AUowinj;-  interest  at  G'/p,  and  oGO  ds.  as  au 
interest  year,  what  does  Jones  owe  Smith  on  final  settlement'/  Aus.  $877.57. 

FIRST    SOLUTION, 
OPERATION 
To  fiud  the  Equated  Date  of  Sales  and  the  time 
limit  of  the  discount  privilege. 
1004. 

$  500  X   0=  0 

700x10=     7000 

400X22=      8800 

1000x54=  54000 


May  10, 
>'    20, 

June   1. 
July    3. 


$2600  )  09800  (  27  ds.  af- 

ter May  10. 
27  ds.  after  May  10,  '04. 
60  ds.  Cr.  on  each  sale, 
10  ds,  Dis,  iJiivilege. 

97  ds.  after  May  10,  =  Aug.  15,  '04,  = 
limit  of  dis.  i)rivilege. 


Or  thus  :  2d  OPERATION. 


1904. 

May  10. 

$  500 X  70  = 

35000 

''  20. 

700 X  80  = 

5G000 

.Tune  1. 

400 X  92  = 

3G800 

July  3. 

1000x124  = 

124000 

$2000 


)  251800  (  97  ds. 


after  May  10,  '04,  =  August  15,  '04,  as 
tlie  Equated  Date  of  the  sales,  allowing 
60  days  credit  and  10  days  discount 
l)rivilege. 


OPERATION  TO  EQUATE  THE  ACCOUNT,  BOTH  DEBIT  AND  CREDIT  ITEMS. 


1904. 

May  10. 

$  .500  X  60=  30000 

"  20. 

700  X  70=  49000 

June  1. 

400  X  82=  32800 

July  3. 

1000x114=114000 

$2000      225800 

1700      120300 

1904. 
June,  9.  $800  X  30  = 
Aug.  25.   900x107  = 

24000 
96300 

$1700 

120300 

$  900  )  105500 ■(  117i  ds.  after  May  10,  '04,  =  Sept.   4,  '04,  as  the 

Equated  date  or  time  for  the  payment  of  the  balance. 

By  inspection,  we  see  that  the  $800  paid  June  9,  was  paid  before  the  expi- 
ration of  the  discount  privilege,  August  15,  and  hence  the  debtor  is  entitled  to 
discount  thereon,  which,  at  3%,  is  $24.  This  dedut^ted  from  the  $900  balance  of 
account,  leaves  a  net  balance  of  $876  due  Sept.  4f ,  '04. 

As  settlement  was  not  made  until  Sept.  15,  '04,  the  creditor  is  therefore 
entitled  to  interest  on  the  net  balance,  from  Sept.  4|,  '04,  to  Sept.  15,  '04,  lOJ  days 
at  6%  =  $1.57  interest.     $876  +  $1.57  =  $877.57,  amount  due  Sept.  15,  '04. 

Note  1. — In  practice,  it  is  not  the  custom  to  use  the  exact  fraction  of  a  day,  but  to  state 
the  day  to  the  nearest  unit.  In  this  prol)leni  we  have  used  the  fraction  of  a  day  in  order  to  obtain 
the  exact  interest,  so  as  to  compare  the  same  with  the  interest  produced  by  the  solution  of  the 
problem,  by  the  Account  Current  and  Interest  Account  Method,  which  is  given  ou  the  following 
page. 

Note  2. — Since  the  purchaser  is  allowed  3%  di.^count  on  all  payments  paid  within  the  dis- 
count time  privilege,  his  disccnint  credit  on  the  $300  payment  w(}uld  be  in  strict  accuracy  and 
justice,  $24.74.  i.  e.  in  the  ratio  $100  credit  for  every  $97  paid.  But  this  degree  of  accuracy  is  not 
observed  in  calculations  of  this  character, 

f  Continued.). 


IOI2 


SOULES    PHILOSOPHIC    PRACTICAL    MATHEMATICS. 


Note  3. — Prolilems  coutainiiii;  conditions  of  credit  :ind  of  discount,  like  tlie  five  precedin" 
problems.  :iro  of  quite  recent  origin,  .and  have  given  rise  to  cousideralile  discussion  among  account- 
.-luts.  These  discussions  result  from  the  ditierent  interjiretations  given  to  the  conditions,  and 
these  interpretations  are  often  reached  because  a  solution  in  accordance  therewith  will  give  a 
iinancial  result  in  favor  of  the  party  holding  such  an  interpretation.  As  au  example  of  this  point 
note  the  following  operation  of  the  above  i>roblem. 

VARIATION    OF    PEOBLEM. 

Equation  op  the  Account  allowing  70  days  cuedit  on  each   itbTM  of  puuchase. 


1004. 

May  10. 

$   500  X  70=  35000 

"  l-'O. 

700 X  80=  50000 

June  1. 

400  X  92=  30S00 

July  3. 

1000  X  124= 11.'4000 

$2000      251800 

1700      120300 

1004. 
June    9. 
Aug.  25. 


$800  X 
000  X 


30  =  24000 
107  =  00300 


$1700 


1L'0300 


Out. 


'04,   as   the 


$  000  )  131500  (  140  ds.   after   May  10,  '04, 

Equated  Date  lor  the  payment  of  the  bahmce. 

This  equation  is  made  on  the  interpretation  that  the  debtor  has  really  70 
days  credit  on  each  item  of  purchase,  because  he  had  10  days  discount  privilege. 
This,  we  think,  is  au  erroneous  constructiou  of  the  conditions  of  the  contract. 
The  10  days  discount  privilege  were  conditioned  on  the  payment  of  cash  as  the 
respective  sales  came  due.  But  as  the  cash  ijaymeiits  were  uot  made,  the  10  days 
privilege  cannot,  injustice,  be  claimed. 

A  dittereuce  of  opiniou  exists  among  accountants  and  also  among  business 

men,  regarding  the  proper  manner  of  adjusting  accounts  like,  or  similar,  to  the 

above.     In  some  cases,  special  agreements  between  tlie  i)arties,  and  iu  some  cases 

the  custom  of  the  trade  will  govern  the  method  of  adjustment  or  settlement. 

Second  Solution  of  Problem  5. 

By  the  Partial  Payment  Method,  ou  uy  making  Pkuioiiic  Partial  Settlements 

COUUlSSPONDIXa  WITH   THE   DaTES   OF   PaYMENT.S. 


1904. 

May  10.  $.500x60=  30000 

''     20.  700x70=  40000 

Juue  1.  '400x82=  32800 


$1000 
800 


lllSOO 
24000 


1904. 
Juue  9. 


$800x30=24000 


$  800 
3%  dis't,         24 


)  87S00  (  109^=110  ds.  after  May  10  =  Aug.  28,  '04. 


Balance,  $776 
1904. 

Juue  28  $776x80=02080 

July     3.  1000x84  =  84000 


)1770 
900 


140080 
09300 


$  876 


1904. 
Aug.  25. 


$000x77=09300 


)  70780  (  87  ds.  after  Jn,)e  0,  '04,  =  Sept.  4,  '04. 

Note. — The  .$776  b.alance  from  the  first  equation  is  not  due  until  August  28,  '04.  Hence,  as 
the  Focal  date  is  .June  9,  there  is  80  days  credit  on  this  balance. 

By  this  method  of  solution,  we  obtain  the  same  equated  date  as  in  the  first  solution,  Sept.  4, 
1904.  The  remaining  part  of  the  operation,  to  find  the  balance  due  to  the  creditor  ou  the  day  of 
settlement,  is  performed  iu  the  same  niiinner  as  in  the  iirst  solution. 


*  ACCOUNT  CURRENT  AND  INT.  ACCOUNT  WITH  SPECIAL  CONDITIONS  IOI3 


THIRD  SOLUTION  BY  THE  ACCOUNT  CURRENT  AND  INTEREST  ACC'T  METHOD. 

Jones  in  Account  Current  and  Interest  Account  at  G%  to  Skpt.  15,  '04,  with  Smith.     60  ds.  credit 

is  allowed  on  all  sales,  and  o"o"'l'*"<"i"t  is  allowed  on  all  payments  made  within  10 

days  after  the  exuiration  of  the  60  days  credit. 

Br.  '  '  Cr. 


J904 
Kay 


June 
July 


lOiMdse.   60  ds.  & 
o^'  dis.  in  10 ds. 

20lildse.     "         " 

3      " 

iDr.  Int.  on  $1. 

To  Bal.  of  ds  on 
$1,  which  di 
vided  by  the 
6"|J  int.  divisor 
6000,  gives 
int.  $1,617 

Less  int.  on 
S24  for  105 
ds.  =  .043 

Int.  due  the 
creditor  $1,574 


When 
due, 


July 


Sept 


Ds 


Prodc't 


34000 

40600 
18400 
14000 


107000 
97300 


9700 


50000 

70000 

400,00 

1000  00 


1 

2601 


1904 
June 

Aug. 


Sept 
Sept. 


Cash, 


Ds.  Int.  on  $1. 


By3?^dis.onS800 
By  balance  due 
the  creditor, 


When 
due. 


June 
Aug. 


Ds 


78400 
18900 


97300 


Amt. 


800,00 
90000 


2400 

S7757 


2601 


57 


Explanation — By  this  method  of  -working  the  problem,  interest  is  computed  on  all  debit  and 
credit  items  from  the  time  due  to  date  of  settlement;  hence  interest  has  been  computed  on  the 
S900  balance  of  account,  from  the  EQUATED  DATE  to  the  date  of  payment,  which  is  lOJ  days 
shown  in  the  solution  above  by  the  equation  method.  This  $900  balance  of  account  is  entitled  to 
a,  credit  of  $24,  the  same  being  3,"o  discount  on  $800  paid  within  the  discount  period,  thus  leaving 
a  net  debit  balance  of  $876,  which  bears  interest  from  the  equated  date  to  the  date  of  settlement. 
Now,  as  interest  has  been  computed  on  $900.  instead  of  $876  from  the  equated  date  to  the  date  of 
settlement,  it  is  evident,  therefore,  that  the  interest  produced  by  this  method  of  solution  is  in 
excess  to  a  sum  equal  to  the  interest  on  the  amount  of  Discount  Credit,  (in  this  case  $24)  from 
the  equated  date  to  the  date  of  settlement.  This  interest,  in  this  problem,  amounts  to  $.043,  and 
is  deducted  from  the  full  amount  of  interest,  $1,617,  as  is  shown  in  the  operation. 

Notk  1.— To  find  the  limit  of  time  of  the  discount  privilege,  equate  the  sales  as  .shown  in 
the  jirecedlng  solution. 

Note  2. — For  a  full  elucidation  of  Equation  of  payments,  see  pages  627  to  653.  For  a  full 
elucidation  of  Accounts  Current  and  Interest  Accounts,  see  pages  653  to  670. 


(t)      ^^     G) 


1014 


SOULE  S   PHILOSOPHIC   PRACTICAL   MATHEMATICS. 


1.  The  following  diagram  represents  an  oil  tank,  which  is  25  feet  long  and  7  feet  in  diameter. 
The  line  AB  is  1  inch  from  the  inner  circumference  of  the  tank  and  indicates  the  surface  of  oil  in. 
the  tank,  flow  many  gallons  of  oil  are  there  in  this  segment  which  is  1  inch  deep?  How  many 
gallons  would  there  be  in  the  tank  if  the  oil  were  6  inches  deep,  as  indicated  by  the  line  CD  ? 

Ist  Answer— 15.R.')3  +  gallons. 
2d  Answer— 228.17  +  gallons. 


:is 


.A. 


3- 


X 


Operation  for  the  1st  Problem. 

(1)  Deptli  of  oil  1  inch  -=-  84  inches   diameter   =    .01190-1   =    quotient  of  versed  sine. 

page  353). 

(2)  .011  =  .00153  =  segment  area. 
.012  =  .00175  =  segment  area. 


(See 


Diflerence    .00022  X  .904  =  .00019SSS. 

Then  .00153  +  .00019888  =  .00172888  (practically)  .00173  =  average  segment  area  by 
which  the  square  of  the  diameter  is  to  be  multiplied. 

(3)  84=  =  7056  X  00173  =  12.206SS  sq.  in.  in  the  area  of  the  segment. 

(4)  12.20688  X  300  inches  length  of  tank  =  3662.064  cubic  inches  in   segment  which    H-    231 

cu.  in.  in  a  gallon  =  15,853  +   gallons  in  the  segment  1  inch  deep. 


(1) 
P) 


(3) 


(4) 


Operation  for  the  2d  Problem. 

6  inches  -f-  84  =  .071428  =  quotient  versed  sine. 

.071428  or  .071  =  .02468  =  seg.  area, 
seg.  area. 


.072  =  .02519 
Difference    .00051  X  .428  =  .00021828  (practically  .00022). 


.02468 
.00022 


175.6944   =    square   inches   in   area  of 


.02490  average  segment  area,    which    X   84' 
segment. 

Then  (175.6944  X  300), -^  231  =  228.17  +  gallons,  in  the  segment  6  inches  deep. 


2.     A  room  is  20  feet  long  by  15  feet  wide  and  10  feet  high.     How   long  is   the   diagonal   of  the 
room;  that  is,  from  one  corner  on  the  floor  to  the  opposite  diagonal  corner  on  the  ceiling? 

Solution. 

(1)  The  diagonal  of  a  room  is  the  hypotenuse  of  a  right  angle  triangle  whose  base  is  the 
diagonal  of  the  floor  and  whose  altitude  is  the  height  of  the  wall;  hence,  two  right  angle  triangles 
are  involved.  The  diagonal  of  the  floor  must  be  found  first,  and  with  this  diagonal  as  the  base  of  a 
right  angle  triangle  with  the  height  of  the  room  as  its  altitude  the  diagonal  of  the  room  is  tlien 
readily  determined.     The  operation  is  as  follows: 


(2)  The  diagonal  of  the  floor  =  V  20=  +  15=  =  25  feet. 

(3)  The  diagonal  of  the  room  is  the  hypotenuse  of  a  right  angle  triangle  whose  base  is  the 
diagonal,  25  feet,  as  above,  and  whose  height  is  10  feet. 

(4)  -SI252  +  10'  =  V725  =  26.92  +  feet,  length  of  the  diagonal  required. 

Or  thus :     20=  +  15=  +  10^  =  725;  V  725  =  26.92  +  feet,  length  of  the  diagonal  required. 


MISCELLANEOUS    PROBLEMS. 


1U15 


3. 


/O' 


EEh- 


A  board  in  the  form  of  a  Trapezoid,  as  shown 
in  the  diagram,  is  10  feet  long  through  the 
middle,  12  inches  wide  at  the  wider  end  and 
6  inches  wide  at  the  narrower  end.  How  far 
from  the  narrower  end  must  it  be  sawed  to 
divide  it   into  two  equal  parts? 

Solution 

As  shown  in  the  diagram  ABCD  is  a  board  in  the  form  of  a  trapezoid,  whose 
middle  length  is  10  feet,  wider  end  12  inches  and  narrower  end  6  inches.  To  find 
the  distance  from  the  narrower  end  that  the  board  must  be  sawed,  to  divide  it  into 
two  equal  parts,  proceed  as  follows: 

( 1 1  Continue  the  lines  AB  and  CD  to  the  point  E  and  thus  form  a  triangle.  As 
the  board  tapers  from  12  inches  to  6  inches  in  a  length  of  10  feet,  which  is  /„ 
or  %  of  an  inch  taper  per  foot,  hence  to  taper  or  converge  6  inches  more  to  the 
point  E,  and  form  the  triangle,  will  require  as  many  feet  increase  of  length  as  6 
inches  is  equal  to  |  of  an  inch,  which  is  1(1  feet;  thus  making  the  triangle  EEC 
20  feet  in  altitude. 

(2)  Find  area  of  triangle  EEC.  Thus:  (20  ft.  EH  X  1  ft.  BC)  -=-2=10  sq.  ft. 
in  triangle  EBC. 

(.5)  Find  area  of  trapezoid  ABCD.  Thus:  1  ft.  BC  -1-  J  ft.  AD  =  \\  ft.  ^  2 
=  }  ft.  average  width  of  trapezoid  or  board.  J  ft.  X  10  ft.  length  of  board  IHv  = 
7.5  sq.  ft.  area  of  board  ABCD. 

(4)  Then  10  sq.  ft.  in  triangle  EBC-7.5  sq.  ft.  area  of  board  ABCD  =  2..=>  sq. 
ft.  in  area  of  triangle  li.AD. 

{h't  Find  area  of  \  of  board  .\BCD,  indicated  by  the  division  line  RS  drawn 
parallel  to  BC.     Thus:     7.5  sq.  ft.   -^  2  =  .3.75  sq.  ft.  area  of  board  .\SRD. 

(6)  Find  area  of  tri.angle  ESR.  Thus:  3.75  sq.  ft.  in  board  ASRD  -f  2.5  sq. 
ft.  in  triangle  E.A.D  =  6.25  sq.  ft.  =  area  of  triangle  ESR. 

(7)  Now  having  two  similar  triangles,  EBC  and  ESR,  and  since  similar  sur- 
faces are  to  each  other  as  the  squares  of  their  like  lines,  we  have  the  following 
proportion  to  find  EO^     Thus:     EBC  :  ESR  :  :  EIP  :  EO^      Substituting  values 


we  have  10  :  6.25 


202 


(S)  Then  ^''  250  =  15.81  -f  ft.  altitude  of  triangle  ESR. 

(9)  Then  15.81  -  10  ft.  EK  altitude  of  triangle  EAD  =  5.81  -f  ft.  from  the 
narrow  end  of  the  trapezoid  or  board  for  the  line  RS,  or  where  the  board  must  be 
sawed  to  divide  it  into  two  equal  parts. 


A,  A  poultryman,  with  his  little  push-cart,  containing  a  basket  of  eggs,  was  on  his  way  to  market 
when  an  automobile,  going  at  full  speed,  upset  the  cart  and  destroyed  all  the  eggs.  The  chauffeur 
admitted  his  responsibility  and  offered  to  pay  for  the  eggs  if  the  poultryman  would  tell  him  how 
many  he  had.  The  poultryman  did  not  know  how  many  eggs  were  in  the  basket,  but  said  there  were 
between  50  and  100,  and  when  he  counted  them  by  2's  and  3's  at  a  time,  none  remained.  But  when 
he  counted  them  by  5's  at  a  time,  there  were  three  remaining.     How  many  eggs  were  in  the  basket? 

Solution. 

(1)  Find  what  numbers  between  50  and  100  are  divisiljle  by  two  and  three  without  a  remainder. 
By  inspection  these  numbers  are  as  follows:     54-60-66-72-78-84-90-  and  96. 

(2)  Then  by  the  terms  of  the  question,  one  of  these  numbers  when  divided  by  Swill  give  a 
remainder  of  3;  by  trial,  78  is  found  to  be  that  number.  Therefore,  78  is  the  number  of  eggs  that 
were  in  the  basket  and  were  destroyed. 


5.  Twenty  men  charter  a  car  to  go  on  an  excursion.  Before  leaving  they  take  in  10  additional 
men,  and  thus  reduce  the  expense  Sli  for  each  man  of  the  original  party.  What  was  the  cost  of 
the  car? 

Solution. 

Since  by  taking  in  10  additional  men  the  expense  is  reduced  SlJ  for  each  of  the  20  men, 
hence  the  reduction  for  the  20  men  is  20  times  Sli,  which  is  S26s.  Then,  since  10  additional  men 
pay  S26i,  1  man  will  pay  the  10th  part,  which  is  S2s,  and  as  there  are  30  men  in  all,  and  as  each  man 
pays  the  same,  the  whole  cost  of  the  car  will  be  30  times  S2j,  which  is  S80. 


1016  soule's  philosophic  practical  mathematics. 

6.  Three  nieclianics,  Jones,  Smith  and  Brown,  have  2  bags  of  goods  to  carry  from  the  factory  to 
the  store,  which  is  three  miles  distant.  Each  is  to  carry  a  bag  the  same  distance.  Plow  should  they 
arrange  it? 

Solution. 

Let  Jones  and  Smith  each  take  a  bag;  let  Jones  carry  his  1  mile  and  turn  it  over  to  Brown, 
who  will  carry  it  2  miles;  let  Smith  carry  his  bag  2  miles,  and  turn  it  over  to  Jones,  who  will  carry  it 
1  mile.     Thus  each  carries  1  bag  2  miles. 


7.     The  sum  of  the  first  and  the  second  of  three  numbers  is  55;  of  the  first  and  third,  62;  of  the 
second  and  the  third,  83.     What  are  the  numbers? 

Solution. 
By  the  terms  of  the  problem,  there  are  the  following  conditions: 
(1)     'I'hefii-st  numljer  plus  the  second  nuniber=  55. 
(  2)     'I'he  first  number  plus  the  third  number  =  62. 
Then,  by  subtracting  condition  1  from  condition  2,  the  result  is: 

(3)  The  third  numljer  (62)  minus  the  SECOXU  number  (55)  =  7. 

(4)  The  SECOND  number  plus  the  THIRD  number  =  83. 

Then,  by  aading  conditions  3  and  4,  it  is  seen  that  two  times  the  third  number  =  90. 

Hence,  KO  -^  2  gives  the  THIRD  number,  which  is  45. 

Snbstituting  in  condition  4,  we  get  the  SECOXD  number  plus  45  =  83. 

Hence,  83  —  45  =  the  second  number,  which  is  38. 

Substituting  in  condition  1,  we  get  the  I'IRST  number  +  38  =  .55. 

Hence,  55  —  38  =  the  first  number  =  17. 

Therefore,  the  numbers  are  17,  first  number;  38,  second  number;  and  45,  thinl  number. 


8.  Three  boys  picked  a  basket  full  of  berries  containing  18  quarts.  They  desire  to  divide  the  berries 
equally  among  them.  They  have  a  10,  an  8  and  a  4  quart  basket.  How  can  they  divide  the  berries 
in  equal  shares? 

Solution. 

Since  there  are  18   quarts  of  berries  to  be   divided  equally  among  3   boys,  therefore  each  boy 
will  receive  6  quarts.     This  division  is  made  as  follows; 

(1)  Eill  the  8  and  the  4  quart  baskets.      This  leaves  6  quarts  in  the  18  quart  basket,  and  the 
10  quart  basket  is  empty. 

(2)  Lmpty  berries  now  in  the  8  quart  basket  into  the  10  quart  basket;  also  pour  2  quarts  from 
the  4  quart  basket  into  the  10  quart  basket. 

This  leaves    6  quarts  in  the  18  quart  basket. 

10  quarts  in  the  10  quart  basket. 

2  quarts  in  the    4  quart  basket. 

0  quarts  in  the    8  quart  basket. 

(3)  Empty  the  2  quarts  now  in  the  4  quart  basket  into  the  8  quart  basket,  then  refill  the  4 
quart  basket  from  the  10  quart  basket,  and  empty  the  same  into  the  8  quart  basket. 

The  division  is  now  complete,  and  there  are  6  quarts  in  the  18  quart  basket,  6  quarts  in  the  10 
quart  basket  and  6  quarts  in  the  8  quart  basket,  and  nothing  in  the  4  quart  basket. 

9.  An  indulgent  parent  paid  his  boy  15  cents  a  day  for  prompt  attendance  at  school,  and  charged 
him  12  cents  for  each  day  he  was  tardy.  At  the  close  of  200  school  days  there  was  due  the  boy 
§29.46.     How  many  days  was  he  punctual? 

Solution. 
If  the  boy  had  attended  school  promptly  200  days  at  15c.  per  da}^  there  would  have  been  due 
him  $30.00.  But  because  of  tardiness  there  was  due  him  only  $29.46.  Then,  S30.00  — $29.46=  54 
cents  as  penalty  for  tardiness.  Then,  as  the  student  was  to  receive  15  cents  for  each  day  he  was 
punctual,  and  was  to  be  charged  12  cents  for  each  day  he  was  tardy,  he  therefore  lost  15c  +  12c  =  27c 
for  each  day  he  was  tardy.  Then,  as  his  total  loss  was  54  cents,  he  must  have  been  tardy  as  many 
days  as  .54  cents  is  equal  to  27  cents,  which  is  2  days.  Then,  200  days— 2  days  =  198  days  of  punctual 
attendance. 


10.    A  tailor  offers  you  a  $40  suit  for  $35;  or  a  $35  suit  for  $30.     Which  is  the  better  offer,  and  how 
nmch  per  cent  better? 

Solution. 

$40  -  $3  =  $35  cost  and  $5  gain.  Then,  $.35  ;  5  :  :  100  :  14.28+%  gain.  $35  -  $5  =  $30  cost 
and  $5  gain.  Then  30  :  5  :  :  100  :  $16.67%  gain.  Then  16.67%  -  14.28%  =  2.39%,  that  the  second 
offer  is  better  than  the  first. 


MISCELLANEOUS   PROBLEMS.  1017 

11.   'A  party  of  8  hired  a  coach.     If  there  had  been  4  more  the  expense  would  have  been  reihiced 
$1  for  each  person.     How  much  was  paid  for  the  coach? 

Solution. 

When  a  party  of  8  pays  for  a  coach,  each  pays  |  of  the  cost;  wlien  a  party  of  12  pays  for  the 
coach,  each  pays  j'j  of  the  cost;  then  i  -  j'j  =  j'^  .  which  is  j'i  of  the  cost  Hence  5'/=  $1,  and 
24  —  24  =  S24  cost  of  tlie  coach. 


1  2.  A  teacher  agreed  to  teach  9  months  for  $.562i  and  his  board.  At  the  end  of  the  term,  on  account 
of  two  montlis'  absence  caused  by  sickness,  he  received  only  $WJi.  What  was  his  board  worth  per 
month? 

Solution. 

(1)  Find  the  teacher's  salary  per  month  without  board,  thus;    S,562.,50  — S409.50  =  S15,),  salary 
for  two  months  without  board;  S1534-2  =  S76.50,  for  one  month  without  board. 

(2)  $76.50X9  =  S688..50,  salary  for  9  months  without  board. 

(3)  S6SS. 50- §562.50  =  §126,  cost  of  board  for  9  months;  S126+  9  =  $14,  cost  of  board  for  one 
month. 


13.    A  carpenter  engaged  to  work  for  12  months  for  S600  and  a  horse.     lie  worked  9  months  and 
received  §432  and  the  horse.     What  was  the  \-alue  of  the  horse? 

Solution. 

By  the  conditions  of  the  problem,  if  the  carpenter  had  worked  12  months  he  would  have 
received  S600  and  a  horse.  As  he  worked  9  months  and  received  §432  and  the  horse,  hence  the  dif- 
ference between  $600  and  S432  =  $168,  which  is  the  value  of  the  services  for  the  three  months  he  did 
not  work.  Then  $168 -H  3  =  $56,  which  is  the  value  of  the  services  for  each  month,  under  the  first 
condition;  hence  the  value  of  the  wages  for  12  months  would  hi  $56X12,  $672.  But  as  he  was  to 
receive  only  $600  and  ahorse,  it  is  clear  that  the  value  of  the  horse  is  §672— $600  =  §72. 


14.    A  teacher  was  engaged  to  teach  a  country  school  on  the  following  conditions: 

If  he  had  .50  pupils  his  salary  was  to  be   §60  per  month;  if  he  had  30  pupils  his  salary  was  to 
be  $50  per  month.     The  actual  number  he  had  was  45.     What  should  his  salary  be? 

Solution. 

First  find  the  respective  price  per  pupil  when  50  and  when  30  are  taught. 
Thus:     50  pupils  for  a  monthly  salary  of  §60  =  $1.20    for  each  pupil. 
30  pupils  for  a  monthly  salary  of  $50  =  $1.66f  for  each  pupil. 

20  =  difference  in  number  of  pupils,  S  .46j  =  increase  on  each  pupil  when  the  num- 
ber was  30. 
Then  it  is  evident  that  if  there  had  been  20  pupils  less  than  50,  the  teacher  would  be  entitled 
to  46i  cents  more  per  pupil  than  if  he  had  had  50  pupils.  Then,  since  20  less  pupils  give  46|  cents 
more,  1  pupil  will  give  the  20th  part,  which  is  2J  cents,  and  5  pupils  less,  which  is  (50  —  45)  the  real 
number  taught  less  than  .50,  will  give  5  times  2i  cents,  which  is  II3  cents,  increase  per  pupil  over 
the  §1.20  price  when  there  were  50  pupils. 

Then,   §1.204- §0.113  =  §1.31j;    §1.313X45,  the  number  of  pupils  taught,  =  $59.25    monthly 
salary. 


15.    A  pyramid  is  10  feet  square  at  the  base  and  20  feet  in  altitude,  or  height.     IIow  far  from  the 
base  must  it  be  divided  parallel  to  the  base  to  make  two  equal  parts? 

Solution. 

(1)  The  solidity  of  a  square  pyramid  is  i  of  a  rectangular  solid  of  equal  height  and  area  of 
base.    Hence,  to  find  the  solidity  or  the  cubic  feet  of  this  pyramid  proceed  thus: 

(10^X20)  H-3  =  666j,  cubic  feet  in  pyramid. 

(2)  666} -H 2  =  333i  cubic  feet  in  each  i  part  of  pyramid  after  it  has  been  divided.  Now,  as 
the  cube  of  the  whole  height  is  to  the  whole  solidity,  666j,  so  is  the  cube  of  i  the  solidity  to  333i, 
and  then  the  cube  root  of  this  number  will  be  the  height  of  the  upper  half  of  the  pyramid.  Hence, 
steps  3,  4  and  5  as  follows: 

(3)  20'  :  666j  :  :  x^  :  333J  =  4000,  cube  of  the  height  of  the  upper  half  of  the  pyramid. 

(4)  ^  4000  =  15.S8-ft  height  of  upper  half  of  pyramid. 

(5)  20  ft,  altitude  — 15..S8  — ft.  =  4.12  ft,  height  of  lower  half  and  the  point  where  it  must  be 
divided  to  make  two  equal  parts. 


1018  soule's  philosophic  practical  mathematics. 

16.    A  building  130  feet  loiiff  lias  a  roof,  each  side  of  which  from  the  eaves  to  the  apex  measures  SO 
feet,  anil  the  apex  forms  a  rifjht  angled  triangle. 

During  a  storm  water  fell  to  the  depth  of  2i  inches.     How  many  gallons  fell  on  the  roof? 

Solution. 

(1)  Pinrl  the  h3'potenuse  of  the  right  angled  triangle,  which  will  be  the  width  of  the  roof 
that  receives  the  rainla'll.  Thus:  (50- +•''0')  =  500U;  \  5000  =  70.71+  ft.  =  the  hypotenuse,  and  the 
width  of  the  building  that  receives  the  rainfall. 

(2)  Find  the  square  inches  in  the  rain  receiving  roof.  Thus:  130  feet  longX70.71  feet  wideX 
144  square  inches  in  a  square  foot  =1,323,691.2  square  inches.  Then,  1,323,691.2  square  inchesX2i 
inches  depth  of  rainfall  =  3,309,228  cubic  inches,  which  -^  231,  the  cubic  inches  in  a  gallon  = 
14, 325. 66 +gallons. 


17.  A  mule  and  a  donkey  were  traveling  together  laden  with  bags  of  goods.  The  donkey  com- 
plained to  the  nuile  that  his  load  was  too  heavy.  "Lazy  animal,  "  said  the  uuile;  "you  have  no  cause 
to  complain,  for  if  I  take  one  of  j-our  bags,  I  shall  have  twice  as  many  as  you,  but  if  I  give  you  one 
of  mine,  we  shall  have  only  an  equal  number."     IIow  many  bags  had  each? 

Solution. 

Since  the  mule  and  the  donkey  will  both  have  equal  loads  when  the  mule  gives  one  of  his  bags 
to  the  donkey,  it  is  evident  that  the  difference  between  the  bags  which  they  respectively  carry  is 
equal  to  2.  Hence,  if  the  mule  receives  one  bag  from  the  donkey,  the  difference  will  be  4;  and  in 
that  case  the  mule  will  have  twice  as  many  bags  as  the  donkey  has;  therefore,  the  mule  will  have  8 
and  the  donkey  4.  Then  if  the  mule  gives  one  to  the  donkey,  the  mule  will  have  7  and  the  donkey 
5;  hence,  these  were  the  number  of  bags  with  which  they  were  respectively  loaded. 


18.  A  party  of  2  men  and  3  boys  went  gunning  for  game  for  a  Thanksgiving  dinner.  They  shot 
and  bagged  2  turkeys  and  S  ducks  in  5J  hours.  In  how  many  hours  would  3  men  and  4  boys  bag  5 
turkeys  and  5  ducks,  allowing  that  each  man  is  equal  to  2  boj'S,  and  that  it  is  as  difficult  to  kill  1 
turkey  as  it  is  to  kill  4  ducks,  and  assuming  that  the  hunting  conditions  were  equal  for  both  parties? 

Solution. 

First  Step.  Find  the  equivalent  of  the  2  men  and  3  boys  in  men  or  boys;  and  the  equiva- 
lent of  the  turkeys  and  ducks  in  turkeys  or  ducks.  In  this  solution  the  equivalents  are  made  in  men 
and  in  turkeys.  Thus,  according  to  the  conditions  of  the  problem,  the  3  boys  in  the  first  party  are 
equivalent  to  1 1  men,  thus  making  3i  men.  In  the  second  party  the  4  boys  are  equivalent  to  2  men, 
thus  making  5  men. 

Second  Step.  The  8  ducks  of  the  first  party  of  hunters  are  equivalent  to  2  turkeys,  plus  the  2 
turkeys  shot  =  4  turkeys.  The  5  ducks  of  the  second  party  are  equivalent  to  li  turkeys,  plus  the  5 
turkeys  shot  =  6i  turkeys. 

T/iird  Step.  Classify  thus:  First  party,  3i  men,  kill  4  turkeys  in  5J  hours.  Second  party,  5 
men,  kill  6i  turkeys  in  ?  hours. 

Fourth  .Step.     Solution  statement  and  reason: 

16  hours.  REASONING  for  the  solution  statement. 

7  First  write  the  SJ  hours  ('3' J  on  the  line. 

Then,  since  it  required  |  men  'g"  hours  to  kill  4  turkeys,  it  will  require  J  a 

man  7  times  as  long,  and  |  or  1  man  A  part  of  the  time;  and  since  it  requires 

25  1  man  this  long,  it  will  require  5  men  J  of  the  time.    Then,  since  4  turkeys 

55  hours.  ^'"'^  killed  in  51  hours,  to  kill  1  turkey  it  will  require  i  as  long,  and  to  kill 

*  i  of  a  turkey  will  require  i  of  the  number  of  hours,  and  to  kill  -',"  turkeys 

it  will  require  25  times  as  many  hours,  which  is  5%  hours,  or  5  hours  50 

minutes. 


19.    A  public  square  is  surrounded  by  a  roadway  2  rods  wide.     The  area  of  the  roadway  around  the 
square  is  one  acre.     What  is  the  area  of  the  square? 

Solution. 

Since  the  roadway  contains  one  acre  or  160  sq.  rds.  and  is  2  rds.  wide,  and  since  it  has  4  right 
angles  around  the  square,  hence  there  are  2X2X4  =  16  sq.  rds.  in  the  4  corners  or  angles.  Then, 
160  —  16  =  144  sq.  rds.,  area  of  the  road  on  the  4  sides  of  the  square.  l44-i-4  =  36  sq.  rds.  of  the  road 
on  each  side  of  the  square.  36^2  rds.  wide  =  IS  rds.,  length  of  each  side  of  the  square.  18X18  = 
324  sq.  rds.  in  the  square;  324  -^  160  =  2^'q  acres  in  the  square. 


MISCELLANEOUS   PROBLEMS. 


1019 


/#' 


A  Mathematicai^  Fi,y  Problem. 

20  A  room  is  16  ft.  long-,  14  ft.  wide,  and  it  is  12  ft.  from  floor  to  ceiling.  A 
fly,  on  the  ceiling-  at  the  angle  A,  is  desirous  of  walking  to  angle  B,  which  is 
diagonally  opposite,  on  the  floor  of  the  room.  What  is  the  shortest  distance 
the  flj-  must  walk  on  the  walls  of  the  room  to  accomplish  its  object.'' 


Solution. 


J-fg.JZ. 


1 


/€  ' 

Figure  1  shows  the  room  in  perspective,  with  the  fly  at  A,  and  B  as  his  objective  point. 

The  fly  has  to  consider  first,  what  route  will  be  the  shortest  to  take.  Will  it  travel  from  A  to 
G,  and  then  from  G  diagonally  to  B;  or  from  A  to  I,  and  then  diagonally  from  I  to  B? 

Considering  the  mathematical  fact  that  a  straight  line  is  the  shortest  distance  between  two 
points,  the  fly  decided  to  follow  the  straight  line  route. 

It  reasoned  to  a  conclusion,  that  if  the  walls  of  the  room  be  projected  as  a  plane  as  shown  in 
figure  2,  it  could  more  easily  find  the  straight  line  to  travel  along  the  side  and  end  walls  of  the  room. 
(See  line  A,  C,  B  in  figures  1  &  2).  Now,  as  the  room  could  not  be  made  a  plane,  it  must  first  find 
at  what  height  from  the  floor  it  must  strike  the  line  I,  C,  E,  in  order  to  make  this  straight  line  route 
to  B.  By  referring  to  figure  2,  it  will  be  seen  that  the  two  triangles,  C,  B,  E  and  A,  B,  D,  have  in 
part  a  common  base  and  hypotenuse,  and  their  altitudes  are  therefore  proportional  to  their  bases: 
Hence,  30  :  12  :  :  14  :  5|.  Therefore  5^  feet  is  the  height  above  the  floor  that  the  fly  must  aim  for  on 
the  line  I,  C,  E.  Then,  on  reaching  this  point  a  change  of  direction  is  made  and  a  straight  line  is 
followed  to  B. 

Figure  2  also  shows  that  the  line  A,  C,  B,  is  the  hypotenuse  of  a  right  angled  triangle,  whose 
base  is  30  and  altitude  is  12  feet  Hence,  (12)'  +  (30)2  =  1044,  and  k  1044  =  32.31  +  feet,  which  is 
the  number  of  feet  the  fly  must  walk. 

f  A  JIathem.^tical  Fly  Problem. 

21  A  room  is  16  ft.  long,  14  ft.  wide  and  12  ft.  high.    A  fly  is  on  the  ceiling- 
at  the  corner   A,  and   is  desirous   of    walking  to  the  corner   B,  diagonally 
J>     opposite  on  the  floor  of  the  room.     Using  the  ceiling  and  cua//,  or  floor  and 
wall,  what  is  the  shortest  route  to  take? 

Solution. 

The  fly  first  travels  from  the  corner  A,  across  the 
ceiling  to  the  point  C,  which  is  7. 384  +  ft.  from  the  corner 
F,  and  then  down  the  side  wall  to  corner  B.  By  this  route 
the  fly  walks  only  30.528  +  ft. 

The  calculations  needed  to  find  the  point  C  are  the 
same  as  were  worked  out  in  the  above  problem,  and  hence 
are  not  repeated. 

By  Fig.  2  we  find  

(26)2  +  (i(,)i  =  9,32  and  ;'932  =  30.528  +  ft. 
NOTE: — The  corners  A  and  B  being  the  same  in  all  cases,  there  would 
be  no  difference  in  distance  between  the  ceiling  and  end  wall  route,  and  the 
ceiling  and  side  wall  route,  in  square  rooms.  In  rooms  longer  than  they  are 
wide  (as  in  drawing  of  room  above)  the  ceiling  and  side  wall  route  is  the 
shorter.  In  rooms  wider  than  they  are  long  the  ceiling  and  end  wall  route 
is  the  shorter. 


B     /y 


1U20 


SOULE's  rniLosoPHic  practical  mathematics. 


22.  The  diagram  A,  B,  C,  D  in  the  margin  represents  the  ground  space  of 
a  farmer's  stable,  40xo0  feet,  built  on  the  side  of  a  meadow.  The  farmer, 
wishing  his  horse  to  graze,  ties  him  with  a  line  to  the  upper  corner  of  the 
stable,  at  B.  The  line  B  K  is  .50  feet  long  from  the  mouth  of  tlie  horse  to  the 
corner  of  the  stable.  Thus  the  horse  can  graze  on  all  sides  of  the  stable  as 
far  as  the  length  of  the  line  will  allow.  Over  how  many  square  yards  cau 
the  horse  graze? 


Solution. 


(5) 
C  DI. 

(6) 
A  HE. 

(7) 
gives  698; 


From  the  diagram  it  is  seen: 

(1)  That  the  horse,  grazing  from  point  II  to  point  I,  grazes 
over  J  of  a  circle,  with  a  radius  of  5U  feet. 

(2)  When  he  reaches  the  point  I  and  moves  toward  D,  the  line 
strikes  the  corner  of  the  stable  at  C,  and  he  then  grazes  over  i  of  a 
circle,  with  a  radius  of  20  feet,  C  to  I. 

(■•,)  When  he  reaches  the  point  II  and  moves  toward  E,  the 
line  strikes  the  corner  of  the  stable  at  A,  and  he  then  grazes  over  i  of 
a  circle,  whose  radius  is  10  feet,  .V  to  II. 

From  this  data,  the  sum  of  the  areas  of  i  of  the  larger  circle, 
and  of  i  of  each  of  the  smaller  circles  is  to  be  found. 

(4)     Area  of  larger  circle   50  ft.  radius  X2  =  100  ft.    diameter.. 

IOO'X.7854,  ratio  between  the  area  of  a  square  and  a  circle  whose 

diameter  is  equal  to  one  side  of  the  square,  =7854  sq.  ft.  area  in  circle; 

J  of  which  is5S90..S. 

second  circle  20  ft.  radius  X2=40  ft.  diameter.    40=X.7854Xi=314.16  sq.  ft.  in 

Area  of  third  circle  10  ft.  radius  X  2  =20  ft.  diameter.      20=X.7854Xi  =  78.54  sq.  ft.  in 


Then  ,5890.5  +  314.16  -^  78.54=6283.2  sq.  ft.  which 
133  +  sq.  yds.  grazed  over  by  the  horse. 


9,  the  number  of  sq.  ft.  in  a  sq.  yd.. 


23.    A  grocer  has  150  gallons  of  a  mixture  of  wine  and  water;  72%  of  which  is  wine, 
■wine  must  be  added  to  make  the  mixture  85%  wine? 


How  much 


Solution. 

In  the  150  gallons  of  the  first  mixture  72%  is  w'ine,  which  =  108  gallons.  Then  1.50  —  108=  42 
gallons  of  water. 

According  to  the  conditions  of  the  problem,  enough  wine  must  be  added  to  the  first  mixture 
to  make  the  wine  in  the  second  mixture  eciual  to  85%  of  the  second  mixture.  Since  in  the  second 
mixture  there  is  to  lie  85%  of  wine,  there  nnist  be  15%  of  water.  And  as  the  amount  of  water  was 
unchanged  in  the  first  mixture,  therefore  the  15%  of  water  in  the  second  mixture  equals  the  42  gal- 
lons in  the  first  mixture.  Hence,  since  15%  of  the  first  mixture  is  42  gallons,  the  whole  of  the  sec- 
ond mixture  will  be  (15  :  42  :  :  100  :  )  2Si)  gallons.  This  second  mixture  was  made  by  the  addition 
of  wine  alone:  hence,  there  must  have  been  added  to  the  first  mixture  the  difference  between  280 
gallons  and  150  gallons,  which  is  130  gallons  of  wine. 

Second  Solutio.n  bv  .Vlligation,  by  the  Analytic  Method. 


72+       I       1  —  13         15  XIO  =     ISO  By  this  solution  the   proportional  quantities 

85      I    lUO  —      I      1  —  15        13  XIO  =     130        of  the  seconil  mixture  are  found  to  be  respec- 

. —  lively  15  of  the  first  mixture  and  13  of  pure 
280  wine.  But  since  1.50  gallons  of  the  first 
mixture  instead  of  15  are  to  be  used,  hence,  1.50  -^  15  =  10,  which  is  the  ratio  mnuber  of  times  that 
each  of  the  proportional  parts  is  to  be  increased  in  order  that  the  gain  and  loss  will  be  equal.  Thus, 
15x10  =  150  gallons  of  original  mixture.  13x10  =  loU  gallons  of  wine  to  be  added.  150  +  130=  280 
gallons  in  the  second  mixture. 


MISCELLANEOUS   PROBLEMS. 


1021 


A  Puzzle  Problem. 
24    A  piece  of  board  8  inches  square,  containing-  64  square  inches,  is  cut  into  4  pieces,  as  shown  ia 
Figure  1  of  the  Sohition.     These  pieces  are  then  arranged  so  as  to  form  a  rectangle  5  inches  wide  at 
each  end  and  13  inches  long;  thus  making  65  square  inches.      How  is  the  1  square  inch  increase 
accounted  for? 

Solution. 


8" 


This  problem  has  puzzled  thousands,  and  this  is  the  first  solu- 
tion ever  published.  The  main  difficulty  in  comprehending  it  arises 
from  tlie  erroneous  statement  that  is  made  bj-  parties  presenting  it, 
to  the  effect  that  the  4  pieces  cut  from  the  8  inch  square  board  fit 
perfectly  when  placed  together  to  form  the  rectangular  piece,  13x.5 
inches.  This  is  not  the  case.  The  pieces  do  not  fit  closel)-.  There 
is  an  opening  the  entire  diagonal  length  of  the  rectangle,  as  shown 
in  Figure  2.  The  corners  in  which  the  diagonal  line  terminate  are 
alone  in  contact.  This  opening  has  its  greatest  width  through  the 
middle  part  of  tlie  diagonal  line,  and  averages  a  little  more  than  4  of 
an  inch.  This  open  space,  the  length  of  tlie  diagonal  13.92  +  inches, 
is  just  equal  to  1  square  inch.     Hence  the  apparent  1  inch  increase. 

If  this  opening  were  closed,  tlie  width  of  the  rectangle  would 
not  be  5  inches,  except  at  the  ends.  Hence,  it  is  clear  that  the  board 
does  not  cover  65  sq.  in.  of  space. 

The  cause  of  this  open  space  results  from  the  fact  that 
the  angles  formed  in  cutting  A  &  B,  Fig.  1,  are  different 
from  the  angles  formed  in  cutting  C  &  D.  This  is  easily 
seen  by  noting  that  the  slant  line  in  A  &  B  diverges  2  inches 
from  a  perpendicular,  in  a  piece  5  in.  wide,  while  the  slant 
or  diagonal  line  in  C  &  D,  Fig.  1,  diverges  o  inches  from  a 
perpendicular  in  a  piece  8  incheslong,  which  is  a  less  degree 
of  slant;  or  to  express  it  differently,  the  angles  O.  O.  in  A 
&  B,  Fig.  1,  are  68°,  while  the  angles  P  P  in  C  &  D,  Fig.  1, 
are  69°.  Hence,  when  the  triangular  pieces  C  &  D  are  adjusted  to  the  trapezoidal  pieces  A  &  B,  the 
diagonal  lines  of  each  half  of  the  rectangle  are  conca-\-e,  as  shown  in  diagram.  Fig.  2,  and  thus  form 
the  open  space,  which,  as  stated  above,  is  equal  to  1  sq.  in.,  and  which,  if  closed,  would  reduce  the 
width  of  the  rectangle  except  at  the  two  ends. 


S" 

ol         3' 

i,  A 

3   i. 

5 

fflO 

3" 

/ea" 
/o            S' 

h  J? 

8"        ^^--^ 

35  At  a  primary  election,  in  which  three  candidates,  A,  B  and  C,  were  contesting,  15,424  votes 
were  cast.  Of  this  number  of  votes  A  received  1006  more  than  B  and  1213  more  than  C.  How  many 
votes  did  each  candidate  receive? 

Solution. 

( 1)  Since  there  were  15,424  votes  cast,  and  since  A  received  1006  more  than  B,  and  1213  more 
than  C=  2219  more  than  B  and  C;  hence  15,424+2219  =  17,643  =  the  number  of  votes  that  would  have 
been  cast  in  order  for  each  candidate  to  have  received  as  many  votes  as  A. 

(2)  Then,  17,642  -^  S  =  5S81  votes  received  by  A;  and  by  the  conditions  of  the  problem 
5881  —  1006  =  4875  votes  received  by  B;  and  5SS1  — 1213  =  4668  votes  received  by  C. 

(3)  5S81+4S75+466S  =  15,424=  total  votes  cast 


26 


A  boy  bought  a  squirrel,  a  rabbit  and  a  bird.  The  squirrel  cost  20  cents.  The  squirrel  and 
rabbit  cost  twice  as  much  as  the  bird,  and  the  rabbit  and  bird  cost  three  times  as  much  as  the  squir- 
rel.    What  was  the  cost  of  the  bird  and  the  rabbit? 

Solution. 
2  birds  =  a  squirrel  and  a  rabbit. 

A  bird  and  a  rabbit  =  60c.  .       .      ^,   .       ,        ,  „ 

Multiplying  equation  (2)  by  2  and  transposing,  an  equation  is  obtained  as  follows: 
2  birds=  SI- 20  minus  the  value  of  two  rabbits, 
in  (1)  and  (3)  were  found  2  equations  for  2  birds.      Then  things  which  are  equal  to  the  same 
thing  are  equal  to  each  other.     Hence  the  following: 

(4)  \  squirrel  and  a  rabljit  =  SI. 20  minus  the  value  of  two  rabbits.  By  subtracting  the  squir- 
rel from  one'skle  and  the  value  of  a  squirrel  from  the  other  side,  the  remainder  is, 

(5)  A  rabbit  =  Sl.00-2  rabbits  (value  of) 

By  adding  two  rabbits  to  each  side  we  obtain: 

(6)  3  rabbits  =  Sl.OO.     Then  1  rabbit  =  33i  cents.  t  ,^    ,  •    ,   ■    r        ,,    , 
Substituting  the  value  of  the  rabbit  in  equaUon  (2) ,  the  value  of  the  bird  is  found  to  be 

26S  cents.      (60  cents -loi  cents  =  26i) . 


(1) 
(3) 

(3) 


1023  soule's  philosophic  practical  mathematics. 

27.  a  grocer  has  150  gallons  of  a  mixture  of  wine  and  water.  The  qn.antity  of  wine  is  72%  of  that 
of  the  water.  How  much  wine  must  be  added  to  make  the  quantity  of  wine  equal  to  85%  of  the 
water .' 

Solution. 

(1)  Let  100  represent  the  quantity  of  water;  then  73  will  equal  the  72%  of  wine. 

(2)  100  +  72  =  172  =  the  sum  of  the  respective  proportions  of  water  and  wine. 

(3)  From  these  proportional  numljers  the  following  statement  is  made  to  find  the  proportion 
of  water  in  the  mixture: 

Statement  ■    (4)     87.21  X72%=62. 79  gallons  of  wine  in  the  mixture,  87.21  + 

100  =  water  62.79  =  150  gals,  of  mixture. 

172  150  =  mixture 


87.21  gals,  water 

(5)     87.21X85'?',  the  quantity  of  wine  required  in  the  second  mixture=74,13  gals.  wine.  74.13  — 
62.79  gals,  of  wine  now  in  first  mixture,  =  11. .34  gals,  of  wine  to  be  added  to  first  mixture. 


28.  A  farmer  has  a  field  80  rods  long  and  40  rods  wide,  which  he  wishes  to  fence  with  posts  and 
boards.  The  posts,  which  are  7  feet  long  and  4  by  6  inches,  are  to  be  set  8  feet  apart,  as  far  as  is  pos- 
sil)le.  The  fence  is  to  be  5  boards  high,  each  board  1  inch  thick  by  4  inches  wide.  How  much  will 
the  lumber  cost  to  fence  the  lot  at  $12.50  per  thousand  board  feet,  allowing  2%  waste? 

Solution. 

Of  several  methods  of  solving  this  problem,  the  following  is  given: 

(1)  Find  the  number  of  posts  thus: 

80  rods=1320  feet,  =  length  of  one  side  of  field. 
40  rods=  660  feet,  =  width  of  one  end  of  field. 

1320  -=-  8  ft. ,  the  distance  the  posts  are  apart,  =  165  +  1  =  166,  posts  on  one  side,  X  2  = 
332  posts  on  two  sides. 

660  -i-  8  ft.,  the  distance  the  posts  are  apart,  =  82i,  practically  83,  less  1,  counted  in  the 
length,  =82  posts  on  one  end,  X  2  =  164  posts  on  two  ends. 
332  +  164  =  496  posts. 
NOTE: — It  will  be  seen  that  on  each  end  there  is  but  4  feet  space  between  two  of  the  posts. 

(2)  Find  the  board  feet  in  the  496  posts  thus: 

7  ft,  =  length  of  posts  x  (4  ',\  6  -^  12)  =  14  sq.  ft.  or  board  feet  in  each  post;  496  X  14  = 
6944  board  feet  in  496  posts. 

(3)  Find  the  board  feet  in  the  boards  of  the  fence,  thus: 

1320  +  1320  +  660  +  660  =3960  feet,  distance  around  the  field.  Then,  as  the  fence  is 
to  be  5  boards  high,  there  will  be  (3960  x  5)  =  19S00  lineal  feet  of  boards  4  inches  wide 
and  1  inch  thick;  then,  as  the  boards  are  i  of  a  foot  wide  there  will  be  as  many  board 
feet  as  19800  is  equal  to  3.      19800  -^  3  =  6600  board  feet. 

(4)  6600  +  6944  =1,3544  board  feet  in  the  boards  and  posts. 

(5)  Find  the  2%  waste  thus: 
13544  X  2%  =270.88  sq.  ft. 

13.544  +  270.88=  138L4.88  sq.  ft.  or  board  ft. 

Then  1,3814.88  ft.  at  $12i  per  M,  =  $172,686,  cost  of  lumber  to  build  the  fence. 
NOTE: — The  above  is  the  approximate  practical  method  of  finding  the  2%  waste.   The  accurate 
method  is  thus:   98  :  lOU  :  :  13544  :  13820  +  sq.  ft.;  then  13820  -  13544  =  276  +  sq.  ft.  =  2% 
waste. 


29.  A  cow  and  a  goat  can  eat  the  contents  of  a  pasture  in  40  days;  a  cow  and  a  goose  can  eat  it  in  60 
days;  and  a  goose  and  a  goat  can  eat  it  in  90  days.  How  many  days  will  it  require  the  cow,  the  goat 
and  the  goose  to  eat  it? 

Solution. 

(1)  Since  a  cow  and  a  goat  can  eat  the  contents  of  a  pasture  in  40  daj'S,  in  one  day  they 
would  eat  the  40  part;  since  a  cow  and  a  goose  can  eat  the  contents  of  the  same  pasture  in  60  days, 
in  one  day  they  would  eat  the  gV.  part;  since  a  goose  and  a  goat  can  eat  the  contents  of  the  same 
pasture  in  90  days,  in  one  day  they  would  eat  the  ^g  part. 

(2)  Then  A  +  e'o  +  no  =  s'/o  =  what  2  cows,  2  goats  and  2  geese  eat  in  a  day;  ^ia  ^  2  = 
^^  =what  1  cow,  1  goat  anil  1  goose  eat  in  1  day. 

(3)  Then,  since  the  eating  of  ^'/j  of  the  contents  of  the  pasture  requires  1  day,  ^0  will 
require  the  19th  part  of  1  day,  and  ||",  or  the  whole  of  the  contents  of  the  pasture  will  require  720 
times  as  many  days,  which  is  37i|.     Or  thus:    19  :  1  :  :  720  :  37},J. 


MISCELLANEOUS    PROBLEMS.  1023 

30.    1.     Multiply  MMCCCXII  by  CXXIV,  using  the  Roman  numerals. 

2.  How  did  the  Roman  school  children  and  the  Roman  business  men  perform  the  operations 
of  numbers? 

Solution. 

MilCCCXII  (1)  XII     multiplied  by  IV  =  48  =  XLVIII 

CXXIV  (2)  CCC  multiplied  by  IV  =  1200  =  MCC 

VniMCCXLVIII  =      9248  ^'^  MM   multiplied  by  IV  =  8000  =  VIII 

XXMMMCXX         =    23120  (j)  ^^jj     multiplied  by  X  =  120  =  CXX 

XXMMMCXX         =    23120  (2)  CCC   multiplied  by  X  =  3000  =  MMM 

C'CXXX.MCC  °  231200  (3)  MM    multiplied  by  X  =20000  =  XX 

CCLXXXVIDCLXXXVIII    =  2866S8  Duplicate  the  multiplication  by  the  second  X. 

Then: 

(1)  XII     multiplied  by  C  =    1200  =  MCC 

(2)  CCC  multiplied  by  C  =  30000  =  XXX 

(3)  MM   multiplied  by  C  =  200,000  =  CC 

In  performing  the  operation,  the  multiplicand  and  the  multiplier  may  be  variously  separated 
and  the  same  result  obtaineil.  Without  the  Arabic  system  of  numbers,  which  the  Romans  did  not 
know,  the  partial  products  would  be  added  with  the  Abacus. 

REMARKS:  It  is  readily  seen  by  this  brief  elucidation  that  with  large  numbers  the  operation 
would  be  very  complicated  and  impracticable  for  business  purposes.  In  reply  to  the  question,  "How 
did  the  Roman  school  children  and  the  Roman  business  men  perform  the  operation  of  numbers?" 
we  answer,  by  the  Abacus  or  arithmetical  board  and  counters.  The  Romans  did  not  work  arithme- 
tical problems  as  herein  elucidated.  The  above  operation  shows  simply  how  Roman  numerals  may 
be  used  in  computing  numl)ers.  For  arithmetical  calculations  they  used  the  Abacus  adopted  from 
the  Greeks,  constructed  with  wires  or  grooves,  with  sliding  balls  or  pebbles  as  counters.  To  each 
pebble  they  gave  the  name  of  calculus  (white  stone),  and  applied  the  verb  calculare  to  express  the 
numerical  operation  or  work.  The  use  of  these  instruments  formed  an  essential  part  of  the  educa- 
tion of  every  Roman  youth.  Instead  of  slate  and  pencil,  every  Roman  boy  carried  to  school  a  box 
or  basket  containing  calculi  or  counters,  and  his  Abacus  or  arithmetical  board.  From  this  work  of 
the  Romans  we  derive  the  words  calculate  and  calculation. 

JIany  pages  could  be  written  in  this  connection,  but  space  forbids.  See  Abacus,  Notations 
and  Numerals  in  any  general  encyclopedia. 


31.    What  sum  must  be  set  aside  annually  as  a  sinking  fund  to  pay  a  debt  of  560,000  in  10  years 
at  6%  compound  interest  ? 

Solution. 

This  problem  involves  the  principles  of  compound  interest  and  geometrical  progression.  The 
following  is  the  solution  formula: 

A  X  R  360,000  >:  .06 

S  = = =  $4552,  practically, 

R"  -1  LOG'"-! 

Or  by  Compounrl  Interest  Table  the  compound  amount  of  $1  for  10  years  at  6%  is  $1.79118477  — 
$1  =  .7908477  interest.  Then,  $60,000  ■:  .06  =  $3600  interest,  which  divided  by  .7908477  =$4552+ 
sinking  fund. 

NOTE: — In  addition  to  this  annual  sinking  fund,  the  yearly  interest  of  $3600  must  also  be 
paid. 

Third  Solution: 

By  the  Annuity  Table  the  amount  of  the  annuity  of  SI  for  10  years  at  6%  is  S13. 1S0795;  hence, 
as  often  as  S60,000  is  equal  to  this  amount  will  be  the  annual  investment  in  a  sinking  fund. 
$60,000  *  $13.180795  =  $4552  +  annual  sinking  fund. 


32.  Three  boys,  A.  B  and  C,  went  duck  hunting.  A  shot  and  bagged  }  of  the  whole  number  killed, 
lacking  2  ducks.  B  bagged  5  ducks,  and  C  bagged  as  many  as  A  and  B  together.  How  many  ducks 
did  all  three  bag? 

Solution. 

Since  C  bagged  as  many  ducks  as  A  and  B  together,  he  evidently  bagged  i  of  tlie  whole 
number. 

Since  A  bagged  i  of  the  whole  number,  less  2  ducks,  and  since  B  bagged  5  ducks,  then  the 
sum  of  -A's  +  B's  shares  equals  i  of  the  ducks  — 2  ducks  +  5  ducks  =  i  of  the  ducks  +  3  ducks  over. 

Then,  the  3  ducks  over  represent  the  difference  between  i  (C's  share)  and  i  of  the  ducks, 
which  i.s  ^  of  the  ducks.  Then,  since  J  of  the  ducks  equals  3,  %  or  the  whole  number  will  be  18 
ducks. 


1024  soule's  philosophic  practical  mathematics. 

33.    A  dealer  in  stock  bought  1270  head  of  horses  and  cows;  3  of  the  number  of  horses  is  to  %  of 
the  number  of  cows  as  g  is  to  4 .     How  many  horses  and  how  many  cows  did  he  purchase? 

Solution. 

(1)  Find  the  ratio  of  [i  to  j  by  reducing  them  to  a  common  denominator,  Jg;  J|,  or  40  horses 
to  42  cows. 

(2)  Find  '   of  the  number  of  horses  and  J  of  the  number  of  cows,  according  to  the  above 
ratio,  thus: 

(3)  If  4  of  the  ratio  number  of  horses  is  40,  1  is  the  i  part  and  ;!  is  4  times  as   many,  which 
is  .534  horses. 

In  the  same  manner  the  42  ratio  niimber  of  cows  is  found  to  be  52i  cows. 

(4)  Since  these  numbers  contain  fractions,  and  as  there  can  be  no  fractions  of  live  stock,  they 
are  midtiplied  by  the  lowest  number  that  will  clear  them  of  fractions,  which  is  6,  thus: 

53i  '■>  6  =  320  horses;  S2i  -■;  6  =  315  cows. 

(5)  Now,  having  found  the  proportional  number  of  horses  and  cows,  according  to  the  condi- 
tions of  the  problem,  the  total  number  purchased  is  divided  accordingly,  thus:     320  +  315  =  635. 

635  :  320  :  :  1270  (total  number  purchased)  :  640  horses. 
635  :  315  :  :  1270  :  630  cows. 


34.  A  coffee  dealer  bought  a  quantity  of  coffee  at  12  cents  per  pound.  At  what  price  per  pound 
must  he  mark  it  in  order  to  allow  163%  for  shrinkage  in  roasting,  and  10%  bad  debts,  and  still  gain 
20%  on  cost? 

Solution. 

(1)  20%  gain  on  12  cents  cost  =  2.4  cents;  12c.  +  2.4  =  14.4c.  selling  price  of  the  coffee. 

(2)  Then,  to  allow  for  the  l(,'n9o  shrinkage  in  roasting  the  coSee  and  10%  loss  for  bad  debts, 
the  following  proportional  solution  statement  is  made: 

Statement  Explanation  of  the  Statement. 

$100      assumed  marking  price.  To  allow  16j  for  shrinkage,  the  $100  assumed  price 

R3i              14.4  must  be  reduced  to  83i;  then,   since  83i  reduced 

VO               100  price  requires  f  100  marking  price,  SI  will  require 

the  83i  part,  and  14.4,  the  selling  price  of  the  coffee, 

1').2   cents  marking  price.  will  require  14.4  times  as  much. 

Then  to  allow  li)%  for  1)ad  debts,  assume  another  $100  marking  price,  and  reason  in  the  same 
manner  as  shown  in  the  first  term  for  the  16j%. 

Second  Solution. 

(1)  To  gain  20%,  which  is  J  on  12  cents  cost,  the  selling  price 
will  be  I  of  the  cost  as  indicated  on  the  statement  line. 

(2)  Since  the  shrinkage  in  roasting  the  coffee  is  165%,  which  is 
J,  hence  the  selling  price  is  §  of  the  asking  price.  Therefore 
J  of  asking  price  will  be  the  5th  part,  and  Q  or  asking  price 
will  be  6  times  as  much. 

(3)  Since  the  allowance  for  bad   debts  is  10%,  the  selling  price  is  j"„  of  the  asking  price. 
Therefore  ^'a  of  asking  price  will  be  the  9th  part  and  }JJ  or  the  asking  price  will  be  10  times  as  much. 


5 
5 
9 

12  cents 

6 

6 
10 

191  cents 

35.  A,  B  and  C  each  owned  a  piece  of  land  as  follows:  (Make  diagrams)  A's  land  was  in  the 
form  of  a  square,  and  measured  on  its  four  sides  320  rods.  B's  land  was  in  the  form  of  a  rectangle, 
and  measured  on  each  side  120  rods,  and  on  each  end  40  rods,  thus  320  rods  on  its  four  sides. 

C's  land  was  circular  in  form,  and  the  circumference  or  distance  around  it  was  320  rods,  the 
same  as  the  distance  around  A's  and  B's  land.  The  three  pieces  were  sold  for  $9674.40.  How  many 
acres  were  there  in  each  piece  of  land,  and  what  amount  of  money  was  due  each  owner? 

Solution. 

(1)  As  A's  land  was  square  and  the  distance  around  it  was  320  rods,  hence,  each  side  was  i  of 
320  =  80  rods.  Then  80X80=6400  sq.  rods  in  the  piece  or  tract.  6400  ^  160  sq.  rods  in  an  acre=40 
acres  in  A's  land. 

(2)  B's  land,  measuring  120  rods  on  the  sides  and  40  rods  on  the  ends,  contained  120^40  = 
4800  sq.  rods;  4800  -s-  160  sq.  rods  in  an  acre=30  acres  in  B's  land. 

(3)  C's  land,  having  a  circumference  of  320  rods,  the  number  of  acres  is  found  as  follows: 
3202X.079.58,  the  ratio  between  the  square  of  the  circumference  and  the  area,  =8148.992  sq.  rds.  area. 
8148.992  ^  160  sq.  rds.  in  an  acre,  =  5(».93  +  acres  in  C's  land. 

(4)  40  acres  in  A's  land,  +  30  acres  B's  land,  +  50.93  acres  C's  land  =  120.93  acres  which  sold 
for  $9674.40. 

(40         :  )  $3200.00  =  A's   share. 

(5)  Then    .as   120.93  :  $9674.40  :: -^  30         :   ^  $2400.00  =  B's  share. 
,  (50.93   :   )  $4074.40  =  C's  share. 

MOTE: — This  problem  shows  an  important  principle  in  mathematics,  which  is  that  the  circle 
is  the  greatest  of  all  figures  of  equal  perimeter  or  boundary  lines;  and  that  the  nearer  the  figure 
approximates  a  circle,  tlie  greater  is  its  capacity. 


MISCELLANEOUS    PROBLEMS.  1025 

36.  A  tinsmith  wishes  to  construct  a  vessel  in  the  form  of  a  frustum  of  a  cone,  the  upper  and  lower 
bases  of  which  are  to  be  respectively  30  and  50  inches  in  diameter,  and  the  slant  height  of  which  is 
to  be  SO  inches. 

Allowing  10%  waste  of  material  in  making,  how  many  sheets  of  tin,  each  12-16  inches,  will  be 
required  to  construct  the  convex  sides  of  the  vessel  ? 

Solution. 

(1)  Find  the  average  circumference  of  the  two  bases,  thus:  3.1416,  circumference  of  a  circle 
whose  diameter  is  1X30  in.;  diameter  of  upper  base,  =  94.2480  in.;  3.1416X,50  in.,  diameter  of  lower 
base,  =  157.0800  in.;  94.248  +  157.08  =  251.328  -  2  =  125.664  in.,  average  circumference  or  width 
of  the  vessel. 

(2)  125.664X80  in  ,  slant  height  =  10053.120  sq.  inches  in  the  convex  surface  of  the  vessel. 
,  (3)     10053.12  +  10%  for  loss  of  material  in  making,  ( 1005.312  sq.  in.)  =  11058.432  sq.  in. 

(4)     11058.432  -<-  (12X16),  size  of  the  sheets  of  tin,  =  57.596  sheets  of  tin. 


37.  (1)  Numerate  and  read  the  following  number  by  the  American  or  French  method  and  by  the 
English  method:     9421532644107805991. 

(2)     Write  and  read  the  following  number  according  to  the  Roman  method:     2467619. 

NOTE:— The  Roman  method  of  Notation  is  by  the  use  of  seven  capital  letters,  as  follows: 
The  letter  I  represents  one;  Y,  five;  X,  ten;  L,  fifty;  C,  one  hundred;  D,  five  hundred;  and  11,  one 
thousand. 

The  intermediate  and  succeeding  numbers  are  expressed  according  to  the  following  principles: 

1st.  Every  time  a  letter  is  repeated,  its  value  is  repeated,  thus:  II  represents  two;  XX  rep- 
resents twenty;  CCCC  represents  four  hundred. 

2d.  When  a  letter  is  placed  after  one  of  greater  value,  the  sum  of  their  values  is  the  number 
expressed.     Thus:     VI  represents  six;  XVII  expresses  seventeen. 

3d.  When  a  letter  is  placed  before  one  of  greater  value,  the  difference  of  their  values  is  the 
number  expressed.     Thus:     IV  expresses  four;  XL,  expresses  forty. 

4th.  When  a  letter  is  placed  between  two  letters  of  a  greater  value,  it  is  combined  wi'.h  the 
one  following  it.     Thus:     XIX  expresses  nineteen;  CXLVI  expresses  one  hundred  forty-six. 

5th.  A  dash  or  bar  placed  over  a  letter  multiplies  its  value  by  1000.  Thus:  V  expresses  5000;  "jT 
expresses  10,000;  L  expresses  50,000;  C  expresses  100,000;  D  expresses  500,000;  'wTexpresses 
1,000,000. 

In   like  manner  a  double  djish  placed  over  a  letter  multiplies  its  value  by  1000  two  times. 

Thus:   C  expresses   100,000,000;     M  e.xpresses  1,000,000,000. 

NOTE:— In  writing  numbers  by  the  Roman  method  always  write  the  different  orders  of  units 
successively,  beginning  with  the  higher  orders. 

Solution. 

(1)  By  the  French  or  American  method  of  numeration,  9421532644107805991  would  be  sepa- 
rated into  periods  of  three  figures  each,  and  read  thus:  9  quintillions,  421  quadrillions,  532  trillions, 
644  billions,  107  millions,  805  thousands,  991  units. 

(2)  By  the  English  method  of  numeration,  the  same  numbers  would  be  separated  into 
periods  of  six  figures  each,  and  read  thus:      9  trillions,  421532  billions,  644107  millions,  S05991  units.  . 

(3)  According  to  the  principles  of  the  Roman  system  of  notation  as  set  forth  in  the  note  to 
the  problem,  2,467,619  would  be  written  thus: 

MMCCCCLXVilDCXIX. _    

EXPLANATION:— MM  =  2  millions;  CCCC  =400,000;  L  =  50,000;  X  =  10,000;  V    =  5,000; 

IT  =2,000;  D  =  500;  C  =  100;  XIX  =  19,  =2,467,619. 


38.    1.     How  many  inches  deep  will  1  cubic  yard  of  earth  fill  8  square  yards  of  ground? 
2.     How  many  square  yards  will  1  cubic  yard  fill  5  inches  deep.' 

Solution. 

(1)  Question.     1  cubic  yard=36x36X36=46656  cu.  in. 

8    sq.    yards=36X36X  8  =  10368  sq.  in. 
Then  46656  -^  10368=4.5  inches  =  the  depth  that  1  cu.  yd.  will  fill  8  sq.  yds. 

(2)  Question.  1  side  of  a  cubic  yard  is  the  same  as  the  surface  of  1  stiuare  yard;  hence  for 
each  of  the  36  inches  thickness  of  the  cubic  yard,  1  square  yard  will  be  filled  1  inch  deep.  Therefore 
as  often  as  the  36  inches  depth  of  the  cube  is  equal  to  the  depth  to  be  filled,  just  so  many  square 
yards  can  be  filled.  Then  36  -^  5,  the  depth  required,  =  7J  sq.  yds.  of  ground  maybe  filled  5  inches 
deep  with  1  cu.  yd.  of  earth. 


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